Given below are two statements: Statement I : [ N i ( C N ) 4 ] 2 − is square planar and diamagnetic complex, with dsp2 hybridizatin for Ni but [ N i ( C O ) 4 ] is tetrahedral, paramagnetic and with sp3 hybridization for Ni. Statement II : [ N i C l 4 ] 2 − a n d [ N i ( C O ) 4 ] both have same d-electron configuration have same geometry and are paramagnetic. In light the above statements, choose the correct answer from the options given below

[Ni (CN)4]2− CN is strong field ligand Ni2+=4s03d8 Here; [Ni (CN)4]2− is square planar and diamagnetic. [Ni (CO)4] Ni = 4s23d8Co is strong field ligand. Here ; [Ni (CO)4] is tetrahedral and diamagnetic [Ni (CN)4]2− has 3d8 configuration while [Ni (CO)4] has 3d10 configuration.

Which of the following compounds (a) and (b) will not react with a mixture of NaBr and H2SO4.Explain why?

This is a short answer type question as classified in NCERT Exemplar (a) Br2 gas is produced when NaBr and H2SO4 are combined. Because of the stable molecule created as a result of resonance stabilisation, molecule (b) Will not react with Br2 gas.

Aryl halides are extremely less reactive towards nucleophilic substitution. Predict and explain the order of reactivity of the following compounds towards nucleophilic substitution:

This is a short answer type question as classified in NCERT Exemplar Because of resonance stabilisation, aryl halides are less reactive to nucleophilic substitution. The inclusion of a -NO2, an electron-withdrawing group in the ortho or para position, enhances the aryl halide's sensitivity to substitution. The more electron withdrawing groups there are in the aryl halide, the more reactive it is. The decreasing sequence of reactivity, according to this theory, is III > II > I.

In the given reaction, X + Y + 3Z ⇌ XYZ3 If one mole of each of X and Y with 0.05 mol of Z gives compound XYZ3. (Given: Atomic masses of X, Y and Z are 10, 20 and 30 amu, respectively.) The yield of XYZ3 is_____________g. (Nearest integer)

x + y + 3z = xyz3 1 mole 1 mole 0.05 mole ∴ n x = 1 n y = 1 n z = 0 . 0 5 3 = 0 . 0 1 6 7 here z is limiting reagent. ? 0 . 0 5 3 mole z gives 1 mole xyz3 ∴ mass of xyz3 = n × molecular mass = 0 . 0 5 3 × ( 1 0 + 2 0 + 3 × 3 0 ) a . m . u . = 0.5 × 4 = 2g

The number of paramagnetic species among the following is B2, Li2, C2, C 2 − , O 2 − 2 , O 2 + and H e 2 + .

B 2 = σ 1 s 2 σ 1 s * 2 σ 2 s 2 σ 2 s * 2 π 2 p x 1 = π 2 p y 1 → Paramagnetic L i 2 = σ 1 s 2 σ 1 s * 2 σ 2 s 2 → D i a m a g n e t i c C 2 = σ 1 s 2 σ 1 s * 2 σ 2 s 2 σ 2 s * 2 π 2 p x 2 ≡ π 2 p y 2 → D i a m a g n e t i c C 2 − = σ 1 s 2 σ 1 s * 2 σ 2 s 2 σ 2 s * 2 π 2 p x 2 ≡ π 2 p y 2 σ 2 p z 1 → P a r a m a g n e t i c O 2 − 2 = σ 1 s 2 σ 1 s * 2 σ 2 s 2 σ 2 s * 2 σ 2 p z 2 π 2 p x 2 ≡ π 2 p y 2 π 2 p x * 2 ≡ π 2 p y * 2 Diamagnetic O 2 + = σ 1 s 2 σ 1 s * 2 σ 2 s 2 σ 2 s * 2 σ 2 p z 2 π 2 p x 2 ≡ π 2 p y 2 π 2 p x * 1 ≡ π 2 p y 0 → Paramagnetic H e 2 + = σ 1 s 2 σ 1 s * 1 → Paramagnetic ∴ Paramagnetic molecules are = B 2 , C 2 − , O 2 + , H e 2 +

150g of acetic acid was contaminated with 10.2g ascorbic acid ( C 6 H 8 O 6 ) to lower down its freezing point by (x × 10-1)°C. The value of x is _________. (Nearest integer) (Given Kf = 3.9 K kg mol-1; molar mass of ascorbic acid = 176 g mol-1]

m = w × 1 0 0 0 m o l e c u l a r w t × w s o l v e n t = 1 0 . 2 × 1 0 0 0 1 7 6 × 1 5 0 = 1 0 2 0 0 1 7 6 × 1 5 0 = 0 . 3 8 6 Δ T f = K f × m 3.9 × 0.386 ∴ x × 1 0 − 1 = 1 5 . 0 5 × 1 0 − 1 Ans =15

Ka for butyric acid (C3H7COOH) is 2 × 10-5. The pH of 0.2 M solution of butyric acid is ____________× 10-1 (Nearest integer) [Given log2 = 0.30]

Ka for C3H7COOH = 2 × 10-5 p K a = − l o g ( 2 × 1 0 − 5 ) = 5 − l o g 2 =5 – 0.3 = 4.7 pH of 0.2 (M) solution = p H = p K a − l o g C 2 = 1 2 ( 4 . 7 ) − 1 2 l o g ( 0 . 2 ) p H = 2 7 × 1 0 − 1 ∴ Ans 27

Match List – I with List – II List – I List – II Name of oxo acid Oxidation state of “P” (a) Hypophosphorous acid (i) + 5 (b) Orthophosphoric acid (ii) + 4 (c) Hypophosphoric acid (iii) + 3 (d) Orthophosphorous acid (iv) + 2 (v) + 1 Choose the correct answer from the options given below:

(a) - (v), (b)-(i), (c)-(ii), (d)-(iii)

(a) - (v), (b)-(iv), (c)-(ii), (d)-(iii)

(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)

a) -(iv), (b)-(v), (c)-(ii), (d)-(iii)

Given below are two statements: Statement I : H 2 O 2 can act as both oxidizing and reducing agent in basic medium. Statement II: In the hydrogen economy, the energy is transmitted in the form of dihydrogen. In the light of the above statements, choose the correct answer from the options given below:

Statement I is true but statement II is false

Both statement I and statement II are false

Statement I is false but statement II is true

Both statement I and statement II are true

Which of the following is Lindlar catalyst?

Partially deactivated palladised charcoal

Cold dilute solution of KMnO4

Zinc chloride and HCl

Given below are two statement: one is labelled as Assertion A and the other is labelled as Reason R: Assertion A: Size of Bk³? ion is less than Np³? ion. Reason R: The above is a consequence of the lanthanoid contraction. In the light of the above statements, choose the correct answer from the options given below:

Both A and R are true but R is not the correct explanation of A

A is true but R is false

Both A and R are true and R is the correct explanation of A

Given below are two statements: Statement I: The E° value for Ce?? / Ce³? is + 1.74V Statement II: Ce is more stable in Ce?? state than Ce³? state In the light of the above statements, choose the most appropriate answer from the options given below:

Both statement I and statement II are incorrect

Statement I is incorrect but statement II is correct

Statement I is correct but statement II is incorrect

Both statement I and statement II are correct

In chromatography technique, the purification of compound is independent of:

Physical state of the pure compound

Mobility or flow of solvent system

Length of the column or TLC plate

Assertion A : Enol form of acetone [CH 3 COCH 3 ] exists in <0.1% quantity. However, the enol form of acetyl acetone [CH 3 COCH 2 OCCH 3 ] exists in approximately 15% quantity. Reason R : Enol form of acetylacetone is stabilized by intramolecular hydrogen bonding, which is not possible in enol form of acetone. Choose the correct statement:

Both A and R are true but R is not the correct explanation of A

Both A and R are true and R is the correct explanation of A

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R: Assertion A: The H-O-H bond angle in water molecule is 104.5°. Reason R: The lone pair- lone pair repulsion of electrons is higher than the bond pair- bond pair repulsion. In the light of the above statements, choose the correct answer from the options given below:

Both A and R are true, but R is not the correct explanation of A

A is false but R is true

Both A and R are true, and R is the correct explanation of A

Match List – I with List – II: List – I List – II Industrial process Application (a) Haber’s process (i) HNO 3 synthesis (b) Ostwald’s process (ii) Aluminium extraction (c) Contact process (iii) NH 3 sythesis (d) Hall-Heroult process (iv) H 2 SO 4 synthesis Choose the correct answer from the options given below:

(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)

(a)- (ii), (b)-(iii), (c)-(iv), (d)-(i)

(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)

Which among the following pairs of Vitamins is stored in our body relatively for longer duration

Thiamine and Vitamin A

The functions of antihistamine are:

Antiallergic and antidepressant

Given below are two statements: Statement I : Both CaCl 2 .6H 2 O and MgCl 2 .8H 2 O undergo dehydration on heating. Statement II : BeO is amphoteric whereas the oxides of other elements in the same group are acidic. In the light of the above statements, choose the correct answer from the options given below:

Both statement I and statement II are true

Statement I is false but statement II is true

Statement I is true but statement II is false

Both statement I and statement II are false

Chemistry NCERT Exemplar Solutions Class 12th Chapter Ten: Overview, Questions, Preparation

Q: Identify the compound Y in the following reaction.

This is a multiple choice answer as classified in NCERT Exemplar The correct answer is (i). Sandmeyer's reaction can be used to synthesise haloarenes from amines. A primary aromatic amine that has been dissolved or suspended in cold aqueous mineral acid is treated with sodium nitrite to create a diazonium salt in this procedure. When this freshly produced salt is combined with cuprous chloride, the diazonium group is replaced with - Cl, resulting in aryl chloride. Option I is the chemical Y, which is an aryl chloride.

Q: What is 'A' in the following reaction?

This is a multiple choice answer as classified in NCERT Exemplar The correct answer is Option (iii). Markonikov's rule can be used to explain the results. The hydrogen from HCl is added to the carbon immediately bonded to the most hydrogen atoms, whereas the -Cl is attached to the carbon directly bonded to the least hydrogens, according to the rule. As a result, the right response is (iii).

Q: Arrange the compounds in increasing order of the rate of reaction towards nucleophilic substitution. (i) (a) < (b) < (c) (ii) (c) < (b) < (a) (iii) (a) < (c) < (b) (iv) (c) < (a) < (b)

This is a multiple choice answer as classified in NCERT Exemplar The Correct Answer is Option (iii). Due to the resonance stabilisation of the benzene ring, the reactivity of aryl halides to nucleophilic substitution is exceedingly low. Because of resonance, the -Cl bond gains a partial double bond. The presence of an electron withdrawing -NO2 group at ortho or para positions on the ring enhances the reactivity of aryl halides. The presence of -NO2 near the C-Cl makes the molecule more reactive. The order of reactivity should be (b) > (c) > (a). The right option is (iii).

Q: An element M crystallises in a body centred cubic unit cell with a cell edge of 300 pm. The density of the element is 6.0g cm-3. The number of atoms in 180g of the element is_________× 1023 (Nearest integer)

B.C.C structure a = 300 pm = 300 × 10-12 m d = 6g/cm3 z = 2 d = Z × M a 3 6 = 2 × A ( 3 0 0 × 1 0 − 1 0 ) 3 = 2 × A 2 7 × 1 0 − 2 4 ∴ A t o m s o f M = 3.69 × 6.022 × 1023 = 22.22 × 1023 the nearest integer = 22

Updated on Jul 24, 2025 11:47 IST

By Raj Pandey

Table of content

Haloalkanes and Haloarenes Questions and Answers
28th June 2022(First Shift)
Haloalkanes and Haloarenes Multiple Choice Type Questions
JEE Mains 2022

28th June 2022(First Shift)

Commonly asked questions

Given below are two statements:

Statement I : ${[N i {(C N)}_{4}]}^{2 -}$ is square planar and diamagnetic complex, with dsp² hybridizatin for Ni but $[N i {(C O)}_{4}]$ is tetrahedral, paramagnetic and with sp³ hybridization for Ni.

Statement II : ${[N i C l_{4}]}^{2 -} a n d [N i {(C O)}_{4}]$ both have same d-electron configuration have same geometry and are paramagnetic.

In light the above statements, choose the correct answer from the options given below

${[N i {(C N)}_{4}]}^{2 -}$

CN is strong field ligand

$N i^{2 +} = 4 s^{0} 3 d^{8}$

Here; ${[N i {(C N)}_{4}]}^{2 -}$ is square planar and diamagnetic.

$[N i {(C O)}_{4}]$

Ni = 4s²3d⁸Co is strong field ligand.

Here ; $[N i {(C O)}_{4}]$ is tetrahedral and diamagnetic

${[N i {(C N)}_{4}]}^{2 -}$ has 3d⁸ configuration while $[N i {(C O)}_{4}]$ has 3d¹⁰ configuration.

The major product (P) in the reaction

Consider the image below

The correct structure of product ‘A’ formed in the following reaction, is

Consider the following image

Identify the major product formed in the following sequence of reactions:

Consider the image below

The incorrect statement about the imperfections in solids is

· Density of solid decreases in Schottky defect and vacancy defect.

· Density of solid increases in interstitial defect.

· Density of solid remain unchanged in Frenkel defect.

Dihydrogen reacts with CuO to give

$C u O + H_{2} \to C u + H_{2} O$

A 2.0g sample containing MnO₂ is treated with HCl liberating Cl₂. The Cl₂ gas is passed into a solution of Kl and 60.0mL of 0.1M Na₂S₂O₃ is required to titrate the liberated iodine. The percentage of MnO₂ in the sample is_______. (Nearest integer)

[Atomic masses (in u) Mn = 55; Cl = 35.5; O = 16, I = 127, Na = 23, K = 39, S = 32]

$M n O_{2} + 4 H C l \to M n C l_{2} + C l_{2} + 2 H_{2} O$

$\begin{array}{l} C l_{2} + 2 K l \to 2 K C l + I_{2} \\ I_{2} + 2 N a_{2} S_{2} O_{3} \to 2 N a l + N a_{2} S_{4} O_{6} \end{array}$

Here, meq of MnO₂ = meq of Na₂S₄O₆

$\frac{w}{E} \times 1000 = M \times n - f a c t o r \times V$

$\frac{w}{87 / 2} \times 1000 = 0.1 \times 1 \times 60$

Mass of MnO₂ in sample = 0.261 g

Percentage of MnO₂ in sample = $\frac{0.261}{2} \times 100$

= 13.05%

The quantity of electricity in Faraday needed to reduce 1 mol of $C r_{2} O_{7}^{2 -}$ to Cr³⁺ is________.

$C r_{2} O_{7}^{2 -} + 6 e^{-} + 14 H^{+} \to 2 C r^{3 +} + 7 H_{2} O$

Here, 6F electricity is required to reduce 1 mol $C r_{2} O_{7}^{2 -}$ to Cr³⁺.

For a first order reaction A→B, the rate constant, k = 5.5 × 10^-¹⁴s^-¹. The time required for 67% completion of reaction is x × 10^-¹ times the half life of reaction. The value of x is ________ (Nearest integer)

(Given : log 3 = 0.4771)

Rate constant, k = 5.5 × 10^-14 s^-1.

$t = \frac{2.303}{k} l o g \frac{{[R]}_{0}}{[R]}$

= $\frac{2.303}{k} l o g 3 - (i)$

$t_{50 %} = \frac{2.303}{k} l o g 2 - (i i)$

From (i) & (ii)

$\frac{t_{67 %}}{t_{50 %}} = \frac{l o g 3}{l o g 2}$

= 1.58 t_50%

So; t_67%is 15.8 × 10^-1 times half life.

X = 16 (the nearest integer)

Number of complexes which will exhibit synergic bonding amongst $[C r {(C O)}_{6}], [M n {(C O)}_{5}]$ and $[M n_{2} {(C O)}_{10}]$ is_______________.

All metal carbonyls have synergic bonds.

In the estimation of bromine, 0.5g of an organic compound gave 0.40g of silver bromide. The percentage of bromine in the given compound is____________% (Nearest integer)

(Relative atomic masses of Ag and Br are 18 u and 80 u, respectively).

Organic compound → AgBr (s)

0.5 g 0.40 g

Here; Moles of Br in organic compound = Moles of Br in Ag Br

= Moles of AgBr

= $\frac{0.40}{188} m o l$

Mass of Br in organic compound = $\frac{0.4}{188} \times 80 g$

$\begin{array}{l} % o f B r = \frac{0.4 \times 80}{188 \times 0.5} \times 100 % \\ = 34 % \end{array}$

The Zeta potential is related to which property of colloids

Potential difference between fixed layer and mobile layer of lyophobic colloid is called zeta potential or electro kinetic potential.

This potential difference is related to charge on the surface of colloidal particles.

Element “E” belongs to the period 4 and group 16of the periodic table. The valence shell electron configuration of the element, which is just above “E” in the group is

General valence shell electronic configuration of group 16 elements is ns²np⁴. For 3^rd period element which is above element “E” is 3s²3p⁴

Given are two statements one is labeled as Assertion A and other is labeled as Reason R.

Assertion A : Magnesium can reduce Al₂O₃ at a temperature below 1350°C, while above 1350°C aluminium can reduce MgO

Reason R : The melting and boiling points of magnesium are lower thatn those of aluminium.

In light of the above statements, choose most appropriate answer for the options given below

Below 1350° C, Mg can reduce Al₂O₃ and above 1350°C, Al can reduce MgO (from Ellingham diagram).

Melting and boiling point of Mg are lower than that of Al.

Nitrogen gas is obtained by thermal decomposition of

N₂ gas is obtained by thermal decomposition of Ba (N₃)₂ as,

$B a {(N_{3})}_{2} \to_{}^{Δ} B a + 3 N_{2} ↑$

Given below are two statements:

Statement I : The pentavalent oxide of group-15 element, E₂O₅, is less acidic than trivalent oxide, E₂O₃, of the same element.

Statement II : The acidic character of trivalent oxide of group 15 elements, E₂O₃ decreases down the group.

In light of the above statements, choose most appropriate answer form the options given below

Pentavalent oxides of group – 15 elements, E₂O₅ is more acidic than trivalent oxides, E₂O₃ of the same element.

Acidic strength of trivalent oxides decreases down the group as metallic strength increases.

Which one of the lanthanoids given below is the most stable in divalent form?

Valence shell electronic configuration of Eu in +2 O.S is,

$E u^{2 +} = 6 s^{0} 4 f^{7}$

It has half filled f-sub shell and stable.

Which amongst the following is not a pesticide?

DDT, organophosphates and dieldrin are pesticides.

Which one of the following techniques is not used to spot components of a mixture separated in thin layer chromatographic plate?

In thin layer chromatography, the spots of colourless compounds, which are invisible to eyes can be detected by putting the plate under UV light.

Another detection technique is to place the plate in a covered jar containing I₂ (s). Sometimes an appropriate reagent may also be sprayed on the plate, ninhydrin in case of amino acids

Which one of the following structures are aromatic in nature?

A primary aliphatic amine on reaction with nitrous acid in cold (273K) and there after raising temperature of reaction mixture to room temperature (298K), gives a / an

$R - N H_{2} \to_{273 K}^{H N O_{2} (a q .)} R - \overset{\oplus}{N_{2}} \to_{298 K}^{H_{2} O / - N_{2} ↑} R - O H$

1° aliphatic amine Diazonium salt Alcohol

Which one of the following is NOT a copolymer?

Copolymer is the polymer formed by two or more monomers having multiple bonds.

Buna – S, PHBV and butadiene – styrene are copolymers. Neoprene is homopolymer.

Stability of α- Helix structure of proteins depends upon

$α -$ Helix (2°) structure of protein is stabilized by intermolecular H-bonding.

The formula of the purple colour formed in Laissaigne’s test for sulphur using sodium nitroprusside is

$\underset{\begin{array}{l} S o d i u m e x t r a c t \\ o f s u l p h u r \end{array}}{N a_{2} S} + N a_{2} \underset{S o d i u m n i t r o p r u s s i d e}{[F e {(C N)}_{5} N O]} \to N a_{4} \underset{v i o l e t / p u r p l e}{[F e {(C N)}_{5} N O S]}$

If the work function of a metal is 6.63 × 10^-¹⁹J, the maximum wavelength of the photon required to remove a photoelectron from the metal is _______nm. (Nearest integer)

[Given h = 6.63 × 10^-³⁴ Js, and c = 3 × 10⁸ ms^-¹]

Work function, $?_{0} = 6.63 \times 1 0^{- 19} J$

Threshold wavelength $λ_{0} = ?$

Using ; $?_{0} = \frac{h c}{λ_{0}}$

$λ_{0} = \frac{h c}{?_{0}}$

= 300 × 109 m = 300 nm

The hybridization of P exhibited in PF₅ is sp^xd^y. The value of y is

Hybridization of P in PF₅ is sp³d, so value of y = 1.

4.0 L of an ideal gas is allowed to expand isothermally into vacuum until the total volume is 20L. The amount of heat absorbed in this expansion is _________L atm.

For expansion in vacuum, workdone, w = 0

For isothermal process, $Δ U = 0$

According to first law of thermodynamics,

$\begin{array}{l} Δ U = q + w \\ q = 0 \end{array}$

The vapour pressure of two volatile liquid A and B at 25°C are 50 Torr and 100 Torr, respectively. If the liquid mixture contains 0.3 mole fraction of A, then the mole fraction of liquid B in the vapour phase is $\frac{x}{17} .$ The value of x is____________.

$P_{A}^{0} = 50 t o r r$

$P_{B}^{0} = 100 t o r r$

Mole fraction of A in liquid phase, x_A = 0.3

Mole fraction of B in liquid phase, x_B = 0.7

Now; $P_{A} = P_{A}^{0} x_{A}$

$P_{B} = P_{B}^{0} x_{B}$

= 100 × 0.7 = 70 torr

Mole fraction of B in vapour phase, $y_{B} = \frac{P_{B}}{P_{A} + P_{B}} = \frac{70}{85}$

$\frac{70}{85} = \frac{x}{17}$

X = 14

The solubility product of a sparingly soluble salt A₂X₃ is 1.1 × 10^-²³. If specific conductance of the solution is 3 × 10^-¹, the limiting molar conductivity of the solution is x × 10^-³ S m² mol^-¹. The value of x is_________.

Using ; k_sp = 108 S⁵

1.1 × 10^-23 = 108 S⁵

S = ${(\frac{110}{108} \times 1 0^{- 25})}^{1 / 5} M \approx 1 0^{- 5} M$

Specific conductance,

$λ_{m}^{0} = \frac{κ}{5 \times 1000} S m^{2} m o l^{- 1}$

= 3 × 10^-3 Sm² mol^-1

So; x = 3

Which one of the following compounds in inactive towards S_N1 reaction?

Consider the image below

Haloalkanes and Haloarenes Multiple Choice Type Questions

1. The order of reactivity of following alcohols with halogen acids is

(i) (A) > (B) > (C)

(ii) (C) > (B) > (A)

(iii) (B) > (A) > (C)

(iv) (A) > (C) > (B)

Ans: The Correct Answer is Option (ii). Haloalkanes are made by combining alcohols with halogen acids, in which the hydroxyl group of the alcohol is replaced by the halogen. Primary, secondary, and tertiary alcohols are represented by options (A), (B), and (C). Tertiary alcohols are more reactive than secondary and primary alcohols, and they generate haloalkanes from haloacids without the need of catalysts at ambient temperature. Alcohols have a reactivity order of 3∘>2∘>1∘. As a result, the proper option is (ii).

2. Which of the following alcohols will yield the corresponding alkyl chloride on reaction with concentrated HCl at room temperature?

Ans: The Correct Answer is Option (iv)Because the tertiary carbocation is the most stable, it interacts the most with concentrated HCl. As a result, for tertiary alcohol, room temperature is sufficient for the reaction. However, for primary and secondary alcohols, the presence of a catalyst (ZnCl₂) is required.

As a result, option (iv) is accurate.

3. Identify the compound Y in the following reaction.

Ans: The correct answer is (i). Sandmeyer's reaction can be used to synthesise haloarenes from amines. A primary aromatic amine that has been dissolved or suspended in cold aqueous mineral acid is treated with sodium nitrite to create a diazonium salt in this procedure. When this freshly produced salt is combined with cuprous chloride, the diazonium group is replaced with - Cl, resulting in aryl chloride. Option I is the chemical Y, which is an aryl chloride.

4. Toluene reacts with a halogen in the presence of iron (III) chloride giving ortho and para halo compounds. The reaction is

(i) Electrophilic elimination reaction

(ii) Electrophilic substitution reaction

(iii) Free radical addition reaction

(iv) Nucleophilic substitution reaction

Ans: The Correct Answer is Option (ii). Toluene is an aromatic hydrocarbon (C₆H₅---CH₃) that may be treated with halogens and utilised in an electrophilic substitution process to create aryl halides in the presence of the Lewis acid catalyst iron (III) chloride. In the absence of light, the reaction is carried out using chlorine or bromine, and the products are o- and p- haloarenes.

5. Which of the following is halogen exchange reaction?

Correct Answer: option (i)

Ans: The correct answer is option (i). Option (i) is an example of a Finklestein reaction, which is a halogen exchange reaction that produces alkyl iodide by treating alkyl halides with NaI in dry acetone. Option (ii) is an addition reaction in which an alkene is transformed to the equivalent alkyl halide. Option (iii) is a substitution reaction, whereas option (iv) is an electrophilic substitution reaction.

6.Which reagent will you use for the following reaction?

(i) light

(ii) NaCl+H₂SO₄

(iii) Cl₂ gas in dark

(iv) Cl₂gas in the presence of iron in dark

Ans: The correct Answer is Option (i).

Chlorination in the presence of UV light can be used to produce alkyl chlorides from alkanes. Under the action of UV radiation, the chlorine molecule generates free radicals, which react with alkanes to create a mixture of isomeric mono- and poly haloalkanes.

7. Arrange the following compounds in the increasing order of their densities.

Ans: The Correct Answer is Option (i).

Alkyl halides have a higher density than water. Their densities are determined by the masses of the halogen atoms, the number of halogen atoms, and the number of carbon atoms. Simply expressed, Br has an atomic mass of 79 while Cl has an atomic mass of 35. (d) will be the heaviest of the molecules, followed by (c), (b), and (a). Because density is exactly proportional to mass, the order of decreasing densities will be the same. The right answer is (i).

8. Arrange the following compounds in increasing order of their boiling points.

Ans: The Correct Answer is Option(iii).

Isomeric alkyl halides' boiling points are related to their branching, with a reduction in B.P. as branching increases. As a result, the boiling point of the tertiary isomer is the lowest, while that of the primary isomer is the highest. Thus, the order of lowering boiling points is (b) > (a) > (c).

9. In which of the following molecules carbon atom marked with asterisk (*) is asymmetric?

(i) (a), (b), (c), (d)

(ii) (a), (b), (c)

(iii) (b), (c), (d)

(iv) (a), (c), (d)

Ans: The correct answer is (ii).The mirror copies of the molecules (a), (b), and (c) are non-superimposable. Because all four species linked to the carbon atom are identical, molecule (d) cannot have an asymmetric carbon atom, thus the mirror image of (d) when rotated 180∘ is superimposable on the original picture.

10. Which of the following structures is enantiomeric with the molecule (A) given below :

Ans: The Correct Answer is Option (i).

Enantiomers are stereoisomers of a substance that are not superimposable on each other. The carbon atom in the molecule (A) is asymmetric. The molecule's mirror image (i) and 180∘ rotation cannot be superimposed on each other (A). The molecules (ii), (iii), and (iv) can all be superimposed on one other (A). The molecules (iii) and (iv) have 180∘ rotations that are superimposable on one another (A).

11. Which of the following is an example of vic-dihalide?

(i) Dichloromethane

(ii) 1,2 -dichloroethane

(iii) Ethylidene chloride

(iv) Allyl chloride

Ans: The Correct Answer is Option (iv).

Geminal halides or gem-dihalides and vicinal halides or vic-dihalides are dihaloalkanes with the same halogen. Vic-dihalides are dihaloalkanes with halogen atoms on two adjacent carbon atoms, whereas Gem-dihalides are molecules with halogen atoms on two adjacent carbon atoms. Gem-dihalides are called alkylidene halides in the conventional naming system, whereas vic-dihalides are called alkylene dihalides. Because dichloromethane has just one carbon, it cannot contain neighbouring halogen atoms. Two carbon atoms are sandwiched between two halogen atoms in 1,2-dichloroethane. Ethylidene chloride is a gem-dihalide, as its name implies. There is only one chlorine atom in allyl chloride.

12. The position of −Br in the compound in CH₃CH=CHC(Br)(CH₃)₂ can be classified as

(i) Allyl

(ii) Aryl

(iii) Vinyl

(iv) Secondary

Ans: The correct answer isOption (i).

When the halogen atom is linked to an sp³-hybridised carbon atom next to an allylic carbon, a carbon-carbon double bond is formed.

Aryl halides are produced when a halogen atom is directly linked to an aromatic ring's sp²-hybridised carbon atom.

When a halogen atom is linked to a sp²-hybridised carbon atom in a carbon-carbon double bond, vinylic bonds are produced. Simply defined, it's an ethylene molecule with one hydrogen atom substituted by another --R group.

When the halogen is attached to a secondary carbon atom in the molecule, secondary bonds form. The −Br is linked to a tertiary carbon atom in the supplied molecule, which is then bonded to an sp³-hybridised carbon-carbon double bond. As a result, this −Br has an allylic position.

13. Chlorobenzene is formed by reaction of chlorine with benzene in the presence of .Which of the following species attacks the benzene ring in this reaction?

(i) Cl−

(ii) Cl+

(iii) AlCl₃

(iv) [AlCl₄]⁻

Ans: The correct answer is option ii.

Aluminum chloride is a Lewis acid catalyst that functions similarly to FeCl₃. By chlorinating benzene in the presence of AlCl₃, benzene is transformed to chlorobenzene. Electrophilic substitution is used to carry out the reaction. Cl₂ forms a coordination complex with AlCl₃ called Cl + AlCl₄^- , which has a small positive charge on Cl and is negatively charged on AlCl₄^- . This Cl⁺ then interacts with the benzene ring's aromatic double bond to generate an addition product, followed by deprotonation to produce chlorobenzene, AlCl₃, and HCl as side products. The best choice is (ii).

14. Ethylidene chloride is a/an

(i) vic-dihalide

(ii) gem-dihalide

(iii) allylic halide

(iv) vinylic halide

Ans: The Correct Answer is Option (ii).

Dihaloalkanes with two halogen atoms of the same kind bonded to the same carbon atom are known as gem-dihalides. Alkylidene dihalides is the most frequent naming scheme for gem-dihalides. As a result, ethylidene dichloride is a gem-dihalide. The best choice is (ii).

15. What is 'A' in the following reaction?

Ans: The correct answer is Option (iii).

Markonikov's rule can be used to explain the results. The hydrogen from HCl is added to the carbon immediately bonded to the most hydrogen atoms, whereas the --Cl is attached to the carbon directly bonded to the least hydrogens, according to the rule. As a result, the right response is (iii).

16. A primary alkyl halide would prefer to undergo
(i) SN1 reaction
(ii) SN2 reaction
(iii) 𝛂-Elimination
(iv) Racemisation

Ans: The Correct Answer is Option (ii).

When an alkyl halide with -hydrogen atoms reacts with a base or a nucleophile, depending on the nature of the alkyl halide, the strength of the base or nucleophile, the size of the molecules, and the reaction circumstances, it can undergo a specific kind of reaction. A substitution or elimination process can be followed by an alkyl halide, as well as the two kinds of substitution, S_N¹ and S_N². Because the entering nucleophile cannot engage with the bulky substituents on or near the carbon atom, the main alkyl halide prefers the S_N²reaction. As a result, primary alkyl halides and methyl halides have the greatest proclivity for S_N²reactions.

17. Which of the following alkyl halides will undergo S_N¹ reaction most readily?

(i) (CH₃)₃C−F

(ii) (CH₃)₃C −Cl

(iii) (CH₃)₃C −Br

(iv) (CH₃)₃C −I

Ans: The Correct Answer is Option (iv).

S_N¹reactions occur mostly in polar protic solvents such as H₂O and follow first-order kinetics. This indicates that the reaction rate is solely determined by one reactant. Because of the great stability of the generated carbocation, this reaction favours tertiary alkyl halides. When a molecule is polarised in water, it generates a carbocation as well as a halide ion. The halides' reactivity is R--I > R--Br > R--Cl > > R--F. As a result, (CH₃)₃C---I will be the most likely to undergo the reaction.

18. Which is the correct IUPAC name for :

(i) 1-Bromo-2-ethylpropane

(ii) 1 -Bromo-2-ethyl-2-methylethane

(iii) 1 -Bromo-2-methylbutane

(iv) 2-Methyl-1-bromobutane

Correct Answer: Option (iii)

Ans: The correct answer is option iii.

First, we must determine which carbon chain is the longest. After that, we should have CH₃---CH₂---CH(CH₃)---CH₂---Br as the real structure. --Br, because the functional halide group is linked to the first carbon atom, we begin numbering there. The methyl group branch is joined to the chain's second carbon atom. Butane is named from the number of carbons in the unbranched parent chain, which is four. 1-Bromo-2-methylbutane is the name of the molecule. Option (iii) is the right answer.

19. What should be the correct IUPAC name for diethylbromomethane?

(i) 1 -Bromo- 1,1 -diethylmethane

(ii) 3-Bromopentane

(iii) 1-Bromo-1-ethylpropane

(iv) 1-Bromopentane

Ans: The Correct Answer is Option (ii).

We may conclude from the given compound's common name that there are two --C₂H₅ groups and one --Br group linked to a CH₄ molecule. Bromomethane would be represented by the symbol CH₃---Br. Two ethyl groups can be replaced for the hydrogens in the methyl molecule to make diethylbromomethane (C₂H₅)CH(C₂H₅)Br. The longest parent chain must be identified in order to offer an IUPAC name. The parent chain is CH₃---CH₂---CH(Br)---CH₂---CH₃, with the --Br group from either side connected to the third carbon atom. Because the parent chain has five carbons, the molecule is termed 3-bromopentane.

20. The reaction of toluene with chlorine in the presence of iron and in the absence of light yields:

Ans: The Correct Answer is Option (iv).

By electrophilic substitution, aromatic arenes react with chlorine in the presence of Lewis acid catalysts such as iron (III) chloride, yielding ortho and para isomers of haloarenes. (ii) and (iii) are both products of the reaction. Cl₂ forms a coordination complex with FeCl₃, creating the Cl + FeCl₄ - complex, which has a small positive charge on Cl and a negative charge on FeCl₄ - . This Cl⁺ then interacts with the aromatic double bonds of the toluene molecule to create an addition product, which is subsequently deprotonated to form a mixture of o-, p-, and m- chlorotoluene isomers. Because the m- isomer is highly unstable, the product is not accessible in the o- and p- forms.

21. Chloromethane on treatment with excess of ammonia yields mainly

Ans: The Correct Answer is Option (iii).

Because it possesses unpaired electrons, the ammonia molecule is a nucleophile in nature. By nucleophilic substitution, this nucleophile attacks the chloromethane molecule and forms methyl amine or methanamine. The carbon atom in the molecule is partially positive due to the electronegativity of the linked halide, which is partially negative. The positive ion is attacked by the electron-rich nucleophile, causing the halide ion to be detached from the molecule. The proper response is (iii).

22. Molecules whose mirror image is non super-imposable over them are known as chiral. Which of the following molecules is chiral in nature?

(i) 2-Bromobutane

(ii) 1-Bromobutane

(iii) 2-Bromopropane

(iv) 2-Bromopropan-2-ol

Ans: The Correct Answer is Option (i).

The structure of 2-Bromobutane is depicted in the diagram below. Because all of the groups connected to the central carbon atom differ, the mirror image of the molecule cannot be superimposed on the original molecule.

The structure of 1-Bromobutane is shown in the diagram below. Because there are just three groups that vary from one another, the molecule is not chiral in nature.

The structure of 2-Bromopropan-2-ol, shown below, has two identical species and hence cannot be chiral.

23. Reaction of C₆H₅CH₂Br with aqueous sodium hydroxide follows

(i) S_N¹ mechanism

(ii) S_N²mechanism

(iii) Any of the above two depending upon the temperature of reaction

(iv) Saytzeff rule

Ans: The Correct Answer is Option (i).

When benzyl chloride is treated with aqueous sodium hydroxide, where --OH is the nucleophile, a nucleophilic substitution reaction occurs, resulting in the formation of benzyl alcohol. The benzene ring is resonance stabilised here, and this stability is extended to the connected methylene group, giving a positive charge to --CH₂, making the whole carbocation stable when the link between benzyl and bromide is broken. This is an S_N¹ reaction with two stages that is followed due to the stability of the carbocation. This is a coordinated reaction that takes place in two phases. The halide group first exits the carbocation, and then the nucleophile binds to the cation, producing alcohol. The right answer is (i).

24. Which of the carbon atoms present in the molecule given below are asymmetric?

(i) a, b, c, d

(ii) b, c

(iii) a, d

(iv) a, b, c

Ans: The Correct Answer is Option (ii).

Chiral molecules are made up of one carbon atom surrounded by four different species. Because of the presence of two or more identical groups, such as hydrogens, all straight chain molecules cannot be chiral. Even carbons with double or triple bonds to a group are not considered chiral. Through covalent connections, an asymmetric carbon must be surrounded by four distinct species. As a result, atoms b and c are asymmetric. The correct answer is option (ii).

25. Which of the following compounds will give racemic mixture on nucleophilic substitution by OH⁻ion?

Solution:

Ans: The Correct Answer is Option (i).

A racemic mixture is one that contains two enantiomers in equal quantities but has no optical activity because the opposing optical rotations of the two enantiomers cancel each other out. The optically active reactant undergoes the S_N¹ reaction in order for a racemic mixture to form following nucleophilic substitution. Option (a) is a chiral carbon atom that will undergo the S_N¹ process, resulting in a racemic mixture. Option (b) does not include an asymmetric carbon, but option (c) has a secondary carbon asymmetric atom that is less reactive to an S_N¹ substitution. The right answer is (i).

In the questions, 26 to 29 arrange the compounds in increasing order of the rate of reaction towards nucleophilic substitution.

Ans: The Correct Answer is Option (iii).

Due to the resonance stabilisation of the benzene ring, the reactivity of aryl halides to nucleophilic substitution is exceedingly low. Because of resonance, the --Cl bond gains a partial double bond. The presence of an electron withdrawing --NO₂ group at ortho or para positions on the ring enhances the reactivity of aryl halides. The presence of --NO₂ near the C---Cl makes the molecule more reactive. The order of reactivity should be (b) > (c) > (a). The right option is (iii).

(i) (a) < (b) < (c)

(ii) (a) < (c) < (b)

(iii) (c) < (b) < (a)

(iv) (b) < (c) < (a)

Ans: The Correct Answer is Option (iv).

Because the --CH₃ group is an electron releasing group, it reduces the reactivity of aryl halides in the ortho and para locations. As a result, aryl halides without electron releasing groups are more reactive. As a result, the order of reactivity is (a) > (c) > (d) (b). The right answer is (iv).

(i) (c) < (b) < (a)

(ii) (b) < (c) < (a)

(iii) (a) < (c) < (b)

(iv) (a) < (b) < (c)

Solution:

Ans: The Correct Answer is option (iv).

( - NO₂) is an electron-withdrawing group that induces nucleophilic substitution when it is in the ortho and para positions. When this identical electron-drawing group is in the meta position, its impact is much reduced. The reactivity of aryl halides rises as the number of electron-withdrawing groups increases. As a result, the greater the number of electron-drawing groups, the greater the rate of nucleophilic substitution.

(i) (a) < (b) < (c)

(ii) (b) < (a) < (c)

(iii) (c) < (b) < (a)

(iv) (a) < (c) < (b)

Ans: The Correct Answer is Option (iii).

The presence of electron releasing groups increases the reactivity of aryl halides; the fewer the electron releasing groups, the slower the rate of nucleophilic substitution. As a result, an increase in methyl groups decreases reactivity. The sequence of reactivity is (a) > (b) > (c). The right answer is (iii).

30. Which is the correct increasing order of boiling points of the following compounds?

1-Iodobutane, 1-Bromobutane, 1-Chlorobutane, Butane

(i) Butane < 1-Chlorobutane < 1-Bromobutane < 1-Iodobutane

(ii) 1-Iodobutane < 1-Bromobutane < 1-Chlorobutane < Butane

(iii) Butane < 1-Iodobutane < 1-Bromobutane < 1-Chlorobutane

(iv) Butane < 1-Chlorobutane < 1-Iodobutane < 1-Bromobutane

Ans: The Correct Answer is Option (i).

The intermolecular force of attraction will grow as the surface area increases. As a result, the boiling point rises as well. The boiling point of a comparable alkyl halide rises with increasing molecular mass. Iodine has the greatest atomic mass. As a result, 1-iodobutane has the greatest boiling point.

31. Which is the correct increasing order of boiling points of the following compounds?

1-Bromoethane, 1-Bromopropane, 1-Bromobutane, Bromobenzene

(i) Bromobenzene < 1-Bromobutane < 1-Bromopropane < 1-Bromoethane

(ii) Bromobenzene < 1-Bromoethane < 1-Bromopropane < 1-Bromobutane

(iii) 1-Bromopropane < 1-Bromobutane < 1-Bromoethane < Bromobenzene

(iv) 1-Bromoethane < 1-Bromopropane < 1-Bromobutane < Bromobenzene

Ans: The Correct Answer is Option (iv).

If the halogen atom in all of the supplied alkyl halides is the same kind, the boiling point of a compound rises as the number of alkyl groups added increases. As the quantity and size of molecules and electrons grow, so do their attractions. As a result, the greater the boiling point, the larger the molecular weight of the alkyl halide. The boiling points of the following compounds are listed in ascending order: 1-Bromoethane 1-Bromopropane 1-Bromobutane Bromobenzene The right answer is (iv).

In the following questions two or more options may be correct.

Consider the following reaction and answer the questions no. 32–34.

32. Which of the statements are correct about the above reaction?

(i) (a) and (e) both are nucleophiles.

(ii) In (c) carbon atom is sp³ hybridised.

(iii) In (c) carbon atom is sp² hybridised.

(iv) (a) and (e) both are electrophiles.

Ans: An S_N² nucleophilic substitution reaction occurs when CH³Cl interacts with the hydroxide ion to produce CH³OH. Both OH - and Cl - are nucleophiles with an excess of electrons in this reaction. Because the binding of OH - and the leaving of Cl - occurs simultaneously in the transition state (c), the carbon atom is linked to just three hydrogens. As a result, the carbon atom becomes sp² hybridised. Aand (C) are the right statements.

Correct Answer: Option (A) and (C)

33. Which of the following statements are correct about this reaction?

(i) The given reaction follows the SN2 mechanism.

(ii) (b) and (d) have the opposite configuration.

(iii) (b) and (d) have the same configuration.

(iv) The given reaction follows the SN1 mechanism.

Ans: An S_N² reaction, also known as a nucleophilic substitution, is the reaction in question. The entering nucleophile causes the groups surrounding the carbon atom to migrate in the opposite direction of the nucleophile, causing the configuration of alkyl halides to invert. As a result, (b) and (d) will be in the opposite order. This reaction is also S_N² since there is just one phase in which OH^- is added and Cl^- is removed at the same time. (i) and (ii) are the right answers.

Correct Answer: Option (A) and (B)

34. Which of the following statements are correct about the reaction intermediate?

(i) Intermediate (c) is unstable because in this carbon is attached to 5 atoms.

(ii) Intermediate (c) is unstable because carbon atom is sp2 hybridised.

(iii) Intermediate (c) is stable because carbon atom is sp2 hybridised.

(iv) Intermediate (c) is less stable than the reactant (b).

Ans: Because the carbon atom is transiently linked to five atoms, the intermediate (iii) is less stable than reactant (ii), which is coupled to four groups.

Answer Q. No. 35 and 36 based on the following reaction.

35. Which of the following statements are correct about the mechanism of this reaction?

(i) A carbocation will be formed as an intermediate in the reaction.

(ii) OH–will attach the substrate (b) from one side and Cl- will leave it simultaneously from the other side.

(iii) An unstable intermediate will be formed in which OH– and Cl– will be attached by weak bonds.

(iv) The reaction proceeds through an SN1 mechanism.

Ans: Because it is a tertiary alkyl halide, the described reaction is a nucleophilic substitution that follows the S_N¹ mechanism. It's a two-step process. The first step is the gradual breaking of a polarised C---Br bond, which results in the formation of a carbocation and a bromide ion. To complete the substitution, the stable intermediate carbocation is attacked by the nucleophile OH⁻. (i) and (iv) are the right statements.

Correct Answer: option (A) and (iv)

36. Which of the following statements are correct about the kinetics of this reaction?

(i) The rate of reaction depends on the concentration of only (b).

(ii) The rate of reaction depends on the concentration of both (a) and (b).

(iii) Molecularity of reaction is one.

(iv) Molecularity of reaction is two.

Ans: The reaction is of the type S_N¹ nucleophilic substitution. This sort of reaction is governed by second order kinetics, in which the rate of reaction is determined only by the concentration of the reactant. A reaction's molecularity is defined as the number of molecules involved in the rate-determining phase. Because of second order kinetics, the moleculeity of this reaction is one. (i) and (iii) are the right answers .

Correct Answer: Option (A) and (C)

37. Haloalkanes contain halogen atom (s) attached to the sp3 hybridised carbon atom of an alkyl group. Identify haloalkane from the following compounds.

(i) 2-Bromopentane

(ii) Vinyl chloride (chloroethene)

(iii) 2-chloroacetophenone

(iv) Trichloromethane

Ans: Option (i) and (iv) have a halogen atom linked to an sp³ hybridised carbon that is single-bonded to other groups. The other possibilities are sp² hybridised because they are linked to carbon atoms, which are coupled to other groups in double bonds. Option I and (iv) are both correct.

Correct Answer: Option (i) and (iv)

38. Ethylene chloride and ethylidene chloride are isomers. Identify the correct statements.

(i) Both the compounds form the same product on treatment with alcoholic KOH.

(ii) Both the compounds form the same product on treatment with aq.NaOH.

(iii) Both the compounds form the same product on reduction.

(iv) Both the compounds are optically active.

Ans: Isomers are ethylene chloride ClCH₂---CH₂Cl and ethylidene dichloride CH₃---CHCl₂. Because the carbon atoms are not surrounded by distinct groups, neither molecule is optically active. When haloalkanes are treated with alcoholic KOH, they undergo an elimination process in which a hydrogen atom from the -carbon atom and a halogen atom from the -carbon atom are eliminated. To form the ethyne molecule, each of these chemicals lose hydrogen and chlorine atoms. When exposed to aqueous NaOH, the molecules undergo nucleophilic substitution, with the –Cl groups being replaced by --OH molecules. The results will differ due to the locations of the halides on various carbon atoms.

Correct Answer: Option (i) and (iii)

39. Which of the following compounds are gem-dihalides?

(i) Ethylidene chloride

(ii) Ethylene dichloride

(iii) Methylene chloride

(iv) Benzyl chloride

Ans:Dihaloalkanes with two halogen atoms linked to the same carbon atom are known as gem-dihalides. Alkylidene halides are another name for them. The chemical ethylene dichloride is a vicinal dihalide, meaning that chlorine atoms are linked to neighbouring carbon atoms. There is only one chlorine atom in benzoyl chloride. According to the molecule's nomenclature, ethylidene chloride is a gem-dihalide. The IUPAC terminology for methylene chloride is dichloromethane. Two chlorine atoms are joined to a single carbon atom in this compound.

Correct Answer: Option (i) and (iii)

40. Which of the following are secondary bromides?

(i) (CH₃)₂ CHBr

(ii) (CH₃)₃CCH₂Br

(iii) CH₃CH(Br)CH₂CH₃

(iv) (CH₃)₂CBrCH₂CH₃

Ans: Checking with the carbon atom it is connected to is one way to see if a compound has a secondary bromide. In a molecule, a secondary carbon group is linked to one hydrogen atom and three non-hydrogen groups. As a result of the foregoing possibilities, secondary bromides are formed when the --Br group is linked to the --CH group. (i) and (iii) are the right answers.

Correct Answer: Option (i) and (iii)

41. Which of the following compounds can be classified as aryl halides?

(i) p-ClC₆H₄CH₂CH(CH₃)₂

(ii) p-CH₃CHCl(C₆H₄ )CH₂CH₃

(iii) o-BrH₂C-C₆H₄CH(CH₃)CH₂CH₃

(iv) C₆H₅ -Cl

Ans: Aryl halides are halides in which the halogen atom is directly linked to the aromatic ring's sp² hybridised carbon atom. Only (i) and (iv) contain halogen atoms directly linked to aryl rings among the choices. As a result, they are known as aryl halides. (i) and (ii) are the right answers (iv).

Correct Answer: Option (i) and (iv)

42. Alkyl halides are prepared from alcohols by treating with

(i) HCl + ZnCl₂

(ii) Redp+Br2

(iii) H₂SO₄ + KI

(iv) All the above

Ans:

Alcohols produce matching alkyl halides when they are treated with various reagents. Alcohols, red phosphorous, and bromine treated with alcohol produce alkyl chlorides and alkyl bromide, respectively, when treated with HCl+ZnCl₂. H₂SO₄+ KI cannot be used to convert an alcohol to alkyl iodide because they will convert KI to the equivalent acid, HI, which will then be oxidised to I₂. As a result, the right statements are (i) and (ii).

Correct Answer: Option (i) and (ii)

43. Alkyl fluorides are synthesised by heating an alkyl chloride/bromide in presence of ____________ or ____________.

(i) CaF₂

(ii) CoF₂

(iii) Hg₂F₂

(iv) Na

Ans: Heat an alkyl chloride/bromide in the presence of a metallic fluoride such as AgF, Hg₂F₂, CoF₂, or SbF₃ to produce alkyl fluorides. Swarts reaction is the name given to this reaction. CoF₂ and Hg₂F₂ are the only choices that work as reagents. NaF is not suitable because the end products of NaF + RCl→ RF + NaCl are soluble in the solvent, i.e. water, making separation difficult. As a result, the right alternatives are (ii) and (iii).

Correct Answer: Option (ii) and (iii)

JEE Mains 2022

Commonly asked questions

In the given reaction,

X + Y + 3Z $⇌$ XYZ₃

If one mole of each of X and Y with 0.05 mol of Z gives compound XYZ₃. (Given: Atomic masses of X, Y and Z are 10, 20 and 30 amu, respectively.) The yield of XYZ₃ is_____________g.

(Nearest integer)

x + y + 3z = xyz₃

1 mole 1 mole 0.05 mole

$∴ n_{x} = 1$

$n_{y} = 1$

$n_{z} = \frac{0.05}{3} = 0.0167$ here z is limiting reagent.

$? \frac{0.05}{3}$ mole z gives 1 mole xyz₃

$∴$ mass of xyz₃ = n × molecular mass

= $\frac{0.05}{3} \times (10 + 20 + 3 \times 30) a . m . u .$

= 0.5 × 4 = 2g

An element M crystallises in a body centred cubic unit cell with a cell edge of 300 pm. The density of the element is 6.0g cm^-3. The number of atoms in 180g of the element is_________× 10²³ (Nearest integer)

B.C.C structure

a = 300 pm = 300 × 10^-12 m

d = 6g/cm³

z = 2

$d = \frac{Z \times M}{a^{3}}$

$6 = \frac{2 \times A}{{(300 \times 1 0^{- 10})}^{3}} = \frac{2 \times A}{27 \times 1 0^{- 24}}$

$∴ A t o m s o f M$ = 3.69 × 6.022 × 10²³

= 22.22 × 10²³

the nearest integer = 22

The number of paramagnetic species among the following is

B₂, Li₂, C₂, $C_{2}^{-}$ , $O_{2}^{- 2}$ , $O_{2}^{+}$ and $H e_{2}^{+}$ .

$B_{2} = σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p x}^{1} = π_{2 p y}^{1} \to$ Paramagnetic

$\begin{array}{l} L i_{2} = σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} \to D i a m a g n e t i c \\ C_{2} = σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p x}^{2} \equiv π_{2 p y}^{2} \to D i a m a g n e t i c \\ C_{2}^{-} = σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p x}^{2} \equiv π_{2 p y}^{2} σ_{2 p z}^{1} \to P a r a m a g n e t i c \end{array}$

$O_{2}^{- 2} = σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} σ_{2 p z}^{2} π_{2 p x}^{2} \equiv π_{2 p y}^{2} π_{2 p x}^{* 2} \equiv π_{2 p y}^{* 2}$ Diamagnetic

$O_{2}^{+} = σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} σ_{2 p z}^{2} π_{2 p x}^{2} \equiv π_{2 p y}^{2} π_{2 p x}^{* 1} \equiv π_{2 p y}^{0} \to$ Paramagnetic

$H e_{2}^{+} = σ_{1 s}^{2} σ_{1 s}^{* 1} \to$ Paramagnetic

$∴$ Paramagnetic molecules are $_{}_{}^{}_{}^{}_{}^{}$ $= B_{2}, C_{2}^{-}, O_{2}^{+}, H e_{2}^{+}$

150g of acetic acid was contaminated with 10.2g ascorbic acid $(C_{6} H_{8} O_{6})$ to lower down its freezing point by (x × 10^-¹)°C. The value of x is _________. (Nearest integer)

(Given K_f = 3.9 K kg mol^-¹; molar mass of ascorbic acid = 176 g mol^-¹]

$m = \frac{w \times 1000}{m o l e c u l a r w t \times w_{s o l v e n t}} = \frac{10.2 \times 1000}{176 \times 150}$

$= \frac{10200}{176 \times 150} = 0.386$

$Δ T_{f} = K_{f} \times m$

3.9 × 0.386

$∴ x \times 1 0^{- 1} = 15.05 \times 1 0^{- 1}$

Ans =15

Ka for butyric acid (C₃H₇COOH) is 2 × 10^-⁵. The pH of 0.2 M solution of butyric acid is ____________× 10^-¹ (Nearest integer)

[Given log2 = 0.30]

Ka for C₃H₇COOH = 2 × 10^-5

$p K a = - l o g (2 \times 1 0^{- 5}) = 5 - l o g 2$

=5 – 0.3 = 4.7

pH of 0.2 (M) solution =

$p H = \frac{p K a - l o g C}{2}$

$= \frac{1}{2} (4.7) - \frac{1}{2} l o g (0.2)$

$p H = 27 \times 1 0^{- 1}$

$∴$ Ans 27

For the given first order reaction

A -> B

The half life of the reaction is 0.3010 min. The ration of the initial concentration of reactant to the concentration of reactant at time 2.0 min will be equal to________________. (Nearest integer)

t1/2 = 0.301 min

t = 2 min

$K = \frac{2.303}{t} l o g (\frac{C o}{C t})$

$\frac{0.693}{0.301} = \frac{2.303}{2} l o g (\frac{C o}{C t})$

$\frac{2.303 \times 0.301}{0.301} = \frac{2.303}{2} l o g (\frac{C o}{C t})$

$∴ 2 = l o g (\frac{C o}{C t})$

$\frac{C o}{C t} = 1 0^{2} = 100$

Ans. 100

The number of interhalogens from the following having square pyramidal structure is:

CIF₃, IF₇, BrF₅, BrF₃, I₂Cl₆, IF₅, ClF, ClF₅

$C l F_{3}$ -> T-shaped (sp³d)

IF₇ -> Pentagonal bipyramidal (sp³d³)

BrF₅ -> Square pyramidal (sp³d²)

BrF₃ -> T-Shaped (sp³d)

I₂Cl₆ -> Triangular bipyramidal (sp³d)

$I F_{5} \to$ Square Pyramidal (sp³d²)

ClF -> (sp³)

ClF₅ -> Square Pyramidal (sp³d²)

Br₅, IF₅ & ClF₅ -> Square Pyramidal

The disproportionation of $M n O_{4}^{2 -}$ in acidic medium resultant in the formation of two manganese compounds A and B. If the oxidation state of Mn is B is smaller than that of A, then the spin-only magnetic moment $(μ)$ value of B in BM is___________.

(Nearest integer)

$M n O_{4}^{- 2} \to A + B$

Oxidation state of Mn in B < A

$M n O_{4}^{- 2} + H^{+} \to M n O_{4}^{-} + M n O_{2}$

B is MnO₂

$∴$ Oxidation state of Mn = +4

$∴_{25} M n^{+ 4} = 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{0} 3 d^{3}$

$∴$ unpaired electron = 3

Spin only magnetic moment $(μ)$

$= \sqrt[]{n (n + 2)} = \sqrt[]{3 (3 + 2)} = \sqrt[]{15}$

$= 3.87 \approx 4$

Total number of relatively more stable isomer(s) possible for octahedral complex $[C u {(e n)}_{2} {(S C N)}_{2}]$ will be________________.

$[C u {(e n)}_{2} {(S C N)}_{2}]$

More stable isomers = 3 (trans isomers)

On complete combustion of 0.492g of an organic compound containing C, H and O, 0.7938 g of CO₂ and 0.4428 g of H₂O was produced. The % composition of oxygen in the compounds is____________.

% of C in organic compound

$= \frac{12}{14} \times \frac{W C O_{2}}{W . O . C .} \times 100$

$= \frac{952.56}{21.648} = 44 %$

$% o f H = \frac{2}{18} \times \frac{W H_{2} O}{W . O . C .} \times 100$

= $\frac{20}{18} \times \frac{0.4428}{0.492} \times 100$ $\frac{}{} \frac{}{}$

= $\frac{88.56}{8.856} = 10 %$ $\frac{}{}$

= 100 – 54 = 46%

Chemistry NCERT Exemplar Solutions Class 12th Chapter Ten Exam

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