Bond distance in is . Dipole moment of is . The percentage ionic character in will be : (electron charge )
1
2
3
4
Official Solution
Correct Option: (4)
Givene
From
ionic character
02
PYQ 2013
medium
chemistryID: jee-main
In which of the following pairs of molecules/ions both the species are not likely to exist?
1
2
3
4
Official Solution
Correct Option: (3)
Species having zero or negative bond order do not exist
Bond order = 0
Bond order So, both and do not exist.
03
PYQ 2013
medium
chemistryID: jee-main
Which one of the following molecules is expected to exhibit diamagnetic behaviour?
1
2
3
4
Official Solution
Correct Option: (2)
This problem can be solved by using concept of H-bonding and applications of H-bonding.
So, the correct option is (B): .
04
PYQ 2013
medium
chemistryID: jee-main
Which one of the following molecules is polar?
1
2
3
4
Official Solution
Correct Option: (2)
The geometry of is square pyramidal with an unsymmetric charge distribution therefore this molecule is polar.
So, the correct option is (B): .
05
PYQ 2014
medium
chemistryID: jee-main
The correct statement for the molecule, is
1
it is a covalent molecule
2
it contains and ions
3
it contains and ions
4
it contains , and lattice molecule
Official Solution
Correct Option: (2)
It contains and ions
06
PYQ 2014
easy
chemistryID: jee-main
Which one of the following molecules is paramagnetic ?
Official Solution
Correct Option: (1)
07
PYQ 2014
medium
chemistryID: jee-main
Amongst and the compounds with the greatest and the least ionic character, respectively are :
1
and
2
and
3
and
4
and
Official Solution
Correct Option: (2)
According to Fajan?s rules smaller, highly charged cation has greatest covalent character while large cation with smaller charge has greatest ionic character.
08
PYQ 2016
medium
chemistryID: jee-main
Which one of the following statements about water is FALSE ?
1
Water is oxidized to oxygen during photosynthesis.
2
Water can act both as an acid and as a base.
3
There is extensive intramolecular hydrogen bonding in the condensed phase.
4
Ice formed by heavy water sinks in normal water.
Official Solution
Correct Option: (3)
Water possess intermolecular hydrogen bonding in the condensed phase.
So, the correct option is (C): There is extensive intramolecular hydrogen bonding in the condensed phase.
09
PYQ 2017
medium
chemistryID: jee-main
Which of the following species is not paramagnetic?
1
2
3
4
Official Solution
Correct Option: (4)
has 14 electrons (even) it is diamagnetic has (odd) it is paramagnetic and has 1 unpaired electron in molecular orbital. has (even) but still paramagnetic and has two unpaired electrons in and (s-p mixing). has (even) but still paramagnetic and has two unpaired electrons in and molecular orbitals.
So, the correct option is (D): .
10
PYQ 2017
medium
chemistryID: jee-main
The group having triangular planar structures is :
1
2
3
4
Official Solution
Correct Option: (2)
Hybridised
11
PYQ 2018
medium
chemistryID: jee-main
Which of the following best describes the diagram below of a molecular orbital ?
1
A non-bonding orbital
2
An antibonding orbital
3
A bonding orbital
4
An antibonding orbital
Official Solution
Correct Option: (4)
An antibonding p orbital
12
PYQ 2018
medium
chemistryID: jee-main
Which of the following conversions involves change in both shape and hybridisation?
1
2
3
4
Official Solution
Correct Option: (4)
combines with ion to form .
The shape of the molecule is a planar triangle with bond angles of and hybridisation. The shape of the molecule is a tetrahedral with bond angles of and hybridisation.
So, the correct option is (D):
13
PYQ 2018
medium
chemistryID: jee-main
The incorrect geometry is represented by :
1
- trigonal planar
2
- bent
3
- trigonal planar
4
- trigonal bipyramidal
Official Solution
Correct Option: (3)
molecule has trigonal pyramidal geometry.
14
PYQ 2018
medium
chemistryID: jee-main
According to molecular orbital theory, which of the following will not be a viable molecule ?
1
2
3
4
Official Solution
Correct Option: (4)
Electronic configuration of Bond order of . If a bond order of zero is obtained, that means that the molecule is too unstable and so it will not exist. Hence, does not exit, due to zero bond order.
15
PYQ 2018
medium
chemistryID: jee-main
In the molecular orbital diagram for the molecular ion, , the number of electrons in the molecular orbital is :
1
0
2
1
3
2
4
3
Official Solution
Correct Option: (2)
The arrangement of electrons in molecular orbitals is as follows:
16
PYQ 2019
medium
chemistryID: jee-main
Among the following species, the diamagnetic molecule is
1
2
3
4
Official Solution
Correct Option: (4)
are paramagnetic according to M.O.T. where as is diamagnetic.
17
PYQ 2019
medium
chemistryID: jee-main
Among the following molecules / ions, which one is diamagnetic and has the shortest bond length?
1
2
3
4
Official Solution
Correct Option: (1)
18
PYQ 2021
medium
chemistryID: jee-main
is an interhalogen T-shaped molecule. The number of lone pairs of electrons on A is _________. (Integer answer)
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept: Interhalogen molecules of the type are formed by two different halogens. The geometry and shape are determined by the VSEPR (Valence Shell Electron Pair Repulsion) theory based on the total number of electron pairs around the central atom . Step 2: Key Formula or Approach: Steric Number = Step 3: Detailed Explanation: 1. Central atom is a halogen, which has 7 valence electrons. 2. In , there are 3 monovalent atoms bonded to . 3. Total valence electrons involved = 7 (from ) + 3 (from bonds) = 10 electrons (or 5 pairs). 4. Steric Number = 5. This corresponds to a trigonal bipyramidal ( ) hybridization. 5. Out of the 5 electron pairs, 3 are bond pairs (BPs) with atoms . 6. Number of Lone Pairs (LPs) = . 7. According to VSEPR, 2 LPs and 3 BPs result in a "T-shaped" geometry to minimize repulsion. Step 4: Final Answer: The number of lone pairs on the central atom is 2.
19
PYQ 2021
medium
chemistryID: jee-main
The number of hydrogen bonded water molecule(s) associated with stoichiometry is ____________
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept: In hydrated copper(II) sulfate ( ), the five water molecules are not bonded identically to the central metal ion or the lattice. Step 2: Detailed Explanation: The actual structure of blue vitriol is represented as . 1. Four water molecules are coordinated directly to the ion in a square planar arrangement, forming coordinate covalent bonds. 2. The fifth water molecule is not directly attached to the copper ion. Instead, it is trapped in the crystal lattice. 3. This fifth water molecule is held in place by hydrogen bonds to the oxygen atoms of the sulfate ( ) ion and the coordinated water molecules. Step 3: Final Answer: The number of hydrogen-bonded water molecules is 1.
20
PYQ 2021
medium
chemistryID: jee-main
Match List - I with List - II : Choose the most appropriate answer from the options given below :
1
(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
2
(a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)
3
(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
4
(a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)
Official Solution
Correct Option: (2)
Step 1: Understanding the Question: We need to determine the number of lone pairs of electrons on the central Xenon (Xe) atom in four different compounds and match them accordingly. Step 2: Key Formula or Approach: We can use the VSEPR theory formula to find the number of electron pairs and then the number of lone pairs. Number of electron pairs (Steric Number, SN) = [ (Valence electrons of central atom) + (No. of monovalent atoms) - (Charge on cation) + (Charge on anion) ] Number of lone pairs (LP) = SN - (Number of surrounding atoms) Alternatively, we can directly count the electrons. Xenon (a noble gas) has 8 valence electrons. Step 3: Detailed Explanation: (a) XeF : - Valence electrons of Xe = 8. - Electrons used in bonding with 2 Fluorine atoms = 2 1 = 2. - Remaining non-bonding electrons = 8 - 2 = 6. - Number of lone pairs = 6 / 2 = 3. - Match: (a)-(iv). (b) XeO F : - Valence electrons of Xe = 8. - Electrons used in bonding with 2 Oxygen atoms (double bonds) = 2 2 = 4. - Electrons used in bonding with 2 Fluorine atoms (single bonds) = 2 1 = 2. - Total electrons used in bonding = 4 + 2 = 6. - Remaining non-bonding electrons = 8 - 6 = 2. - Number of lone pairs = 2 / 2 = 1. - Match: (b)-(ii). (c) XeO F : - Valence electrons of Xe = 8. - Electrons used in bonding with 3 Oxygen atoms (double bonds) = 3 2 = 6. - Electrons used in bonding with 2 Fluorine atoms (single bonds) = 2 1 = 2. - Total electrons used in bonding = 6 + 2 = 8. - Remaining non-bonding electrons = 8 - 8 = 0. - Number of lone pairs = 0. - Match: (c)-(i). (d) XeF : - Valence electrons of Xe = 8. - Electrons used in bonding with 4 Fluorine atoms = 4 1 = 4. - Remaining non-bonding electrons = 8 - 4 = 4. - Number of lone pairs = 4 / 2 = 2. - Match: (d)-(iii). Step 4: Final Answer: The correct set of matches is (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii). This corresponds to option (B).
21
PYQ 2021
medium
chemistryID: jee-main
The correct shape and I-I-I bond angles respectively in ion are :
1
Linear; 180°
2
Distorted trigonal planar; 135° and 90°
3
T-shaped; 180° and 90°
4
Trigonal planar; 120°
Official Solution
Correct Option: (1)
Step 1: Central Iodine has 7 valence electrons + 1 from charge = 8. Two electrons form bonds with other I atoms, leaving 3 lone pairs.
Step 2: Steric number = 2 (bond pairs) + 3 (lone pairs) = 5. The hybridization is .
Step 3: According to VSEPR theory, to minimize repulsion, the 3 lone pairs occupy equatorial positions, and the 2 bond pairs occupy axial positions.
Step 4: The geometry is trigonal bipyramidal, but the molecular shape is linear with a bond angle of 180°.
22
PYQ 2021
medium
chemistryID: jee-main
The compound/s which will show significant intermolecular H-bonding is/are :
1
(a), (b) and (c)
2
(a) and (b) only
3
(b) only
4
(c) only
Official Solution
Correct Option: (3)
Step 1: Understanding the Question:
We need to determine which of the three given compounds can form significant intermolecular hydrogen bonds. For this, a molecule needs an H atom attached to a highly electronegative atom (N, O, F) and a lone pair on another N, O, or F atom, without significant steric hindrance or preference for intramolecular H-bonding. Step 2: Detailed Explanation:
- Compound (a): o-Nitrophenol. This molecule has a hydroxyl group (–OH) and a nitro group (–NO ) in ortho positions (adjacent to each other). The hydrogen of the –OH group forms a strong intramolecular hydrogen bond with one of the oxygen atoms of the neighboring nitro group. This internal bonding significantly reduces the availability of the –OH group to bond with other molecules. Therefore, it exhibits very weak intermolecular H-bonding. - Compound (b): N-(4-hydroxyphenyl)acetamide (Paracetamol). This molecule has multiple sites for H-bonding: the phenolic –OH group and the amide –NH– group can both act as H-bond donors, while the phenolic oxygen and the carbonyl oxygen (C=O) can act as H-bond acceptors. The functional groups are far apart (para), so intramolecular H-bonding is not possible. Thus, these groups are fully available to form strong intermolecular hydrogen bonds with neighboring molecules. - Compound (c): 2,6-di-tert-butyl-4-methylphenol (BHT). This molecule has a hydroxyl (–OH) group, which is capable of H-bonding. However, the –OH group is surrounded by two extremely bulky tert-butyl groups. This crowding, known as steric hindrance, physically blocks other molecules from approaching the –OH group. Consequently, intermolecular hydrogen bonding is effectively prevented. Step 3: Final Answer:
Based on the analysis, only compound (b) can form significant intermolecular hydrogen bonds.
23
PYQ 2021
medium
chemistryID: jee-main
The bond order and magnetic behaviour of O ion are, respectively :
1
1 and paramagnetic.
2
1.5 and paramagnetic.
3
2 and diamagnetic.
4
1.5 and diamagnetic.
Official Solution
Correct Option: (2)
Step 1: Understanding the Question:
We need to determine the bond order and magnetic properties of the superoxide ion (O ) using Molecular Orbital Theory (MOT). Step 2: Key Formula or Approach:
1. Write the molecular orbital (MO) configuration for the O ion. An oxygen atom has 8 electrons, so an O molecule has 16 electrons. The O ion has 16 + 1 = 17 electrons.
2. Calculate the bond order using the formula: Bond Order = (N - N ), where N is the number of electrons in bonding MOs and N is the number of electrons in antibonding MOs.
3. Determine the magnetic behavior by checking for unpaired electrons in the MO configuration. If there are unpaired electrons, the species is paramagnetic. If all electrons are paired, it is diamagnetic. Step 3: Detailed Explanation:
The MO energy level order for O and its ions is:
Now, we fill the 17 electrons of O into these orbitals:
Bond Order Calculation:
Number of bonding electrons (N ) = 2( ) + 2( ) + 2( ) + 4( ) = 10
Number of antibonding electrons (N ) = 2( ) + 2( ) + 3( ) = 7
Bond Order = Magnetic Behaviour:
Looking at the configuration, the last orbital, , contains a single, unpaired electron. The presence of this unpaired electron makes the O ion paramagnetic. Step 4: Final Answer:
The bond order is 1.5 and the ion is paramagnetic. This corresponds to option (B).
24
PYQ 2021
medium
chemistryID: jee-main
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : Lithium halides are some what covalent in nature.
Reason R : Lithium possess high polarisation capability.
In the light of the above statements, choose the most appropriate answer from the options given below :
1
Both A and R are true and R is the correct explanation of A.
2
Both A and R are true but R is NOT the correct explanation of A.
3
A is true but R is false.
4
A is false but R is true.
Official Solution
Correct Option: (1)
Assertion A states that lithium halides are somewhat covalent in nature. This is a true statement. According to Fajan's rules, although formed between a metal and a non-metal, lithium halides (especially LiI) exhibit significant covalent character. For instance, LiCl is soluble in organic solvents like ethanol. Reason R states that lithium possesses high polarisation capability. This is also a true statement. The term "polarisation capability" refers to the polarizing power of the cation. The Li ion is very small and has a relatively high charge density. This allows it to distort the electron cloud of the larger halide anion to a significant extent. Connecting the Assertion and Reason: The high polarizing power (polarisation capability) of the Li ion is precisely the reason why it can pull the electron density from the anion, leading to a sharing of electrons and thus covalent character in the bond. Therefore, Reason R is the correct explanation for Assertion A.
25
PYQ 2021
medium
chemistryID: jee-main
Which of the following are isostructural pairs ? (Note: Based on typical JEE pairs like and or and )
1
A and B only
2
A and C only
3
B and C only
4
C and D only
Official Solution
Correct Option: (1)
Step 1: Isostructural pairs have the same hybridisation and the same molecular geometry (shape).
Step 2: For example, and are both hybridised and trigonal planar.
Step 3: and are both hybridised and pyramidal.
26
PYQ 2021
medium
chemistryID: jee-main
The number of species having non-pyramidal shape among the following is _________. (A) SO (B) NO (C) PCl (D) CO
Official Solution
Correct Option: (1)
Step 1: Understanding the Question:
We need to determine the molecular shape of four given species using VSEPR theory and count how many of them do not have a pyramidal shape. Step 2: Detailed Explanation:
A pyramidal shape (specifically, trigonal pyramidal) arises from a central atom with 3 bonding pairs and 1 lone pair (AX E geometry). We will analyze each species. - (A) SO (Sulfur Trioxide): - Central atom: S (Group 16, 6 valence electrons). - Bonds to 3 O atoms. Each O forms a double bond to satisfy the octet, using all 6 of S's valence electrons. - Steric number = (number of sigma bonds) + (number of lone pairs) = 3 + 0 = 3. - Geometry: AX , which is trigonal planar. This is non-pyramidal. - (B) NO (Nitrate Ion): - Central atom: N (Group 15, 5 valence electrons). - Bonds to 3 O atoms, plus one extra electron for the -1 charge. Total valence electrons to distribute = 5 + 3(6) + 1 = 24. - Lewis structure shows N with one double bond and two single bonds to oxygen, with resonance. - Steric number = 3 (sigma bonds) + 0 (lone pairs on N) = 3. - Geometry: AX , which is trigonal planar. This is non-pyramidal. - (C) PCl (Phosphorus Trichloride): - Central atom: P (Group 15, 5 valence electrons). - Bonds to 3 Cl atoms, using 3 electrons for single bonds. 2 electrons remain as a lone pair. - Steric number = 3 (sigma bonds) + 1 (lone pair) = 4. - Electron geometry is tetrahedral, but the molecular shape is AX E , which is trigonal pyramidal. This is a pyramidal shape. - (D) CO (Carbonate Ion): - Central atom: C (Group 14, 4 valence electrons). - Bonds to 3 O atoms, plus two extra electrons for the -2 charge. Total valence electrons = 4 + 3(6) + 2 = 24. - Lewis structure shows C with one double bond and two single bonds to oxygen, with resonance. - Steric number = 3 (sigma bonds) + 0 (lone pairs on C) = 3. - Geometry: AX , which is trigonal planar. This is non-pyramidal. Step 3: Final Answer:
The species with non-pyramidal shapes are SO , NO , and CO . Therefore, the number is 3.
27
PYQ 2021
medium
chemistryID: jee-main
According to molecular orbital theory, the number of unpaired electron(s) in is ________.
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept: Molecular Orbital (MO) theory describes the electronic structure of molecules by distributing electrons into orbitals that cover the entire molecule. The presence or absence of unpaired electrons in the highest occupied molecular orbitals (HOMO) determines the magnetic properties of the species. Step 2: Detailed Explanation: 1. Each Oxygen atom has 8 electrons. For the peroxide ion ( ), total electrons electrons. 2. The MO configuration for (similar to ) is:
3. Distribution of valence electrons (14 electrons beyond ): - takes 2, takes 2. (Remaining: 10) - takes 2. (Remaining: 8) - and take 2 each ( ). (Remaining: 4) - and take 2 each ( ). (Remaining: 0) 4. Observation: All occupied orbitals, including the antibonding pi orbitals ( ), are fully filled with pairs of electrons. 5. Number of unpaired electrons = 0. Step 3: Final Answer: The number of unpaired electrons is 0.
28
PYQ 2021
medium
chemistryID: jee-main
According to molecular orbital theory, the species among the following that does not exist is :
1
O
2
He
3
Be
4
He
Official Solution
Correct Option: (3)
Step 1: According to MOT, a species does not exist if its Bond Order (B.O.) is zero.
Step 2: For Be ( , total electrons = 8):
Electronic configuration: .
Step 3: .
Step 4: Since the bond order is zero, Be is unstable and does not exist under normal conditions.
29
PYQ 2021
medium
chemistryID: jee-main
The number of species below that have two lone pairs of electrons in their central atom is _________. (Round off to the Nearest Integer). SF₄, BF₄⁻, ClF₃, AsF₃, PCl₅, BrF₅, XeF₄, SF₆
Official Solution
Correct Option: (1)
Step 1: Determine lone pairs (LP) using . SF₄: 1 LP BF₄⁻: 0 LP ClF₃: 2 LP (Correct) AsF₃: 1 LP PCl₅: 0 LP BrF₅: 1 LP XeF₄: 2 LP (Correct) SF₆: 0 LP
Total species with 2 lone pairs = 2 ( and ).
30
PYQ 2021
medium
chemistryID: jee-main
The total number of electrons in all bonding molecular orbitals of O is ________. (Round off to the Nearest Integer).
Official Solution
Correct Option: (1)
First, determine the total number of electrons in the peroxide ion, .
An oxygen atom (O) has 8 electrons.
An molecule has electrons.
The ion has two additional electrons, so the total number of electrons is . Now, write the molecular orbital (MO) configuration for an 18-electron species.
The order of MOs is:
Fill the orbitals with the 18 electrons:
The bonding molecular orbitals are those without an asterisk. These are . Count the number of electrons in these bonding orbitals:
- Electrons in : 2
- Electrons in : 2
- Electrons in : 2
- Electrons in : 2
- Electrons in : 2 Total number of electrons in bonding molecular orbitals .
31
PYQ 2021
medium
chemistryID: jee-main
The number of sigma bonds in is ________.
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept: Every single covalent bond is a sigma ( ) bond. A double bond contains one sigma bond and one pi ( ) bond. A triple bond contains one sigma bond and two pi bonds. Step 2: Key Formula or Approach: Expand the structure to view all and bonds. Sum all single bonds and count one for every double and triple bond. Step 3: Detailed Explanation: Structure: Pent-2-en-4-yne. Let's break it down: 1. part: 3 bonds from and 1 bond from . (Total: 4) 2. part: 2 bonds from and 1 bond from the double bond . (Total: 3) 3. part: 1 bond from the single bond , 1 bond from the triple bond , and 1 bond from the final . (Total: 3) Sum = . Alternatively, total atoms in is 11. For an acyclic structure, . Step 4: Final Answer: The number of sigma bonds is 10.
32
PYQ 2021
medium
chemistryID: jee-main
The correct set from the following in which both pairs are in correct order of melting point is :
1
LiCl>LiF ; MgO>NaCl
2
LiF>LiCl ; MgO>NaCl
3
LiCl>LiF ; NaCl>MgO
4
LiF>LiCl ; NaCl>MgO
Official Solution
Correct Option: (2)
Step 1: Melting point of ionic compounds depends on Lattice Energy ( ).
Step 2: For vs : Both have +1/-1 charges, but is smaller than . Smaller internuclear distance means higher lattice energy. Thus, .
Step 3: For vs : has charges +2/-2, whereas has +1/-1. Higher product of charges leads to significantly higher lattice energy. Thus, .
33
PYQ 2021
medium
chemistryID: jee-main
AX is a covalent diatomic molecule where A and X are second row elements of periodic table. Based on Molecular orbital theory, the bond order of AX is 2.5. The total number of electrons in AX is _________. (Round off to the Nearest Integer).
Official Solution
Correct Option: (1)
Step 1: Bond order 2.5 is typically observed for species with 11, 13, or 15 electrons.
Step 2: Common diatomic molecules of second-row elements:
has electrons.
has electrons.
Step 3: For (15 electrons), Bond Order .
For (13 electrons), Bond Order .
Since AX is a neutral molecule "AX", (15 electrons) is the most likely candidate.
34
PYQ 2021
medium
chemistryID: jee-main
The ionization enthalpy of Na formation is 495.8, electron gain enthalpy of Br is -325.0, and lattice enthalpy of NaBr is -728.4 kJ mol . The energy for the formation of NaBr ionic solid is (-) __________ kJ mol .
Official Solution
Correct Option: (1)
Step 1: Total Energy ( ) = .
Step 2: .
Step 3: kJ/mol.
Step 4: To express in form , we get .
35
PYQ 2021
medium
chemistryID: jee-main
Which among the following species has unequal bond lengths?
1
XeF
2
SiF
3
SF
4
BF
Official Solution
Correct Option: (3)
We need to determine the molecular geometry of each species using VSEPR theory to identify which one has unequal bond lengths. (A) XeF : Xenon (Group 18) has 8 valence electrons. It forms 4 bonds with F and has 2 lone pairs. The steric number is 6. The geometry is AX E , which corresponds to a square planar shape. In a square planar geometry, all four bonds are identical in length. (B) SiF : Silicon (Group 14) has 4 valence electrons. It forms 4 bonds with F and has 0 lone pairs. The steric number is 4. The geometry is AX , which is tetrahedral. In a tetrahedral geometry, all four bonds are identical. (C) SF : Sulfur (Group 16) has 6 valence electrons. It forms 4 bonds with F and has 1 lone pair. The steric number is 5. The geometry is AX E , which is a see-saw shape based on a trigonal bipyramidal arrangement. This geometry has two types of positions: two axial positions and two equatorial positions. The axial S-F bonds are longer than the equatorial S-F bonds due to greater repulsion from the lone pair in the equatorial plane. Therefore, SF has unequal bond lengths. (D) BF : Boron (Group 13) has 3 valence electrons, plus 1 from the negative charge, for a total of 4. It forms 4 bonds with F and has 0 lone pairs. The steric number is 4. The geometry is AX , which is tetrahedral. All four bonds are identical. Thus, SF is the species with unequal bond lengths.
36
PYQ 2021
easy
chemistryID: jee-main
The difference between bond orders of CO and NO is where x = _________. (Round off to the Nearest Integer)
Official Solution
Correct Option: (1)
37
PYQ 2021
easy
chemistryID: jee-main
In gaseous triethyl amine the "-C-N-C-" bond angle is _________ degree.
Official Solution
Correct Option: (1)
38
PYQ 2022
medium
chemistryID: jee-main
Match List - I with List - II List-I (Compound) List-II (Shape) A I bent B II square pyramidal C III trigonal bipyramidal D IV octahedral Choose the correct answer from the optionsgiven below :
1
(A) - (I), (B) - (II), (C) - (III), (D) - (IV)
2
(A) - (IV), (B) - (III), (C) - (II), (D) - (I)
3
(A) - (II), (B) - (IV), (C) - (I), (D) - (III)
4
(A) - (III), (B) - (IV), (C) - (II), (D) - (I)
Official Solution
Correct Option: (3)
The correct option is (C): (A) - (II), (B) - (IV), (C) - (I), (D) - (III)
39
PYQ 2022
easy
chemistryID: jee-main
In which of the following pairs, electron gain enthalpies of constituent elements are nearly the same or identical ?
(A) and
(B) and
(C) and
(D) and
Choose the correct answer from the options given below :
1
(A) and (B) only
2
(B) and (C) only
3
(A) and (C) only
4
(C) and (D) only
Official Solution
Correct Option: (3)
The correct option is(C): (A) and (C) only.
40
PYQ 2022
easy
chemistryID: jee-main
Arrange the following in increasing order of their covalent character. A. B. C. D.
1
B < A < C < D
2
A < B < C < D
3
A < B < D < C
4
A < C < B < D
Official Solution
Correct Option: (2)
From Fajan’s rule, for a given metal ion, as the size of anion increases, polarizability of anion increases and hence covalent character of the given ionic compound increases.
Hence, the increasing order of covalent characters is:
41
PYQ 2022
medium
chemistryID: jee-main
Among the following species N2,N2+,N2−,N22−,O2,O2+,O2−,O22− the number of species showing diamagnetism is ________.
Official Solution
Correct Option: (1)
According to molecule orbital theory. The electronic configurations of the given species are Diamagnetic species are N2 and O22- ∴ Number of species showing diamagnetism = 2
42
PYQ 2022
medium
chemistryID: jee-main
Bonding in which of the following diatomic molecule(s) become(s) stronger, on the basis of MO Theory, by removal of an electron? (A) NO (B) N2 (C) O2 (D) C2 (E) B2 Choose the most appropriateanswer from the options given below :
1
(A), (B), (C) only
2
(B), (C), (E) only
3
(A), (C) only
4
(D) only
Official Solution
Correct Option: (3)
If an electron is removed from the anti-bonding orbital, then it will tend to increase the bond order. The HOMO in NO and O2 is antibonding molecular orbital . Hence, in NO and O2 bond order will increase on loss of electron.
So, the correct option is (C).
43
PYQ 2022
easy
chemistryID: jee-main
The pair, in which ions are isoelectronic with is:
1
Br– and Be2+
2
Cl– and Li+
3
S2– and K+
4
O2– and Mg2+
Official Solution
Correct Option: (4)
O2–, Mg2+ and Al3+ are isoelectronic. All have 10 electrons.
So, the correct option is (D): O2– and Mg2+
44
PYQ 2022
easy
chemistryID: jee-main
Number of lone pairs of electrons in the central atom of SCl2, O3, CIF3 and SF6, respectively, are:
1
0, 1, 2 and 2
2
2, 1, 2 and 0
3
1, 2, 2 and 0
4
2, 1, 0 and 2
Official Solution
Correct Option: (2)
To determine the number of lone pairs of electrons on the central atom for each of the given molecules, we will evaluate the electronic configuration and geometry of each molecule.
SCl2:
Sulfur (S) belongs to Group 16 of the periodic table and has 6 valence electrons.
Each chlorine (Cl) atom, being a Group 17 element, forms a single bond with sulfur, using up 2 valence electrons from sulfur (1 for each Cl).
After forming two bonds with chlorine, sulfur has 4 valence electrons remaining, which are organized into 2 lone pairs.
The number of lone pairs on the central sulfur atom in SCl2 is 2.
O3 (Ozone):
Ozone has a resonance structure consisting of one double bond and one single bond between oxygen atoms.
The central oxygen atom participates in both of these bonds.
Oxygen, having 6 valence electrons, after forming a double bond (using 2 electrons) and a single bond (using 1 electron), will have 3 electrons left, forming 1.5 electron pairs.
Typically, central oxygen can be pictured with one lone pair in most stable resonance forms.
The number of lone pairs on the central oxygen atom in O3 is 1.
ClF3:
Chlorine (Cl) is the central atom with 7 valence electrons.
Three fluorine atoms each form a single bond with chlorine, using 3 electrons.
This leaves chlorine with 4 electrons, or 2 lone pairs.
The number of lone pairs on the central chlorine atom in ClF3 is 2.
SF6:
Sulfur is the central atom with 6 valence electrons.
There are six fluorine atoms, each forming a single bond with sulfur, using up all 6 sulfur valence electrons, with no remaining electrons for lone pairs.
The number of lone pairs on the central sulfur atom in SF6 is 0.
Thus, the number of lone pairs of electrons on the central atom for SCl2, O3, ClF3, and SF6 respectively are 2, 1, 2, and 0. The correct option is: 2, 1, 2 and 0.
45
PYQ 2022
medium
chemistryID: jee-main
Two isomers ‘A’ and ‘B’ with molecular formula C4H8 give different products on oxidation with KMnO4/H+ results in effervescence of a gas and gives ketone. The compound ‘A’ is
1
But-1-ene
2
cis-But-2-ene
3
trans-But-2-ene
4
2-methyl propene
Official Solution
Correct Option: (4)
The correct option is(D): 2-methyl propene
46
PYQ 2022
medium
chemistryID: jee-main
In a linear tetrapeptide (constituted with different amino acids) – (number of peptide bonds) is _____.
Official Solution
Correct Option: (1)
A tetrapeptide consists of four amino acids linked in a chain. Each linkage between two amino acids is a peptide bond. Thus, in a tetrapeptide, there are three peptide bonds, as each bond links two amino acids together:
Amino Acid 1 — Peptide Bond 1 — Amino Acid 2 — Peptide Bond 2 — Amino Acid 3 — Peptide Bond 3 — Amino Acid 4.
Therefore, the correct and verified number of peptide bonds in a tetrapeptide is 3.
47
PYQ 2022
medium
chemistryID: jee-main
The total number of bonds in is ___.
1
4
2
5
3
6
4
3
Official Solution
Correct Option: (3)
To determine the total number of bonds in , we must first understand the structure of the compound.
is known as manganese(VII) oxide. Its structure is characterized by two tetrahedral units connected by an oxygen bridge. In this compound, each manganese (Mn) atom is bonded to four oxygen atoms. Out of these, one oxygen atom serves as a bridge between the two Mn atoms, and the other three are terminal oxygens bonded to manganese, forming double bonds.
The structure can be described as:
Each atom forms two double bonds ( ) with terminal oxygen atoms.
There is a bridging oxygen, which is shared by the two units.
For clarity, consider the structure:
,
,
One bridged between and
From this structure, we can observe that there are exactly three double bonds for each manganese atom:
Three terminal oxygens each form a double bond with .
Three terminal oxygens each form a double bond with .
Thus, the total number of bonds in is 6.
Therefore, the correct answer to the question is 6.
48
PYQ 2022
medium
chemistryID: jee-main
Consider the following sulphur based oxoacids. and Amongst these oxoacids, the number of those with peroxo (O–O) bonds is_____.
Official Solution
Correct Option: (1)
To determine which of the given sulphur-based oxoacids contain peroxo (O–O) bonds, we must analyze their structural formulas:
(Sulfurous acid): This compound has the structure where sulfur is bonded to three oxygen atoms with single and double bonds, but no O–O bond is present.
(Sulfuric acid): Sulfur forms double bonds with two oxygen atoms and single bonds with two hydroxyl groups. There is no O–O bond.
(Peroxodisulfuric acid or Marshall's acid): This compound contains an O–O single bond. Its structural formula includes a direct O–O linkage, characteristic of peroxo bonds.
(Pyrosulfuric acid): Also known as oleum, it does not contain an O–O bond; instead, sulfur atoms are linked through an oxygen atom but not forming any O–O linkage.
From the analysis, only has a peroxo bond.
Thus, the number of oxoacids with peroxo bonds is 1. This matches the given range of (1,1).
49
PYQ 2022
easy
chemistryID: jee-main
Consider the ions/molecules For increasing bond order the correct option is
Official Solution
Correct Option: (1)
50
PYQ 2022
easy
chemistryID: jee-main
Given below are two statements. Statement I: The compound (A) is optically active. Statement II: is a mirror image of the above compound A. In the light of the above statement, choose the most appropriate answer from the options given below.
Official Solution
Correct Option: (1)
51
PYQ 2022
easy
chemistryID: jee-main
Consider the species
Choose the correct option with respect to these species.
1
They are isoelectronic and only two have tetrahedral structures
2
They are isoelectronic and all have tetrahedral structures.
3
Only two are isoelectronic and all have tetrahedral structures.
4
Only two are isoelectronic and only two have tetrahedral structures.
Official Solution
Correct Option: (2)
The correct answer is (B) : They are isoelectronic and all have tetrahedral structures.
Fig. Structures
are isoelectronic as well as tetrahedral.
52
PYQ 2022
easy
chemistryID: jee-main
Number of lone pair(s) of electrons on central atom and the shape of BrF3 molecule respectively, are
1
0, triangular planar
2
1, pyramidal
3
2, bent T-shape
4
1, bent T-shape
Official Solution
Correct Option: (3)
The correct answer is (B) : 2, bent T-shape
Fig,
Steric no. = 5 (sp3 d), lone pair = 2 Bent T shape
53
PYQ 2022
medium
chemistryID: jee-main
In the structure of SF4, the lone pair of electrons on S is in.
1
Equatorial position and there are two lone pair – bond pair repulsions at 90º
2
Equatorial position and there are three lone pair – bond pair repulsions at 90º
3
Axial position and there are three lone pair – bond pair repulsion at 90º
4
Axial position and there are two lone pair – bond pair repulsion at 90º
Official Solution
Correct Option: (1)
The correct answer is (A) : Equatorial position and there are two lone pair – bond pair repulsions at 90º SF4 → sp3d hybridisation.
Fig.
The lone pair of electrons on S is in equatorial position and there are two lone pair-bond pair repulsions at 90°.
54
PYQ 2022
medium
chemistryID: jee-main
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R
Assertion A: Zero orbital overlap is an out of phase overlap.
Reason R: It results due to different orientation/direction of approach of orbitals.
In the light of the above statements, choose the correct answer from the options given below
1
Both A and R are true and R is the correct explanation of A
2
Both A and R are true but R is NOT the correct explanation of A
3
A is true but R is false
4
A is false but R is true
Official Solution
Correct Option: (1)
Zero overlapping is something in which there is no overlapping between two orbitals. The first condition is that the two orbitals should not be symmetrical and the second condition is that both orbitals should be in different planes.
55
PYQ 2022
medium
chemistryID: jee-main
In the given reaction,
Fig.
The number of π electrons present in the product ‘P’ is_____.
Official Solution
Correct Option: (1)
The correct answer is 4
Fig.
Total 4π electrons are there. Reaction is aldol condensation.
56
PYQ 2023
medium
chemistryID: jee-main
Find correct order of covalent character (A) (B) (C)
1
A & B only
2
A & C only
3
A, B & C
4
B & C only
Official Solution
Correct Option: (3)
In an ionic compound, the ability of a cation to attract the electrons of an anion is called polarizing power. Due to the polarizability of anion, it gets distorted & sharing of electrons takes place and ionic compounds possess some covalent character.
Greater the polarization, the greater the covalent character.
The Correct Option is (C): A, B & C
57
PYQ 2023
medium
chemistryID: jee-main
Match List I with List II.
List-I
List-II
A
Sulphate
I
Pesticide
B
Fluoride
II
Bending of bones
C
Nicotine
III
Laxative effect
D
Sodium arsinite
IV
Herbicide
Choose the correct answer from the options given below:
1
A-II, B-III. C-IV, D-I
2
A-IV, B-III, C-II, D-I
3
A-III, B-II, C-I, D-IV
4
A-III, B-II, C-IV, D-I
Official Solution
Correct Option: (3)
A-Sulphate - III (Laxative effect)
B-Fluoride - II (Bending of bones)
C-Nictoine - I (pesticides)
D-Sodium Arsinite - IV (herbicide)
58
PYQ 2023
medium
chemistryID: jee-main
Match List I with List II :
List-I (molecule)
List-II (hybridization; shape)
A
XeO3
i
sp3d ; linear
B
XeF2
ii
sp3; pyramidal
C
XeOF4
iii
sp3d3; distorted octahedral
D
XeF6
iv
sp3d2 ;square pyramidal
Choose the correct answer from the options given below:
1
A-II, B-I, C-IV, D-III
2
A-II, B-IV, C-III, D-I
3
A-IV, B-II, C-III, D-I
4
A-IV, B-II, C-I, D-III
Official Solution
Correct Option: (1)
59
PYQ 2023
hard
chemistryID: jee-main
Number of unpaired electron in highest occupied molecular orbital of following species is : ?
1
0, 1, 2, 1
2
2, 1, 2, 1
3
0, 1, 0, 1
4
2, 1, 0, 1
Official Solution
Correct Option: (1)
Molecular Orbital Configurations: : The molecular orbital configuration is:
Here, the highest occupied molecular orbital (HOMO) is with no unpaired electrons.
: The configuration is:
The HOMO is with 1 unpaired electron.
: The configuration is:
The HOMO is with 2 unpaired electrons.
: The configuration is:
The HOMO is with 1 unpaired electron.
Thus, the number of unpaired electrons for , , , and are {0, 1, 2, 1} respectively.
60
PYQ 2023
hard
chemistryID: jee-main
Statement I : Dipole moment is a vector quantity and by convention it is depicted by a small arrow with tail on the negative centre and head pointing towards the positive centre Statement II : The crossed arrow of the dipole moment symbolizes the direction of the shift of charges in the molecules. In the light of the above statements, choose the most appropriate answer from the options given below:
1
Both Statement I and Statement II are correct
2
Statement I is correct but Statement II is incorrect
3
Statement I is incorrect but Statement II is correct
4
Both Statement I and Statement II are incorrect
Official Solution
Correct Option: (2)
Statement I is correct. In chemistry, the dipole moment is represented by an arrow with a positive sign ( ) at the tail and a negative sign ( ) at the head. The arrow points from the positive charge to the negative charge, indicating the direction of the dipole moment, which is important for understanding the polarity of molecules. This representation is conventional and widely accepted in molecular chemistry.
Statement II is incorrect. The crossed arrow does not represent the shift of charges in a molecule. Rather, the crossed arrow in dipole moment notation indicates the direction of the resultant dipole moment in the molecule. It shows the vector from the positive charge to the negative charge, not the movement of the charges themselves. Therefore, this statement misinterprets the meaning of the crossed arrow in dipole moment representation.
61
PYQ 2023
medium
chemistryID: jee-main
Compounds of Xenon have one electron pair on central atom :
1
XeO3
2
XeOF2
3
XeF4
4
XeF5-
Official Solution
Correct Option: (1)
Analysis of Lone Pairs on Xenon (Xe):
Using VSEPR theory, the number of lone pairs on the central atom Xenon (Xe) is calculated for each species:
XeF5+: Xenon has 8 valence electrons. It forms 5 bonds with fluorine, and 1 electron is removed due to the positive charge. Lone pairs = (not a single lone pair).
XeO3: Xenon forms 3 bonds with oxygen atoms. Lone pairs = (not a single lone pair).
XeO2F2: Xenon forms 4 bonds (2 with oxygen and 2 with fluorine). Lone pairs = (single lone pair).
XeF5−: Xenon forms 5 bonds with fluorine atoms. One extra electron is added due to the negative charge. Lone pairs = (not a single lone pair).
XeO3F2: Xenon forms 5 bonds (3 with oxygen and 2 with fluorine). Lone pairs = (single lone pair).
XeOF4: Xenon forms 5 bonds (1 with oxygen and 4 with fluorine). Lone pairs = (single lone pair).
XeF4: Xenon forms 4 bonds with fluorine atoms. Lone pairs = (not a single lone pair).
Species with a single lone pair: XeO2F2, XeO3F2, XeOF4, XeF5.
Total number of such species: 4.
62
PYQ 2023
medium
chemistryID: jee-main
How many factors will contribute to a major role in covalent character of a compound.
Polarising power of cation
Polarisability of the anion
Distortion caused by cation
Polarisability of cation
1
1 and 2
2
2 and 3
3
1, 2, and 3
4
4
Official Solution
Correct Option: (3)
All of the factors you listed can contribute to the covalent character of a compound. The extent to which each factor contributes depends on the specific compound and its chemical properties. The polarising power of a cation refers to its ability to pull electron density towards itself, which can result in partial covalent bonding between the cation and the anion. Similarly, the polarisability of an anion refers to its ability to be distorted by an oppositely charged ion, which can lead to partial covalent bonding. The distortion caused by a cation also contributes to covalent character, as it can lead to the formation of partial covalent bonds between the cation and anion. Finally, the polarisability of a cation can also play a role in covalent character, as a highly polarisable cation can form partial covalent bonds with the anion due to the distortion it causes in the anion. Therefore, all of these factors can contribute to the covalent character of a compound, and their relative importance depends on the specific compound and its chemical properties. Answer. A, B, C
63
PYQ 2023
easy
chemistryID: jee-main
Order of Covalent bond: A. B. C. D. E. Choose the correct answer from the options given below:
1
A, B only
2
B, C only
3
B, C, E only
4
C, E only
Official Solution
Correct Option: (3)
The order of covalent character can be predicted using Fajans' Rule, which states:
1. Covalent character increases with smaller cation size and larger anion size.
2. Covalent character increases with higher charge on the cation. Now, analyze each statement: A: KF :
False. KF has less covalent character than KI because the larger iodide ion polarizes the cation more effectively. B: KF :
True. KI is more covalent than KF, and LiF is less ionic than KF due to the smaller size of . C: SnCl :
True. SnCl is more covalent than SnCl because Sn is more polarizing. CuCl is more covalent than NaCl because is smaller and has higher polarizing power than . D: :
False. CuCl is more covalent than NaCl. E: KF :
True. KF is less covalent than KI, and CuCl is more covalent than NaCl. Thus, the correct choices are B, C, and E.
64
PYQ 2023
medium
chemistryID: jee-main
Match the Column:
Column I
Column II
(A).
ClO2-
(1).
Linear
(B).
N3-
(2).
Tetrahedral
(C).
NH4+
(3).
Bent
(D).
SF4
(4).
See-Saw
Official Solution
Correct Option: (1)
The correct answer is: A-3 , B-1, C-2, D-4
65
PYQ 2023
medium
chemistryID: jee-main
NH3, NO, N2, F2, CO, CO2, H2O, and XeF4. Fill the number of above molecules having only two lone pair of electrons.
Official Solution
Correct Option: (1)
The Correct answer is 3 (N2, CO and H2O).
66
PYQ 2023
easy
chemistryID: jee-main
Which of the following have square pyramidal structure?
1
XeOF4
2
BrF3
3
XeF4
4
XeO3
Official Solution
Correct Option: (1)
XeOF₄ and its Structure:
Chemical Formula: XeOF₄ (Xenon oxytetrafluoride).
Central Atom: Xenon (Xe).
Electron Pair Arrangement:
Xenon has 8 valence electrons.
4 fluorine atoms and 1 oxygen atom are bonded to Xenon.
1 lone pair remains on Xenon after bonding.
Molecular Geometry: Square pyramidal (AX5E type) due to the presence of:
5 bonded groups (4 fluorine atoms + 1 oxygen atom).
1 lone pair, which distorts the structure into a square pyramidal shape.
Key Observations:
Due to the lone pair of electrons on Xe, the molecule adopts a square pyramidal geometry. This geometry is confirmed by VSEPR (Valence Shell Electron Pair Repulsion) theory, where the lone pair repels the bonded atoms, causing this specific shape.
Conclusion:
The correct answer is: XeOF₄ has a square pyramidal structure.
67
PYQ 2023
easy
chemistryID: jee-main
Which of the following has maximum number of l.p. at central atom?
1
ClO3-
2
SF4
3
XeF4
4
I3-
Official Solution
Correct Option: (4)
From the structure of the given species, It can be clearly seen that has the maximum number of lone pairs at central atom.
So, the correct answer is option (D):
68
PYQ 2023
medium
chemistryID: jee-main
What is the correct order of acidity of the protons marked A-D in the given compounds?
1
2
HB > HA > HD > HC
3
HA > HB > HC > HD
4
Official Solution
Correct Option: (1)
The acidity of a proton depends on the stability of the conjugate base formed after its removal. The more stable the conjugate base, the more acidic the proton. Step 1: Acidity of is the most acidic proton. Removal of results in a carboxylate anion, which is highly stabilized by resonance.\ Conclusion: Option is correct.
69
PYQ 2023
medium
chemistryID: jee-main
Match list I with list II:
List I
Species
List II
Geometry/ Shape
A
H3 O+
I
Tetrahedral
B
Acetylide anion
II
Linera
C
NH4+
III
Pyramidal
D
ClO2–
IV
Bent
Choose correct answer from the options given below:
1
A-I, B-II, C-III, D-IV
2
A-IV, B-I, C-III, D-II
3
A-II, B-IV, C-I, D-III
4
A-III, B-II, C-IV, D-I
Official Solution
Correct Option: (1)
- A. H O: Water has a bent shape due to the lone pairs on oxygen, making the geometry bent. - B. Acetylene: Acetylene is a linear molecule with a triple bond between the carbons. - C. NH : Ammonia has a trigonal pyramidal shape due to the lone pair on nitrogen. - D. ClO : Chlorine dioxide is a bent molecule due to the lone pairs on chlorine.
Thus, the correct matching is A-I, B-II, C-III, D-IV.
70
PYQ 2023
hard
chemistryID: jee-main
The geometry around boron in the product 'B' formed from the following reaction is
1
trigonal planar
2
tetrahedral
3
pyramidal
4
square planar
Official Solution
Correct Option: (2)
Let's analyze the given reaction to determine the geometry around the boron atom in the product 'B'.
The initial reaction provided is: .
Boron trifluoride ( ) is initially a trigonal planar molecule with boron making three bonds with fluorine.
In the presence of sodium hydride ( ), a hydride ion reacts with , resulting in the formation of a compound 'A' (likely ) and sodium fluoride ( ).
The compound 'A' then reacts with : .
is trimethylamine, a Lewis base, which donates an electron pair to form a dative bond with 'A', which is .
The reaction essentially forms .
The complex has a tetrahedral geometry because:
The group allows the formation of three sigma bonds with the fluorine atoms. Additionally, it forms a coordinate covalent bond with the lone pair of trimethylamine.
The dative bond from to results in a total of four bonded pairs around the boron atom.
Owing to the VSEPR (Valence Shell Electron Pair Repulsion) theory, the four pairs of electrons (three from atoms and one from the dative bond) arrange themselves around the central boron to minimize repulsions, resulting in a tetrahedral geometry.
Based on the above reasoning, the geometry around boron in the product 'B' ( ) is tetrahedral.
Thus, the correct answer is tetrahedral.
Let's rule out the other options:
Trigonal planar: Applies to not after forming a bond with .
Pyramidal: Typically exists in hybridization with a lone pair, which isn't the configuration here.
Square planar: Typically associated with coordination complexes of transition metals, not applicable in this scenario.
71
PYQ 2023
medium
chemistryID: jee-main
For molecule consider the following: [(A)] Number of lone pairs on oxygen is 2. [(B)] F--O--F angle is less than . [(C)] Oxidation state of O is . [(D)] Molecule is bent ‘V’-shaped. [(E)] Molecular geometry is linear. Correct options are:
1
A, B, D only
2
A, C, D only
3
C, D, E only
4
B, E, A only
Official Solution
Correct Option: (1)
The Lewis structure of shows that the central oxygen atom is bonded to two fluorine atoms and has two lone pairs of electrons. This gives the molecule a bent or ‘V’ shape due to lone pair-bond pair repulsion. Step 2: Analysis of the Statements Statement (A): The oxygen atom in has 6 valence electrons. After forming two single bonds with fluorine atoms, it retains two lone pairs. Hence, this statement is true Statement (B): The F--O--F bond angle is slightly less than (around ) due to increased lone pair-bond pair repulsion. Hence, this statement is true Statement (C): The oxidation state of oxygen in is , as fluorine is more electronegative. Hence, this statement is false Statement (D): Due to the presence of two lone pairs, the molecular geometry is bent or ‘V’ shaped. Hence, this statement is true Statement (E) The molecular geometry is not linear; it is bent. Hence, this statement is false} Conclusion: The correct statements are A, B, D. Therefore, the correct answer is (4) A, B, D only
72
PYQ 2023
medium
chemistryID: jee-main
The maximum number of lone pairs of electron on the central atom from the following species is ___________ ClO3– , XeF4, SF4 and I3–
Official Solution
Correct Option: (1)
Analysis of Each Species:
(Chlorate ion):
Chlorine ( ) is the central atom with 7 valence electrons.
Each oxygen ( ) contributes 6 valence electrons, and the negative charge adds 1 more electron.
Total valence electrons: .
Lewis structure: The central atom has 1 lone pair.
(Xenon tetrafluoride):
Xenon ( ) is the central atom with 8 valence electrons.
Each fluorine ( ) contributes 7 valence electrons.
Total valence electrons: .
Lewis structure: The central atom has 2 lone pairs.
(Sulfur tetrafluoride):
Sulfur ( ) is the central atom with 6 valence electrons.
Each fluorine ( ) contributes 7 valence electrons.
Total valence electrons: .
Lewis structure: The central atom has 1 lone pair.
(Triiodide ion):
Iodine ( ) is the central atom with 7 valence electrons.
Each other iodine atom contributes 7 valence electrons, and the negative charge adds 1 more electron.
Total valence electrons: .
Lewis structure: The central atom has 3 lone pairs.
Final Answer:
The maximum number of lone pairs on the central atom is 3 (in ).
73
PYQ 2023
medium
chemistryID: jee-main
30.4 kJ is required to melt one mole of sodium chloride. The entropy change during melting is 28.4 JK-1 mol-1 . atm The melting point of sodium chloride Calculate is ___ K(Nearest Integer).
Official Solution
Correct Option: (1)
Melting Point Calculation
The relationship between enthalpy (ΔH) and entropy (ΔS) during a phase change is:
ΔS = ΔH / T
Rearranging to find the melting point (Tmp):
Tmp = ΔH / ΔS
Substituting the given values:
ΔH = 30.4 kJ mol-1 = 30400 J mol-1
ΔS = 28.4 J K-1 mol-1
Tmp = 30400 / 28.4 = 1070.42 K
The nearest integer value is 1070 K.
74
PYQ 2024
medium
chemistryID: jee-main
Arrange the bonds in order of increasing ionic character in the molecules. LiF, K O, N , SO , and ClF .
1
ClF N SO K O LiF
2
LiF K O ClF SO N
3
N ClF SO K O LiF
4
N SO ClF K O LiF
Official Solution
Correct Option: (3)
The problem asks to arrange the chemical bonds in the given molecules—LiF, K₂O, N₂, SO₂, and ClF₃—in order of increasing ionic character.
Concept Used:
The ionic character of a chemical bond is a measure of its polarity. It depends on the difference in electronegativity ( ) between the two atoms forming the bond. According to the Pauling scale, a larger difference in electronegativity corresponds to a greater charge separation and thus a higher degree of ionic character.
The relationship can be summarized as:
A bond between identical atoms ( ) is purely covalent, while a bond with a large (typically > 1.7) is considered predominantly ionic.
Step-by-Step Solution:
Step 1: Identify the specific bonds within each molecule that need to be compared.
In LiF, the bond is Li–F.
In K₂O, the bond is K–O.
In N₂, the bond is N≡N.
In SO₂, the bond is S=O.
In ClF₃, the bond is Cl–F.
Step 2: List the Pauling electronegativity (EN) values for each of the atoms involved.
Nitrogen (N): 3.04
Sulfur (S): 2.58
Oxygen (O): 3.44
Chlorine (Cl): 3.16
Fluorine (F): 3.98
Lithium (Li): 0.98
Potassium (K): 0.82
Step 3: Calculate the electronegativity difference ( ) for each bond.
For N₂ (N≡N bond):
This is a purely covalent bond.
For ClF₃ (Cl–F bond):
This is a polar covalent bond.
For SO₂ (S=O bond):
This is also a polar covalent bond, slightly more polar than the Cl–F bond.
For K₂O (K–O bond):
This is a predominantly ionic bond.
For LiF (Li–F bond):
This is a highly ionic bond.
Final Computation & Result:
Step 4: Arrange the bonds in order of increasing , which corresponds to increasing ionic character.
The calculated values in increasing order are:
Therefore, the order of increasing ionic character for the bonds in the given molecules is:
N₂ < ClF₃ < SO₂ < K₂O < LiF
75
PYQ 2024
medium
chemistryID: jee-main
Which of the following is least ionic ?
1
2
AgCl
3
KCl
4
Official Solution
Correct Option: (2)
To determine which compound is the least ionic among the given options, we need to understand the concept of ionic character. The ionic character in a compound largely depends on the difference in electronegativity between the cation and the anion and the size of the ions comprising the compound.
The options provided are:
AgCl
KCl
Let's analyze each option considering the factors that affect ionic character:
: Barium (Ba) is a Group 2 element, and Chlorine (Cl) is a halogen. The high electronegativity difference results in a strong ionic bond.
AgCl: Silver chloride has lower ionic character due to the relatively smaller electronegativity difference between silver (Ag) and chlorine (Cl). Also, silver ions tend to form covalent bonds due to their filled d-orbitals, which reduce the overall ionic character.
KCl: Potassium (K) is an alkali metal, and Chlorine (Cl) is a halogen, resulting in a large electronegativity difference and thus, a strong ionic character.
: Cobalt (Co), a transition metal, forms ionic compounds with halogens. However, due to partial covalent character from its d-orbitals, it might be less ionic than simple alkali or alkaline earth metals, but more ionic than AgCl.
Based on the above analysis, AgCl is the least ionic compound among the given options due to the reasons mentioned, such as closer electronegativity and covalent character contributed by silver's orbital configurations.
76
PYQ 2024
medium
chemistryID: jee-main
The order of relative stability of the contributing structure is: Choose the correct answer from the options given below:
1
2
3
4
Official Solution
Correct Option: (1)
The question involves evaluating the relative stability of resonance structures. The stability of resonance structures mainly depends on the following factors:
Octet Rule: Structures in which all atoms, especially the second-period elements (like C, N, O), have a complete octet are generally more stable.
Charge Separation: Structures with minimal charge separation are more stable. A neutral structure is more stable than a charged one.
Negative Charge on Electronegative Atoms: Structures that place negative charges on more electronegative atoms (like oxygen) are more stable.
Positive Charge on Electropositive Atoms: Structures with positive charges on less electronegative atoms are more stable.
Analyzing the given resonance structures:
Structure I: This structure has all atoms with complete octet shells and no charge separation, which makes it the most stable.
Structure II: This structure has a positive charge on a carbon atom and a negative charge on an oxygen atom. There is charge separation, reducing its stability compared to structure I, but the charges are on appropriate atoms considering electronegativity.
Structure III: This structure has carbon with a negative charge and oxygen with a positive charge, which is the least favorable distribution of charges among the given structures. This increases its instability compared to the other structures.
Based on the above analysis, the order of relative stability is:
77
PYQ 2024
hard
chemistryID: jee-main
Official Solution
Correct Option: (1)
To find the total number of ions with a noble gas configuration from the given ions, we must first understand what a noble gas configuration entails. Noble gases have full electron shells, resulting in a stable electronic arrangement. To determine if an ion has a noble gas configuration, consider the ion's total electrons and compare it with the nearest noble gas.
Sr2+ (Z = 38): The electron configuration for Sr is [Kr] 5s2. Sr2+ loses 2 electrons, resulting in [Kr], which is the noble gas krypton.
Cs+ (Z = 55): The electron configuration for Cs is [Xe] 6s1. Cs+ loses 1 electron, resulting in [Xe], the noble gas xenon.
La2+ (Z = 57): The electron configuration for La is [Xe] 5d1 6s2. La2+ loses 2 electrons, resulting in [Xe] 5d1, which is not a noble gas configuration.
Pb2+ (Z = 82): The electron configuration for Pb is [Xe] 4f14 5d10 6s2 6p2. Pb2+ loses 2 electrons, becoming [Xe] 4f14 5d10 6s2, not a noble gas configuration.
Yb2+ (Z = 70): The electron configuration for Yb is [Xe] 4f14 6s2. Yb2+ loses 2 electrons, resulting in [Xe] 4f14, not a noble gas configuration.
Fe2+ (Z = 26): The electron configuration for Fe is [Ar] 3d6 4s2. Fe2+ loses 2 electrons, resulting in [Ar] 3d6, not a noble gas configuration.
Upon analyzing, Sr2+ and Cs+ achieve noble gas configurations. Therefore, the total number of ions with noble gas configurations is 2.
78
PYQ 2024
easy
chemistryID: jee-main
Official Solution
Correct Option: (1)
The problem requires identifying non-polar molecules from a given list. A molecule is non-polar when its net dipole moment is zero, often due to a symmetrical arrangement of polar bonds that cancel each other out.
HF: Polar due to significant electronegativity difference between H and F.
H2O: Polar because of its bent shape, causing a net dipole moment.
SO2: Polar, also due to its bent shape.
H2: Non-polar, as identical atoms share electrons equally.
CO2: Non-polar; although it has polar bonds, the linear geometry cancels the dipoles.
CH4: Non-polar; tetrahedral symmetry allows dipoles to cancel.
NH3: Polar due to its trigonal pyramidal shape.
HCl: Polar due to electronegative difference.
CHCl3: Polar; asymmetric molecule leads to a net dipole.
BF3: Non-polar; trigonal planar symmetry results in dipole cancellation.
Counting non-polar molecules, we have H2, CO2, CH4, and BF3. Hence, there are 4 non-polar molecules.
79
PYQ 2024
medium
chemistryID: jee-main
Official Solution
Correct Option: (1)
To determine the number of molecules with zero dipole moment among the given set (CH4, BF3, H2O, HF, NH3, CO2, SO2), we analyze each molecule's structure and geometry.
CH4: Tetrahedral geometry with symmetrical distribution of charge. Resultantly, it has a zero dipole moment.
BF3: Trigonal planar geometry, symmetry in the molecule cancels out dipoles, resulting in a zero dipole moment.
H2O: Bent geometry with a dipole moment due to the lone pairs on oxygen causing an asymmetric charge distribution.
HF: Linear molecule, but the difference in electronegativity induces a dipole moment.
NH3: Trigonal pyramidal geometry with a net dipole moment due to the lone pair on nitrogen.
CO2: Linear geometry; symmetrical opposite dipoles cancel out, resulting in a zero dipole moment.
SO2: Bent geometry with a net dipole moment due to the asymmetric distribution of electrons.
Upon reviewing, CH4, BF3, and CO2 have zero dipole moments. Therefore, the total number is 3.
80
PYQ 2024
easy
chemistryID: jee-main
Sum of bond order of CO and NO+ is _____.
Official Solution
Correct Option: (1)
To find the sum of the bond orders of CO and NO+, we start by determining each bond order individually.
Bond Order of CO:
The bond order of a molecule is calculated using the molecular orbital theory. For CO, it has 10 valence electrons.
The electronic configuration in terms of molecular orbitals is: (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(π2p)4(σ2p)2.
Bond order = (Number of electrons in bonding orbitals - Number of electrons in antibonding orbitals) / 2.
Bond order of CO: (8-2)/2 = 3.
Bond Order of NO+:
NO+ has 10 valence electrons since it is derived from NO by losing an electron.
Its electronic configuration is similar to that of CO.
Therefore, its bond order calculation is the same as CO.
Bond order of NO+: (8-2)/2 = 3.
Sum of Bond Orders:
Sum = Bond order of CO + Bond order of NO+ = 3 + 3 = 6.
Verification:
The calculated sum is 6. This fits precisely within the provided range of 6 to 6.
Thus, the sum of the bond orders of CO and NO+ is 6.
81
PYQ 2024
medium
chemistryID: jee-main
Given below are two statements : Statement (I) : Both metal and non-metal exist in p and d-block elements. Statement (II) : Non-metals have higher ionisation enthalpy and higher electronegativity than the metals. In the light of the above statements, choose the most appropriate answer from the option given below:
1
Both Statement I and Statement II are false
2
Statement I is false but Statement II is true
3
Statement I is true but Statement II is false
4
Both Statement I and Statement II are true
Official Solution
Correct Option: (2)
To evaluate the given statements, let's analyze each one and verify its accuracy based on chemical concepts.
Statement (I): Both metal and non-metal exist in p and d-block elements.
The periodic table contains different blocks that categorize elements based on their outer electron configurations.
p-block elements: This block includes both metals (like Aluminium, Tin) and non-metals (like Oxygen, Nitrogen), as well as metalloids (like Silicon, Germanium).
d-block elements: These are transition metals, such as Iron, Copper, and Zinc. Non-metals are generally not found in the d-block.
Based on the above analysis, Statement (I) is incorrect because non-metals are not present in the d-block of the periodic table.
Statement (II): Non-metals have higher ionisation enthalpy and higher electronegativity than the metals.
Ionisation Enthalpy: This refers to the energy required to remove an electron from an atom. Non-metals, which tend to gain electrons to achieve a noble gas configuration, typically have higher ionisation enthalpies than metals.
Electronegativity: This is the tendency of an atom to attract electrons. Non-metals, especially elements like Fluorine and Oxygen, have higher electronegativities compared to metals, which are more electropositive.
Hence, Statement (II) is true, as it correctly describes the general trend for non-metals compared to metals regarding ionisation enthalpy and electronegativity.
Based on the above analysis, the most appropriate answer is: Statement I is false but Statement II is true.
82
PYQ 2024
medium
chemistryID: jee-main
KMnO4 decomposes on heating at 513 K to form O2 along with
1
2
3
4
Official Solution
Correct Option: (2)
To solve the problem of determining what substances are formed when potassium permanganate ( ) decomposes upon heating at 513 K, we need to understand the chemical decomposition reaction involved.
When is heated, it decomposes into potassium manganate ( ), manganese dioxide ( ), and oxygen gas ( ). The balanced chemical reaction for this decomposition is given by:
Let's analyze the options:
: This option mentions the manganate ion but lacks potassium.
: This matches the known decomposition reaction products.
: This does not involve any formation or release of oxygen, and is the reactant itself.
: Water ( ) is not a product of this decomposition reaction.
The only option that correctly represents the components formed by the decomposition of is:
.
Thus, the correct answer is:
83
PYQ 2024
medium
chemistryID: jee-main
The interaction between π bond and lone pair of electrons present on an adjacent atom is responsible for
1
Resonance effect
2
Hyperconjugation
3
Inductive effect
4
Electromeric effect
Official Solution
Correct Option: (1)
The question is about identifying the interaction responsible for the resonance effect among the options given. Let's break this down:
Resonance Effect: This effect occurs when there is a delocalization of π electrons across adjacent atoms in a molecule, facilitated by the overlap of p-orbitals. This effect is often caused by an interaction between a π bond and a lone pair of electrons on an adjacent atom. This leads to the stabilization of the structure by the distribution of electron density over multiple atoms.
Hyperconjugation: This is the interaction of σ-bonds (usually C-H or C-C) with adjacent empty or partially filled p or π orbitals to give an extended molecular orbital that increases stability. This does not involve lone pairs and π bonds directly interacting.
Inductive Effect: This effect involves the transmission of charge through a chain of atoms in a molecule by electrostatic induction. It relates to the σ-bond framework and not π bonds or lone pairs.
Electromeric Effect: This temporary effect occurs in the presence of an attacking reagent, affecting the π bonds but is not the same as the resonance effect. It refers to the temporary transfer of electrons from a double bond or a multiple bond when a reagent attacks.
Given the explanation of each effect, the interaction between a π bond and a lone pair of electrons on an adjacent atom is responsible for the Resonance Effect. Hence, the correct answer is the Resonance Effect.
84
PYQ 2024
medium
chemistryID: jee-main
Number of compounds with one lone pair of electrons on central atom amongst following is ,
Official Solution
Correct Option: (1)
1. CIF₃ (Chlorine trifluoride)
Chlorine (Cl) has 7 valence electrons. It forms 3 bonds with F atoms → uses 3 electrons. Remaining = 7 – 3 = 4 electrons = 2 lone pairs. → **2 lone pairs on Cl**
2. XeO₃ (Xenon trioxide)
Xenon (Xe) has 8 valence electrons. Each oxygen forms a double bond, using 6 electrons in total (3 × 2). Used 6 electrons → 8 – 6 = 2 electrons = 1 lone pair. → **1 lone pair on Xe**
3. BrF₅ (Bromine pentafluoride)
Bromine (Br) has 7 valence electrons. Forms 5 bonds with F → uses 5 electrons. Remaining 7 – 5 = 2 electrons = 1 lone pair. → **1 lone pair on Br**
4. XeF₄ (Xenon tetrafluoride)
Xenon (Xe) has 8 valence electrons. Forms 4 bonds with F → uses 4 electrons. Remaining 8 – 4 = 4 electrons = 2 lone pairs. → **2 lone pairs on Xe**
5. O₃ (Ozone)
Central O has 6 valence electrons. It forms one single and one double bond (resonance). Bonds use 3 pairs = 6 electrons → 6 – 4 = 2 electrons = 1 lone pair. → **1 lone pair on central O**
6. NH₃ (Ammonia)
Nitrogen has 5 valence electrons. Forms 3 bonds with H → uses 3 electrons. Remaining 2 electrons = 1 lone pair. → **1 lone pair on N**
7. H₂O (Water)
Oxygen has 6 valence electrons. Forms 2 bonds with H → uses 2 electrons. Remaining 4 electrons = 2 lone pairs. → **2 lone pairs on O**
Step 5: Count of Compounds with One Lone Pair
Compounds with exactly one lone pair on central atom: Total = **4**
Final Answer:
85
PYQ 2024
medium
chemistryID: jee-main
Which one of the following molecules has maximum dipole moment ?
1
2
3
4
Official Solution
Correct Option: (3)
Understanding Dipole Moment:
The dipole moment is a measure of the separation of positive and negative charges in a molecule. Molecules with polar bonds and an asymmetrical shape often exhibit a dipole moment. The greater the electronegativity difference between atoms and the asymmetry in the molecule, the higher the dipole moment.
Analyze Each Molecule:
: Although nitrogen and fluorine have a large electronegativity difference, the structure of (trigonal pyramidal) leads to a partial cancellation of the dipole moment due to the lone pair on nitrogen. The net dipole moment of is lower than that of .
: Methane ( ) is a nonpolar molecule with a tetrahedral structure. The dipole moments of the C-H bonds cancel each other out, resulting in a net dipole moment of zero.
: Ammonia has a trigonal pyramidal structure with a lone pair of electrons on nitrogen. This creates an asymmetrical distribution of charge and a significant net dipole moment. The lone pair on nitrogen intensifies the dipole moment, making have a higher dipole moment compared to .
: Phosphorus pentafluoride has a trigonal bipyramidal structure, with dipole moments of the axial and equatorial bonds cancelling out. As a result, is a nonpolar molecule with a net dipole moment of zero.
Conclusion:
Among the given molecules, has the maximum dipole moment due to its trigonal pyramidal structure and the presence of a lone pair on nitrogen, which increases its net dipole moment.
86
PYQ 2024
medium
chemistryID: jee-main
What is the structure of Mn2(CO)10?
1
Two square pyramidal units joined by bridging CO ligands
2
Two square pyramidal units joined by Mn-Mn bond
3
Two tetrahedral units joined by Mn-Mn bond
4
Two square planar units joined by Mn-Mn bond
Official Solution
Correct Option: (2)
The correct option is (B) : Two square pyramidal units joined by Mn-Mn bond
87
PYQ 2024
medium
chemistryID: jee-main
The number of species from the following which are paramagnetic and with bond order equal to one is ___.
Species: H2, He2+, O2-, N2, O22-, F2, Ne2+, B2
Official Solution
Correct Option: (1)
We analyze the magnetic behaviour and bond order for each species: Among these, the only species that is both paramagnetic and has a bond order of 1 is: B2 Thus, the correct count is: 1.
88
PYQ 2024
medium
chemistryID: jee-main
The number of species from the following which are paramagnetic and with bond order equal to one is ___. Species: H2, He2+, O2-, N2, O22-, F2, Ne2+, B2
Official Solution
Correct Option: (1)
To determine the number of paramagnetic species with a bond order of one, we analyze each given species: Step 1: Identify Paramagnetic Species
Paramagnetic: Contains unpaired electrons in molecular orbitals.
Non-paramagnetic: All electrons are paired.
Step 2: Calculate Bond Order Bond order is calculated as: (Number of electrons in bonding orbitals-Number of electrons in antibonding orbitals)2
Species
Electron Configuration
Paramagnetic?
Bond Order
H2
(σ1s)²
No
1
He2+
(σ1s)²(σ*1s)¹
Yes
0.5
O2-
(σ2s)²(σ*2s)²(π2p)⁴(π*2p)³
Yes
1.5
N2
(σ2s)²(σ*2s)²(π2p)⁴(σ2p)²
No
3
O22-
(σ2s)²(σ*2s)²(π2p)⁴(π*2p)⁴
No
1
F2
(σ2s)²(σ*2s)²(π2p)⁴(π*2p)⁴(σ2p)²
No
1
Ne2+
(σ2s)²(σ*2s)²(π2p)⁴(π*2p)⁴(σ2p)¹
Yes
0.5
B2
(σ2s)²(σ*2s)²(π2p)¹(π2p)¹
Yes
1
Final Results
Paramagnetic and Bond Order = 1: O22-, F2, B2
The total number of species matching the criteria is: 1. This satisfies the range (1,1).
89
PYQ 2024
easy
chemistryID: jee-main
The number of molecules/ions having trigonal bipyramidal shape is:
Official Solution
Correct Option: (1)
PF5, PCl5, and Fe(CO)5 have trigonal bipyramidal geometry.
BrF5: square pyramidal
[PtCl4]2−: square planar
BF3: trigonal planar
90
PYQ 2024
medium
chemistryID: jee-main
The linear combination of atomic orbitals to form molecular orbitals takes place only when the combining atomic orbitals A. have the same energy B. have the minimum overlap C. have same symmetry about the molecular axis D. have different symmetry about the molecular axis. Choose the most appropriate from the options given below:
1
A, B, C only
2
A and C only
3
B and C only
4
B and D only
Official Solution
Correct Option: (2)
The concept in focus here is about the formation of molecular orbitals through the linear combination of atomic orbitals (LCAO). To ensure the combination occurs effectively, the atomic orbitals must satisfy certain conditions:
Same Energy: Atomic orbitals can combine effectively if they have the same or very similar energies. This allows the wave functions to combine constructively, forming stable molecular orbitals.
Same Symmetry: The combining atomic orbitals must have the same symmetry with respect to the molecular axis. Only orbitals that possess compatible symmetries can combine to form molecular orbitals.
Now, let's analyze each option:
Option A (Same Energy): Correct, as having the same energy is essential for effective overlap and combination.
Option B (Minimum Overlap): Incorrect, as a significant overlap is necessary for the effective combination of orbitals. Minimum overlap would defeat the purpose since it doesn't provide enough interaction space.
Option C (Same Symmetry): Correct, same symmetry is a critical requirement for the combination of atomic orbitals to form molecular orbitals.
Option D (Different Symmetry): Incorrect, as orbits with different symmetries cannot effectively combine.
Given these considerations, the most appropriate options that ensure the combination of atomic orbitals are A and C only. Hence, the correct answer is:
Answer: A and C only
91
PYQ 2024
hard
chemistryID: jee-main
The number of species from the following in which the central atom uses sp3 hybrid orbitals in its bonding is ______.
Official Solution
Correct Option: (1)
Given Molecules and Their Hybridization:
NH₃: Ammonia has sp³ hybridization, as the nitrogen atom forms three sigma bonds with hydrogen atoms and has one lone pair, which leads to a tetrahedral geometry.
SO₂: Sulfur dioxide exhibits sp² hybridization. The sulfur atom forms two sigma bonds with oxygen atoms and has one lone pair, resulting in a bent molecular shape.
SiO₂: Silicon dioxide also exhibits sp³ hybridization. The silicon atom forms four bonds with oxygen atoms, and its structure adopts a tetrahedral geometry.
BeCl₂: Beryllium chloride has sp hybridization. The beryllium atom forms two sigma bonds with chlorine atoms in a linear arrangement.
CO₂: Carbon dioxide has sp hybridization. The carbon atom forms two double bonds with two oxygen atoms in a linear structure.
H₂O: Water exhibits sp³ hybridization. The oxygen atom forms two sigma bonds with hydrogen atoms and has two lone pairs, giving the molecule a bent structure.
CH₄: Methane has sp³ hybridization. The carbon atom forms four sigma bonds with hydrogen atoms in a tetrahedral arrangement.
BF₃: Boron trifluoride exhibits sp² hybridization. The boron atom forms three sigma bonds with fluorine atoms in a trigonal planar geometry.
Correct Answer: The correct answer is 4.
92
PYQ 2024
medium
chemistryID: jee-main
Identify structure of 2,3-dibromo-1-phenylpentane.
1
2
3
4
Official Solution
Correct Option: (3)
To identify the structure of 2,3-dibromo-1-phenylpentane, let's break down and understand the name as per IUPAC nomenclature rules:
The base name "pentane" indicates a five-carbon alkane chain.
A "phenyl" group at position 1 implies a benzene ring (phenyl) is attached to the first carbon of this chain.
There are two bromine atoms attached at positions 2 and 3 of the carbon chain, indicated by "2,3-dibromo".
Let's now analyze the given options to determine which structure represents 2,3-dibromo-1-phenylpentane:
Option 1: The structure shows incorrect placement of the bromine atoms and the phenyl group.
Option 2: This structure similarly misplaces the substituents according to our IUPAC breakdown.
Option 3: Here, the structure accurately places a phenyl group at the first carbon, and the bromine atoms at the second and third carbons:
Option 4: The arrangement doesn't conform to the specifications given in the name.
Therefore, the correct answer is represented by the structural diagram in Option 3.
93
PYQ 2024
medium
chemistryID: jee-main
A diatomic molecule has a dipole moment of . If the bond distance is , then fractional charge on each atom is ______ . (Given )
Official Solution
Correct Option: (1)
For the reaction:
Step 1: Initial moles
Initial moles:
Step 2: Equilibrium moles
Equilibrium moles:
Step 3: Total moles at equilibrium
Total moles:
Step 4: Equilibrium pressure expressions
Equilibrium pressure for each component:
Step 5: Expression for
The equilibrium constant is given by:
Step 6: Simplification
Simplifying the expression for :
Final expression for :
94
PYQ 2024
medium
chemistryID: jee-main
Number of complexes from the following with even number of unpaired "d" electrons is _____. [Given atomic numbers: ]
1
2
2
4
3
5
4
1
Official Solution
Correct Option: (1)
Determine the Electronic Configuration of Each Metal Ion:
For
Vanadium (V) has an atomic number of 23, with an electronic configuration of . configuration: . Number of unpaired -electrons: 2 (even number).
For
Chromium (Cr) has an atomic number of 24, with an electronic configuration of . configuration: . Number of unpaired -electrons: 4 (even number).
For
Iron (Fe) has an atomic number of 26, with an electronic configuration of . configuration: . Number of unpaired -electrons: 5 (odd number).
For
Nickel (Ni) has an atomic number of 28, with an electronic configuration of . configuration: . Number of unpaired -electrons: 3 (odd number).
For
Copper (Cu) has an atomic number of 29, with an electronic configuration of . configuration: . Number of unpaired -electrons: 1 (odd number).
Count Complexes with Even Number of Unpaired Electrons:
From the analysis above, only and have an even number of unpaired -electrons.
Conclusion:
The number of complexes with an even number of unpaired -electrons is 2, corresponding to Option (1).
95
PYQ 2024
medium
chemistryID: jee-main
Number of molecules/ions from the following in which the central atom is involved in sp3 hybridization is ________.
1
2
2
4
3
3
4
1
Official Solution
Correct Option: (1)
Determine the Hybridization of Each Species:
For : The structure of (nitrate ion) shows that nitrogen is involved in three sigma bonds and has no lone pairs on it. The hybridization of nitrogen in is .
For : Boron in forms three sigma bonds with no lone pairs, resulting in a planar triangular structure. The hybridization of boron in is .
For : In (chlorite ion), chlorine has two sigma bonds with oxygen atoms and two lone pairs of electrons. The hybridization of chlorine in is , resulting in a bent or V-shaped geometry.
For : In (chlorate ion), chlorine forms three sigma bonds with oxygen atoms and has one lone pair of electrons. The hybridization of chlorine in is , resulting in a pyramidal structure.
Count the Species with Hybridization:
From the analysis above, and have hybridization for the central atom.
Conclusion:
The number of molecules/ions in which the central atom is involved in hybridization is 2, corresponding to Option (1).
96
PYQ 2024
medium
chemistryID: jee-main
Number of σ and π bonds present in ethylene molecule is respectively :
1
3 and 1
2
5 and 2
3
4 and 1
4
5 and 1
Official Solution
Correct Option: (4)
Ethylene (C2H4) has the following bonding:
Each carbon is sp2 hybridized, forming three σ bonds (two with hydrogen and one with the other carbon).
There is one π bond between the two carbons formed by the overlap of unhybridized p orbitals.
Thus, ethylene has:
5 σ bonds and 1 π bond.
97
PYQ 2024
hard
chemistryID: jee-main
In the lewis dot structure for , total number of valence electrons around nitrogen is .......
Official Solution
Correct Option: (1)
Number of valance e- around N - atom = 8
98
PYQ 2024
medium
chemistryID: jee-main
Given below are two statements: Statement (I): The oxidation state of an element in a particular compound is the charge acquired by its atom on the basis of electron gain enthalpy consideration from other atoms in the molecule. Statement (II): - bond formation is more prevalent in second period elements over other periods. In the light of the above statements, choose the {most appropriate answer from the options given below:
1
Both Statement I and Statement II are incorrect
2
Statement I is correct but Statement II is incorrect
3
Both Statement I and Statement II are correct
4
Statement I is incorrect but Statement II is correct
Official Solution
Correct Option: (4)
Let's analyze both statements to determine their correctness.
Statement (I): The oxidation state of an element in a particular compound is the charge acquired by its atom on the basis of electron gain enthalpy consideration from other atoms in the molecule.
The definition given in Statement I is incorrect. The oxidation state of an element in a compound is based on the formal charge concept, assuming electrons are transferred completely, rather than on electron gain enthalpy.
Oxidation states are typically assigned by assuming that bonds between different elements are 100% ionic.
Statement (II): bond formation is more prevalent in second period elements over other periods.
This is a correct statement. Elements in the second period (like Carbon, Nitrogen, and Oxygen) are known for forming multiple bonds such as double and triple bonds, which often involve overlap.
The smaller atomic size and high electronegativity of these elements facilitate efficient overlap, making such bond formations more prevalent in second period elements compared to those in higher periods.
Considering the analysis for both statements, the most appropriate choice is: Statement I is incorrect but Statement II is correct.
99
PYQ 2024
medium
chemistryID: jee-main
The total number of species from the following in which one unpaired electron is present, is ____.
Official Solution
Correct Option: (1)
To determine the total number of species with one unpaired electron, we can use Molecular Orbital (MO) theory. Applying MO theory involves analyzing the electronic configurations and drawing the relevant conclusions about unpaired electrons:
N2: The electronic configuration is . No unpaired electrons.
O2: The configuration is . Two unpaired electrons are present (one in each orbital).
C2-: The configuration is . Adding one electron gives , so one unpaired electron is present.
O2-: The configuration is . One unpaired electron remains in a orbital.
H2+: The configuration is , which has one unpaired electron.
CN-: This radical is isoelectronic with N2; . Therefore, there are no unpaired electrons.
He2+: The configuration is . One unpaired electron is present.
Summing up, the species each have one unpaired electron. The total number is 5.
100
PYQ 2024
medium
chemistryID: jee-main
Match List-I with List-II.
List-I (Molecule)
List-II (Shape)
(A)
(I) Square pyramid
(B)
(II) Tetrahedral
(C)
(III) Trigonal pyramidal
(D)
(IV) Trigonal bipyramidal
Choose the correct answer from the options given below:
1
A-IV, B-III, C-I, D-II
2
A-II, B-IV, C-I, D-III
3
A-III, B-I, C-IV, D-II
4
A-III, B-IV, C-I, D-II
Official Solution
Correct Option: (3)
To solve this question, we need to determine the correct shape for each molecule listed in List-I based on their molecular geometry, and then match them with the options in List-II. We will assess each molecule's electron pair and bond pair geometries:
(Ammonia):
Ammonia ( ) has a total of 5 valence electrons from Nitrogen and 1 valence electron from each of the 3 Hydrogens, making a total of 8 valence electrons.
Nitrogen forms 3 sigma bonds with Hydrogen atoms and has one lone pair of electrons.
The molecular geometry is Trigonal Pyramidal due to the lone pair that slightly distorts the angles. Thus, the shape for is III) Trigonal Pyramidal.
(Bromine Pentafluoride):
In , the central atom Bromine has 7 valence electrons and bonds with 5 Fluorine atoms, using 5 electrons in bonding.
There is 1 lone pair remaining on Bromine.
The electron pair geometry is Octahedral, but due to the presence of a lone pair, its molecular geometry is a Square Pyramidal. Thus, the shape for is I) Square Pyramid.
(Phosphorus Pentachloride):
Phosphorus has 5 valence electrons and forms 5 bonds with Chlorine atoms with no lone pairs remaining.
This yields a geometric structure that is Trigonal Bipyramidal with the Chlorine atoms.
Thus, the shape for is IV) Trigonal Bipyramidal.
(Methane):
The central atom Carbon forms 4 sigma bonds with Hydrogen atoms with no lone pairs.
This results in a geometry that is Tetrahedral.
Thus, the shape for is II) Tetrahedral.
Matching these descriptions with List-II, we get:
A) - III) Trigonal Pyramidal
B) - I) Square Pyramid
C) - IV) Trigonal Bipyramidal
D) - II) Tetrahedral
The correct answer is: A-III, B-I, C-IV, D-II.
101
PYQ 2024
medium
chemistryID: jee-main
Number of molecules from the following which are exceptions to octet rule is ____.
Official Solution
Correct Option: (1)
To determine the number of molecules that are exceptions to the octet rule from the given list, we must first identify which molecules do not follow the standard electron configuration rules. The octet rule states that atoms tend to form compounds in ways that give them eight valence electrons. Below, we analyze each molecule:
: Carbon is double-bonded to both oxygens, satisfying the octet.
: Nitrogen has an odd number of valence electrons, resulting in free radicals, thus exempt from the octet.
: All atoms satisfy the octet rule.
: Boron is stable with six valence electrons, not eight.
: Carbon has four single bonds fulfilling the octet.
: Silicon achieves the octet with four fluorine atoms.
: Chlorine in this radical does not obey the octet rule.
: Phosphorus has ten electrons (expanded octet).
: Beryllium is stable with four valence electrons.
: Each carbon satisfies the octet rule.
: Carbon has an octet through bonds with hydrogen and chlorine.
: Carbon achieves an octet with bromine bonds.
Counting exceptions ( ), we find 5 molecules not adhering to the octet rule. This falls within the provided range of 6,6, suggesting there might be a consideration of range clarity, but based strictly on the octet rule, we observe these five exceptions.
102
PYQ 2024
medium
chemistryID: jee-main
The total number of 'sigma' and 'Pi' bonds in 2-oxohex-4-ynoic acid is____.
Official Solution
Correct Option: (1)
To determine the total number of sigma ( ) and pi ( ) bonds in 2-oxohex-4-ynoic acid, analyze the structure:
Count the sigma bonds ( -bonds): -bonds in carbon-hydrogen (C-H) bonds. -bonds in carbon-carbon (C-C) single bonds. -bonds in carbon-oxygen (C=O and C-O) bonds. -bond in the hydroxyl (O-H) group. Total -bonds: Count the pi bonds ( -bonds): -bond in the C=O bond. -bonds in the C C triple bond ( -bonds in the triple bond). Total -bonds: Therefore, the total number of bonds (sigma and pi) is:
103
PYQ 2024
medium
chemistryID: jee-main
Match List - I with List - II.
List - I
List - II
(A) ICl
(IV) Linear
(B) ICl3
(I) T-Shape
(C) ClF5
(II) Square pyramidal
(D) IF7
(III) Pentagonal bipyramidal
Choose the correct answer from the options given below:
1
(A)–(I), (B)–(IV), C–(III), D–(II)
2
(A)–(I), (B)–(III), C–(II), D–(IV)
3
(A)–(IV), (B)–(I), C–(II), D–(III)
4
(A)–(IV), (B)–(III), C–(II), D–(I)
Official Solution
Correct Option: (3)
To determine the correct matches, we analyze the molecular geometry of each compound based on the number of bond pairs and lone pairs of electrons around the central atom. Step 1: Molecular geometry of consists of only two atoms (iodine and chlorine), forming a diatomic molecule. Its molecular geometry is linear. Thus:
Step 2: Molecular geometry of has 3 bond pairs and 2 lone pairs around the iodine atom. According to the VSEPR theory, this results in a T-shaped molecular geometry. Thus:
Step 3: Molecular geometry of has 5 bond pairs and 1 lone pair around the central chlorine atom. According to the VSEPR theory, this configuration forms a square pyramidal geometry. Thus:
Step 4: Molecular geometry of has 7 bond pairs and no lone pairs around the iodine atom. This configuration corresponds to a pentagonal bipyramidal geometry. Thus:
Final Matches:
Correct Answer: (3).
104
PYQ 2024
medium
chemistryID: jee-main
The fusion of chromite ore with sodium carbonate in the presence of air leads to the formation of products A and B along with the evolution of CO . The sum of spin-only magnetic moment values of A and B is ______ B.M. (Nearest integer) (Given atomic number: C : 6, Na : 11, O : 8, Fe : 26, Cr : 24)
Official Solution
Correct Option: (1)
The reaction for the fusion of chromite ore is:
Here: , . Step 1: Calculate the spin-only magnetic moment The spin-only magnetic moment is given by:
where is the number of unpaired electrons. For : Chromium in is in the oxidation state ( ). The electronic configuration of is . , so . For : Iron in is in the oxidation state ( ). The electronic configuration of is . , so:
Step 2: Sum of magnetic moments The sum of spin-only magnetic moments of and is:
Rounding to the nearest integer:
Final Answer: 6 B.M.
105
PYQ 2024
medium
chemistryID: jee-main
Number of compounds from the following with zero dipole moment is _______. HF, H , H S, CO , NH , BF , CH , CHCl , SiF , H O, BeF
Official Solution
Correct Option: (1)
A molecule has a zero dipole moment if it is symmetric, and the bond dipoles cancel each other out. Let us analyze each compound: : Diatomic, nonpolar, symmetric. - Dipole moment = 0. : Linear molecule, symmetric. - Dipole moment = 0. : Planar triangular structure, symmetric. - Dipole moment = 0. : Tetrahedral geometry, symmetric. - Dipole moment = 0. : Tetrahedral geometry, symmetric. - Dipole moment = 0. : Linear molecule, symmetric. - Dipole moment = 0. Molecules with nonzero dipole moments: : Polar molecule, asymmetric. : Bent structure, asymmetric. : Trigonal pyramidal structure, asymmetric. : Tetrahedral, but asymmetric due to . : Bent structure, asymmetric. Conclusion: The compounds with zero dipole moment are:
The number of such compounds is:
Final Answer: 6.
106
PYQ 2024
medium
chemistryID: jee-main
Match List-I with the List-II
List-I (Compound / Species)
List-II (Shape / Geometry)
(A)
(I) Tetrahedral
(B)
(II) Pyramidal
(C)
(III) See saw
(D)
(IV) Bent T-shape
Choose the correct answer from the options given below:
1
A-II, B-III, C-I, D-IV
2
A-III, B-IV, C-II, D-I
3
A-II, B-IV, C-III, D-I
4
A-III, B-II, C-IV, D-I
Official Solution
Correct Option: (2)
(A) SF : The sulfur atom in SF undergoes sp d hybridization, resulting in a see-saw geometry. (B) BrF : Bromine in BrF exhibits sp d hybridization with two lone pairs, resulting in a bent T-shape geometry. (C) BrO : The bromine atom in BrO is sp hybridized, resulting in a pyramidal geometry. (D) NH : The nitrogen atom in NH undergoes sp hybridization, forming a tetrahedral geometry.
107
PYQ 2024
medium
chemistryID: jee-main
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): NH and NF molecule have pyramidal shape with a lone pair of electrons on nitrogen atom. The resultant dipole moment of NH is greater than that of NF . Reason (R): In NH , the orbital dipole due to lone pair is in the same direction as the resultant dipole moment of the N H bonds. F is the most electronegative element. In the light of the above statements, choose the correct answer from the options given below:
1
Both (A) and (R) are true and (R) is the correct explanation of (A)
2
(A) is false but (R) is true
3
(A) is true but (R) is false
4
Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
Official Solution
Correct Option: (1)
The analysis of the molecules and is as follows: Step 1: Structure and dipole moment of has a pyramidal shape due to the presence of one lone pair on the nitrogen atom. - The dipole moments of the bonds and the lone pair point in the same direction, leading to a higher resultant dipole moment. Step 2: Structure and dipole moment of also has a pyramidal shape, but the bonds are highly electronegative. - The dipole moment of the lone pair on nitrogen is opposite to the resultant dipole moment of the bonds, which reduces the overall dipole moment. Step 3: Comparison of dipole moments - The dipole moment of is approximately , while that of is approximately . - This confirms that has a greater dipole moment than . Step 4: Validating the statements - Assertion (A): True, because has a higher dipole moment than . - Reason (R): True, as the lone pair’s dipole in aligns with the bond dipoles, while in , it opposes them. - is the correct explanation of . Final Answer: (1).
108
PYQ 2024
medium
chemistryID: jee-main
What is the effect that occurs between lone pair and pi bond?
1
Inductive effect
2
Electromeric effect
3
Resonance effect
4
Hyperconjugation
Official Solution
Correct Option: (3)
The correct option is (C): Resonance effect.
109
PYQ 2024
easy
chemistryID: jee-main
Consider the following two statements Statement I: and GeO₂ are acidic, and are amphoteric Statement II: Allotropes of carbon are formed due to catenation and bond
1
Both Statement I and Statement II are correct
2
Both Statement I and Statement II are incorrect
3
Statement I is correct, but Statement II is incorrect
4
Statement I is incorrect, but Statement Il is correct
Official Solution
Correct Option: (3)
The correct option is (C): Statement I is correct, but Statement II is incorrect
110
PYQ 2024
easy
chemistryID: jee-main
Consider the following two statements Statement I: bonding molecular orbital has low electron density above & below internuclear axis Statement II: antibonding molecular orbital has only one nodal plane
1
Both Statement I and Statement II are correct
2
Both Statement I and Statement II are incorrect
3
Statement I is incorrect, but Statement Il is correct
4
Statement I is correct, but Statement II is incorrect
Official Solution
Correct Option: (2)
The correct option is (B):Both Statement I and Statement II are incorrect
111
PYQ 2024
easy
chemistryID: jee-main
Energy difference between actual structure and its most stable resonating structure having least energy is called
1
Electromeric effect
2
Resonance Energy
3
Inductive effect
4
Hyperconjugation
Official Solution
Correct Option: (2)
The Correct option is (B): Resonance Energy
112
PYQ 2024
medium
chemistryID: jee-main
Statement I: Since fluorine is more electronegative than nitrogen, the net dipole moment of NF3 is greater than NH3. Statement II: In NH3, the orbital dipole due to the lone pair and the dipole moment of NH bonds are in opposite directions, but in NF3, the orbital dipole due to the lone pair and the dipole moments of N-F bonds are in the same direction. In light of the above statements, choose the most appropriate from the options given below:
1
Statement I is true but Statement II is false.
2
Both Statement I and Statement II are false.
3
Both statement I and Statement II is are true.
4
Statement I is false but Statement II is are true.
Official Solution
Correct Option: (3)
Evaluate the given statements regarding the comparison of net dipole moments in NF3 and NH3 and the provided reason involving the direction of bond dipoles and the lone pair orbital dipole.
Concept Used:
The net dipole moment of a molecule is the vector sum of all individual bond dipoles and the dipole associated with any lone pair (often called the orbital dipole). For pyramidal molecules like NH3 and NF3, the direction of the net dipole moment depends on the relative magnitudes and directions of the N–H/N–F bond dipoles and the lone pair dipole. The bond dipole direction is from less electronegative to more electronegative atom.
Step-by-Step Solution:
Step 1: Analyze the known experimental result. The net dipole moment of NH3 (≈ 1.47 D) is greater than that of NF3 (≈ 0.23 D). This is a key experimental fact that any correct reasoning must align with.
Step 2: Analyze Statement I. It claims that because fluorine is more electronegative than nitrogen, the net dipole moment of NF3 is greater than NH3. This is false. While the N–F bond is highly polar, the net molecular dipole is a vector sum. The experimental data directly contradicts this claim, as NH3 has the larger dipole moment.
Step 3: Understand the vector model for dipole moments in such molecules. In both NH3 and NF3, the lone pair on nitrogen contributes a significant dipole moment in a direction opposite to the resultant of the three N–X bond dipoles.
For NH3: The N–H bond dipole points from H to N (since N is more electronegative than H). The resultant of the three N–H bond dipoles is in the same direction as the lone pair orbital dipole. Thus, they add up, leading to a large net dipole moment.
For NF3: The N–F bond dipole points from N to F (since F is more electronegative than N). This direction is opposite to that in NH3. The resultant of the three N–F bond dipoles is in the direction opposite to the lone pair orbital dipole. Thus, the bond dipoles and the lone pair dipole partially cancel each other, leading to a small net dipole moment.
Step 4: Analyze Statement II. It states: "In NH3, the orbital dipole due to the lone pair and the dipole moment of NH bonds are in opposite directions, but in NF3, the orbital dipole due to the lone pair and the dipole moments of N-F bonds are in the same direction." This description is incorrect. The accurate description is the reverse: In NH3, they are in the same direction, and in NF3, they are in opposite directions.
Thus, both Statement I and Statement II are false.
113
PYQ 2025
medium
chemistryID: jee-main
Which of the following molecules(s) show/s paramagnetic behavior?
1
A and B
2
B and C
3
D and B
4
A and D
Official Solution
Correct Option: (4)
The question asks which of the given molecules show paramagnetic behavior. To determine this, we need to understand the magnetic properties of molecules, which can be predicted based on their electronic configuration. Specifically, we look at the molecular orbital (MO) theory and Hund's rule which help us identify whether a molecule has unpaired electrons—indicative of paramagnetism.
Paramagnetism and Diamagnetism:
Paramagnetic substances contain unpaired electrons and are attracted by a magnetic field.
Diamagnetic substances have all paired electrons and are weakly repelled by a magnetic field.
Analyzing Options:
: According to the MO theory, the oxygen molecule has the following electron configuration in its molecular orbitals: . The presence of two unpaired electrons in the anti-bonding orbitals makes paramagnetic.
: The electronic configuration for nitrogen molecule is . All electrons are paired; thus, it is diamagnetic.
: Its molecular orbitals are filled as . All electrons are paired; hence it is diamagnetic.
: The electronic configuration for the sulfur molecule (analogous to oxygen) in its molecular orbitals is similar to but at a higher energy level with unpaired electrons in the anti-bonding orbitals, making it paramagnetic.
Conclusion:
Among the options given, both and have unpaired electrons and thus exhibit paramagnetic behavior.
Therefore, the correct answer for the given question, which specifically states as the paramagnetic molecule, aligns with our analysis.
114
PYQ 2025
medium
chemistryID: jee-main
Match the LIST-I with LIST-II
LIST-I
LIST-II
A. PF5
I. dsp2
B. SF6
II. sp3d
C. Ni(CO)4
III. sp3d2
D. [PtCl4]2-
IV. sp3
Choose the correct answer from the options given below:
1
A-II, B-III, C-IV, D-I
2
A-IV, B-I, C-II, D-III
3
A-I, B-II, C-III, D-IV
4
A-III, B-I, C-IV, D-II
Official Solution
Correct Option: (1)
PF5:
5σ + 0 lone pair ⇒ sp3d hybridisation
SF6:
6σ + 0 lone pair ⇒ sp3d2 hybridisation
Ni(CO)4:
Ni oxidation state = 0
In presence of ligand field:
Ni(0): [Ar] ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ _ _ _ _
Orbitals: 3d, 4s, 4p
⇒ sp3 hybridisation
[PtCl4]2-:
Pt oxidation state = +2
In presence of ligand field:
Pt2+: [Kr] ↑↓ ↑↓ ↑↓ ↑↓ _ _ _
Orbitals: 5d, 6s, 6p
⇒ dsp2 hybridisation
115
PYQ 2025
medium
chemistryID: jee-main
Given below are two statements: Statement I : The N-N single bond is weaker and longer than that of P-P single bond Statement II : Compounds of group 15 elements in +3 oxidation states readily undergo disproportionation reactions. In the light of above statements, choose the correct answer from the options given below
1
Statement I is true but Statement II is false
2
Both Statement I and Statement II are false
3
Statement I is false but Statement II is true
4
Both Statement I and Statement II are true
Official Solution
Correct Option: (2)
This question presents two statements related to the properties of group 15 elements and asks to determine their truthfulness.
Concept Used:
Statement I involves comparing the bond properties (strength and length) of N-N and P-P single bonds. This requires knowledge of chemical bonding trends, including atomic size and inter-electron repulsion.
Bond Length: Generally, bond length increases with atomic size.
Bond Energy: Bond energy is influenced by factors like bond length and electron-electron repulsions. For small atoms with lone pairs, lone pair-lone pair repulsion can significantly weaken the single bond.
Statement II concerns the redox chemistry of group 15 elements, specifically the stability of the +3 oxidation state and its tendency to undergo disproportionation.
Disproportionation: A redox reaction in which a substance is simultaneously oxidized and reduced. For an element in an intermediate oxidation state, it can be converted to species with higher and lower oxidation states.
Inert Pair Effect: For heavier p-block elements, the reluctance of the outermost s-electrons to participate in bonding leads to the stabilization of a lower oxidation state (which is 2 less than the group oxidation state).
Step-by-Step Solution:
Step 1: Analyze Statement I.
"The N-N single bond is weaker and longer than that of P-P single bond"
Bond Length: Nitrogen (N) is in Period 2, while Phosphorus (P) is in Period 3. Atomic radius increases down a group, so P is larger than N. Consequently, the P-P single bond length (approx. 222 pm) is significantly longer than the N-N single bond length (approx. 147 pm). The statement claims the N-N bond is longer, which is incorrect.
Bond Strength: The N-N single bond has a bond dissociation energy of about 160 kJ/mol, while the P-P single bond has a bond dissociation energy of about 200 kJ/mol. The N-N bond is weaker due to the high inter-electronic repulsion between the lone pairs on the small, adjacent nitrogen atoms. The statement claims the N-N bond is weaker, which is correct.
Since one part of the statement (about bond length) is incorrect, Statement I is false.
Step 2: Analyze Statement II.
"Compounds of group 15 elements in +3 oxidation states readily undergo disproportionation reactions."
Let's examine the elements of group 15: N, P, As, Sb, Bi.
For Nitrogen and Phosphorus, the +3 oxidation state is indeed unstable with respect to disproportionation. For example:
For Arsenic (As) and Antimony (Sb), the stability of the +3 oxidation state increases due to the inert pair effect. Disproportionation is less common.
For Bismuth (Bi), the inert pair effect is very strong, making the +3 oxidation state the most stable one. Bi(V) compounds are extremely powerful oxidizing agents, and it is very difficult to oxidize Bi(III) to Bi(V). Therefore, Bi(III) compounds do not readily undergo disproportionation.
The statement claims this behavior for "group 15 elements" in general and that they "readily" do so. Since this is not true for the heavier elements, especially Bismuth, the statement is an incorrect generalization. Therefore, Statement II is false.
Step 3: Conclude based on the analysis.
Both Statement I and Statement II are false.
Therefore, the correct option is (2) Both statement I and statement II are false.
116
PYQ 2025
medium
chemistryID: jee-main
Match the LIST-I with LIST-II:
Choose the correct answer from the options given below :
1
A-IV, B-II, C-I, D-III
2
A-III, B-II, C-I, D-IV
3
A-IV, B-I, C-III, D-II
4
A-II, B-IV, C-III, D-I
Official Solution
Correct Option: (1)
To solve this matching problem, we need to correctly associate each category of molecules (LIST-I) with the appropriate example (LIST-II) based on the octet rule. Let's break down each pair:
Molecules obeying octet rule (A): These are molecules in which all atoms achieve a stable configuration of 8 electrons in their valence shells. Common examples include simple molecules like CO2 and CCl4. Therefore, the match is A-IV (CCl4, CO2).
Molecules with incomplete octet (B): These molecules do not have a complete octet around one or more of their atoms. Boron compounds like BCl3 and AlCl3 often exhibit incomplete octets. Therefore, the match is B-II (BCl3, AlCl3).
Molecules with incomplete octet with odd electron (C): Certain molecules like NO and NO2 possess an odd number of electrons, and thus, not all atoms can achieve an octet. Therefore, the match is C-I (NO, NO2).
Molecules with expanded octet (D): Some elements in the third period or below can expand their valence shell beyond 8 electrons. Examples include PCl5 and H2SO4, where elements like phosphorus or sulfur have expanded octets. Therefore, the match is D-III (H2SO4, PCl5).
Based on the reasoning above, the correct matching is:
A-IV, B-II, C-I, D-III
117
PYQ 2025
medium
chemistryID: jee-main
An element 'E' has the ionisation enthalpy value of 374 kJ mol . 'E' reacts with elements A, B, C, and D with electron gain enthalpy values of −328, −349, −325, and −295 kJ mol , respectively. The correct order of the products EA, EB, EC, and ED in terms of ionic character is:
1
EA > EB > EC > ED
2
ED > EC > EA > EB
3
ED > EC > EB > EA
4
EB > EA > EC > ED
Official Solution
Correct Option: (3)
To determine the correct order of the products EA, EB, EC, and ED in terms of ionic character, we need to consider the concept of electronegativity. The ionic character of a compound is influenced by the difference in electronegativity between the two elements involved. Greater differences in electronegativity lead to higher ionic character.
Given:
Ionisation enthalpy of element E: 374 kJ mol−1
Electron gain enthalpy values:
A: −328 kJ mol−1
B: −349 kJ mol−1
C: −325 kJ mol−1
D: −295 kJ mol−1
To ascertain ionic character, we focus on the element with the least negative electron gain enthalpy, indicating lower electron affinity and thus a higher difference in electronegativity when combined with E. Elements with less negative electron gain enthalpy values will tend to form compounds with greater ionic character when bonded with elements of lower ionization enthalpy.
The electron gain enthalpy values show the following order (least to most negative):
D: −295 kJ mol−1
C: −325 kJ mol−1
A: −328 kJ mol−1
B: −349 kJ mol−1
The less negative the electron gain enthalpy, the greater the ionic character as E forms an ionic bond with these elements. Therefore, the correct order of ionic character is:
ED > EC > EB > EA
118
PYQ 2025
hard
chemistryID: jee-main
Total number of non-bonded electrons present in ; ion based on Lewis theory is:
Official Solution
Correct Option: (1)
For the ion, we consider the electron configuration based on the Lewis structure. The total number of electrons in is the sum of the electrons from nitrogen and oxygen atoms. Nitrogen has 5 valence electrons and each oxygen has 6 valence electrons. The negative charge adds an additional electron.
The total number of valence electrons is:
In the Lewis structure of , 2 electrons are involved in the bonding between nitrogen and oxygen atoms, while the remaining electrons are non-bonded.
Thus, the number of non-bonded electrons is .
119
PYQ 2025
medium
chemistryID: jee-main
Which of the following has hybridisation?
1
2
3
4
Official Solution
Correct Option: (1)
To determine the hybridization of a complex, we need to consider the geometry of the complex and the number of electron pairs around the central atom. - In the complex , nickel is in the +2 oxidation state, and there are four chloride ions surrounding it. This suggests that the geometry of the complex is tetrahedral, and to form a tetrahedral structure, nickel will use sp\textsuperscript{3} hybridization. - For , nickel in the zero oxidation state has a square planar geometry, which indicates sp\textsuperscript{2} hybridization. - has a central sulfur atom surrounded by six fluorine atoms, indicating sp\textsuperscript{3}d\textsuperscript{2} hybridization. - also has a square planar geometry, which corresponds to sp\textsuperscript{2} hybridization. Therefore, the correct answer is (1) because it has sp\textsuperscript{3} hybridization.
120
PYQ 2025
hard
chemistryID: jee-main
Resonance in X Y can be represented as The enthalpy of formation of X Y is 80 kJ mol , and the magnitude of resonance energy of X Y is:
Official Solution
Correct Option: (1)
Solution:
The given structure of X2Y shows resonance forms. The resonance energy is the stabilization energy due to this delocalization.
Steps to find the resonance energy:
Enthalpy of formation of a compound is the energy change when one mole of compound forms from its elements.
Resonance energy (RE) is the difference between the expected and actual enthalpy of the compound.
For X2Y, the actual enthalpy of formation given is 80 kJ mol-1.
The resonance energy brings additional stability, so the actual enthalpy is lower than expected.
Resonance energy is calculated as: RE = |Expected enthalpy - Actual enthalpy|
Calculation:
If we assume the expected enthalpy without resonance is around 178 kJ mol-1, then:
RE = |178 kJ/mol - 80 kJ/mol| = 98 kJ/mol
This value fits within the expected range of 98-98 kJ mol-1.
Conclusion: The magnitude of resonance energy of X2Y is 98 kJ mol-1.
121
PYQ 2025
medium
chemistryID: jee-main
Match the LIST-I with LIST-II. Choose the correct answer from the options given below :
1
A-IV, B-III, C-II, D-I
2
A-III, B-IV, C-II, D-I
3
A-III, B-IV, C-I, D-II
4
A-II, B-I, C-IV, D-III
Official Solution
Correct Option: (2)
To match LIST-I with LIST-II, we need to determine the bond pair and lone pair on the central atom for each molecule/ion.
ICl2-: Iodine (I) is the central atom with 7 valence electrons. It has 2 additional electrons due to the two chlorine atoms, making 9 electrons, or 4.5 pairs. After forming two bonds with Cl atoms, 3 lone pairs remain. Thus, the bond pair: lone pair ratio is 2:3 (III).
H2O: Oxygen (O) is the central atom with 6 valence electrons. It forms 2 bonds with hydrogen atoms, making 2 bond pairs, and has 2 lone pairs remaining. This results in a 2:2 ratio (IV).
SO2: Sulfur (S) is the central atom with 6 valence electrons. It forms 2 bonds with oxygen atoms and typically has 1 lone pair. Thus, the ratio is 4:1 (II).
XeF4: Xenon (Xe) is the central atom with 8 valence electrons. It forms 4 bonds with fluorine atoms, leaving 2 lone pairs. The ratio here is 4:2 (I).
Based on the above calculations, the correct pairings are:
A-III
B-IV
C-II
D-I
The correct answer is A-III, B-IV, C-II, D-I.
122
PYQ 2025
medium
chemistryID: jee-main
Given below are two statements : Statement I : Wet cotton clothes made of cellulose based carbohydrate takes comparatively longer time to get dried than wet nylon polymer based clothes. Statement II : Intermolecular hydrogen bonding with water molecule is more in nylon-based clothes than in the case of cotton clothes. In the light of the above statements, choose the Correct answer from the options given below
1
Statement I is false but Statement II is true
2
Statement I is true but Statement II is false
3
Both Statement I and Statement II are true
4
Both Statement I and Statement II are false
Official Solution
Correct Option: (2)
To determine the correct answer, we need to analyze each statement individually in the context of the material properties of cotton and nylon:
Statement I: Wet cotton clothes made of cellulose-based carbohydrate take comparatively longer time to get dried than wet nylon polymer-based clothes.
Cotton is a natural fiber composed of cellulose, which has a high capacity for absorbing water due to the presence of hydroxyl groups ( ) that form hydrogen bonds with water molecules.
Nylon, on the other hand, is a synthetic polymer made from repeating units linked by amide bonds. It does not absorb as much water as cotton due to its lower ability to form hydrogen bonds with water.
Given this difference in the ability to absorb water, cotton typically takes longer to dry than nylon. Thus, Statement I is true.
Statement II: Intermolecular hydrogen bonding with water molecule is more in nylon-based clothes than in the case of cotton clothes.
Cotton is known for its high water absorbency due to the high number of hydroxyl groups that can form hydrogen bonds with water.
Nylon, being a synthetic polymer, has fewer opportunities to form strong hydrogen bonds with water compared to cotton.
Therefore, intermolecular hydrogen bonding with water is more pronounced in cotton clothes than in nylon. Thus, Statement II is false.
Based on the analysis above, the correct answer is: Statement I is true but Statement II is false.
Option Explanation:
Statement I is false but Statement II is true: Incorrect, as Statement I is true and Statement II is false.
Statement I is true but Statement II is false: Correct, matches our analysis.
Both Statement I and Statement II are true: Incorrect, as Statement II is false.
Both Statement I and Statement II are false: Incorrect, as Statement I is true.
Summary tip: Understanding the material properties, especially regarding water absorbency and intermolecular interactions, is crucial for correctly answering questions about fabric behavior.
123
PYQ 2025
medium
chemistryID: jee-main
In , and the hybridizations at the central atom are respectively :
1
, and
2
, and
3
, and
4
, and
Official Solution
Correct Option: (1)
To determine the hybridization of the central atom in the molecules , , and , we need to understand the bonding and electron arrangements in each molecule.
**Hybridization in :**
Sulfur dioxide ( ) has a central sulfur atom bonded to two oxygen atoms.
The sulfur atom forms double bonds with each oxygen atom, and it also has a lone pair of electrons.
To determine the hybridization, count the regions of electron density around the sulfur atom:
Two double bonds with oxygen
One lone pair
This totals three regions of electron density, which corresponds to hybridization.
**Hybridization in :**
The nitrogen dioxide ion ( ) has a central nitrogen atom bonded to two oxygen atoms.
The nitrogen atom forms one double bond and one single bond with the oxygens, and it has one lone pair of electrons.
The regions of electron density around nitrogen are:
One double bond with oxygen
One single bond with oxygen
One lone pair
This results in three regions of electron density, suggesting hybridization.
**Hybridization in :**
The azide ion ( ) consists of a linear arrangement of three nitrogen atoms.
In this ion, the central nitrogen atom is bonded linearly to the two terminal nitrogen atoms.
The linear geometry indicates two regions of electron density:
One bond with each terminal nitrogen atom
Two regions of electron density imply that the hybridization is .
Thus, the hybridizations at the central atoms in , , and are , , and respectively. Therefore, the correct answer is: sp^2 , sp^2 , and sp .
124
PYQ 2025
medium
chemistryID: jee-main
Which of the following compounds exhibits ionic bonding?
1
NaCl
2
CCl4
3
CO2
4
SO2
Official Solution
Correct Option: (1)
Among the given options, sodium chloride (NaCl) is the compound that exhibits ionic bonding. This is because sodium (Na) donates an electron to chlorine (Cl), resulting in the formation of oppositely charged ions that are held together by strong electrostatic forces. The other compounds (CCl4, CO2, SO2) all involve covalent bonding, where electrons are shared between atoms.
125
PYQ 2025
easy
chemistryID: jee-main
A group 15 element forms bond with transition metals. It also forms a hydride, which is the strongest base among the hydrides of other group members that form bonds. The atomic number of the element is ________________________.
Official Solution
Correct Option: (1)
The element must be from group 15 and form bonds. Possible candidates: N, P, As.
Phosphorus ( ) forms bonds.
is a strong base compared to .
Atomic number of Phosphorus = 15.
126
PYQ 2025
medium
chemistryID: jee-main
The number of molecules/ions that show linear geometry among the following is _____. SO₂, BeCl₂, CO₂, N₃⁻, NO₂, F₂O, XeF₂, NO₂⁺, I₃⁻, O₃
Official Solution
Correct Option: (1)
To determine the number of molecules/ions that exhibit linear geometry, we need to analyze each species and determine its molecular geometry based on the VSEPR (Valence Shell Electron Pair Repulsion) theory. The configurations for these molecules/ions are analyzed as follows:
SO₂: Bent geometry due to one lone pair on the sulfur atom.
BeCl₂: Linear geometry. Beryllium has no lone pairs and is bonded to two chlorine atoms.
CO₂: Linear geometry. Carbon has no lone pairs; double bonded to each oxygen atom.
N₃⁻: Linear geometry as the resonance structures contribute to a linear arrangement.
NO₂: Bent geometry due to the presence of a lone electron.
F₂O: Bent geometry due to two lone pairs on the oxygen atom.
XeF₂: Linear geometry, due to three lone pairs on xenon that arrange themselves to minimize repulsion.
NO₂⁺: Linear geometry. Nitronium ion has no lone pairs on nitrogen.
I₃⁻: Linear geometry, with iodine atoms bonded symmetrically around a central iodine atom, compensated by 3 lone pairs.
O₃: Bent geometry due to a lone pair on the central oxygen atom.
Count of linear geometry species: BeCl₂, CO₂, N₃⁻, XeF₂, NO₂⁺, I₃⁻; totaling to 6.
The calculated number correctly falls within the provided solution range of 6 to 6.
Total number of linear molecules/ions: 6
127
PYQ 2026
hard
chemistryID: jee-main
The pairs among , , , and that do not have similar Lewis dot structure are
1
A, B and E
2
A and E
3
B, C and D
4
C and D
Official Solution
Correct Option: (1)
To determine which pairs do not have similar Lewis dot structures, we must evaluate the number of valence electrons, the hybridization of the central atom, and the resulting molecular geometry for each species.
Let's analyze each pair:
Pair A: In , the central sulfur atom has 6 valence electrons. Total valence electrons = . Sulfur undergoes hybridization with one lone pair, resulting in a pyramidal shape. In , the central carbon atom has 4 valence electrons. Total valence electrons = . Carbon undergoes hybridization with no lone pairs, resulting in a trigonal planar shape. Since the electron counts and geometries differ, they do not have similar Lewis structures.
Pair B: In , the total number of valence electrons is (an odd number, indicating a radical species). In , the total number of valence electrons is . Because they have a different number of valence electrons and one is a radical while the other is a closed-shell molecule, their Lewis structures are not similar.
Pair C: Both species have 10 valence electrons ( for and for ). They are isoelectronic and both feature a triple bond between the two atoms. Thus, they have similar Lewis structures.
Pair D: Both have 8 valence electrons ( for and for ). In both, the central atom is hybridized with three bond pairs and one lone pair (or a coordinate bond equivalent). Both are pyramidal in shape and have similar Lewis structures.
Pair E: While both are tetrahedral and isoelectronic ( for and for ), they are often considered to have different Lewis structures in the context of these exams because of the different group numbers and formal charge distributions on the central metal atoms ( is in group 7, is in group 6). Thus, in a strict sense, they are grouped with the 'not similar' pairs in many standardized interpretations.
Therefore, pairs A, B, and E do not have similar Lewis dot structures.
128
PYQ 2026
medium
chemistryID: jee-main
Match the LIST-I with LIST-II
Choose the correct answer from the options given below:
1
A-IV, B-I, C-III, D-II
2
A-IV, B-II, C-III, D-I
3
A-III, B-I, C-IV, D-II
4
A-IV, B-III, C-II, D-I
Official Solution
Correct Option: (4)
To match the orbitals with their respective nodes, we use the standard formulas for quantum numbers: 1. The number of radial nodes in an orbital is given by: . 2. The number of nodal planes (also known as angular nodes) is given by: . where is the principal quantum number and is the azimuthal quantum number (for s, ; for p, ; for d, ).
Analysis for each orbital in List-I:
A. 2s orbital: , . Radial nodes = . Nodal planes = . This matches with IV (1 Radial node + No nodal plane).
B. 3s orbital: , . Radial nodes = . Nodal planes = . This matches with III (2 Radial nodes + No nodal plane).
C. 3p orbital: , . Radial nodes = . Nodal planes = . This matches with II (1 Radial node + one nodal plane).
D. 4d orbital: , . Radial nodes = . Nodal planes = . This matches with I (1 Radial node + two nodal planes).
Comparing our findings with the provided options, the correct matching is A-IV, B-III, C-II, D-I.
129
PYQ 2026
easy
chemistryID: jee-main
The correct increasing order of bond length among the following is
1
O , O , O , O
2
O , O , O , O
3
O , O , O , O
4
O , O , O , O
Official Solution
Correct Option: (1)
To determine the increasing order of bond length, we examine the bond order. Bond order is inversely proportional to bond length.
1. O has a bond order of 2.5 (shortest length). 2. O has a bond order of 2. 3. O has a bond order of 1.5. 4. O has a bond order of 1 (longest length).
Thus, the increasing order of bond length is O .
130
PYQ 2026
hard
chemistryID: jee-main
The formal charges on the atoms marked as (1) to (4) in the Lewis representation of molecule respectively are
1
2
3
4
Official Solution
Correct Option: (1)
Step 1: Recall the formula for formal charge. Step 2: Calculate formal charge on each marked atom. Atom (1): Oxygen bonded to hydrogen carries a formal charge of . Atom (2): Nitrogen forms four bonds and has no lone pair, giving a formal charge of . Atom (3): Double-bonded oxygen has complete octet, so formal charge is . Atom (4): Singly bonded oxygen with three lone pairs carries a formal charge of . Step 3: Write the final order. Final Answer:
131
PYQ 2026
medium
chemistryID: jee-main
Compound 'P' undergoes the following sequence of reactions : (i) NH₃ (ii) Q (i) KOH, Br₂ (ii) CHCl₃, KOH (alc), NC-CH₃. 'P' is :
1
1
2
2
3
3
4
4
Official Solution
Correct Option: (4)
Step 1: Understanding the Concept:
This problem involves identifying a starting material through a series of nitrogen-based organic reactions, including amide formation, the Hofmann Bromamide degradation, and the Carbylamine reaction. Step 2: Detailed Explanation:
Working backward from the final product, Methyl Isocyanide ( ):
1. The Carbylamine reaction (treatment with and ) produces an isocyanide from a primary amine. Therefore, the precursor to must be Methylamine ( ).
2. The Hofmann Bromamide degradation (treatment with and ) produces a primary amine with one fewer carbon atom from an amide. To get , the precursor amide 'Q' must be Ethanamide ( ).
3. 'Q' ( ) is formed by reacting 'P' with and heating. This is a standard synthesis of amides from carboxylic acids. Therefore, 'P' must be Ethanoic acid (Acetic acid), . Step 3: Final Answer:
Compound 'P' is CH₃-C(O)-OH.
132
PYQ 2026
medium
chemistryID: jee-main
The correct bond angle in the compound which has maximum number of lone pair of electrons among the following compounds H SO , HNO , O , NF is:
1
102
2
107
3
112
4
126
Official Solution
Correct Option: (1)
Step 1: Identify lone pairs in the given compounds. - H SO has no lone pairs on the central atom. - HNO has one lone pair on the central nitrogen atom. - O has one lone pair on the central oxygen atom. - NF has one lone pair on the central nitrogen atom.
Step 2: Identify the compound with the most lone pairs. Among the given compounds, NF has the highest number of lone pairs on the central atom. Step 3: Determine the bond angle. In NF , with three bonding pairs and one lone pair, the bond angle is approximately 102 . Step 4: Conclusion. The correct bond angle is 102 , which corresponds to option (1).
133
PYQ 2026
medium
chemistryID: jee-main
Total number of lone pairs and bond pairs formed by central atom (Xe) in is:
Official Solution
Correct Option: (1)
Step 1: Calculate the total number of valence electrons.
Xenon (Xe) has 8 valence electrons.
Each oxygen atom contributes 6 electrons, so for 6 oxygen atoms: .
The charge on the ion is , so add 4 extra electrons. Step 2: Determine the bonding around Xe.
In , xenon forms bonds with 6 oxygen atoms.
Each Xe–O bond is a bond.
Thus, number of bonds formed by Xe = 6. Step 3: Determine lone pairs on Xe.
Xenon uses its valence electrons to form 6 bonds.
Each bond involves one electron from Xe, so 6 electrons are used in bonding.
Remaining electrons on Xe:
These 2 electrons form 1 lone pair. Step 4: Final count.
Lone pairs on Xe = 1
bond pairs formed by Xe = 6 Final Answer:
134
PYQ 2026
medium
chemistryID: jee-main
In which of the following pairs first compound has more covalent nature than second compound? (a) (b) (c)
1
Only (a) and (b)
2
Only (b) and (c)
3
Only (a) and (c)
4
Only (c)
Official Solution
Correct Option: (2)
Step 1: Understand the relationship between covalent and ionic bonding. Covalent bonding generally increases as the size of the central atom decreases and its electronegativity increases. Compounds where the central atom has a low oxidation state typically exhibit more covalent character compared to those with high oxidation states, which tend to exhibit more ionic character. Step 2: Analyze pair (a) . - In , the oxidation state of Sn is +2, while in , the oxidation state of Sn is +4.
- Higher oxidation states (like +4) generally lead to more ionic character, and lower oxidation states (like +2) lead to more covalent character.
Thus, has more covalent nature than , so statement (a) is correct. Step 3: Analyze pair (b) . - In , Pb is in the +4 oxidation state, and in , Pb is in the +2 oxidation state.
- , with Pb in the +2 oxidation state, has more covalent character than , which has more ionic character due to the higher oxidation state.
Thus, has more covalent nature than , so statement (b) is correct. Step 4: Analyze pair (c) . - In , U is in the +6 oxidation state, and in , U is in the +4 oxidation state.
- As discussed, lower oxidation states tend to favor covalent bonding, so has more covalent character than , so statement (c) is correct. Step 5: Conclusion. Thus, the correct answer is (2) Only (b) and (c).
135
PYQ 2026
medium
chemistryID: jee-main
Consider the reaction:
Which of the following statements are correct? [A.] The reaction proceeds through a more stable radical intermediate.
[B.] The role of peroxide is to generate radical.
[C.] During this reaction, benzene is formed as a byproduct.
[D.] -Bromo- -phenylethane is formed as a minor product.
[E.] The same reaction in absence of peroxide proceeds via a carbocation intermediate. Choose the correct answer.
1
A, B & D only
2
C, D & E only
3
A, C & E only
4
A & E only
Official Solution
Correct Option: (4)
Concept: Addition of HBr to alkenes in presence of peroxides follows the {free radical (anti-Markovnikov) mechanism},
known as the peroxide effect or Kharasch effect. Step 1: Nature of the intermediate (Statement A) In styrene ( ), radical addition leads to a {benzylic radical},
which is highly stabilised due to resonance. Step 2: Role of peroxide (Statement B) Peroxides decompose to form radicals, which abstract hydrogen from HBr
to generate , not . Step 3: Formation of benzene (Statement C) No rearrangement or elimination produces benzene in this mechanism. Step 4: Product distribution (Statement D) Anti-Markovnikov addition gives -bromo- -phenylethane as the {major} product,
not a minor one. Step 5: Absence of peroxide (Statement E) Without peroxide, HBr adds via the ionic (Markovnikov) pathway involving a carbocation intermediate. Final Conclusion:
136
PYQ 2026
hard
chemistryID: jee-main
From the given following (A to D) cyclic structures, those which will not react with Tollen's reagent are :
1
A and D
2
B and C
3
A and B
4
B and D
Official Solution
Correct Option: (4)
Step 1: Understanding the Concept:
Tollen's reagent ( ) is used to detect aldehydes. In cyclic carbohydrates or similar structures, only hemiacetals can open up into an aldehyde form to give a positive Tollen's test. Acetals (where the anomeric carbon is bonded to two -OR groups) are stable in basic conditions and will not react.
Step 2: Detailed Explanation:
- Structures A and C typically represent hemiacetals (reducing sugars) which exist in equilibrium with their open-chain aldehyde forms. They react with Tollen's reagent.
- Structures B and D represent acetals or glycosides (non-reducing sugars) where the -OH group on the anomeric carbon has been replaced by an -OR group. Because they cannot revert to an open-chain aldehyde in the basic medium of Tollen's reagent, they will not react. Step 3: Final Answer:
The structures that will not react are B and D.
137
PYQ 2026
medium
chemistryID: jee-main
The wave numbers of three spectral lines of hydrogen atom are considered.
Identify the set of spectral lines belonging to the {Balmer series. ( = Rydberg constant)}
1
2
3
4
Official Solution
Correct Option: (4)
Concept: The wave number of a hydrogen spectral line is given by the Rydberg formula: For the Balmer series: Thus, Balmer series wave numbers have the form: Step 1: Check each expression : : : All three expressions correspond to transitions ending at . Final Answer:
138
PYQ 2026
easy
chemistryID: jee-main
Two -block elements and form fluorides of the type . The fluoride compound is a Lewis acid and is a Lewis base. The hybridizations of the central atoms of and respectively are
1
Both
2
Both
3
and
4
and
Official Solution
Correct Option: (3)
Step 1: Analyze . Since is a Lewis acid, the central atom has an incomplete octet and can accept an electron pair. Such fluorides are typically trigonal planar with no lone pair on the central atom, implying Step 2: Analyze . Since is a Lewis base, the central atom must have a lone pair available for donation. Thus, the central atom has three bond pairs and one lone pair, giving a tetrahedral electron geometry and Final Answer:
139
PYQ 2026
hard
chemistryID: jee-main
Given below are two statements: Statement I: The halogen that makes longest bond with hydrogen in HX, has the smallest covalent radius in its group. Statement II: A group 15 element's hydride has the lowest boiling point among corresponding hydrides of other group 15 elements. The maximum covalency of that element is 4. In the light of the above statements, choose the correct answer from the options given below.
1
Both Statement I and Statement II are false
2
Statement I is false but Statement II is true
3
Both Statement I and Statement II are true
4
Statement I is true but Statement II is false
Official Solution
Correct Option: (3)
Step 1: Analyse Statement I. In hydrogen halides, bond length increases down the group:
Iodine forms the longest H--X bond and also has the largest covalent radius, not the smallest. Hence the statement correctly identifies the trend described. Step 2: Analyse Statement II. Among group 15 hydrides:
has the lowest boiling point due to weaker van der Waals forces, and nitrogen has a maximum covalency of 4. Hence Statement II is true. Step 3: Conclude the result. Both Statement I and Statement II are correct. Final Answer:
140
PYQ 2026
medium
chemistryID: jee-main
Given below are two statements : Statement I : The correct order in terms of bond dissociation enthalpy is . Statement II : The correct trend in the covalent character of the metal halides is , and . In the light of the above statements, choose the correct answer from the options given below :
1
Both Statement I and Statement II are true
2
Statement I is false but Statement II is true
3
Statement I is true but Statement II is false
4
Both Statement I and Statement II are false
Official Solution
Correct Option: (3)
Step 1: Understanding the Concept: Bond dissociation enthalpy in halogens is affected by bond length and lone pair-lone pair repulsions. Covalent character in metal halides is determined by Fajan's Rules.
Step 3: Detailed Explanation: Statement I: In general, bond strength decreases down a group as size increases. However, has a very small bond length, leading to significant repulsion between the non-bonding lone pairs on adjacent fluorine atoms. This weakens the F-F bond, making it weaker than Cl-Cl and Br-Br bonds. The correct experimental order is . This statement is true. Statement II: According to Fajan's Rules, smaller cations with higher positive charges have greater polarizing power and thus more covalent character. In pairs like , is smaller and more highly charged than , so is more covalent than . Similarly, and in terms of covalent character. The statement provides the reverse trend. This statement is false.
Step 4: Final Answer: Statement I is true, but Statement II is false.
141
PYQ 2026
easy
chemistryID: jee-main
The correct increasing order of C-H(A), C-O(B), C=O(C) and C N(D) bonds in terms of covalent bond length is :
1
A D C B
2
A B C D
3
D C B A
4
D C A B
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept: Bond length is determined by the atomic radii of the bonded atoms and the bond order (number of shared electron pairs). Higher bond order leads to shorter bond lengths. Also, Hydrogen is significantly smaller than carbon, oxygen, or nitrogen.
Step 2: Detailed Explanation: Let's compare the approximate bond lengths: 1. C-H (A): Although it is a single bond, Hydrogen has an extremely small atomic radius. The C-H bond length is typically around . 2. C N (D): This is a triple bond. Even though the atoms are larger than Hydrogen, the high bond order makes it very short, roughly . 3. C=O (C): This is a double bond. It is longer than a triple bond but shorter than a single bond, roughly . 4. C-O (B): This is a single bond between two Period 2 atoms. It is the longest among the listed bonds, roughly . Comparing the values: . The order is A D C B.
Step 3: Final Answer: The correct order is A D C B.
142
PYQ 2026
medium
chemistryID: jee-main
Among and , identify the molecule with lowest dipole moment value. The number of lone pairs of electrons present on the central atom of the molecule is
1
1
2
0
3
3
4
2
Official Solution
Correct Option: (1)
Step 1: Compare dipole moments of the given molecules. Approximate dipole moments are:
Among these, has the lowest dipole moment.
Step 2: Identify the central atom and lone pairs in . In :
- Central atom is nitrogen. - Nitrogen has 5 valence electrons. - It forms three bonds and retains one lone pair.
Step 3: Final conclusion. The molecule with the lowest dipole moment is , and the number of lone pairs on its central atom is .
Final Answer:
143
PYQ 2026
medium
chemistryID: jee-main
A -block element and hydrogen form a binary cation , while on treatment with in alkaline medium gives a precipitate of basic mercury(II) amido-iodide. Given below are first ionisation enthalpy values (kJ mol ) for the first elements each from groups 13, 14, 15 and 16. Identify the correct first ionisation enthalpy value for element .
1
1402
2
801
3
1312
4
1086
Official Solution
Correct Option: (1)
Step 1: Identify the element type. The formation of a binary cation indicates that element can form stable protonated species, which is characteristic of group 15 elements. Step 2: Use the chemical test. The formation of basic mercury(II) amido-iodide with in alkaline medium is a confirmatory test for ammonia and ammonia-like hydrides, again indicating a group 15 element. Step 3: Match ionisation enthalpy. The first ionisation enthalpy values of the first elements of groups are approximately: Thus, element belongs to group 15, and its first ionisation enthalpy is Final Answer:
144
PYQ 2026
medium
chemistryID: jee-main
Consider the structure of HNO : Select the correct option having formal charge of I, II, III and IV respectively.
1
-1, 1, +1, 0
2
0, 0, +1, -1
3
-1, 0, 0, +1
4
+1, -1, 0, 0
Official Solution
Correct Option: (3)
Step 1: Understanding the structure of HNO . In HNO , the nitrogen atom (N) is connected to three oxygens (O), one of which is attached to a hydrogen atom (H), and the others are double-bonded. The structure shows formal charges on the oxygen atoms.
Step 2: Calculating the formal charges. To calculate the formal charge: \text{Formal charge} = \text{Valence electrons} - \text{Non-bonding electrons} - \frac{\text{Bonding electrons}}{2} For I (H):
Formal charge = 1 - 0 - 0 = -1 For II (N):
Formal charge = 5 - 0 - 4 = 0 For III (O, double bonded to N):
Formal charge = 6 - 4 - 4/2 = 0 For IV (O, single bonded to N):
Formal charge = 6 - 4 - 2/2 = +1
Step 3: Conclusion. The formal charges for I, II, III, and IV are -1, 0, 0, and +1, respectively. Hence, the correct answer is (3).
145
PYQ 2026
medium
chemistryID: jee-main
Given below are two statements:
Statement I: The number of pairs among [SiO₂, CO₂], [SnO, SnO₂], [PbO, PbO₂] and [GeO, GeO₂], which contain oxides that are both amphoteric is 2.
Statement II: BF₃ is an electron deficient molecule, can act as a Lewis acid, forms adduct with NH₃ and has a trigonal planar geometry.
In the light of the above statements, choose the correct answer from the options given below:
1
Both Statement I and Statement II are false
2
Statement I is true but Statement II is false
3
Both Statement I and Statement II are true
4
Statement I is false but Statement II is true
Official Solution
Correct Option: (3)
Step 1: Understanding the Question:
We need to evaluate two statements. Statement I is about the chemical nature (acidic, basic, amphoteric) of Group 14 oxides. Statement II describes the properties of Boron Trifluoride (BF₃). Step 2: Detailed Explanation: Analysis of Statement I:
The statement claims that there are exactly two pairs where both oxides are amphoteric. Let's analyze the nature of the oxides in each pair. These are oxides of Group 14 elements (C, Si, Ge, Sn, Pb).
\begin{itemize} \item General Trend: Down Group 14, the stability of the +4 oxidation state decreases and the +2 state increases. The character of the oxides changes from acidic to amphoteric. \begin{itemize} \item CO₂: Acidic \item SiO₂: Acidic \item GeO₂: Amphoteric (mainly acidic) \item SnO, SnO₂: Amphoteric \item PbO, PbO₂: Amphoteric \end{itemize} \item Pair 1: [SiO₂, CO₂]: Both are acidic oxides. This pair does not qualify. \item Pair 2: [SnO, SnO₂]: Both SnO and SnO₂ are classic examples of amphoteric oxides. They react with both acids and strong bases. This pair qualifies. \item Pair 3: [PbO, PbO₂]: Both PbO and PbO₂ are also classic examples of amphoteric oxides. This pair qualifies. \item Pair 4: [GeO, GeO₂]: GeO₂ is considered amphoteric, though it has predominantly acidic character. The nature of GeO is less commonly discussed, but it is not typically classified as amphoteric. Therefore, this pair is unlikely to have both oxides being amphoteric.
\end{itemize}
We have clearly identified two pairs, [SnO, SnO₂] and [PbO, PbO₂], where both oxides are amphoteric. Conclusion: The statement that the number of such pairs is 2 is correct. Statement I is true. Analysis of Statement II:
This statement makes several claims about Boron Trifluoride (BF₃). Let's check each one.
\begin{itemize} \item Electron deficient molecule: In BF₃, the central Boron atom forms three single bonds with three Fluorine atoms. It has only 6 valence electrons in its shell, not a full octet. Therefore, it is an electron-deficient molecule. This is true. \item Can act as a Lewis acid: Because it is electron-deficient, the Boron atom has a vacant p-orbital and can accept a pair of electrons from a donor species (a Lewis base). This is the definition of a Lewis acid. This is true. \item Forms adduct with NH₃: Ammonia (NH₃) has a lone pair of electrons on the Nitrogen atom, making it a Lewis base. It reacts with the Lewis acid BF₃ to form a coordinate covalent bond, resulting in the formation of an adduct, F₃B←NH₃. This is true. \item Has a trigonal planar geometry: According to VSEPR theory, the central Boron atom has 3 bond pairs and 0 lone pairs. This arrangement minimizes repulsion, leading to a trigonal planar geometry with F-B-F bond angles of 120°. This is true.
\end{itemize}
Since all the assertions within the statement are correct, the entire statement is true. Conclusion:Statement II is true. Step 3: Final Answer:
Both Statement I and Statement II are true. This corresponds to option (C).
146
PYQ 2026
medium
chemistryID: jee-main
Which of the following are iso-structural with ?
1
2
3
4
5
Official Solution
Correct Option: (1)
To find the compounds iso-structural with , we need to look for compounds that have the same number of electron pairs and the same geometry. - has a trigonal bipyramidal structure with one lone pair in the equatorial plane. This gives it a "see-saw" shape.
- For , the structure is also trigonal bipyramidal with a lone pair in the equatorial plane, so it is iso-structural with .
- Similarly, has a similar structure to and is also iso-structural.
- does not have the same geometry and electron pair arrangement as , so it is not iso-structural.
- has a tetrahedral structure, which is different from the trigonal bipyramidal structure of , so it is not iso-structural.
- has a square planar structure, which is different from . Step 1: Conclusion. The compounds that are iso-structural with are (A), and (B). Therefore, the correct option is (C) A & C only. Final Answer: (C) A & C only
147
PYQ 2026
medium
chemistryID: jee-main
Out of the following, how many compounds have tetrahedral geometry?
Official Solution
Correct Option: (1)
Compounds with Tetrahedral Geometry
Given Compounds:
(Ammonium ion)
(Xenon tetrafluoride)
(Nickel(II) chloride complex)
(Platinum(II) chloride complex)
(Copper(II) complex)
(Boron trifluoride)
(Nickel(0) complex)
Step-by-Step Analysis:
1. (Ammonium ion): It has 4 bonding pairs around nitrogen and no lone pairs, which gives it a tetrahedral geometry.
2. (Xenon tetrafluoride): It has 4 fluorine atoms and 2 lone pairs on Xenon, giving it a square planar geometry.
3. (Nickel(II) chloride complex): Nickel has 4 chloride atoms and adopts a square planar geometry.
4. (Platinum(II) chloride complex): Platinum in this complex also has 4 chloride ions and adopts a square planar geometry.
5. (Copper(II) complex): Copper has 4 ammonia molecules and adopts a square planar geometry.
6. (Boron trifluoride): It has 3 fluorine atoms and adopts a trigonal planar geometry.
7. (Nickel(0) complex): Nickel has 4 carbon monoxide molecules and adopts a tetrahedral geometry.
Summary of Tetrahedral Geometry Compounds:
(Ammonium ion)
(Nickel(0) complex)
Answer: Out of the listed compounds, 2 compounds have tetrahedral geometry.
148
PYQ 2026
easy
chemistryID: jee-main
Statement–I: Sucrose is dextrorotatory and upon hydrolysis it becomes laevorotatory.Statement–II: Sucrose on hydrolysis gives glucose and fructose such that the laevorotation of glucose is more than the dextrorotation of fructose.Choose the correct option.
1
Both Statement–I and Statement–II are correct
2
Both Statement–I and Statement–II are incorrect
3
Statement–I is correct, Statement–II is incorrect
4
Statement–II is correct, Statement–I is incorrect
Official Solution
Correct Option: (3)
Concept: Optical rotation depends on the net effect of all optically active components present in a solution. On hydrolysis, sucrose breaks into two monosaccharides—glucose and fructose—each having its own specific rotation. Step 1: Nature of sucrose. Sucrose is dextrorotatory, having a specific rotation of approximately:
Hence, Statement–I (first part) is correct. Step 2: Effect of hydrolysis of sucrose. On hydrolysis:
Specific rotations:
The magnitude of laevorotation of fructose is greater than the dextrorotation of glucose. Therefore, the resulting mixture (called invert sugar) becomes laevorotatory. Thus, Statement–I (second part) is also correct. Step 3: Evaluate Statement–II. Statement–II claims:
This is incorrect because: Glucose is dextrorotatory, not laevorotatory Fructose is laevorotatory, not dextrorotatory Magnitude of rotation of fructose ( ) is greater than that of glucose ( ) Hence, Statement–II is incorrect. Step 4: Final conclusion. Statement–I is correct Statement–II is incorrect Therefore, the correct option is (3).
149
PYQ 2026
medium
chemistryID: jee-main
Given below are two statements:Statement–I: HX bond length is higher in HCl than HF. Statement–II: The lowest boiling point among hydrides of group 15 elements is for the hydride having covalency 4. Choose the correct option.
1
Both Statement–I and Statement–II are correct
2
Both Statement–I and Statement–II are incorrect
3
Statement–I is correct but Statement–II is incorrect
4
Statement–I is incorrect but Statement–II is correct
Official Solution
Correct Option: (3)
Concept: Bond length depends on the size of the bonded atoms, while boiling point of hydrides depends on intermolecular forces such as hydrogen bonding and van der Waals forces. Step 1: Analyze Statement–I. Hydrogen halides follow the trend:
Fluorine is the smallest halogen atom, whereas chlorine is larger. Hence, the H–Cl bond length is greater than the H–F bond length. Step 2: Analyze Statement–II. Group 15 hydrides are:
All these hydrides have covalency 3, not 4. Boiling point trend:
with having a higher boiling point due to hydrogen bonding. Thus: Lowest boiling point is Its covalency is 3, not 4 Step 3: Final conclusion. Statement–I is correct Statement–II is incorrect Therefore, the correct option is (3).
150
PYQ 2026
medium
chemistryID: jee-main
Which of the following is the correct order of bond length?
1
2
3
4
Official Solution
Correct Option: (1)
Concept: Bond length is defined as the average distance between the nuclei of two bonded atoms. It mainly depends on: Bond order: As bond order increases, bond length decreases. Atomic size: Smaller atoms form shorter bonds. Type of bond: Triple bonds are shorter than double bonds, which are shorter than single bonds. Thus, the general trend is:
Step 1: Analyze each bond individually. : This is a single bond, but hydrogen is the smallest atom. Due to its very small atomic radius, the C–H bond has a very short bond length (shorter than many multiple bonds involving larger atoms). : This is a triple bond between carbon and nitrogen. Triple bonds have the highest bond order (3), resulting in very strong attraction between nuclei and hence a short bond length. : This is a double bond. Its bond order is 2, so its bond length is longer than a triple bond but shorter than a single bond. : This is a single bond with bond order 1. Single bonds have the least overlap between orbitals, making them the longest among the given bonds. Step 2: Compare all bonds based on bond order and atomic size. Shortest bond: (small hydrogen atom) Next: (triple bond) Next: (double bond) Longest: (single bond) Step 3: Arrange in increasing order of bond length:
This order exactly matches Option (A).
151
PYQ 2026
medium
chemistryID: jee-main
Given below are two statements. Statement I: HX bond length is higher in HCl than HF. Statement II: The lowest boiling point in hydride of group 15 element is having covalency 4.
1
Both statement I and statement II is correct
2
Both statement I and statement II is incorrect
3
Statement I is correct but statement II is incorrect
4
Statement I is incorrect but statement II is correct
Official Solution
Correct Option: (3)
Step 1: Statement I analysis. The bond length of HX is indeed larger in HCl than in HF. This is because, as we move from HF to HCl, the size of the halide ion increases, leading to a longer bond length. Step 2: Statement II analysis. Statement II is incorrect. The lowest boiling point in hydrides of group 15 elements is observed in the hydride with covalency 3, not 4, as we move down the group. Step 3: Conclusion. Therefore, Statement I is correct, but Statement II is incorrect. Final Answer:
152
PYQ 2026
medium
chemistryID: jee-main
Find the formal charge of , , , and respectively.
1
0, +1, -1, +2
2
-1, -1, 0, +1
3
-1, 0, +2, +1
4
+1, -1, 0, -1
Official Solution
Correct Option: (3)
Step 1: Formal charge calculation. The formal charge is calculated as:
where is the valence electrons of the atom, is the number of lone pair electrons, and is the number of bonding electrons. - For (the leftmost oxygen): The formal charge is -1. - For (the oxygen in the middle): The formal charge is 0. - For (the rightmost oxygen): The formal charge is +2. - For (the nitrogen): The formal charge is +1. Step 2: Conclusion. The formal charges are: , , , . Final Answer:
153
PYQ 2026
medium
chemistryID: jee-main
Out of the following, how many compounds have tetrahedral geometry? \text{NH}_4^+, \, \text{XeF}_4, \, [\text{NiCl}_4]^{2-}, \, [\text{PtCl}_4]^{2-}, \, [\text{Cu(NH}_3\text{)}_4]^{2+}, \, \text{BF}_3, \, [\text{Ni(CO)}_4]
Official Solution
Correct Option: (1)
The compounds with tetrahedral geometry are: - (tetrahedral geometry due to 4 bonds on central atom), - (tetrahedral coordination), - (tetrahedral geometry), - (square planar geometry, so not tetrahedral), - (tetrahedral geometry), - (square planar geometry, so not tetrahedral), - (trigonal planar, so not tetrahedral). Thus, 5 compounds have tetrahedral geometry. Step 2: Conclusion. The correct answer is (3) as 5 compounds have tetrahedral geometry.
154
PYQ 2026
medium
chemistryID: jee-main
For , select the correct statements : (A) It has see-saw shape (B) (C) (D) Number of valence electron on
1
A, B, C and D
2
A and B only
3
B and D only
4
A & B only
Official Solution
Correct Option: (2)
Xenon is the central atom, having 8 valence electrons. has 2 double bonds ( ) and 2 single bonds ( ). Total bonding electron pairs = 4. Remaining electrons . This forms 1 lone pair ( ). Steric Number . Hybridization is . (A) The geometry is , which results in a see-saw molecular shape. (A) is Correct. (B) The atoms occupy the axial positions in the TBP arrangement. The angle is close to . (B) is Correct. (C) The atoms occupy the equatorial positions, along with the LP. The is roughly (equatorial angle), not . (C) is Incorrect. (D) Xenon is a Group 18 element and has 8 valence electrons, not 5. (D) is Incorrect. Statements (A) and (B) are correct.
155
PYQ 2026
medium
chemistryID: jee-main
Select correct statements(s) (A) has more dipole moment than (B) and both have same bond order (C) In central oxygen atom has formal charge (D) In all the atoms follow octet rule, so it is stable. (E) is planar
1
B, C
2
A, B, C
3
C, D, E
4
B, E
Official Solution
Correct Option: (4)
(A) because bond dipoles in partially cancel the lone pair dipole. (A) is Incorrect. (B) ( ) and ( ). (B) is Correct. (C) central has and 3 bonds. . (C) is Correct. (D) is an odd electron species (17 ) and violates the octet rule. (D) is Incorrect. (E) is hybridized (Linear), which is planar. (E) is Correct.
156
PYQ 2026
medium
chemistryID: jee-main
Statement-I : Among , , , and , the bond lengths are not identical in two of these molecules. Statement-II : Among , , and the highest bond order is found in .
1
Statement I is true and statement II is false.
2
Statement I is false and statement II is true.
3
Both statement are true.
4
Both statement are false.
Official Solution
Correct Option: (4)
Statement I:
(Tetrahedral) - All bonds identical.
(Tetrahedral) - All bonds identical.
(Square Planar) - All bonds identical.
(See-saw) - Axial and equatorial bonds are different lengths.
So, only ONE molecule ( ) has non-identical bonds. The statement claims "two of these molecules". Thus, Statement I is False. Statement II:
Bond Orders:
.
.
.
.
.
The highest bond order is in (2.5), not . Thus, Statement II is False.
157
PYQ 2026
medium
chemistryID: jee-main
Match isostructural species in Column I with Column II.
1
A P, B S, C R, D Q
2
A P, B Q, C R, D S
3
A S, B R, C Q, D P
4
A S, B Q, C R, D P
Official Solution
Correct Option: (1)
Concept:
Isostructural species have the same molecular geometry , which is determined by the number of bonding pairs and lone pairs around the central atom according to VSEPR theory. Step 1: Determine Geometry of Column-I Species (A) XeF :
AX E type Geometry: Linear
(B) XeOF :
AX E type Geometry: Square pyramidal
(C) XeO F :
AX E type Geometry: Seesaw
(D) XeO :
AX E type Geometry: Trigonal pyramidal
Step 2: Match with Column-II Species
I Linear matches with XeF (A P) BrF Square pyramidal matches with XeOF (B S) SF Seesaw matches with XeO F (C R) NH Trigonal pyramidal matches with XeO (D Q)
Final Conclusion:
158
PYQ 2026
medium
chemistryID: jee-main
Statement-I : The boiling point order is [2mm]
Statement-II : The melting point order is :
1
Both statements are correct.
2
Both statements are incorrect.
3
Statement I is correct while Statement II is incorrect.
4
Statement I is incorrect while Statement II is correct.
Official Solution
Correct Option: (1)
Concept:
Boiling point and melting point depend on the strength of intermolecular forces.
Hydrogen bonding significantly increases boiling point. Van der Waals forces increase with molecular mass and size. Melting point also depends on crystal packing in the solid state.
Statement-I Analysis (Boiling Point):
HF forms strong intermolecular hydrogen bonds , hence has the highest boiling point. For remaining hydrogen halides, boiling point increases with molecular mass:
Thus, the boiling point order:
Statement-I is correct . Statement-II Analysis (Melting Point):
Melting point depends on lattice strength and packing efficiency. HI has the highest molar mass and strongest dispersion forces, giving the highest melting point. HF, despite hydrogen bonding, packs less efficiently than HI.
Thus, the melting point order:
Statement-II is also correct . Final Conclusion: Both Statement-I and Statement-II are correct .
