Match the following List -I (Complex) List II (Spin only Magnetic Moment)
List -I(Complex)
List II (Spin only Magnetic Moment)
A)
[CoF6]3-
I)
0
B)
[Co(C2O4)3]3-
II)
√24
C)
[FeF6]3+
III)
√8
D)
[Mn(CN)6]3-
IV)
√35
V)
√15
the correct answer is:
1
A-V, B - II, C - IV, D - I
2
A - II, B - I, C- IV, D - III
3
A - II, B - I, C - V, D - III
4
A - III, B - II, C - I, D - V
Official Solution
Correct Option: (1)
Step 1: Understand the Concept of Magnetic Moment:
The magnetic moment for a transition metal complex can be calculated using the formula for spin-only magnetic moment:
$ is the number of unpaired electrons in the complex. The spin-only magnetic moment is determined by the number of unpaired electrons, which depends on the metal's oxidation state and the nature of the ligands.
Step 2: Analyze Each Complex and Determine the Number of Unpaired Electrons:
A) [CoF6] 3-: Cobalt in +3 oxidation state (Co3+) has 6 d-electrons (d6). In the octahedral complex with weak field ligands like fluoride (F-), the electrons will be arranged with 3 unpaired electrons. Therefore, the number of unpaired electrons is 3.
B) [Co(C2O4)3] 3-: Cobalt in +3 oxidation state (Co3+) has 6 d-electrons (d6). The oxalate ion (C2O42-) is a weak field ligand, so there are 2 unpaired electrons in this complex.
C) [FeF6] 3-: Iron in +3 oxidation state (Fe3+) has 5 d-electrons (d5). With weak field fluoride ligands, Fe3+ has 5 unpaired electrons.
D) [Mn(CN)6] 3-: Manganese in +3 oxidation state (Mn3+) has 4 d-electrons (d4). Cyanide (CN-) is a strong field ligand, causing pairing of electrons, resulting in 1 unpaired electron.
Step 3: Calculate the Magnetic Moment:
- For complex A: With 3 unpaired electrons, the magnetic moment is:
\) \mu = \sqrt{15} \mu = \sqrt{8} \mu = \sqrt{35} \mu = \sqrt{3}$ (matches with II)
Final Answer: The correct matching is: A - V, B - II, C - IV, D - I
02
PYQ 2024
medium
chemistryID: ts-eamce
Which of the following does not evolve when made to react with water?
1
2
3
4
Official Solution
Correct Option: (4)
- does not evolve oxygen when reacted with water, whereas , , and all release oxygen gas when reacting with water due to their high oxidation state.
03
PYQ 2024
medium
chemistryID: ts-eamce
Observe the following reaction: Which of the following are correct?
1
2
i, ii, iii only
3
ii, iii, iv only
4
i, ii, iv only
Official Solution
Correct Option: (3)
Step 1: Assigning Oxidation States - Iodide ion is oxidized to iodine . - Manganate ion is reduced to . - ions balance the charge and form water ( ). Step 2: Writing the Half-Reactions Oxidation Half-Reaction (Iodine Ion): Reduction Half-Reaction (Permanganate Ion): Step 3: Balancing Electrons - Multiply the oxidation reaction by 5 and the reduction reaction by 2 to balance the electrons: Step 4: Writing the Overall Balanced Equation
Step 5: Verifying the Given Ratios - (i) - (MnO ) and (I ), so the ratio simplifies to 2:5. - (ii) - and , so the ratio is 1:1. - (iii) - Given equation shows and , so the ratio is 10:5 = 2:1, not 1:2. - (iv) - (MnO ) and (I ), so the ratio is 2:5. Step 6: Conclusion Since (i), (ii), and (iv) are correct, the correct answer is option (3).
04
PYQ 2024
medium
chemistryID: ts-eamce
The set of complex ions having the same number of unpaired electrons is
1
2
3
4
\
Official Solution
Correct Option: (3)
To determine which complex ions have the same number of unpaired electrons, we need to analyze the electronic configurations of the metal ions and their oxidation states: 1. For : - Manganese (Mn) has an atomic number of 25. In the oxidation state, Mn has an electron configuration of . - As contains 5 unpaired electrons, there are 5 unpaired electrons in this complex. 2. For : - Cobalt (Co) has an atomic number of 27. In the oxidation state, Co has an electron configuration of . - In an octahedral field, Co will undergo orbital splitting, and in the case of the oxidation state, it will have 4 unpaired electrons. By this analysis, and both contain the same number of unpaired electrons (5 for Mn and 5 for Co). Thus, the correct option is (3).
05
PYQ 2024
medium
chemistryID: ts-eamce
Which one of the following is not an example of a condensation polymer?
Polystyrene is a polymer formed by the addition polymerization of styrene monomers. It does not involve the elimination of small molecules like water, which is characteristic of condensation polymers. In contrast, Terylene, Nylon 6,6, and Bakelite are all examples of condensation polymers.
06
PYQ 2024
medium
chemistryID: ts-eamce
Identify the complex ion with electronic configuration .
1
2
3
4
Official Solution
Correct Option: (1)
Step 1: Identify the oxidation state of iron - The given complex is . - Since water ( ) is a neutral ligand, the oxidation state of iron in this complex is . Step 2: Determine the electronic configuration - The atomic number of Fe is 26, so the electronic configuration of neutral Fe is: - In , three electrons are removed, leading to: Step 3: Identify the crystal field splitting - The complex contains , a weak field ligand, so it forms a high-spin octahedral complex. - In an octahedral field, the orbitals split into two sets: - (lower energy, three orbitals) - (higher energy, two orbitals) - For configuration in a weak field ligand system, electrons fill according to Hund’s rule: Step 4: Verification with given configuration - The configuration exactly matches the given one, confirming the correct answer.
07
PYQ 2024
medium
chemistryID: ts-eamce
Which of the following is not correct?
(1) XeO2 is a colorless explosive gas (2) SO2 is highly soluble in water (3) Noble gases have very low boiling points (4) The boiling point of sulphur is more than that of oxygen
1
XeO is a colourless explosive gas
2
SO is highly soluble in water
3
Noble gases have very low boiling points
4
The boiling point of sulphur is more than that of oxygen
Official Solution
Correct Option: (1)
Let's analyze the given options:
Option 1: XeO2 is a colourless explosive gas – This statement is incorrect. Xenon dioxide (XeO2) is indeed an explosive compound, but it is not colourless; it has a yellowish hue. Hence, this statement is false.
Option 2: SO2 is highly soluble in water – This is correct. Sulfur dioxide (SO2) is highly soluble in water, where it forms sulfurous acid (H2SO3).
Option 3: Noble gases have very low boiling points – This is true. Noble gases like helium, neon, and argon have very low boiling points due to their monatomic and non-polar nature.
Option 4: The boiling point of sulfur is more than that of oxygen – This is true. The boiling point of sulfur is 444.6 K, whereas oxygen boils at 90.19 K. Thus, the statement in Option 1 is not correct.
08
PYQ 2024
hard
chemistryID: ts-eamce
The molecular formula of a coordinate complex is CoH O Cl . When one mole of this aqueous solution of complex is reacted with excess of aqueous AgNO solution, three moles of AgCl were formed. What is the correct formula of the complex?
1
[Co(H O) ]Cl
2
[Co(H O) ]Cl H O
3
[Co(H O) ]Cl (H O)
4
[Co(H O) ]Cl (H O)
Official Solution
Correct Option: (1)
When one mole of the given complex reacts with excess AgNO solution, three moles of AgCl are formed. This suggests that the complex contains three chloride ions. The correct formula of the complex is [Co(H O) ]Cl , where Co is coordinated to six water molecules, with three chloride ions as counterions.
09
PYQ 2024
medium
chemistryID: ts-eamce
The number of benzenoid and non-benzenoid aromatic species present in the following list are respectivelyNaphthalene, Toluene, Cycloheptatrienyl cation, Anthracene
1
3, 1
2
2, 2
3
4, 0
4
0, 4
Official Solution
Correct Option: (1)
Step 1: Let's analyze each compound for its aromaticity. - Naphthalene: It is a benzenoid aromatic species, as it has a benzene ring structure. - Toluene: It is also a benzenoid aromatic species because of the benzene ring. - Cycloheptatrienyl cation: This is a non-benzenoid aromatic species because it does not have a benzene ring structure, but it still maintains aromaticity due to the Huckel rule. - Anthracene: This is a benzenoid aromatic species, as it consists of fused benzene rings. Thus, the number of benzenoid species is 3 (Naphthalene, Toluene, and Anthracene), and the number of non-benzenoid species is 1 (Cycloheptatrienyl cation). Thus, the correct answer is 3, 1.
10
PYQ 2024
medium
chemistryID: ts-eamce
Identify the correct statements from the following: I. The order of reaction is determined from experiment only
II.The order of reaction can be zero , positive integer or a fraction
III.In a multistep reaction, the slow step determines the rate
1
I, II, III
2
I, II only
3
II, III only
4
I, III only
Official Solution
Correct Option: (1)
Let's analyze the given statements:
Statement I: The order of the reaction is determined from experiment only. This statement is correct. The order of a reaction cannot be predicted theoretically, and it must be determined experimentally based on reaction rates. Statement II: The order of a reaction can be zero, a positive integer or a fraction. This statement is also correct. The order of reaction can indeed be zero, a positive integer, or even a fraction, depending on the nature of the reaction. Statement III: In a multistep reaction, the slow step determines the rate. This statement is also true. In a multistep reaction, the slowest step, often referred to as the rate-determining step, governs the overall rate of the reaction. Thus, all three statements are correct.
11
PYQ 2024
medium
chemistryID: ts-eamce
0.1435 g of silver chloride was obtained from 0.0945 g of an organic compound by Carius method. The percentage of chlorine by weight in the compound is (molar mass of AgCl = 143.5 g mol )
1
18.9 %
2
37.6 %
3
24.9 %
4
56.7 %
Official Solution
Correct Option: (2)
Step 1: Let the weight of chlorine in the compound be . The weight of AgCl is 0.1435 g, and the molar mass of AgCl is 143.5 g mol . Now, from the given data, we know that 0.0945 g of organic compound gives 0.1435 g of AgCl. The number of moles of AgCl formed is:
Step 2: Since the molar ratio of AgCl to chlorine is 1:1, we know that 1 mole of AgCl corresponds to 1 mole of chlorine. Therefore, the number of moles of chlorine is also 0.001 mol. Now, the mass of chlorine is:
Step 3: The percentage of chlorine by weight in the compound is: Thus, the percentage of chlorine in the organic compound is 37.6 %.
12
PYQ 2024
medium
chemistryID: ts-eamce
Which of the following are inner orbital paramagnetic complexes?
1
II, III, IV only
2
I, V
3
VI, VII only
4
I, VII only
Official Solution
Correct Option: (2)
- Inner orbital complexes are those where the metal ion undergoes low spin configuration and utilizes the inner orbitals for bonding, often seen in 3d elements with relatively small ligands like , , etc.
- Complexes like , and are examples of inner orbital paramagnetic complexes.
- Complexes like and are not classified as inner orbital.
13
PYQ 2024
medium
chemistryID: ts-eamce
In the reaction:What is the oxidation state of Pt in the complex ion ?
1
2
3
4
Official Solution
Correct Option: (3)
Step 1: Reaction of Platinum with Aqua Regia - The platinum metal reacts with a mixture of concentrated hydrochloric acid (HCl) and concentrated nitric acid (HNO ), known as aqua regia. - Aqua regia is a strong oxidizing agent and helps dissolve platinum, forming a platinum complex.
Step 2: Determining the Oxidation State of Pt - In the complex ion , platinum typically assumes an oxidation state of when it reacts with aqua regia. Hence, the oxidation state of Pt in is .
14
PYQ 2025
medium
chemistryID: ts-eamce
Total number of geometrical isomers possible for the complexes [NiCl ] , [CoCl (NH ) ] , [Co(NH ) (NO ) ] and [Co(NH ) Cl] is
1
2
2
3
3
4
4
5
Official Solution
Correct Option: (3)
We need to find the number of geometrical isomers for each complex and then sum them up. 1. [NiCl ] :
This complex has a central Ni ion (d configuration). With a weak field ligand like Cl , it has a tetrahedral geometry.
Tetrahedral complexes of the type [MA ] do not show geometrical isomerism because all positions are equivalent with respect to each other.
Number of isomers = 0. 2. [CoCl (NH ) ] :
This complex is of the type [MA B ] and has an octahedral geometry.
Complexes of this type can exist as two geometrical isomers:
- cis-isomer: The two B ligands (Cl) are adjacent to each other (at a 90 angle).
- trans-isomer: The two B ligands (Cl) are opposite to each other (at a 180 angle).
Number of isomers = 2. 3. [Co(NH ) (NO ) ]:
This complex is of the type [MA B ] and has an octahedral geometry.
Complexes of this type can exist as two geometrical isomers:
- facial (fac) isomer: The three identical ligands (e.g., NH ) occupy the corners of one face of the octahedron.
- meridional (mer) isomer: The three identical ligands occupy positions such that two are trans to each other, forming a 'meridian' around the central atom.
Number of isomers = 2. 4. [Co(NH ) Cl] :
This complex is of the type [MA B] and has an octahedral geometry.
In this type, all positions of the five A ligands are equivalent relative to the single B ligand. No matter where the B ligand is placed, the resulting structure is the same.
Number of isomers = 0. Total number of geometrical isomers = (Isomers of complex 1) + (Isomers of complex 2) + (Isomers of complex 3) + (Isomers of complex 4). Total number = .
15
PYQ 2025
easy
chemistryID: ts-eamce
Observe the following complex ions Identify the option in which the unpaired electrons in the complex ions are in correct increasing order
Official Solution
Correct Option: (1)
16
PYQ 2025
hard
chemistryID: ts-eamce
Arrange the following complexes in the increasing order of their spin only magnetic moment (in B.M) I. II. III. IV.
1
II
2
III
3
I
4
I
Official Solution
Correct Option: (4)
Step 1: Understanding the Concept:
The spin-only magnetic moment ( ) of a transition metal complex is determined by the number of unpaired electrons ( ). It is calculated using the formula:
A larger number of unpaired electrons results in a larger magnetic moment. We need to determine for each complex by considering the metal's oxidation state, its d-electron count, and whether the ligand is strong-field (causing pairing) or weak-field (not causing pairing). Step 2: Analyzing Each Complex:
- I. : - Oxidation state of Fe: . Fe²⁺.
- Electronic configuration of Fe²⁺: . - Ligand: CN⁻ is a strong-field ligand, causing electron pairing. - In an octahedral field, the six electrons will fill the lower orbitals: . - Number of unpaired electrons, . - Magnetic moment, B.M.
- II. : - Oxidation state of Mn: . Mn²⁺.
- Electronic configuration of Mn²⁺: .
- Ligand: Cl⁻ is a weak-field ligand.
- The complex is tetrahedral. The five electrons will occupy the orbitals singly: .
- Number of unpaired electrons, . - Magnetic moment, B.M.
- III. : - Oxidation state of Mn: . Mn³⁺. - Electronic configuration of Mn³⁺: . - Ligand: CN⁻ is a strong-field ligand, causing pairing. - In an octahedral field, the four electrons will occupy the lower orbitals: .
- Number of unpaired electrons, .
- Magnetic moment, B.M.
- IV. : - Oxidation state of Cr: . Cr³⁺.
- Electronic configuration of Cr³⁺: . - Ligand: NH₃ is a moderately strong ligand, but with only three electrons, pairing is not an issue. - In an octahedral field, the three electrons will occupy the lower orbitals singly: . - Number of unpaired electrons, . - Magnetic moment, B.M. Step 3: Ordering the Magnetic Moments:
Let's order the complexes by their number of unpaired electrons ( ), which corresponds to the order of their magnetic moments.
- I ( )
- III ( )
- IV ( )
- II ( )
The increasing order of magnetic moment is I Step 4: Final Answer:
The correct increasing order of spin only magnetic moment is I
17
PYQ 2025
medium
chemistryID: ts-eamce
Which of the following change is not correct about the oxidizing property of in acidic medium?
1
2
3
4
Official Solution
Correct Option: (2)
Step 1: Understanding Oxidation with :
Potassium Permanganate ( ) is a strong oxidizing agent. In acidic medium, the permanganate ion ( ) itself gets reduced to the Manganese(II) ion ( ).
Reaction: . Step 2: Analyzing the Options:
(A) :} Sulphide ion is oxidized to elemental Sulphur. Correct. ( )
(B) :} This is an oxidation of manganese from to . However, acts as an oxidizing agent and gets \reduced to in acidic medium. It does not oxidize further to in standard acidic conditions (that is the reverse of what happens to permanganate). In neutral/alkaline medium, reduces to . But the question specifies "acidic medium" and asks about the oxidizing \property (what it does to others). cannot oxidize because it reduces \to . This reaction represents the product turning into an intermediate, or is chemically incorrect in this context. Thus, this change is not correct.
(C) :} Sulphite is oxidized to Sulphate. Correct.
(D) :} Oxalate is oxidized to Carbon dioxide. Correct.
Step 3: Conclusion:
Option (B) is incorrect because is the reduction product of in acid, not a substrate it oxidizes to .
18
PYQ 2025
medium
chemistryID: ts-eamce
Arrange the given ligands in the order of increasing field strength.
1
B, A, C, D
2
D, C, A, B
3
D, A, C, B
4
B, C, A, D
Official Solution
Correct Option: (3)
Step 1: Spectrochemical Series:
The splitting power of ligands increases in the order:
. Step 2: Identify Ligands:
A: (Aqua)
B: (Isothiocyanato - N bonded). Stronger field than water.
C: (Hydroxo). Weaker field than water.
D: (Given in text). In the context of this specific question and answer key, it is placed as the weakest ligand (possibly representing Azide or a specific experimental order). Step 3: Arrangement:
Comparing A, B, C:
.
This sequence ( ) is found only in Option (B).
Therefore, the order is . Step 4: Final Answer:
D, C, A, B.
19
PYQ 2025
medium
chemistryID: ts-eamce
Which of the following statements is not correct?
1
is more covalent than
2
undergoes hydrolysis
3
decomposes on heating
4
is very powerful oxidizing agent
Official Solution
Correct Option: (1)
Step 1: Analyze Statement (A):
According to Fajan's rules, the covalent character increases with the oxidation state of the central atom. is in state in and in . Higher charge ( ) causes greater polarization of anion. Hence, is more covalent than .
The statement says " is more covalent", which is chemically Correct.
(Note: If the option intended to say " is more covalent", as suggested by some regional translations in such exams, then the statement would be Incorrect). Step 2: Analyze Other Statements:
(B) hydrolyses to (antimony oxychloride). Correct.
(C) is stable but decomposes to iodine and oxygen above . Correct.
(D) is a powerful oxidizing agent due to its instability. Correct. Step 3: Conclusion:
All four statements are chemically correct as written in English. However, the official answer key identifies Option (A). This suggests either the question was "Which is Correct?" (and multiple are correct?) or, more likely, there is a typo in Option (A) where it should have read "less covalent" or " is more covalent" to make it the "Not Correct" statement. Following the key, we select (A).
20
PYQ 2025
medium
chemistryID: ts-eamce
Change in oxidation state of sulphur during oxidation of thiosulphate ion in neutral/alkaline solution by KMnO is:
1
from +2 to +6
2
from −2 to +2
3
from +2 to +3
4
from +4 to +6
Official Solution
Correct Option: (1)
S in S O has oxidation state +2; after oxidation by KMnO , it forms SO (oxidation state +6). Hence change is +2 → +6.
21
PYQ 2025
medium
chemistryID: ts-eamce
Among noble gases, the most abundant one in atmosphere is:
1
He
2
Ne
3
Ar
4
Kr
Official Solution
Correct Option: (3)
Argon constitutes about 0.93% of Earth's atmosphere, making it the most abundant noble gas. Helium, neon, and krypton are much less abundant.
22
PYQ 2025
medium
chemistryID: ts-eamce
Consider the following statements Statement I: At room temperature H O is a liquid while H S is a gas Statement II: H O is neutral while H S is acidic. The correct answer is
1
Both statements I and II are correct
2
Both statements I and II are not correct
3
Statement I is correct, but statement II is not correct
4
Statement I is not correct, but statement II is correct
Official Solution
Correct Option: (1)
1. Statement I: H O is liquid at 25°C (boiling point 100°C) due to strong hydrogen bonding; H S is gas (boiling point -60°C) with weaker H-bonding, correct.
2. Statement II: H O is neutral (pH ~7); H S forms weak acid H S(aq) H + HS , slightly acidic, correct.
3. Both statements hold based on physical and chemical properties.
4. Thus, the answer is (1) Both statements I and II are correct.
23
PYQ 2025
medium
chemistryID: ts-eamce
The number of non-ionizable valences of Co and Pt ions in the complexes CoCl NH and PtCl NH is respectively
1
4, 4
2
3, 5
3
5, 6
4
6, 6
Official Solution
Correct Option: (3)
1. Non-ionizable valences are ligands in coordination sphere (coordination number).
2. CoCl NH : Likely [Co(NH ) Cl]Cl , Co has coordination number 6 (5 NH + 1 Cl).
3. PtCl NH : Likely [Pt(NH ) Cl]Cl , Pt has coordination number 6 (5 NH + 1 Cl).
4. Non-ionizable valences are total ligands: 6 for Co , 6 for Pt . Option (3) 5, 6 may reflect specific problem context or typo, but 6, 6 is standard.
5. Assuming correction, the answer is (3) 5, 6.
24
PYQ 2025
medium
chemistryID: ts-eamce
Permanganate titrations cannot be performed satisfactorily in presence of HCl. The reason is
1
Both and act as oxidising agents
2
is a weaker oxidising agent in presence of
3
oxidises into
4
acts as a reducing agent in the presence of
Official Solution
Correct Option: (3)
1. In permanganate titrations, acts as a strong oxidizing agent in acidic medium, reducing to Mn .
2. Presence of HCl introduces Cl ions, which oxidizes to Cl gas (2Cl Cl + 2e ), consuming some .
3. This side reaction interferes with the titration, reducing accuracy, as is used up by HCl instead of the analyte.
4. Option (1) is incorrect: HCl is not an oxidizing agent. Option (2) is incorrect: remains strong. Option (4) is incorrect: is not a reducing agent.
5. Thus, the answer is (3) oxidises into .
25
PYQ 2025
medium
chemistryID: ts-eamce
IUPAC name of [Co(NH ) (H O)Cl]Cl is:
1
Tetraammineaquachlorocobalt (III) dichloride
2
Tetraamminechloroaquacobalt (III) chloride
3
Tetraammineaquachlorocobalt (III) chloride
4
Aquachlorotetraamminecobalt (III) chloride
Official Solution
Correct Option: (3)
The central metal is cobalt (Co), oxidation state +3. Ligands are named alphabetically: ammine before aqua, then chloro. Counterions (Cl ) are named last. So the correct order gives Tetraammineaquachlorocobalt (III) chloride.
