CBSE 12 Chemistry Question Paper-2022 Set-4 by Pavan | Practice Test to Test Your Knowledge
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  3. CBSE 12 Chemistry Question Paper-2022 Set-4
CBSE 12 Chemistry Question Paper-2022 Set-4

CBSE 12 Chemistry Question Paper-2022 Set-4

This mock test includes actual CBSE Class 12 Chemistry board exam questions from the year 2022 Set-4, helping students understand exam trends and practice real paper format

2025-08-14
Pavan Pavan
CBSE Chemistry 2022 Class 12 Grade 12

Duration

26 min

Questions

26

Marking

Negative

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Questions Preview

How will you carry out the following conversions: Propanal to Propane, Ethanal to But-2-enal, Ethanoic acid to ethanamide?

A
Propanal to Propane: Wolf-Kishner Reduction; Ethanal to But-2-enal: Aldol Condensation; Ethanoic acid to ethanamide: Reaction with ammonia followed by heating
B
Propanal to Propane: Clemmensen Reduction; Ethanal to But-2-enal: Friedel-Crafts reaction; Ethanoic acid to ethanamide: Reaction with ammonia
C
Propanal to Propane: Wurtz reaction; Ethanal to But-2-enal: Cannizzaro reaction; Ethanoic acid to ethanamide: Reaction with PCl₅
D
Propanal to Propane: Aldol Condensation; Ethanal to But-2-enal: Clemmensen Reduction; Ethanoic acid to ethanamide: Reaction with LiAlH₄

In the given reaction $N_2(g)+3H_2(g) \to 2NH_3(g)$ the rate of formation of $NH_3$ is $3.6 \times 10^{-4}$ mol $L^{-1}s^{-1}$. Calculate the rate of reaction, and the rate of disappearance of $H_2(g)$.

A
Rate of reaction: 1.8 \times 10⁻⁴ mol L⁻¹s⁻¹; Rate of disappearance of H₂: 5.4 \times 10⁻⁴ mol L⁻¹s⁻¹
B
Rate of reaction: 3.6 \times 10⁻⁴ mol L⁻¹s⁻¹; Rate of disappearance of H₂: 1.2 \times 10⁻⁴ mol L⁻¹s⁻¹
C
Rate of reaction: 5.4 \times 10⁻⁴ mol L⁻¹s⁻¹; Rate of disappearance of H₂: 1.8 \times 10⁻⁴ mol L⁻¹s⁻¹
D
Rate of reaction: 1.2 \times 10⁻⁴ mol L⁻¹s⁻¹; Rate of disappearance of H₂: 3.6 \times 10⁻⁴ mol L⁻¹s⁻¹

State Kohlrausch's law of independent migration of ions. If $\wedge_m$ and $\wedge_m^o$ for $CH_3COOH$ are 48 S cm² mol⁻¹ and 400 S cm² mol⁻¹ respectively, calculate the degree of dissociation ($\alpha$) of $CH_3COOH$.

A
Law states that limiting molar conductivity is the sum of limiting ionic conductivities of cation and anion. \alpha = 0.12.
B
Law states that molar conductivity increases with dilution. \alpha = 0.08.
C
Law states that molar conductivity is directly proportional to concentration. \alpha = 0.24.
D
Law states that conductivity depends on the solvent. \alpha = 0.15.

Write the hybridisation and magnetic character of the following complexes: $[NiCl_4]^{2-}$, $[Co(NH_3)_6]^{3+}$, and $[FeF_6]^{3-}$.

A
$[NiCl_4]^{2-}$: $sp^3$, paramagnetic; $[Co(NH_3)_6]^{3+}$: $d^2sp^3$, diamagnetic; $[FeF_6]^{3-}$: $sp^3d^2$, paramagnetic
B
$[NiCl_4]^{2-}$: $dsp^2$, diamagnetic; $[Co(NH_3)_6]^{3+}$: $sp^3d^2$, paramagnetic; $[FeF_6]^{3-}$: $d^2sp^3$, diamagnetic
C
$[NiCl_4]^{2-}$: $sp^3$, diamagnetic; $[Co(NH_3)_6]^{3+}$: $d^2sp^3$, paramagnetic; $[FeF_6]^{3-}$: $sp^3d^2$, diamagnetic
D
$[NiCl_4]^{2-}$: $dsp^2$, paramagnetic; $[Co(NH_3)_6]^{3+}$: $sp^3d^2$, diamagnetic; $[FeF_6]^{3-}$: $d^2sp^3$, paramagnetic

What is the difference between an Ambidentate ligand and a Bidentate ligand?

A
A bidentate ligand has two donor atoms, but only one can form a coordinate bond. An ambidentate ligand has a single donor atom that can form two coordinate bonds.
B
A bidentate ligand has one donor atom, while an ambidentate ligand has two donor atoms.
C
An ambidentate ligand has two different donor atoms, but only one is used to form a coordinate bond. A bidentate ligand has two donor atoms that both form coordinate bonds.
D
A bidentate ligand can form a chelate ring, while an ambidentate ligand cannot.

Out of $[Fe(NH_3)_6]^{3+}$ and $[Fe(C_2O_4)_3]^{3-}$, which complex is more stable and why?

A
$[Fe(C_2O_4)_3]^{3-}$ because of the chelate effect, where the bidentate oxalate ligand forms stable five-membered rings.
B
$[Fe(C_2O_4)_3]^{3-}$ because the oxalate ligand is a stronger ligand.
C
$[Fe(NH_3)_6]^{3+}$ because it has a higher coordination number.
D
$[Fe(NH_3)_6]^{3+}$ because ammonia is a neutral ligand.

Calculate the emf for the cell: $Zn(s)|Zn^{2+}(0.01 M)||(0.001 M)Ag^{+}|Ag(s)$ Given: $E^{\ominus}{Zn^{2+}/Zn} = -0.76 V$ and $E^{\ominus}{Ag^{+}/Ag} = +0.80 V$.

A
1.56 V
B
1.54 V
C
1.50 V
D
1.59 V

Predict the products formed when $CH_3CH_2CHO$ reacts with the following reagents: PhMgBr and then $H_3O^{+}$, $LiAlH_4$, and HCN.

A
PhMgBr: 1-phenylpropan-1-ol; LiAlH₄: propan-1-ol; HCN: 2-hydroxybutanenitrile
B
PhMgBr: 1-phenylpropan-1-ol; LiAlH₄: propanoic acid; HCN: 2-hydroxybutanoic acid
C
PhMgBr: 1-phenylpropan-1-ol; LiAlH₄: propan-1-ol; HCN: 2-hydroxybutanenitrile
D
PhMgBr: 2-phenylpropan-2-ol; LiAlH₄: propanoic acid; HCN: 2-hydroxybutanoic acid

Define molar conductivity for the solution of an electrolyte. How does it vary with concentration?

A
Molar conductivity is the conductivity of a solution containing one mole of an electrolyte placed between two electrodes. It increases with dilution.
B
Molar conductivity is the conductivity of a solution of an electrolyte. It decreases with dilution.
C
Molar conductivity is the measure of the concentration of an electrolyte. It remains constant with dilution.
D
Molar conductivity is the reciprocal of resistivity. It increases with concentration.

In a reaction $2N_2O_5(g) \to 4NO_2(g)+O_2(g)$, the concentration of $N_2O_5$ decreases from 0.5 mol $L^{-1}$ to 0.4 mol $L^{-1}$ in 10 minutes. Calculate the average rate of this reaction and rate of production of $NO_2$ during this period.

A
Average rate: 5 \times 10⁻³ mol L⁻¹min⁻¹; Rate of production of $NO_2$: 2 \times 10⁻² mol L⁻¹min⁻¹
B
Average rate: 1 \times 10⁻² mol L⁻¹min⁻¹; Rate of production of $NO_2$: 4 \times 10⁻³ mol L⁻¹min⁻¹
C
Average rate: 5 \times 10⁻³ mol L⁻¹min⁻¹; Rate of production of $NO_2$: 1 \times 10⁻² mol L⁻¹min⁻¹
D
Average rate: 2 \times 10⁻³ mol L⁻¹min⁻¹; Rate of production of $NO_2$: 5 \times 10⁻³ mol L⁻¹min⁻¹

Account for the following: Transition metals and their compounds show catalytic activities. Zn, Cd and Hg are non-transition elements. Zr and Hf are of almost identical atomic radii.

A
Catalytic activities due to variable oxidation states; Zn, Cd, Hg have filled d-orbitals; Zr and Hf have similar radii due to lanthanoid contraction.
B
Catalytic activities due to high ionization enthalpy; Zn, Cd, Hg are p-block elements; Zr and Hf have similar radii due to relativistic effect.
C
Catalytic activities due to completely filled d-orbitals; Zn, Cd, Hg have variable oxidation states; Zr and Hf have similar radii due to screening effect.
D
Catalytic activities due to low enthalpy of atomisation; Zn, Cd, Hg are non-transition elements because they are in the d-block; Zr and Hf have different radii due to electronic configuration.

From the given $E^{\ominus}$ values of the first row transition elements, answer the following questions: Why is $E^{\ominus}{Mn^{2+}/Mn}$ value highly negative as compared to other elements? What is the reason for the irregularity in the above E values? Why is $E^{\ominus}{Cu^{2+}/Cu}$ value exceptionally positive?

A
Mn²⁺/Mn is negative due to the stability of d⁵ configuration; irregularity is due to stability of d⁵ and d¹⁰; Cu²⁺/Cu is positive due to its high enthalpy of hydration.
B
Mn²⁺/Mn is negative due to high hydration enthalpy; irregularity is due to half-filled and full-filled d-orbitals; Cu²⁺/Cu is positive due to high enthalpy of atomisation.
C
Mn²⁺/Mn is negative due to high enthalpy of atomisation; irregularity is due to irregular trends in ionization enthalpies; Cu²⁺/Cu is positive due to its low enthalpy of hydration.
D
Mn²⁺/Mn is negative due to low ionization enthalpy; irregularity is due to the small size of the atoms; Cu²⁺/Cu is positive due to its negative enthalpy of atomisation.

Write three differences between Lyophobic sol and Lyophilic sol. State the reason for Brownian motion in colloidal particles.

A
Lyophobic sols are solvent-attracting, stable, and reversible. Lyophilic sols are solvent-repelling, less stable, and irreversible. Brownian motion is due to continuous bombardment by dispersion medium molecules.
B
Lyophilic sols are solvent-attracting, stable, and reversible. Lyophobic sols are solvent-repelling, less stable, and irreversible. Brownian motion is due to continuous bombardment by dispersion medium molecules.
C
Lyophilic sols are prepared by simple mixing. Lyophobic sols require special methods. Brownian motion is due to gravitational forces.
D
Lyophilic sols are unstable. Lyophobic sols are stable. Brownian motion is due to electrostatic repulsion.

Write the dispersed phase and dispersion medium of 'milk'. What is the reason for Brownian motion in colloidal particles? Why does physisorption decrease with an increase in temperature?

A
Dispersed phase: Fat, Dispersion medium: Water. Brownian motion is due to continuous bombardment of the dispersed phase particles by the molecules of the dispersion medium. Physisorption decreases with temperature because it is an exothermic process.
B
Dispersed phase: Water, Dispersion medium: Fat. Brownian motion is due to the electric charge on colloidal particles. Physisorption decreases because the activation energy is high.
C
Dispersed phase: Protein, Dispersion medium: Water. Brownian motion is due to gravitational forces. Physisorption decreases with temperature because the kinetic energy of gas molecules decreases.
D
Dispersed phase: Fat, Dispersion medium: Water. Brownian motion is due to the mutual attraction between particles. Physisorption decreases with temperature because the process is irreversible.

Predict the products formed when $CH_3CHO$ reacts with the following reagents: $CH_3MgBr$ and then $H_3O^{+}$, $Zn(Hg)/Conc. HCl$, and Tollens' reagent.

A
Propan-2-ol with $CH_3MgBr$; Ethane with $Zn(Hg)/Conc. HCl$; Silver mirror with Tollens' reagent.
B
Ethanol with $CH_3MgBr$; Propane with $Zn(Hg)/Conc. HCl$; Silver mirror with Tollens' reagent.
C
Propan-2-ol with $CH_3MgBr$; Ethane with $Zn(Hg)/Conc. HCl$; Red precipitate with Tollens' reagent.
D
Propan-2-ol with $CH_3MgBr$; Propane with $Zn(Hg)/Conc. HCl$; Silver mirror with Tollens' reagent.

The resistance and conductivity of a conductivity cell containing 0.001 M KCl solution at 298 K are 1200 Ω and $1.5 \times 10^{-4}$ S $cm^{-1}$. Calculate its cell constant and molar conductivity.

A
Cell constant: 0.18 cm⁻¹; Molar conductivity: 150 S cm² mol⁻¹
B
Cell constant: 0.15 cm⁻¹; Molar conductivity: 120 S cm² mol⁻¹
C
Cell constant: 1.2 cm⁻¹; Molar conductivity: 1500 S cm² mol⁻¹
D
Cell constant: 1.8 cm⁻¹; Molar conductivity: 15 S cm² mol⁻¹

A reaction is first order in X and second order in Y. How is the rate affected on increasing the concentration of Y three times? How is the rate affected when the concentrations of both X and Y are doubled?

A
Increasing Y by three times increases the rate by a factor of nine. Doubling both X and Y increases the rate by a factor of eight.
B
Increasing Y by three times increases the rate by a factor of three. Doubling both X and Y increases the rate by a factor of four.
C
Increasing Y by three times increases the rate by a factor of nine. Doubling both X and Y increases the rate by a factor of twelve.
D
Increasing Y by three times increases the rate by a factor of six. Doubling both X and Y increases the rate by a factor of six.

Predict the products in the following reactions: $CH_3MgBr + CO_2 \to (a) H_3O^{+}$; $CH_3CN \to (a) DIBAL-H, (b) H_2O$; Benzamide \to H_3O^{+}/\Delta$.

A
Propanoic acid from $CH_3MgBr$; Ethanal from $CH_3CN$; Benzoic acid from Benzamide.
B
Propanoic acid from $CH_3MgBr$; Ethanoic acid from $CH_3CN$; Benzoic acid from Benzamide.
C
Acetic acid from $CH_3MgBr$; Ethanal from $CH_3CN$; Benzaldehyde from Benzamide.
D
Butanoic acid from $CH_3MgBr$; Ethanoic acid from $CH_3CN$; Benzaldehyde from Benzamide.

Arrange the following compounds in increasing order of their property as indicated: $CH_3CHO$, $CH_3CH_2CHO$, $CH_3-C-CH_2-CH_3$, $CH_3-C-CH_3$ (Reactivity towards HCN).

A
$CH_3-C-CH_2-CH_3 < CH_3-C-CH_3 < CH_3CH_2CHO < CH_3CHO$
B
$CH_3CHO < CH_3CH_2CHO < CH_3-C-CH_3 < CH_3-C-CH_2-CH_3$
C
$CH_3CH_2CHO < CH_3-C-CH_3 < CH_3-C-CH_2-CH_3 < CH_3CHO$
D
$CH_3-C-CH_3 < CH_3-C-CH_2-CH_3 < CH_3CHO < CH_3CH_2CHO$

Arrange the following compounds in increasing order of their acidic strength: $CH_3COOH$, $ClCH_2COOH$, $(CH_3)_2CH-COOH$.

A
$(CH_3)_2CH-COOH < CH_3COOH < ClCH_2COOH$
B
$CH_3COOH < ClCH_2COOH < (CH_3)_2CH-COOH$
C
$ClCH_2COOH < (CH_3)_2CH-COOH < CH_3COOH$
D
$CH_3COOH < (CH_3)_2CH-COOH < ClCH_2COOH$

Write the major products in the following: Benzaldehyde \to HNO_3 + H_2SO_4 / 273-283 K; Sodium benzoate \to CaO + NaOH / \Delta; Acetophenone \to NH_2OH.

A
m-nitrobenzaldehyde; Benzene; Acetophenone oxime
B
o-nitrobenzaldehyde; Benzoic acid; Acetophenone oxime
C
p-nitrobenzaldehyde; Benzene; Acetamide
D
m-nitrobenzaldehyde; Benzene; Benzene

Why is the oxidation of Propanal easier than Propanone?

A
Propanone has an alpha-hydrogen atom, which makes it less reactive.
B
Propanal has a hydrogen atom attached to the carbonyl carbon.
C
Propanal is a ketone, which is more easily oxidized.
D
The carbonyl group in propanone is sterically hindered.

How will you distinguish between acetophenone and benzophenone?

A
Acetophenone gives a positive iodoform test, while benzophenone does not.
B
Benzophenone gives a positive Tollens' test, while acetophenone does not.
C
Acetophenone gives a positive Fehling's test, while benzophenone does not.
D
Both give a positive iodoform test, but only benzophenone gives a positive Tollens' test.

Draw the structure of the 2,4-dinitrophenylhydrazone of propanone.

A
The structure shows a propanone molecule bonded to the nitrogen of 2,4-dinitrophenylhydrazine, with the oxygen of propanone and two hydrogens of the hydrazine removed as a water molecule.
B
The structure shows propanone with a hydroxyl group and a phenyl group attached.
C
The structure shows 2,4-dinitrophenylhydrazine with a propene group attached to the nitrogen.
D
The structure shows propanone with two nitro groups attached to the phenyl ring.

What are the common oxidation states of the first-row transition elements? Give a reason for the same.

A
All elements show only +2 oxidation state. Reason: loss of two s-electrons.
B
Variable oxidation states. Reason: The small energy difference between the (n-1)d and ns orbitals allows for the use of electrons from both subshells in bonding.
C
Only the +3 oxidation state. Reason: loss of one s-electron and two d-electrons.
D
Fixed oxidation states. Reason: The d-orbitals are completely filled.

Write the IUPAC name of the following coordination compounds: $[Co(NH_3)_5Cl]Cl_2$ and $[Ni(CO)_4]$.

A
Pentaamminechloridocobalt(III) chloride and Tetracarbonylnickel(0).
B
Chloropentaamminecobalt(II) chloride and Tetracarbonylnickel(0).
C
Pentaamminechloridocobalt(II) chloride and Nickel tetracarbonyl.
D
Pentaamminechloridocobalt(III) dichloride and Tetracarbonylnickel(II).