Inorganic Chemistry

01

PYQ 2022

medium

chemistry ID: ap-eapce

The empirical formula of Calgon is

1

2

3

4

02

PYQ 2022

medium

chemistry ID: ap-eapce

Identify the correct statements from the following: The formula of Calgon is (II) Exhausted permutit is regenerated by aqueous KCl solution. (III) Calcium stearate is soluble in water. (IV) In ice crystals, each oxygen is surrounded by four oxygen atoms.

1

III & IV only

2

I & IV only

3

I & III only

4

II & III only

03

PYQ 2022

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chemistry ID: ap-eapce

Identify the species which does **not** exist:

1

2

3

4

04

PYQ 2022

medium

chemistry ID: ap-eapce

What are X and Y respectively in the following reactions?

1

AlCl\textsubscript{3}, CaCl\textsubscript{2}

2

Be\textsubscript{2}C, Al\textsubscript{4}C\textsubscript{3}

3

Al\textsubscript{4}C\textsubscript{3}, CaC\textsubscript{2}

4

CaC\textsubscript{2}, Al\textsubscript{4}C\textsubscript{3}

05

PYQ 2022

medium

chemistry ID: ap-eapce

Which of the following conditions are not suitable for a spontaneous reaction?

1

and at low temperature

2

and

3

and at low temperature

4

and at high temperature

06

PYQ 2022

medium

chemistry ID: ap-eapce

The compound that is used in smoke screens is:

1

2

3

4

07

PYQ 2022

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chemistry ID: ap-eapce

Identify the correct statements:

1

BeSO is soluble in water.

2

BeO is an amphoteric oxide.

3

CO can be obtained by heating CaCO at 1070-1270 K.

4

a, b, c

08

PYQ 2022

medium

chemistry ID: ap-eapce

Identify P and Q in the following reaction: $ $

1

2

3

4

09

PYQ 2023

medium

chemistry ID: ap-eapce

The decomposition products of [FeO] are:

1

Fe O, O

2

FeO, O

3

Fe O, O

4

FeO, Fe O

10

PYQ 2023

easy

chemistry ID: ap-eapce

Which of the following oxides is highly basic?

11

PYQ 2023

easy

chemistry ID: ap-eapce

Identify the correct statements from the following. i. Melting points and boiling points of group 15 elements are much higher than those of corresponding group 14 elements ii. SiO only exists at high temperatures iii. PbI does not exist iv. Buckminster fullerene contain twelve 6-membered carbon rings and twenty 5 - membered carbon rings.

12

PYQ 2023

medium

chemistry ID: ap-eapce

How many of the following are neutral oxides? CO, NO, Na O, N O, Cl O, Al O, As O

1

4

2

5

3

3

4

2

13

PYQ 2023

medium

chemistry ID: ap-eapce

Identify the correct statement

1

is an oxidant

2

is paramagnetic

3

is basic

4

Brass is an alloy of Cu, Sn

14

PYQ 2023

medium

chemistry ID: ap-eapce

The metal purified by Mond pross is X and metal purified by van Arkel method is Y. X and Y respectively are:

1

Ni, Zr

2

Mn, Ti

3

Zn, Ti

4

Mn, Ni

15

PYQ 2023

medium

chemistry ID: ap-eapce

The spin-only magnetic moment of is 5.9 BM. The geometry of the complex and x respectively are

1

tetrahedral, 1

2

square planar, 1

3

square planar, 2

4

tetrahedral, 2

16

PYQ 2023

medium

chemistry ID: ap-eapce

Among the following the correct statements are I. LiH, BeH and MgH are saline hydrides with significant covalent character II. Saline hydrides are volatile III. Electron - precise hydrides are Lewis bases IV. The formula for chromium hydride is CrH The correct option is

1

I, III only

2

II, IV only

3

I, IV only

4

III, IV only

Official Solution

Correct Option:

(3)

17

PYQ 2023

medium

chemistry ID: ap-eapce

Correct order of basic strength of metallic hydroxides

1

2

3

4

18

PYQ 2023

medium

chemistry ID: ap-eapce

Which of the following fluorides of Xenon does not exist?

1

2

3

4

19

PYQ 2024

medium

chemistry ID: ap-eapce

Hydrated sodium aluminum silicate is called:

1

Calgon

2

Zeolite

3

Dead burnt plaster

4

Kaolinite

20

PYQ 2024

medium

chemistry ID: ap-eapce

Which one of the following statements is NOT correct about the compounds of alkaline earth metals?

1

Basic nature increases from Mg(OH)₂ to Ba(OH)₂

2

Thermal stability decreases from BeCO₃ to BaCO₃

3

Solubility of sulphates in water decreases from BeSO₄ to BaSO₄

4

Nitrates of these on heating give oxides

21

PYQ 2025

medium

chemistry ID: ap-eapce

Which of the following elements has the highest first ionization energy?

1

Sodium

2

Magnesium

3

Aluminum

4

Silicon

22

PYQ 2025

medium

chemistry ID: ap-eapce

Which of the following sets are correctly matched? Molecule Hybridization Geometry I. BrF 5 sp 3 d 2 square pyramidal II. XeF 6 sp 3 d 3 Distorted octahedral III. SF 4 dsp 2 square planar IV. PbCl 2 sp linear

Molecule	Hybridization	Geometry
I. BrF₅	sp³d²	square pyramidal
II. XeF₆	sp³d³	Distorted octahedral
III. SF₄	dsp²	square planar
IV. PbCl₂	sp	linear

1

I & IV

2

II & III

3

III & IV

4

I & II

23

PYQ 2025

medium

chemistry ID: ap-eapce

Vessel (A) contains 1 mole each of H and I, and vessel (B) contains 2 moles each of H and I . Both vessels are at the same temperature and reach equilibrium for the reaction H (g) + I (g) 2HI(g). What is the ratio of the equilibrium constants for vessels A and B?

1

for A and B are in the ratio of 1 : 2

2

for A and B are in the ratio of 1 : 1

3

for A and B are equal

4

for A and B are in the ratio of 2 : 1

24

PYQ 2025

medium

chemistry ID: ap-eapce

The reactions which give as one of the products are: I) Complete hydrolysis of II) Complete hydrolysis of III) Complete hydrolysis of

1

I& II only

2

II& III only

3

I& III only

4

I, II, III

25

PYQ 2025

medium

chemistry ID: ap-eapce

The buffer system that helps in maintaining pH of blood between 7.26 and 7.42 is

1

HCN / CN

2

H CO / HCO

3

CH COOH / CH COO

4

H PO / H PO

26

PYQ 2025

medium

chemistry ID: ap-eapce

Which of the following statements are correct regarding lithium and magnesium? I) They react slowly with water II) Their bicarbonates are solids III) Their chlorides are not soluble in ethanol IV) Their nitrates decompose easily on heating

1

II & III only

2

I & III only

3

I & IV only

4

III & IV only

27

PYQ 2025

medium

chemistry ID: ap-eapce

At T(K), value for is 16. In a closed 1 L flask, one mole each of,,, and are taken and heated to T(K). What is the concentration (in mol L) of at equilibrium?

1

0.4

2

0.6

3

1.6

4

1.4

28

PYQ 2025

medium

chemistry ID: ap-eapce

Among,, and, the oxides which react with alkalies are:

1

, only

2

, only

3

, only

4

,, and

29

PYQ 2025

medium

chemistry ID: ap-eapce

In which of the following intramolecular H-bonding is absent? Options: - Salicylic acid - Salicylaldehyde - Quinol - Catechol

1

Salicylic acid

2

Salicylaldehyde

3

Quinol

4

Catechol

30

PYQ 2025

medium

chemistry ID: ap-eapce

Identify the air pollutant which in high concentration leads to stiffness of flower buds?

1

CO

2

SO

3

CO

4

CH

31

PYQ 2025

medium

chemistry ID: ap-eapce

Which of the following will give a white precipitate with solution?

1

CH COOH

2

NaCl

3

C H OH

4

Na CO

32

PYQ 2025

medium

chemistry ID: ap-eapce

Among the following compounds, which one is not responsible for the depletion of ozone layer?

1

CH

2

CFC

3

NO

4

Cl

Official Solution

Correct Option:

(1)

Ozone layer depletion is primarily caused by substances that release halogen radicals (like chlorine and bromine) or certain nitrogen oxides in the stratosphere. - CH (Methane): Methane is a greenhouse gas and contributes to global warming. In the stratosphere, it can react with chlorine radicals to form HCl (), which acts as a reservoir for chlorine, temporarily deactivating it from ozone depletion cycles. Methane itself is not directly considered a primary ozone-depleting substance in the same way as CFCs or halons, though its atmospheric chemistry is complex and can indirectly affect ozone concentrations. However, compared to the others, it's the least direct cause of depletion. - CFCl (Trichlorofluoromethane, a CFC - specifically CFC-11): Chlorofluorocarbons (CFCs) are highly stable in the lower atmosphere but break down in the stratosphere under UV radiation to release chlorine radicals (Cl). These chlorine radicals catalytically destroy ozone molecules.; . CFCs are major ozone-depleting substances. - NO (Nitric Oxide): Nitric oxide and nitrogen dioxide (NO) can participate in catalytic cycles that destroy ozone in the stratosphere. For example,; . Sources of stratospheric NO include natural processes and emissions from high-flying aircraft. - Cl (Chlorine gas): While molecular chlorine itself might not be as stable as CFCs to reach the stratosphere in large quantities from ground-level emissions, if it does reach the stratosphere, UV radiation can dissociate it into chlorine radicals (), which then deplete ozone. The primary concern is substances that transport chlorine to the stratosphere. Comparing the options, CH is the one that is not directly responsible for ozone layer depletion in the same way as CFCs and NO are. While it interacts with ozone chemistry, its primary role isn't catalytic destruction. Therefore, CH is the compound not primarily responsible for ozone depletion among the choices. This matches option (1).

33

PYQ 2025

medium

chemistry ID: ap-eapce

Identify the correct set of molecules with zero dipole moment.

1

CO, NH, H O

2

NH, NF, BF

3

PF, NH, CH

4

CH, BF, CO

34

PYQ 2025

medium

chemistry ID: ap-eapce

Which of the following orders is not correct about the property shown against it?

1

- First ionisation enthalpy

2

- Negative electron gain enthalpy

3

- Negative electron gain enthalpy

4

- Size

35

PYQ 2025

medium

chemistry ID: ap-eapce

The incorrect statement from the following is

1

Classical smog is also called reducing smog.

2

Common components of classical smog are O₃, NO, HCHO.

3

Photochemical smog leads to cracking of rubber and corrosion of metals.

4

Photochemical smog occurs in warm, dry, and sunny climate.

36

PYQ 2025

medium

chemistry ID: ap-eapce

The sum of number of antibonding electrons present in O, O, O is:

1

15

2

9

3

21

4

6

37

PYQ 2025

medium

chemistry ID: ap-eapce

Match the following List-I (Molecule) List-II (Dipole moment in D) A) I) B) II) C) III) D) IV)

List-I (Molecule)	List-II (Dipole moment in D)
A) $%5Ctext%7BHCCl%7D$	I) $1.85%20%5C%2C%20%5Ctext%7BD%7D$
B) $%5Ctext%7BNH%7D_3$	II) $1.07%20%5C%2C%20%5Ctext%7BD%7D$
C) $%5Ctext%7BH%7D_2%5Ctext%7BO%7D$	III) $0.23%20%5C%2C%20%5Ctext%7BD%7D$
D) $%5Ctext%7BNF%7D_3$	IV) $1.47%20%5C%2C%20%5Ctext%7BD%7D$

1

A-II, B-IV, C-I, D-III

2

A-IV, B-III, C-I, D-II

3

A-II, B-I, C-IV, D-I

4

A-III, B-II, C-IV, D-I

38

PYQ 2025

medium

chemistry ID: ap-eapce

Among the following compounds, which one is not responsible for the depletion of ozone layer?

1

CH

2

CFC

3

NO

4

Cl

Official Solution

Correct Option:

(1)

Ozone layer depletion is primarily caused by substances that release halogen radicals (like chlorine and bromine) or certain nitrogen oxides in the stratosphere. - CH (Methane): Methane is a greenhouse gas and contributes to global warming. In the stratosphere, it can react with chlorine radicals to form HCl (), which acts as a reservoir for chlorine, temporarily deactivating it from ozone depletion cycles. Methane itself is not directly considered a primary ozone-depleting substance in the same way as CFCs or halons, though its atmospheric chemistry is complex and can indirectly affect ozone concentrations. However, compared to the others, it's the least direct cause of depletion. - CFCl (Trichlorofluoromethane, a CFC - specifically CFC-11): Chlorofluorocarbons (CFCs) are highly stable in the lower atmosphere but break down in the stratosphere under UV radiation to release chlorine radicals (Cl). These chlorine radicals catalytically destroy ozone molecules.; . CFCs are major ozone-depleting substances. - NO (Nitric Oxide): Nitric oxide and nitrogen dioxide (NO) can participate in catalytic cycles that destroy ozone in the stratosphere. For example,; . Sources of stratospheric NO include natural processes and emissions from high-flying aircraft. - Cl (Chlorine gas): While molecular chlorine itself might not be as stable as CFCs to reach the stratosphere in large quantities from ground-level emissions, if it does reach the stratosphere, UV radiation can dissociate it into chlorine radicals (), which then deplete ozone. The primary concern is substances that transport chlorine to the stratosphere. Comparing the options, CH is the one that is not directly responsible for ozone layer depletion in the same way as CFCs and NO are. While it interacts with ozone chemistry, its primary role isn't catalytic destruction. Therefore, CH is the compound not primarily responsible for ozone depletion among the choices. This matches option (1).

39

PYQ 2025

medium

chemistry ID: ap-eapce

Match the following pollutant concentration limits in water with their effects: List-I (Pollutant Concentration Limit in Water) List-II (Effect) Pollutant 1 Effect 1 Pollutant 2 Effect 2 Pollutant 3 Effect 3 Pollutant 4 Effect 4

List-I (Pollutant Concentration Limit in Water)	List-II (Effect)
Pollutant 1	Effect 1
Pollutant 2	Effect 2
Pollutant 3	Effect 3
Pollutant 4	Effect 4

1

Pb ppb & (IV) Liver damage

2

ppm & (III) Laxative effect

3

ppm & (II) 'Blue baby' syndrome

4

ppm & (I) Brown mottling of teeth \hline \end{tabular} \end{center}

40

PYQ 2025

medium

chemistry ID: ap-eapce

Nature of two oxides of nitrogen X and Y formed in the reaction of sodium nitrite with hydrochloric acid is

1

Both X and Y are acidic in nature

2

X is acidic and Y is neutral in nature

3

Both X and Y are neutral in nature

4

X is amphoteric and Y is neutral in nature

41

PYQ 2025

medium

chemistry ID: ap-eapce

Compound 'X' is prepared commercially by the electrolysis of brine solution. Which of the following is not the use of 'X'?

1

Manufacture of paper

2

Petroleum refining

3

Antichlor

4

Mercirising cotton fabrics

42

PYQ 2025

medium

chemistry ID: ap-eapce

Which of the following orders is not correct for the given property?

1

Li

2

Br

3

C

4

Mg - ionic radius

43

PYQ 2025

medium

chemistry ID: ap-eapce

Select the correct statements from the following A) Aluminium liberates gas with dil. but not with aqueous . B) Formula of sodium metaborate is C) Boric acid is a weak monobasic acid D) For thallium, +1 state is more stable than +3 state.

1

A & B

2

B & C

3

C & D

4

A & D

44

PYQ 2025

medium

chemistry ID: ap-eapce

The order of dipole moments of H\textsubscript{2}O (A), CHCl\textsubscript{3} (B) and NH\textsubscript{3} (C) is

1

B < A < C

2

B < C < A

3

C < B < A

4

C < A < B

45

PYQ 2025

medium

chemistry ID: ap-eapce

Consider the sets I, II and III. Identify the set(s) which is (are) correctly matched. [I.] Staggered ethane>eclipsed ethane \dotfill torsional strain [II.] 2,2-Dimethylbutane>2-methylpentane \dotfill boiling point [III.] cis-But-2-ene>trans-But-2-ene \dotfill dipole moment

1

I, II only

2

II, III only

3

III only

4

I, II, III

46

PYQ 2025

medium

chemistry ID: ap-eapce

Identify the correct graph for an ideal gas (-axis = compressibility factor; -axis = pressure)

1

2

3

4

47

PYQ 2025

medium

chemistry ID: ap-eapce

The increasing order of number of lone pair electrons on the central atom of the following molecules is which of the following? I) II) III) IV)

1

IV < III < II < I

2

I < II < III < IV

3

IV < III < I < II

4

IV < III < II < I

48

PYQ 2025

medium

chemistry ID: ap-eapce

$Match the following: List-I (Atomic number, Z) List-II (Block) A) 112 I) B) 116 II) C) 88 III) D) 100 IV)$

1

$A-III, B-I, C-II, D-IV$

2

$A-III, B-II, C-I, D-IV$

3

$A-IV, B-II, C-III, D-I$

4

$A-II, B-III, C-IV, D-I$

Official Solution

Correct Option: $(2)$

$Step 1: Determine blocks - Atomic number 112: belongs to -block (transition metal). - Atomic number 116: belongs to -block (post-transition metal). - Atomic number 88: belongs to -block (alkaline earth metal). - Atomic number 100: belongs to -block (actinides). Step 2: Match accordingly A-III, B-II, C-I, D-IV is correct.$

List-I (Atomic number, Z)	List-II (Block)
A) 112	I) $s$
B) 116	II) $p$
C) 88	III) $d$
D) 100	IV) $f$

49

PYQ 2025

medium

chemistry ID: ap-eapce

$Identify the correct statements from the following: A) Among alkali metal ions, Li⁺ has the highest hydration enthalpy. B) Boiling point of alkali metals increases from Li to Cs. C) Density of K is less than that of Na and Rb. D) Li has strong tendency to form superoxide.$

1

$A & B$

2

$B & C$

3

$A & C$

4

$A & D$

Official Solution

Correct Option: $(3)$

$- A) Li⁺ has the highest hydration enthalpy due to its small ionic size. - B) Boiling point increases from Li to Cs due to the size and ionization energy of alkali metals. - C) The density of K is less than Na and Rb. - D) Li does not have a strong tendency to form superoxide, unlike other alkali metals. Thus, the correct answer is A & C.$

50

PYQ 2025

medium

chemistry ID: ap-eapce

$Which of the following sets are correctly matched? Molecule Number of lone pairs of electrons on central atom Hybridization I) 1 II) 2 III) 2 IV) 2$

1

$I & II$

2

$II & III$

3

$II & IV$

4

$I & IV$

Official Solution

Correct Option: $(1)$

$- For PCF₃, the central atom (C) has 1 lone pair and forms 3 bonds with fluorine atoms, leading to an sp³ hybridization. - For SO₂, the central atom (S) has 2 lone pairs and forms 2 bonds with oxygen atoms, leading to an sp² hybridization. Therefore, the correct matching is I & II.$

Molecule	Number of lone pairs of electrons on central atom	Hybridization
I) $%5Ctext%7BPCF%7D_3$	1	$%5Ctext%7Bsp%7D%5E3$
II) $%5Ctext%7BSO%7D_2$	2	$%5Ctext%7Bsp%7D%5E2$
III) $%5Ctext%7BSF%7D_4$	2	$%5Ctext%7Bsp%7D%5E3d%5E2$
IV) $%5Ctext%7BCF%7D_4$	2	$%5Ctext%7Bsp%7D%5E3$

51

PYQ 2025

medium

chemistry ID: ap-eapce

$The correct order of the non-metallic character among the elements B, C, N, F and Si is$

1

$B>C>Si>N>F$

2

$Si>C>B>N>F$

3

$F>N>C>B>Si$

4

$F>N>C>Si>B$

Official Solution

Correct Option: $(3)$

$Non-metallic character generally increases across a period (from left to right) and decreases down a group in the periodic table. This is related to electronegativity and ionization energy. Elements that readily gain electrons to form negative ions or share electrons in covalent bonds are more non-metallic. The given elements are: - Boron (B): Group 13, Period 2 - Carbon (C): Group 14, Period 2 - Nitrogen (N): Group 15, Period 2 - Fluorine (F): Group 17, Period 2 - Silicon (Si): Group 14, Period 3 Comparing elements in Period 2 (B, C, N, F): Non-metallic character increases from left to right: B So, F is the most non-metallic, followed by N, then C, then B. Now compare with Silicon (Si). Silicon is in the same group as Carbon (Group 14) but in Period 3. Carbon is in Period 2. Non-metallic character decreases down a group. So, Carbon (C) is more non-metallic than Silicon (Si). Silicon is a metalloid, having properties intermediate between metals and non-metals. Boron is also considered a metalloid but is generally less metallic than Silicon. Combining these trends: Fluorine (F) is the most non-metallic among the given elements. Nitrogen (N) is next. Carbon (C) is next. Between Boron (B) and Silicon (Si): - Boron is in Period 2, Group 13. - Silicon is in Period 3, Group 14. Carbon is more non-metallic than Boron. Silicon is less non-metallic than Carbon. Comparing B and Si: Electronegativity values (Pauling scale): B: 2. 04 C: 2. 55 N: 3. 04 F: 3. 98 Si: 1. 90 Higher electronegativity generally corresponds to higher non-metallic character. Based on electronegativity: F>N>C>B>Si. So the order of decreasing non-metallic character is: F>N>C>B>Si. This matches option (3).$

52

PYQ 2025

medium

chemistry ID: ap-eapce

$How many of the following molecules have two lone pairs of electrons on central atom? SF, BF, ClF, PCl, BrF, XeF, H O, SF$

1

$5$

2

$4$

3

$3$

4

$2$

Official Solution

Correct Option: $(3)$

$Let's determine the number of lone pairs on the central atom for each molecule. Valence electrons: S=6, B=3, Cl=7, P=5, Br=7, Xe=8, O=6, F=7, H=1. 1. SF : Central atom S. Valence e of S = 6. Bonds = 6 (with 6 F). Total e pairs around S = 6. Bonding pairs = 6. Lone pairs = 6 - 6 = 0. 2. BF : Central atom B. Valence e of B = 3. Bonds = 3 (with 3 F). Total e pairs around B = 3. Bonding pairs = 3. Lone pairs = 3 - 3 = 0. (Incomplete octet) 3. ClF : Central atom Cl. Valence e of Cl = 7. Bonds = 3 (with 3 F). Electrons used in bonding = 3. Remaining valence e on Cl = 7 - 3 = 4. Lone pairs = 4/2 = 2. (Total e pairs = (7+3)/2 = 5. Bonding pairs = 3. Lone pairs = 5-3=2). Shape: T-shaped. 4. PCl : Central atom P. Valence e of P = 5. Bonds = 5 (with 5 Cl). Total e pairs around P = 5. Bonding pairs = 5. Lone pairs = 5 - 5 = 0. 5. BrF : Central atom Br. Valence e of Br = 7. Bonds = 5 (with 5 F). Electrons used in bonding = 5. Remaining valence e on Br = 7 - 5 = 2. Lone pairs = 2/2 = 1. (Total e pairs = (7+5)/2 = 6. Bonding pairs = 5. Lone pairs = 6-5=1). Shape: Square pyramidal. 6. XeF : Central atom Xe. Valence e of Xe = 8. Bonds = 4 (with 4 F). Electrons used in bonding = 4. Remaining valence e on Xe = 8 - 4 = 4. Lone pairs = 4/2 = 2. (Total e pairs = (8+4)/2 = 6. Bonding pairs = 4. Lone pairs = 6-4=2). Shape: Square planar. 7. H O: Central atom O. Valence e of O = 6. Bonds = 2 (with 2 H). Electrons used in bonding = 2. Remaining valence e on O = 6 - 2 = 4. Lone pairs = 4/2 = 2. (Total e pairs = (6+2)/2 = 4. Bonding pairs = 2. Lone pairs = 4-2=2). Shape: Bent. 8. SF : Central atom S. Valence e of S = 6. Bonds = 4 (with 4 F). Electrons used in bonding = 4. Remaining valence e on S = 6 - 4 = 2. Lone pairs = 2/2 = 1. (Total e pairs = (6+4)/2 = 5. Bonding pairs = 4. Lone pairs = 5-4=1). Shape: See-saw. Molecules with two lone pairs on the central atom are: - ClF - XeF - H O There are 3 such molecules. This matches option (3).$

53

PYQ 2025

medium

chemistry ID: ap-eapce

$The pair of molecules / ions with the same bond order value is$

1

$B, C$

2

$O, C$

3

$O, O$

4

$H, Li$

Official Solution

Correct Option: $(2)$

$We use Molecular Orbital Theory (MOT) to determine bond orders. Bond Order (B. O.) = . 1. B : Boron (Z=5), 2 B atoms = 10 electrons. MO configuration (up to): . (For B, fills before). Bonding e (N) = 2 (from) + 2 (from) + 2 (from) = 6. Antibonding e (N) = 2 (from) + 2 (from) = 4. B. O. for B = . 2. C : Carbon (Z=6), 2 C atoms = 12 electrons. MO configuration: . N = 2+2+4 = 8. N = 2+2 = 4. B. O. for C = . 3. O : Oxygen (Z=8), 2 O atoms = 16 electrons. MO configuration (for O, fills before in terms of energy, but filling order after is): . N = 2+2+2+4 = 10. N = 2+2+2 = 6. B. O. for O = . 4. O : 15 electrons. Remove one e from of O . N = 10. N = 5. B. O. for O = . 5. O : 17 electrons. Add one e to of O . One will be full, other half. MO for O ends with . N = 10. N = 7. B. O. for O = . 6. H : Hydrogen (Z=1). 1 electron. MO configuration: . N = 1. N = 0. B. O. for H = . 7. Li : Lithium (Z=3). 2 Li atoms = 6 electrons. MO configuration: . N = 2 (from) + 2 (from) = 4. (Core electrons usually ignored for valence B. O, but here total B. O.). N = 2 (from). B. O. for Li = . Comparing bond orders: % Option (1) B (B. O. =1), C (B. O. =2). Different. % Option (2) O (B. O. =2), C (B. O. =2). Same. % Option (3) O (B. O. =2. 5), O (B. O. =1. 5). Different. % Option (4) H (B. O. =0. 5), Li (B. O. =1). Different. The pair with the same bond order is O and C . This matches option (2).$

54

PYQ 2025

medium

chemistry ID: ap-eapce

$The correct order of density of Be, Mg, Ca, Sr is:$

1

2

3

4

Official Solution

Correct Option: $(2)$

$Step 1: Density Trend in Alkaline Earth Metals Density in Group 2 (Alkaline Earth Metals) typically increases down the group, but Beryllium is an exception due to its compact atomic structure. Step 2: Density Values Measured densities (g/cm³): Conclusion Thus, the correct order is:$

55

PYQ 2025

medium

chemistry ID: ap-eapce

$Which of the following is not correct?$

1

$Potassium permanganate on heating gives potassium manganate and manganese dioxide only$

2

$Phosphine is used in smoke screens$

3

$Bleaching action of chlorine is due to oxidation$

4

$Noble gases have very low boiling points$

Official Solution

Correct Option: $(1)$

$Step 1: Understand the thermal decomposition of KMnO . When potassium permanganate (KMnO) is heated, it decomposes to give potassium manganate (K MnO), manganese dioxide (MnO), and oxygen (O), not just K MnO and MnO : Step 2: Verify other options. - Phosphine is indeed used in smoke screens. - Chlorine bleaches by oxidation. - Noble gases have very low boiling points due to weak Van der Waals forces. Hence, option (1) is incorrect.$

56

PYQ 2025

easy

chemistry ID: ap-eapce

$Which of the following statements is incorrect with reference to alkaline earth metals?$

Official Solution

Correct Option: $(1)$

57

PYQ 2025

easy

chemistry ID: ap-eapce

$Consider the following: Assertion (A): CO is poisonous to living beings Reason (R): CO binds to hemoglobin forming carboxyhaemoglobin, which is less stable than oxygen-hemoglobin complex.$

Official Solution

Correct Option: $(1)$

58

PYQ 2025

easy

chemistry ID: ap-eapce

$The electronic configurations of four elements A, B, C, D are given below: (A) 1s22s22p63s1 (B) 1s22s22p63s23p1 (C) 1s22s22p63s2 (D) 1s22s22p63s23p2$

Official Solution

Correct Option: $(1)$

59

PYQ 2025

medium

chemistry ID: ap-eapce

$A molecule has a T-shape. The total number of electron pairs in the valence shell of its central atom is:$

1

2

3

4

Official Solution

Correct Option: $(2)$

$Step 1: Understanding Molecular Geometry A T-shaped molecule forms when: - The central atom has 5 electron pairs, but only three are bonding pairs. - The remaining two are lone pairs, distorting the molecular shape. Step 2: Example of a T-Shaped Molecule A common example is ClF (Chlorine trifluoride). The central atom (Cl) has five electron pairs: - Three bonding pairs (with fluorine atoms). - Two lone pairs, causing repulsion and leading to the T-shape. Conclusion Thus, the correct answer is:$

60

PYQ 2025

easy

chemistry ID: ap-eapce

$The sum of bond order values of and is, which is equal to the sum of bond order values of and . What are and ?$

Official Solution

Correct Option: $(1)$

61

PYQ 2025

medium

chemistry ID: ap-eapce

$Which chloride does not exist as hydrate ?$

1

$MgCl$

2

$CaCl$

3

$LiCl$

4

$KCl$

Official Solution

Correct Option: $(4)$

62

PYQ 2025

medium

chemistry ID: ap-eapce

$The correct order of density of Be, Mg, Ca, Sr is:$

1

2

3

4

Official Solution

Correct Option: $(2)$

$Step 1: Density Trend in Alkaline Earth Metals Density in Group 2 (Alkaline Earth Metals) typically increases down the group, but Beryllium is an exception due to its compact atomic structure. Step 2: Density Values Measured densities (g/cm³): Conclusion Thus, the correct order is:$

63

PYQ 2025

medium

chemistry ID: ap-eapce

$Match the following pollutant concentration limits in water with their effects:$

1

$Pb ppb & (IV) Liver damage$

2

$ppm & (III) Laxative effect$

3

$ppm & (II) 'Blue baby' syndrome$

4

$ppm & (I) Brown mottling of teeth \hline \end{tabular} \end{center}$

Official Solution

Correct Option: $(1)$

$Step 1: Understanding the Effects of Pollutants - Lead (Pb>50 ppb): Liver damage, as lead toxicity affects major organs. - Sulfate (ppm): Laxative effect, leading to digestive disturbances. - Nitrate (ppm): Causes 'Blue baby' syndrome, where oxygen transport in infants is impaired. - Fluoride (ppm): Leads to brown mottling of teeth (dental fluorosis). Conclusion Thus, the correct answer is:$

64

PYQ 2025

medium

chemistry ID: ap-eapce

$The sum of bond order values of and is, which is equal to the sum of bond order values of and . What are and ?$

1

2

3

4

Official Solution

Correct Option: $(2)$

$Step 1: Bond Order Calculation Bond order is determined using the formula: Using Molecular Orbital Theory, the bond orders are: Summing: Step 2: Finding Looking at the bond orders: Summing: Conclusion Thus, the correct answer is:$

65

PYQ 2025

medium

chemistry ID: ap-eapce

$How many of the following lanthanide elements exhibit +4 oxidation state? Ce, Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy$

1

$5$

2

$4$

3

$3$

4

$6$

Official Solution

Correct Option: $(1)$

$Step 1: Recall lanthanides that show +4 oxidation state. Only a few lanthanides can show the +4 oxidation state under specific conditions. The known ones include: Step 2: Count these elements. The total number is 5: Ce, Pr, Nd, Tb, Dy.$

66

PYQ 2025

medium

chemistry ID: ap-eapce

$The incorrect statement from the following is$

1

$Aluminium dissolves in conc. HNO and liberates H gas$

2

$Borazole contains 12 and 3 bonds$

3

$Gallium oxide is amphoteric in nature$

4

$BF is a Lewis acid$

Official Solution

Correct Option: $(1)$

$Step 1: Examine Aluminium's reaction with concentrated HNO Aluminium forms a passive oxide layer (Al O) in concentrated nitric acid due to strong oxidizing nature of HNO, preventing further reaction. So, it does not liberate hydrogen gas. Hence, Statement 1 is incorrect. Step 2: Analyze Borazole bonding Borazole (B N H) has a structure similar to benzene and contains 12 bonds (6 B-N, 3 B-H, and 3 N-H) and 3 bonds (due to resonance). So, Statement 2 is correct. Step 3: Amphoteric nature of Ga O Gallium oxide reacts with both acids and bases, confirming its amphoteric nature. So, Statement 3 is correct. Step 4: BF as Lewis acid Boron trifluoride is an electron-deficient compound and readily accepts a lone pair, making it a classic Lewis acid. So, Statement 4 is correct.$

67

PYQ 2025

medium

chemistry ID: ap-eapce

$The incorrect statement in the following is:$

1

$Ionic hydrides are crystalline in nature$

2

$Group 14 elements form electron precise hydrides$

3

$Covalent hydrides are non-volatile compounds$

4

$Generally, saline hydrides react violently with water$

Official Solution

Correct Option: $(3)$

$Step 1: Understanding hydride classifications. Hydrides are broadly categorized into: Ionic (saline) hydrides — formed with s-block metals; they are crystalline and react violently with water. Covalent hydrides — formed with p-block elements, and they are generally volatile due to their molecular nature. Metallic hydrides — formed with d- and f-block elements. Step 2: Evaluating the options. (1) True — Ionic hydrides like NaH, CaH 2 are crystalline. (2) True — Group 14 elements form electron-precise hydrides like CH 4, SiH 4 . (3) False — Covalent hydrides (e.g., CH 4, NH 3, H 2 O) are typically volatile. (4) True — Saline hydrides react with water to produce hydrogen gas explosively. Thus, the incorrect statement is (3).$

68

PYQ 2025

medium

chemistry ID: ap-eapce

$At 25 C, the percentage of ionization of 'x' M acetic acid is 4.242. What is the value of x?$

1

$0.05$

2

$0.04$

3

$0.02$

4

$0.01$

Official Solution

Correct Option: $(4)$

$Step 1: Use the formula for percent ionization. Given: Step 2: Rearranging and solving for :$

69

PYQ 2025

medium

chemistry ID: ap-eapce

$Which of the following orders are correct against the stated property? I) NaO_2 < RbO_2_2 < Ca(OH)_2_3 < CaCO_3$ — thermal stability$

1

$I & III only$

2

$II & III only$

3

$I & II only$

4

$I, II & III$

Official Solution

Correct Option: $(4)$

$I) Stability of superoxides increases down the group due to increase in cation size which stabilizes the larger anion O better. Hence, NaO_2 < RbO_2_2 < Ca(OH)_2_3 < CaCO_3$.$

70

PYQ 2025

medium

chemistry ID: ap-eapce

$Compound 'X' is prepared commercially by the electrolysis of brine solution. Which of the following is not the use of 'X'?$

1

$Manufacture of paper$

2

$Petroleum refining$

3

$Antichlor$

4

$Mercirising cotton fabrics$

Official Solution

Correct Option: $(3)$

$Compound 'X' prepared by the electrolysis of brine (concentrated NaCl solution) is Sodium Hydroxide (NaOH). The overall reaction is . So, X is NaOH. Now let's examine the uses: 1. Manufacture of paper: NaOH is used in the pulping process to break down wood fibers (Kraft process). This is a use of NaOH. 2. Petroleum refining: NaOH is used to remove acidic impurities (like sulfur compounds, phenols) from petroleum products. This is a use of NaOH. 3. Antichlor: An antichlor is a substance used to remove excess chlorine after bleaching, typically in the textile or paper industry. Common antichlors are sodium thiosulfate (Na S O), sulfur dioxide (SO), sodium bisulfite (NaHSO). NaOH is a strong base and is not used as an antichlor; in fact, it reacts with chlorine but not in the typical antichlor role for removing excess already used chlorine in that context. While NaOH can react with Cl (e. g., to form NaClO and NaCl), its primary function here is not "antichlor". 4. Mercerising cotton fabrics: Mercerization is a treatment for cotton fabric/yarn with a caustic soda (NaOH) solution to improve its luster, strength, and dye affinity. This is a use of NaOH. Therefore, "Antichlor" is not a typical use of NaOH (Compound X). This matches option (3).$

71

PYQ 2025

medium

chemistry ID: ap-eapce

$The pair of molecules / ions with the same bond order value is$

1

$B, C$

2

$O, C$

3

$O, O$

4

$H, Li$

Official Solution

Correct Option: $(2)$

$We use Molecular Orbital Theory (MOT) to determine bond orders. Bond Order (B. O.) = . 1. B : Boron (Z=5), 2 B atoms = 10 electrons. MO configuration (up to): . (For B, fills before). Bonding e (N) = 2 (from) + 2 (from) + 2 (from) = 6. Antibonding e (N) = 2 (from) + 2 (from) = 4. B. O. for B = . 2. C : Carbon (Z=6), 2 C atoms = 12 electrons. MO configuration: . N = 2+2+4 = 8. N = 2+2 = 4. B. O. for C = . 3. O : Oxygen (Z=8), 2 O atoms = 16 electrons. MO configuration (for O, fills before in terms of energy, but filling order after is): . N = 2+2+2+4 = 10. N = 2+2+2 = 6. B. O. for O = . 4. O : 15 electrons. Remove one e from of O . N = 10. N = 5. B. O. for O = . 5. O : 17 electrons. Add one e to of O . One will be full, other half. MO for O ends with . N = 10. N = 7. B. O. for O = . 6. H : Hydrogen (Z=1). 1 electron. MO configuration: . N = 1. N = 0. B. O. for H = . 7. Li : Lithium (Z=3). 2 Li atoms = 6 electrons. MO configuration: . N = 2 (from) + 2 (from) = 4. (Core electrons usually ignored for valence B. O, but here total B. O.). N = 2 (from). B. O. for Li = . Comparing bond orders: % Option (1) B (B. O. =1), C (B. O. =2). Different. % Option (2) O (B. O. =2), C (B. O. =2). Same. % Option (3) O (B. O. =2. 5), O (B. O. =1. 5). Different. % Option (4) H (B. O. =0. 5), Li (B. O. =1). Different. The pair with the same bond order is O and C . This matches option (2).$

72

PYQ 2025

medium

chemistry ID: ap-eapce

$How many of the following molecules have two lone pairs of electrons on central atom? SF, BF, ClF, PCl, BrF, XeF, H O, SF$

1

$5$

2

$4$

3

$3$

4

$2$

Official Solution

Correct Option: $(3)$

$Let's determine the number of lone pairs on the central atom for each molecule. Valence electrons: S=6, B=3, Cl=7, P=5, Br=7, Xe=8, O=6, F=7, H=1. 1. SF : Central atom S. Valence e of S = 6. Bonds = 6 (with 6 F). Total e pairs around S = 6. Bonding pairs = 6. Lone pairs = 6 - 6 = 0. 2. BF : Central atom B. Valence e of B = 3. Bonds = 3 (with 3 F). Total e pairs around B = 3. Bonding pairs = 3. Lone pairs = 3 - 3 = 0. (Incomplete octet) 3. ClF : Central atom Cl. Valence e of Cl = 7. Bonds = 3 (with 3 F). Electrons used in bonding = 3. Remaining valence e on Cl = 7 - 3 = 4. Lone pairs = 4/2 = 2. (Total e pairs = (7+3)/2 = 5. Bonding pairs = 3. Lone pairs = 5-3=2). Shape: T-shaped. 4. PCl : Central atom P. Valence e of P = 5. Bonds = 5 (with 5 Cl). Total e pairs around P = 5. Bonding pairs = 5. Lone pairs = 5 - 5 = 0. 5. BrF : Central atom Br. Valence e of Br = 7. Bonds = 5 (with 5 F). Electrons used in bonding = 5. Remaining valence e on Br = 7 - 5 = 2. Lone pairs = 2/2 = 1. (Total e pairs = (7+5)/2 = 6. Bonding pairs = 5. Lone pairs = 6-5=1). Shape: Square pyramidal. 6. XeF : Central atom Xe. Valence e of Xe = 8. Bonds = 4 (with 4 F). Electrons used in bonding = 4. Remaining valence e on Xe = 8 - 4 = 4. Lone pairs = 4/2 = 2. (Total e pairs = (8+4)/2 = 6. Bonding pairs = 4. Lone pairs = 6-4=2). Shape: Square planar. 7. H O: Central atom O. Valence e of O = 6. Bonds = 2 (with 2 H). Electrons used in bonding = 2. Remaining valence e on O = 6 - 2 = 4. Lone pairs = 4/2 = 2. (Total e pairs = (6+2)/2 = 4. Bonding pairs = 2. Lone pairs = 4-2=2). Shape: Bent. 8. SF : Central atom S. Valence e of S = 6. Bonds = 4 (with 4 F). Electrons used in bonding = 4. Remaining valence e on S = 6 - 4 = 2. Lone pairs = 2/2 = 1. (Total e pairs = (6+4)/2 = 5. Bonding pairs = 4. Lone pairs = 5-4=1). Shape: See-saw. Molecules with two lone pairs on the central atom are: - ClF - XeF - H O There are 3 such molecules. This matches option (3).$

73

PYQ 2025

medium

chemistry ID: ap-eapce

$The correct order of the non-metallic character among the elements B, C, N, F and Si is$

1

$B>C>Si>N>F$

2

$Si>C>B>N>F$

3

$F>N>C>B>Si$

4

$F>N>C>Si>B$

Official Solution

Correct Option: $(3)$

$Non-metallic character generally increases across a period (from left to right) and decreases down a group in the periodic table. This is related to electronegativity and ionization energy. Elements that readily gain electrons to form negative ions or share electrons in covalent bonds are more non-metallic. The given elements are: - Boron (B): Group 13, Period 2 - Carbon (C): Group 14, Period 2 - Nitrogen (N): Group 15, Period 2 - Fluorine (F): Group 17, Period 2 - Silicon (Si): Group 14, Period 3 Comparing elements in Period 2 (B, C, N, F): Non-metallic character increases from left to right: B So, F is the most non-metallic, followed by N, then C, then B. Now compare with Silicon (Si). Silicon is in the same group as Carbon (Group 14) but in Period 3. Carbon is in Period 2. Non-metallic character decreases down a group. So, Carbon (C) is more non-metallic than Silicon (Si). Silicon is a metalloid, having properties intermediate between metals and non-metals. Boron is also considered a metalloid but is generally less metallic than Silicon. Combining these trends: Fluorine (F) is the most non-metallic among the given elements. Nitrogen (N) is next. Carbon (C) is next. Between Boron (B) and Silicon (Si): - Boron is in Period 2, Group 13. - Silicon is in Period 3, Group 14. Carbon is more non-metallic than Boron. Silicon is less non-metallic than Carbon. Comparing B and Si: Electronegativity values (Pauling scale): B: 2. 04 C: 2. 55 N: 3. 04 F: 3. 98 Si: 1. 90 Higher electronegativity generally corresponds to higher non-metallic character. Based on electronegativity: F>N>C>B>Si. So the order of decreasing non-metallic character is: F>N>C>B>Si. This matches option (3).$

74

PYQ 2025

medium

chemistry ID: ap-eapce

$In which of the following reactions is not formed?$

1

$Acidifying an aqueous solution of$

2

$Heating of Boron in air$

3

$Burning of in oxygen$

4

$On heating above$

Official Solution

Correct Option: $(1)$

$\textbf{Step 1: Understand the Formation of }} is a common oxide of boron and is typically formed when boron or its compounds react with oxygen under certain conditions. Let's analyze each option to determine whether is formed. Step 2: Analyze Each Option Option (1): Acidifying an aqueous solution of is sodium tetraborate, also known as borax. When acidified, the reaction produces boric acid ( ), not . The reaction is: $ \text{B}_2\text{O}_3 \text{B}_2\text{O}_3 \text{B}_2\text{O}_3 \text{B}_2\text{H}_6 \text{B}_2\text{H}_6 \text{B}_2\text{O}_3 \text{B}_2\text{O}_3 \text{H}_3\text{BO}_3 370 \, \text{K} \text{H}_3\text{BO}_3 370 \, \text{K} \text{B}_2\text{O}_3 \text{B}_2\text{O}_3 \text{B}_2\text{O}_3 $ is not formed.$

75

PYQ 2025

medium

chemistry ID: ap-eapce

$A gaseous mixture at temperature T(K) consists of dihydrogen and dioxygen, with a total pressure of 2 2 bar. If the mixture is 33.33% dihydrogen and 66.67% dioxygen, what is the approximate partial pressure (in bar) of dihydrogen? (T(K), 2 bar total pressure)$

1

$1.778$

2

$0.222$

3

$1.5$

4

$0.5$

Official Solution

Correct Option: $(4)$

$The total pressure of the mixture is 2 bar. The mole fraction of dihydrogen is 33.33% = . Partial pressure of dihydrogen = mole fraction total pressure = bar. The closest option to 0.67 bar is 0.5, so the approximate partial pressure of dihydrogen is 0.5 bar.$

76

PYQ 2025

medium

chemistry ID: ap-eapce

$For the reactions: NaBH + I A + NaI + H, and A + LiH B, what are A and B, respectively? (Reactions are not balanced)$

1

$B H, LiBH$

2

$B H, (BN)$

3

$BH, CO, LiBH$

4

$BH, CO, B N H$

Official Solution

Correct Option: $(1)$

$First reaction: NaBH + I A + NaI + H . NaBH reacts with I to form diborane (B H). The balanced reaction is: 2NaBH + I B H + 2NaI + H . So, A = B H . Second reaction: A + LiH B. B H + 2LiH 2LiBH . So, B = LiBH . Thus, A = B H, B = LiBH .$

77

PYQ 2025

medium

chemistry ID: ap-eapce

$One mole of an ideal gas is contained in a 1 L vessel at T(K) with a pressure of p atm. The vessel is split into two equal parts. What are the pressure and volume in each part?$

1

2

3

4

Official Solution

Correct Option: $(1)$

$Initially, 1 mol of gas occupies 1 L at pressure and temperature . Using the ideal gas law, : . The vessel is divided into two equal parts, so each part has a volume of L, and the gas splits equally (0.5 mol per part). For each part, apply the ideal gas law: . Thus, each part has a pressure of atm and a volume of L.$

78

PYQ 2025

medium

chemistry ID: ap-eapce

$Which of the following metals is used for measuring high temperatures?$

1

$Al$

2

$Ga$

3

$In$

4

$Sn$

Official Solution

Correct Option: $(2)$

$Gallium (Ga) has a low melting point and a stable liquid range which makes it suitable for high-temperature thermometers. Unlike mercury or alcohol, gallium can be used in higher temperature ranges without decomposing.$

79

PYQ 2025

medium

chemistry ID: ap-eapce

$Match the following: List - I (Metal) List - II (Flame colour) A) Li I) Crimson red B) Na II) Yellow C) Ca III) Brick red D) Ba IV) Apple green$

1

$A-III, B-V, C-IV, D-II$

2

$A-II, B-III, C-IV, D-I$

3

$A-III, B-I, C-IV, D-II$

4

$A-II, B-IV, C-V, D-III$

Official Solution

Correct Option: $(3)$

$Different metals impart characteristic colours to flame: Lithium gives crimson red, Sodium gives yellow, Calcium gives brick red, and Barium gives apple green. These flame colours help in identifying the elements in flame tests.$

List - I (Metal)		List - II (Flame colour)
A)	Li	I)	Crimson red
B)	Na	II)	Yellow
C)	Ca	III)	Brick red
D)	Ba	IV)	Apple green

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Chapter Questions 79 MCQs

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Chapter Questions
79 MCQs