KCET 2021 Chemistry Questions with Answers Key Solutions

Q1. The compound $A$ (major product) is

A

B

C

D

Solution:

Q2. Bond enthalpies of $A_2$ , $B_2$ and $AB$ are in the ratio $2 : 1 : 2$ . If bond enthalpy of formation of $AB$ is $-100 \,KJ \,mol ^{-1}$ .The bond enthalpy of $B _{2}$ is

A

$100\, KJ\, mol ^{-1}$

B

$50\, KJ\, mol ^{-1}$

C

$200\, KJ\, mol ^{-1}$

D

$150\, KJ\, mol ^{-1}$

Solution:

Let bond enthalpy of $A _{2}$ be $x , B _{2}$ be $\frac{ x }{2},AB$ be $x$
$A _{2}+ B _{2} \rightarrow 2 AB$
$x +\frac{ x }{2}\,\,\,2x$
$\Delta H _{\text {form }}=\frac{1}{2} A _{2}+\frac{1}{2} B _{2} \rightarrow AB$
$-100=\left(\frac{ x }{2}+\frac{ x }{4}\right)- x$
$x =400$
$B _{2}=\frac{ x }{2}=200 \,kJ$

Q3. The order of reactivity of the compounds $C _{6} H _{5} CH _{2} Br , C _{6} H _{5} CH \left( C _{6} H _{5}\right) Br , C _{6} H _{5} CH \left( CH _{3}\right) Br$ and $C _{6} H _{5} C \left( CH _{3}\right)\left( C _{6} H _{5}\right) Br \text { in } S _{ N }^{2}$ reaction is

A

$C_6H_5 - \underset{\overset{|}{C_6H_5}}{\overset{\underset{|}{CH_3}}{C}} - Br < C_6H_5 - \underset{\overset{|}{C_6H_5}}{\overset{\underset{|}{H}}{C}} - Br < C_6H_5 - \underset{\overset{|}{C}H_3}{\overset{\underset{|}{H}}{C}} - Br < C_6H_5 - \underset{\overset{|}{H}}{\overset{\underset{|}{H}}{C}} - Br$

B

$C_6H_5 - \underset{\overset{|}{H}}{\overset{\underset{|}{H}}{C}} - Br < C_6H_5 - \underset{\overset{|}{CH_3}}{\overset{\underset{|}{H}}{C}} - Br < C_6H_5 - \underset{\overset{|}{C_6H_5}}{\overset{\underset{|}{H}}{C}} - Br < C_6H_5 - \underset{\overset{|}{C_6H_5}}{\overset{\underset{|}{CH_3}}{C}} - Br$

C

$C_6H_5 - \underset{\overset{|}{CH_3}}{\overset{\underset{|}{H}}{C}} - Br < C_6H_5 - \underset{\overset{|}{H}}{\overset{\underset{|}{H}}{C}} - Br < C_6H_5 - \underset{\overset{|}{C_6H_5}}{\overset{\underset{|}{H}}{C}} - Br < C_6H_5 - \underset{\overset{|}{C_6H_5}}{\overset{\underset{|}{CH_3}}{C}} - Br$

D

$C_6H_5 - \underset{\overset{|}{C_6H_5}}{\overset{\underset{|}{H}}{C}} - Br < C_6H_5 - \underset{\overset{|}{H}}{\overset{\underset{|}{H}}{C}} - Br < C_6H_5 - \underset{\overset{|}{CH_3}}{\overset{\underset{|}{H}}{C}} - Br < C_6H_5 - \underset{\overset{|}{C_6H_5}}{\overset{\underset{|}{CH_3}}{C}} - Br$

Solution:

Lesser the steric hindrance, easier and higher the rate of reaction towards $SN ^{2}$

Q4. The major product of the following reaction is $CH _{2}= CH - CH _{2}- OH \xrightarrow[\text{Excess} ]{HBr}$ product

A

$CH _{3}- CHBr - CH _{2} Br$

B

$CH _{2}= CH - CH _{2} Br$

C

$CH _{3}- CHBr - CH _{2}- OH$

D

$CH _{3}- CHOH - CH _{2} OH$

Solution:

$CH _{2}= CH - CH _{2}OH \xrightarrow[\text{Excess} ]{HBr} CH_3 - \underset{\overset{|}{B}r}{CH} - CH_2Br$

Q5. The product 'A' gives white precipitate when treated with bromine water. The product ' $B$ ' treated with Barium hydroxide to give the product $C$ . The compound $C$ is heated strongly form product $D$ . The product $D$ is

A

4-Methylpent-3-en-2-one

B

But-2 enal

C

3-Methylpent-3-en-2-one

D

2-Methylbut-2-enal

Solution:

Q6. For the reaction $A ( g )+ B ( g ) \rightleftharpoons C ( g )+ D ( g ) ; \Delta H =- QKJ$
The equilibrium constant cannot be disturbed by

A

Addition of A

B

Addition of D

C

Increasing of pressure

D

none of the above

Solution:

Equilibrium constant is directly proportional to temperature

Q7. An organic compound ' $X$ ' on treatment with $PCC$ in dichloromethane gives the compound $Y$ . Compound ' $Y$ ' reacts with $I_2$ and alkali to form yellow precipitate of triiodomethane. The compound X is

A

$CH _{3} CHO$

B

$CH _{3} COCH _{3}$

C

$CH _{3} CH _{2} OH$

D

$CH _{3} COOH$

Solution:

Q8. A compound ' $A$ ' $\left( C _{7} H _{8} O \right)$ is insoluble in $NaHCO _{3}$ solution but dissolve in $NaOH$ and gives a characteristic colour with neutral $FeCl _{3}$ solution. When treated with Bromine water compound ' $A$ ' forms the compound $B$ with the formula $C _{7} H _{5} OBr _{3} .$ ' $A$ ' is

A

B

C

D

Solution:

Q9. In set of reactions, identify $D$
$CH _{3} COOH \xrightarrow{ SOCl _{2}}A \xrightarrow[\text{Anh} \,AlCl_3]{\text { Benzene }} B \xrightarrow{ HCN } C \xrightarrow{ H _{2} O }D$

A

B

C

D

Solution:

Q10. $K _{ u }$ values for acids $H _{2} SO _{3}, HNO _{2}, CH _{3} COOH$ and $HCN$ are respectively $1.3 \times 10^{-2}, 4 \times 10^{-4}, 1.8 \times 10^{-5}$ and $4 \times 10^{-10}$ , which of the above acids produces stronger conjugate base in aqueous solution?

A

$H_2SO_3$

B

$HNO_2$

C

$CH_3COOH$

D

$HCN$

Solution:

$K _{ a }( HCN )=4 \times 10^{-10}$
$HCN \rightleftharpoons H ^{+}+ CN ^{-}$
Weaker the acid, stronger the conjugate base.

Q11. $A, B$ and $C$ respectively are

A

ethanol, ethane nitrile and ethyne

B

ethane nitrile, ethanol and ethyne

C

ethyne, ethanol and ethane nitrile

D

ethyne, ethane nitrile and ethanol

Solution:

Q12. The reagent which can do the conversion $CH _{3} COOH \longrightarrow CH _{3}- CH _{2}- OH$ is

A

$LiAIH_4$ / ether

B

$H_2, Pt$

C

$NaBH_4$

D

$Na$ and $C_2H_5OH$

Solution:

$\underset{\text{Ethanoic acid}}{CH _{3} COOH} \xrightarrow{\text { LiAlH }_{4} / \text { ether }} \underset{\text{Ethanol}}{CH _{3} CH _{2} OH}$

Q13. $CH _{3} CHO \xrightarrow[(ii)H_3O^+]{(i) CH _{3} MgBr } A \xrightarrow[\Delta]{\text{ Conc} H _{2} SO _{4}} B \xrightarrow[(ii)H_2O,OH^-]{(i) B _{2} H _{6}} CA$ and $C$ are

A

Identical

B

Position isomers

C

Functional isomers

D

Optical isomers

Solution:

Q14. Which of the following is not true for oxidation?

A

addition of oxygen

B

addition of electronegative element

C

removal of hydrogen

D

removal of electronegative element

Solution:

Oxidation is addition of oxygen or electronegative element;
Oxidation is removal of hydrogen or electropositive element.

Q15. Which is the most suitable reagent for the following conversion?
$CH_3 - CH = CH - CH_2 -\overset{\underset{||}{O}}{C}-CH_3 \to CH_3 - CH = CH- CH_2 - \overset{\underset{||}{O}}{C}-OH$

A

Tollen's reagent

B

Benzoyl peroxide

C

$I_2$ and $NaOH$ solution with subsequent acidification

D

$Sn$ and $NaOH$ solution

Solution:

Q16. $C_6H_5CH_2Cl \ce{->[\text{alc}.NH_3]} A \ce{->[2CH_3Cl]}B$ . The product $B$ is

A

N, N - Dimethyl phenyl methanamine

B

N, N - Dimethyl benzenamine

C

N - Benzyl - N - methyl methanamine

D

phenyl - N, N - dimethyl methanamine

Solution:

Q17. The method by which aniline cannot be prepared is

A

Nitration of benzene followed by reduction with Sn and con HCI

B

Degradation of benzamide with bromine in alkaline solution

C

Reduction of nitrobenzene with $H_2 / Pd$ is ethanol

D

Potassium salt of phthalimide treated with chlorgbenzene followed by the hydrolysis with aqueous NaOH solution

Solution:

Gabriel phthalimide synthesis not preferred for preparing aniline because amine formation involves nucleophilic substitution of alkyl halides by the anion formed by phthalimide. But aryl halides do not undergo nucleophilic substitution.

Q18. Permanent hardness cannot be removed by

A

Using washing soda

B

Calgon's method

C

Clark's method

D

Ion exchange method

Solution:

Clark’s method is for the removal of temporary hardness.

Q19. A hydrocarbon $A \left( C _{4} H _{8}\right)$ on reaction with HCl gives a compound $B \left( C _{4} H _{9} Cl \right)$ which on reaction with 1 mol of $NH _{3}$ gives compound $C \left( C _{4} H _{10} N \right) .$ On reacting with $NaNO _{2}$ and $HCl$ followed by treatment with water, compound $C$ yields an optically active compound $D$ . The $D$ is

A

B

C

D

Solution:

Q20. RNA and DNA are chiral molecules, their chirality is due to the presence of

A

D-Sugar component

B

L-Sugar component

C

Chiral bases

D

Chiral phosphate ester unit

Solution:

Deoxyribose and ribose sugars are D-chiral sugars in DNA and RNA.

Q21. The property of the alkaline earth metals that increases with their atomic number is

A

Ionisation enthalpy

B

Electronegativity

C

Solubility of their hydroxide in water

D

Solubility of their sulphate in water

Solution:

The solubility of alkaline earth metal hydroxides increases with increase in atomic number.

Q22. Primary structure in a nucleic acid chain contains bases as G A T G C ........The chain which is complementary to this chain is

A

G G T G A .......

B

T G A A G .........

C

C T A C G .......

D

T T T A G ........

Solution:

Q23. In the detection of II group acid radical, the salt containing chloride is treated with concentrated sulphuric acid, the colourless gas is liberated. The name of the gas is

A

Hydrogen chloride gas

B

Chlorine gas

C

Sulphur dioxide gas

D

Hydrogen gas

Solution:

$MCl + H _{2} SO _{4} \longrightarrow MHSO _{4}+ HCl _{( g )}$

Q24. The number of six membered and five membered rings in Buckminster Fullerene respectively is

A

20,12

B

12,20

C

14,18

D

14,11

Solution:

$12 : 5$ Membered ring
$20 : 6$ Membered ring

Q25. In Chrysoberyl, a compound containing Beryllium, Aluminium and oxygen, oxide ions form cubic close packed structure. Aluminium ions occupy $1/4^\text{ th}$ of tetrahedral voids and Beryllium ions occupy $1/4^\text{ th}$ of octahedral voids. The formula of the compound is

A

$BeAlO _{4}$

B

$BeAl _{2} O _{4}$

C

$Be _{2} AlO _{2}$

D

$BeAlO _{2}$

Solution:

$CCP = N =4$
Octahedral void $= N =1 \times 4=4 \times \frac{1}{4}=1: Be$
Tetrahedral void $=2 N =2 \times 4=2 \times \frac{1}{4}=2: A l$

Q26. The correct statement regarding defects in solids is

A

Frenkel defect is a vacancy defect

B

Schottky defect is a dislocation defect

C

trapping of an electron in the lattices leads to the formation of F-centre

D

Schottky defect has no effect on density.

Solution:

Q27. A metal crystallises in $BCC$ lattice with unit cell edge length of $300\, pm$ and density $6.15\, g\, cm^{-3}$ . The molar mass of the metal is

A

$50\, g\, mol ^{-1}$

B

$60\, g\, mol ^{-1}$

C

$40\, g\, mol ^{-1}$

D

$70\, g\, mol ^{-1}$

Solution:

$d =\frac{ ZM }{ a ^{3} N _{ A }}$
$6.15=\frac{2 \times M }{\left(3 \times 10^{-8}\right)^{3} \times 6.02 \times 10^{23}}$
$M =\frac{6.15 \times 27 \times 10^{-24} \times 6.02 \times 10^{23}}{2}$
$M =50\, g / mol$

Q28. Henry's law constant for the solubility of $N_2$ gas in water at $298 \,K$ is $1.0 \times 10^5$ atm . The mole fraction of $N_2$ in air is $0.8$ .The number of moles of $N_2$ from air dissolved in $10$ moles of water $298\, K$ and $5$ atm pressure is

A

$4.0 \times 10^{-4}$

B

$4.0 \times 10^{-5}$

C

$5.0 \times 10^{-4}$

D

$4.0 \times 10^{-6}$

Solution:

Total pressure is $5 atm$ . The mole fraction of nitrogen is $0.8$ .
Hence, the partial pressure of nitrogen $= P _{ T } \times X =0.8 \times 5=4 atm$
According to Henry's law, $P = KX$
$x$ is the mole fraction
K is the henry's law constant
$P$ is the pressure in atm.
Substitute values in the above expression.
$X =\frac{4 atm }{1 \times 10^{5} atm }=4 \times 10^{-5}=\frac{ n }{10}$
$n =4.0 \times 10^{-4} mol$

Q29. A pure compound contains $2.4\, g$ of $C$ , $1.2 \times 10^{23}$ atoms of $H, 0.2$ moles of oxygen atoms. Its empirical formula is

A

$C _{2} HO$

B

$C _{2} H _{2} O _{2}$

C

$CH _{2} O$

D

$CHO$

Solution:

Number of moles of $C =\frac{2.4}{12}=0.2$
Number of moles of $H =\frac{1.2 \times 10}{6.0 \times 10}=0.2$
Number of moles of $O =0.2 \,mol$
Empirical formula is $CHO$ .

Q30. Choose the correct statement

A

$K _{ H }$ value is same for a gas in any solvent

B

Higher the $K_{H}$ value more the solubility of gas

C

$K _{ H }$ value increases on increasing the temperature of the solution

D

none of the above

Solution:

option 3

Q31. The $K _{ H }$ value (K bar) of Argon (I), Carbondioxide (II) formuldehyde (III) and methane (IV) are respectively $40.3,1,67,1.83 \times 10^{-5}$ and $0.413$ at $298\, K$ . The increasing order of solubility of gas in liquid is

A

$I < II < IV < III$

B

$III < IV < II < I$

C

$I < III < II < IV$

D

$I < IV < II < III$

Solution:

$K _{ H }$ is inversely proportional to solubility of gas in liquid.

Q32. The vapour pressure of pure liquids $A$ and $B$ are $450$ and $700\, mm$ of $Hg$ at $350 \,K$ respectively. If the total vapour pressure of the mixture is $600 \,mm$ of $Hg$ , the composition of the mixture in the solution is

A

$x_{A}=0.4, x_{B}=0.6 \quad$

B

$x_{A}=0.6, x_{B}=0.4$

C

$x_{A}=0.3, x_{B}=0.7$

D

$x_{A}=0.7, x_{B}=0.3$

Solution:

$p _{\text {tot }}= p _{ A }^{0} x _{ A }+ p _{ B }^{0} x _{ B }$
$600=450 x _{ A }+700\left(1- x _{ A }\right)$
$600=450 x _{ A }+700-700 x _{ A }$
$-100=-250 x _{ A }$
$x _{ A }=\frac{100}{250}=\frac{2}{5}$
$=0.4$
$x _{ B }=1- x _{ A }=1-0.4$
$=0.6$

Q33. Consider the following electrodes $P = Zn ^{2+}(0.0001 M ) / Zn \quad Q = Zn ^{2+}(0.1 M ) / Zn$ $R = Zn ^{2+}(0.01 M ) / Zn \quad S = Zn ^{2+}(0.001 M ) / Zn$ $E ^{\circ} Zn / Zn ^{2+}=-0.76\, V$ Electrode potentials of the above electrodes in volts are in the order

A

$P > S > R > Q$

B

$S>R>Q>P$

C

$Q > R > S > P$

D

$P > Q > R > S$

Solution:

Electrode potential is directly proportional to concentration of $Zn ^{2}$ ions.

Q34. The number of angular and radial nodes in $3 p$ orbital respectively are

A

3,1

B

1,1

C

2,1

D

2,3

Solution:

Number of angular nodes $=l$
Number of radial nodes $= n -l-1$
$\therefore$ For $3p$ orbital, angular nodes $=1$
For $3 p$ orbital, radial nodes $=3-1-1=1$

Q35. The resistance of $0.01 \,m \,KCl$ solution at $298\, K$ is $1500\, \Omega$ . If the conductivity of $0.01\, m \,KCl$ solution at $298\, K$ is $0.146 \,\times \,10^{-3}\, S \,cm ^{-1} .$ The cell constant of the conductivity cell in $cm ^{-1}$ is

A

0.219

B

0.291

C

0.301

D

0.194

Solution:

$\kappa=\frac{1}{ R }\left(\frac{l}{ a }\right)$
$\therefore \frac{l}{ a }=\kappa \cdot R$
$=0.146 \times 10^{-3} \times 1500$
$=219 \times 10^{3}=0.219\, cm ^{-1}$

Q36. $H _{2(g)}+2 AgCl _{(s)} \rightleftharpoons 2 Ag _{(s)}+2 HCl _{(a q)}$
$E _{\text {cell }}^{\circ}$ at $25^{\circ} C$ for the cell is $0.22\, V$ . The equilibrium constant at $25^{\circ} C$ is

A

$2.8 \times 10^{7}$

B

$5.2 \times 10^{8}$

C

$2.8 \times 10^{5}$

D

$5.2 \times 10^{4}$

Solution:

$E _{\text {cell }}^{0}=\frac{0.059}{ n } \log K _{ C }$
$0.22=\frac{0.059}{2} \log K _{ c }$
$\log K _{ c }=\frac{0.22 \times 2}{0.059}=7.457$
$K _{ C }=2.8 \times 10^{7}$

Q37. For a reaction $A+2 B \rightarrow$ Products, when concentration of $B$ alone is increased half life remains the same. If concentration of $A$ alone is doubled, rate remains the same. The unit of rate constant for the reaction is

A

$S ^{-1}$

B

$L\, mol ^{-1} \, S ^{-1}$

C

mol $L^{-1} S^{-1}$

D

$atm ^{-1}$

Solution:

For first-order reaction, half-life $t _{1 / 2}=\frac{0.693}{ k }$
Since half-life does not depend on the concentration of reactants.
According to the given data, the order of reaction with respect to $B$ is one, hence, does not change half-life.

Similarly, an order of reaction with respect to $A$ is also one since the rate increase twice on doubling the concentration.
Hence, total order $=1+1=2$
Second order reaction and its unit is $Lmol ^{-1} s ^{-1}$ .

Q38. The third ionisation enthalpy is highest in

A

Alkali metals

B

Alkaline earth metals

C

Chalcogens

D

Pnictogens

Solution:

Alkaline earth metals have $ns ^{2}$ configuration.
After the removal of $2^{\text {nd }}$ electron, they get stable noble gas configuration.

Q39. If the rate constant for a first order reaction is $k$ , the time $(t)$ required for the completion of $99 \%$ of the reaction is given by

A

$t=\frac{4.606}{k}$

B

$t=\frac{2.303}{k}$

C

$t=\frac{0.693}{k}$

D

$t=\frac{6.909}{k}$ .

Solution:

$t =\frac{2.303}{ k } \log \frac{100}{1}$
$=\frac{2.303}{ k } \times 2$
$t =\frac{4.606}{ k }$

Q40. The rate of a gaseous reaction is given by the expression $k [ A ][ B ]^{2} .$ If the volume of vessel is reduced to one half of the initial volume, the reaction rate as compared to original rate is

A

$\frac{1}{16}$

B

$\frac{1}{8}$

C

8

D

16

Solution:

$r = K [ A ][ B ]^{2}$
When volume is reduced to half, pressure increases by two times.
$r = k [2 A ]^{1}[2 B ]^{2}=2 \times 4=8$ times

Q41. The correct IUPAC name of

A

4-Ethyl-1-Fluoro-2-nitrobenzene

B

1-Ethyl-4-Fluoro-3-nitrobenzene

C

3-Ethyl-6-Fluoronitrobenzene

D

5-Ethyl-2-Fluoronitrobenzene

Solution:

Based on lowest sum of locants,

4-Ethyl-1-fluoro-2-nitrobenzene

Q42. Higher order $(>3)$ reactions are rare due to

A

Shifting of equilibrium towards reactants due to elastic collisions

B

Loss of active species on collision

C

Low probability of simultaneous collision of all reacting species

D

Increase in entropy as more molecules are involved

Solution:

Collisions between more than three reactant molecules is rare.

Q43. Arrange benzene, $n$ -hexane and ethyne in decreasing order of their acidic behaviour

A

Benzene $>n$ -hexane $>$ ethyne

B

n-hexane > Benzene > ethyne

C

ethyne $>$ n-hexane $>$ Benzene

D

ethyne $>$ Benzene $>n$ -hexane

Solution:

Higher the percentage s-character, higher the acidic character.

Q44. A colloidal solution is subjected to an electric field than colloidal particles more towards anode. The amount of electrolytes of $BaCl _{2}, AlCl _{3}$ and $NaCl$ required to coagulate the given colloid is in the order

A

$NaCl > BaCl _{2}> AlCl _{3}$

B

$BaCl _{2}> AlCl _{3}> NaCl$

C

$AlCl _{3}= NaCl = BaCl _{2}$

D

$AlCl _{3}> BaCl _{2}> NaCl$

Solution:

Hardy Schulze Law: The effective ions of the electrolytes in bringing about coagulation are those which carry a charge opposite to that of colloidal particles. Greater the valency of coagulating or fluctuating ion, greater is its power to bring about coagulation.

When colloidal solution is subjected to electrical field, coagulating power should be in the order of $AlCl _{3}> BaCl _{2}> NaCl$ as thier valency is increasing.

Q45. Which of the following is an incorrect statement?

A

Hydrogen bonding is stronger than dispersion forces

B

Sigma bonds are stronger than $\pi$ -bonds

C

lonic bonding is non-directional :

D

$\sigma$ -electrons are referred to as mobile electrons

Solution:

$\pi$ -electrons are referred to as mobile electrons (with respect to graphite).

Q46. Zeta potential is

A

Potential required to bring about coagulation of a colloidal sol.

B

Potential required to give the particle a speed of $1 \,cm\, S ^{-1} .$

C

Potential difference between fixed charged layer and the diffused layer having opposite charges

D

Potential energy of the colloidal particles.

Solution:

The potential difference between fixed charged layer and the diffused layer having opposite charges is known as zeta potential

Q47. Which of the following compound on heating gives $N _{2} O$ ?

A

$Pb \left( NO _{3}\right)_{2}$

B

$NH _{4} NO _{3}$

C

$NH _{4} NO _{2}$

D

$NaNO _{3}$

Solution:

$NH _{4} NO _{3} \xrightarrow{\Delta}N _{2} O +2 H _{2} O$

Q48. Which of the following property is true for the given sequence $NH _{3}> PH _{3}> AsH _{3}> SbH _{3}> BiH _{3} ?$

A

Reducing property

B

Thermal stability

C

Bond. angle

D

none of the above

Solution:

Thermal stability and bond angle decreases down the group due to increase in bond length.

Q49. The correct order of boiling point in the following compounds is

A

$HF > H _{2} O > NH _{3}$

B

$H _{2} O > HF > NH _{3}$

C

$NH _{3}> H _{2} O > HF$

D

$NH _{3}> HF > H _{2} O$

Solution:

Boiling point is directly proportional to molecular mass and strength of hydrogen bond

Q50. $XeF _{6}$ on partial hydrolysis gives a compound $X$ , which has square pyramidal geometry ' $X$ ' is

A

$XeO _{3}$

B

$XeO _{4}$

C

$XeOF _{4}$

D

$XeO _{2} F _{2}$

Solution:

$XeF _{6}+ H _{2} O \longrightarrow XeOF _{4}+2 HF$

Q51. A colourless, neutral, paramagnetic oxide of Nitrogen $'P'$ on oxidation gives reddish brown gas $Q$ . $Q$ on cooling gives colourless gas $R$ . $R$ on reaction with P gives blue solid $S$ . Identify $P , Q , R , S$ , respectively

A

$N _{2} O\, NO\, NO _{2} \,N _{2} O _{5}$

B

$N _{2} O\, NO _{2}\, N _{2} O _{4} \,N _{2} O _{3}$

C

$NO \,NO _{2} \,N _{2} O _{4}\, N _{2} O _{3}$

D

$NO \,NO\, N _{2} O _{4} \,N _{2} O _{5}$

Solution:

$\underset{ P }{ NO } \xrightarrow{[ O ]} \underset{ Q }{ NO _{2}} \xrightarrow{\text { cooling }} \underset{ R }{ N _{2} O _{4}} \xrightarrow{[ H ]} \underset{S}{ N _{2} O _{3}}$

Q52. Which of the following does not represent property stated against it?

A

$CO ^{+2}< Fe ^{+2}< Mn ^{+2}$ - lonic size

B

$Ti < V < Mn$ - Number of oxidation states

C

$Cr ^{+2}< Mn ^{+2}< Fe ^{+2}$ - Paramagnetic behaviour

D

none of the above

Solution:

$\underset{24}{ Cr }[ Ar ] 3 d ^{5} 4 s ^{1} \,\,\, Cr ^{2+} 3 d ^{4} 4 s ^{0}=4$ unpaired electrons
$Mn [ Ar ] 3 d ^{5} 4 s ^{2} Mn ^{2+} 3 d ^{5} 4 s ^{0}=5$ unpairedelectrons
$Fe [ Ar ] 3 d ^{6} 4 s ^{2} \,\,\,Fe ^{2+} 3 d ^{6} 4 s ^{0}=4$ unpaired electrons
Since both $Cr ^{2+}$ and $Fe ^{2+}$ contain $4$ unpaired electrons, the correct option is $C$ .
Option D is also correct because density order given is not correct.

Q53. Which one of the following is correct for all elements from $Sc$ to $Cu$ ?

A

The lowest oxidation state shown by them is +2

B

4 S orbital is completely filled in the ground state

C

3 d orbital is not completely filled in the ground state

D

The ions in +2 oxidation states are paramagnetic

Solution:

Paramagnetism is due to the presence of one or more unpaired electrons

Q54. When the absolute temperature of ideal gas is doubled and pressure is halved, the volume of gas

A

will be half of original volume

B

will be 4 times the original volume

C

will be 2 times the original volume

D

will be 1 / 4 th times the original volume

Solution:

From the combined gas equation,
$\frac{p_{1} V_{1}}{T_{1}}=\frac{p_{2} V_{2}}{T_{2}}$
$\frac{p_{1} V_{1}}{T_{1}}=\frac{\frac{1}{2} p_{1} V_{2}}{2 T_{1}}$
$V_{1}=\frac{1}{4} V_{2}$
$V_{2}=4 V_{1}$

Q55. Which of the following pairs has both the ions coloured in aqueous solution?
[Atomic numbers of $S c=21, T i=22, N i=28, C u=29, M n=25$ ]

A

$Sc ^{3+}, Mn ^{2+}$

B

$Ni ^{2+}, Ti ^{4+}$

C

$Ti ^{3+}, Cu ^{+}$

D

$Mn ^{2+}, Ti ^{3+}$

Solution:

$Sc =21, Sc ^{3+}=[ Ar ] 3 d ^{0} 4 s ^{0}=$ no unpaired electrons
$Ti =22, Ti ^{3+}=[ Ar ] 3 d ^{1} 4 s ^{0}=1$ unpaired electron
$Ti ^{4+}=[ Ar ] 3 d ^{\circ} 4 s ^{\circ}$ no unpaired electrons
$Ni =28, Ni ^{2+}=[ Ar ] 3 d ^{8} 4 s ^{2}=2$ unpaired electrons
$Cu =29, Cu ^{+}=[ Ar ] 3 d ^{10} 4 s ^{0}=$ no unpaired electrons
$Mn =25, Mn ^{2+}=[ Ar ] 3 d ^{5} 4 s ^{2}=5$ unpaired electrons

Q56. For the crystal field splitting in octahedral complexes,

A

the energy of the $e _{ g }$ orbitals will decrease by $(3 / 5) \Delta_{0}$ and that of the $t _{2 g }$ will increase by $(2 / 5) \Delta_{0}$

B

the energy of the $e _{ g }$ orbitals will increase by $(3 / 5) \Delta_{0}$ and that of the $t_{2 g}$ will decrease by $(2 / 5) \Delta_{0}$

C

the energy of the $e_{g}$ orbitals will increase by $(3 / 5) \Delta_{0}$ and that of the $t_{2 g}$ will increase by $(2 / 5) \Delta_{0}$

D

the energy of the $e _{ g }$ orbitals will decrease by $(3 / 5) \Delta_{0}$ and that of the $t_{2 g}$ will decrease by $(2 / 5) \Delta_{0}$

Solution:

Q57. Peroxide effect is observed with the addition of $HBr$ but not with the addition of $HI$ to unsymmetrical alkene because

A

H-l bond is stronger that H-Br and is not cleaved by the free radical

B

H-l bond is weaker than H-Br bond so that iodine free radicals combine to form iodine molecules

C

Bond strength of HI and HBr are same but free radicals are formed in HBr

D

All of these

Solution:

Due to free radical mechanism (secondary free radical)

Q58. The IUPAC name of $\left[ Co \left( NH _{3}\right)_{5}\left( CO _{3}\right)\right] Cl$ is

A

Pentaamminecarbonatocobalt (III) Chloride

B

Carbonatopentamminecobalt (III) Chloride

C

Pentaamminecarbonatocobaltate (III) Chloride

D

Pentaammine cobalt (III) Carbonate Chloride

Solution:

Pentaamminecarbonatocobalt(III) chloride

Q59. Homoleptic complexes among the following are
(A) $K _{3}\left[ Al \left( C _{2} O _{4}\right]_{3}\right.$ ,
(B) $\left[ CoCl _{2}( en )_{2}\right]^{+}$
(C) $K _{2}\left[ Zn ( OH )_{4}\right]$

A

A only .

B

(A) and (B) only

C

(A) and (C) only

D

(C) only

Solution:

Complex compounds having only one type of ligands are homoleptic complexes

Q60. The correct order for wavelengths of light absorbed in the complex ions $\left[ CoCl \left( NH _{3}\right)_{5}\right]^{2+},\left[ Co \left( NH _{3}\right)_{6}\right]^{3+}$ and $\left[ Co ( CN )_{6}\right]^{3-}$ is

A

$\left[ CoCl \left( NH _{3}\right)_{5}\right]^{2+}>\left[ Co \left( NH _{3}\right)_{6}\right]^{3+}>\left[ Co ( CN )_{6}\right]^{3-}$

B

$\left[ Co \left( NH _{3}\right)_{6}\right]^{3+}>\left[ Co ( CN )_{6}\right]^{3-}>\left[ CoCl \left( NH _{3}\right)_{5}\right]^{2+}$

C

$\left[ Co ( CN )_{6}\right]^{3-}>\left[ CoCl \left( NH _{3}\right)_{5}\right]^{2+}>\left[ Co ( CN )_{6}\right]^{3-}$

D

$\left[ Co \left( NH _{3}\right)_{6}\right]^{3+}>\left[ CoCl \left( NH _{3}\right)_{5}\right]>\left[ Co ( CN )_{6}\right]^{3-}$

Solution:

KCET 2021 Chemistry Questions with Answers Key Solutions

Q1. The compound AA (major product) is

Q2. Bond enthalpies of A2A_2, B2B_2 and ABAB are in the ratio 2:1:22 : 1 : 2. If bond enthalpy of formation of ABAB is −100KJmol−1-100 \,KJ \,mol ^{-1} .The bond enthalpy of B2B _{2} is

100KJmol−1100\, KJ\, mol ^{-1}

50KJmol−150\, KJ\, mol ^{-1}

200KJmol−1200\, KJ\, mol ^{-1}

150KJmol−1150\, KJ\, mol ^{-1}

Lesser the steric hindrance, easier and higher the rate of reaction towards SN2SN ^{2}

Q4. The major product of the following reaction is CH2=CH−CH2−OHHBr→ExcessCH _{2}= CH - CH _{2}- OH \xrightarrow[\text{Excess} ]{HBr} product

CH3−CHBr−CH2BrCH _{3}- CHBr - CH _{2} Br

CH2=CH−CH2BrCH _{2}= CH - CH _{2} Br

CH3−CHBr−CH2−OHCH _{3}- CHBr - CH _{2}- OH

CH3−CHOH−CH2OHCH _{3}- CHOH - CH _{2} OH

CH2=CH−CH2OHHBr→ExcessCH3−CH|Br−CH2BrCH _{2}= CH - CH _{2}OH \xrightarrow[\text{Excess} ]{HBr} CH_3 - \underset{\overset{|}{B}r}{CH} - CH_2Br

Q5. The product 'A' gives white precipitate when treated with bromine water. The product ' BB ' treated with Barium hydroxide to give the product CC. The compound CC is heated strongly form product DD. The product DD is

4-Methylpent-3-en-2-one

But-2 enal

3-Methylpent-3-en-2-one

2-Methylbut-2-enal

Q6. For the reaction A(g)+B(g)⇌C(g)+D(g);ΔH=−QKJ A ( g )+ B ( g ) \rightleftharpoons C ( g )+ D ( g ) ; \Delta H =- QKJ The equilibrium constant cannot be disturbed by

Addition of A

Addition of D

Increasing of pressure

none of the above

Equilibrium constant is directly proportional to temperature

Q7. An organic compound 'XX' on treatment with PCCPCC in dichloromethane gives the compound YY. Compound 'YY' reacts with I2I_2 and alkali to form yellow precipitate of triiodomethane. The compound X is

CH3CHOCH _{3} CHO

CH3COCH3CH _{3} COCH _{3}

CH3CH2OHCH _{3} CH _{2} OH

CH3COOHCH _{3} COOH

Q9. In set of reactions, identify DD CH3COOHSOCl2→A Benzene →AnhAlCl3BHCN→CH2O→DCH _{3} COOH \xrightarrow{ SOCl _{2}}A \xrightarrow[\text{Anh} \,AlCl_3]{\text { Benzene }} B \xrightarrow{ HCN } C \xrightarrow{ H _{2} O }D

H2SO3H_2SO_3

HNO2HNO_2

CH3COOHCH_3COOH

HCNHCN

Ka(HCN)=4×10−10K _{ a }( HCN )=4 \times 10^{-10} HCN⇌H++CN−HCN \rightleftharpoons H ^{+}+ CN ^{-} Weaker the acid, stronger the conjugate base.

Q11. A,BA, B and CC respectively are

ethanol, ethane nitrile and ethyne

ethane nitrile, ethanol and ethyne

ethyne, ethanol and ethane nitrile

ethyne, ethane nitrile and ethanol

Q12. The reagent which can do the conversion CH3COOH⟶CH3−CH2−OHCH _{3} COOH \longrightarrow CH _{3}- CH _{2}- OH is

LiAIH4LiAIH_4 / ether

H2,PtH_2, Pt

NaBH4NaBH_4

NaNa and C2H5OHC_2H_5OH

CH3COOHEthanoic acid LiAlH 4/ ether →CH3CH2OHEthanol\underset{\text{Ethanoic acid}}{CH _{3} COOH} \xrightarrow{\text { LiAlH }_{4} / \text { ether }} \underset{\text{Ethanol}}{CH _{3} CH _{2} OH}

Q13. CH3CHO(i)CH3MgBr→(ii)H3O+A ConcH2SO4→ΔB(i)B2H6→(ii)H2O,OH−CACH _{3} CHO \xrightarrow[(ii)H_3O^+]{(i) CH _{3} MgBr } A \xrightarrow[\Delta]{\text{ Conc} H _{2} SO _{4}} B \xrightarrow[(ii)H_2O,OH^-]{(i) B _{2} H _{6}} CA and CC are

Identical

Position isomers

Functional isomers

Optical isomers

Q14. Which of the following is not true for oxidation?

addition of oxygen

addition of electronegative element

removal of hydrogen

removal of electronegative element

Oxidation is addition of oxygen or electronegative element; Oxidation is removal of hydrogen or electropositive element.

Q15. Which is the most suitable reagent for the following conversion? CH3−CH=CH−CH2−O||C−CH3→CH3−CH=CH−CH2−O||C−OHCH_3 - CH = CH - CH_2 -\overset{\underset{||}{O}}{C}-CH_3 \to CH_3 - CH = CH- CH_2 - \overset{\underset{||}{O}}{C}-OH

Tollen's reagent

Benzoyl peroxide

I2I_2 and NaOHNaOH solution with subsequent acidification

SnSn and NaOHNaOH solution

Q16. C6H5CH2Clalc.NH3→A2CH3Cl→BC_6H_5CH_2Cl \ce{->[\text{alc}.NH_3]} A \ce{->[2CH_3Cl]}B. The product BB is

N, N - Dimethyl phenyl methanamine

N, N - Dimethyl benzenamine

N - Benzyl - N - methyl methanamine

phenyl - N, N - dimethyl methanamine

Q17. The method by which aniline cannot be prepared is

Nitration of benzene followed by reduction with Sn and con HCI

Degradation of benzamide with bromine in alkaline solution

Reduction of nitrobenzene with H2/PdH_2 / Pd is ethanol

Potassium salt of phthalimide treated with chlorgbenzene followed by the hydrolysis with aqueous NaOH solution

Gabriel phthalimide synthesis not preferred for preparing aniline because amine formation involves nucleophilic substitution of alkyl halides by the anion formed by phthalimide. But aryl halides do not undergo nucleophilic substitution.

Q18. Permanent hardness cannot be removed by

Using washing soda

Calgon's method

Clark's method

Ion exchange method

Clark’s method is for the removal of temporary hardness.

Q1. The compound $A$ (major product) is

Q2. Bond enthalpies of $A_2$ , $B_2$ and $AB$ are in the ratio $2 : 1 : 2$ . If bond enthalpy of formation of $AB$ is $-100 \,KJ \,mol ^{-1}$ .The bond enthalpy of $B _{2}$ is

$100\, KJ\, mol ^{-1}$

$50\, KJ\, mol ^{-1}$

$200\, KJ\, mol ^{-1}$

$150\, KJ\, mol ^{-1}$

Lesser the steric hindrance, easier and higher the rate of reaction towards $SN ^{2}$

Q4. The major product of the following reaction is $CH _{2}= CH - CH _{2}- OH \xrightarrow[\text{Excess} ]{HBr}$ product

$CH _{3}- CHBr - CH _{2} Br$

$CH _{2}= CH - CH _{2} Br$

$CH _{3}- CHBr - CH _{2}- OH$

$CH _{3}- CHOH - CH _{2} OH$

$CH _{2}= CH - CH _{2}OH \xrightarrow[\text{Excess} ]{HBr} CH_3 - \underset{\overset{|}{B}r}{CH} - CH_2Br$

Q5. The product 'A' gives white precipitate when treated with bromine water. The product ' $B$ ' treated with Barium hydroxide to give the product $C$ . The compound $C$ is heated strongly form product $D$ . The product $D$ is

Q6. For the reaction $A ( g )+ B ( g ) \rightleftharpoons C ( g )+ D ( g ) ; \Delta H =- QKJ$
The equilibrium constant cannot be disturbed by

Q7. An organic compound ' $X$ ' on treatment with $PCC$ in dichloromethane gives the compound $Y$ . Compound ' $Y$ ' reacts with $I_2$ and alkali to form yellow precipitate of triiodomethane. The compound X is

$CH _{3} CHO$

$CH _{3} COCH _{3}$

$CH _{3} CH _{2} OH$

$CH _{3} COOH$

Q9. In set of reactions, identify $D$
$CH _{3} COOH \xrightarrow{ SOCl _{2}}A \xrightarrow[\text{Anh} \,AlCl_3]{\text { Benzene }} B \xrightarrow{ HCN } C \xrightarrow{ H _{2} O }D$

$H_2SO_3$

$HNO_2$

$CH_3COOH$

$HCN$

$K _{ a }( HCN )=4 \times 10^{-10}$
$HCN \rightleftharpoons H ^{+}+ CN ^{-}$
Weaker the acid, stronger the conjugate base.

Q11. $A, B$ and $C$ respectively are

Q12. The reagent which can do the conversion $CH _{3} COOH \longrightarrow CH _{3}- CH _{2}- OH$ is

$LiAIH_4$ / ether

$H_2, Pt$

$NaBH_4$

$Na$ and $C_2H_5OH$

$\underset{\text{Ethanoic acid}}{CH _{3} COOH} \xrightarrow{\text { LiAlH }_{4} / \text { ether }} \underset{\text{Ethanol}}{CH _{3} CH _{2} OH}$

Q13. $CH _{3} CHO \xrightarrow[(ii)H_3O^+]{(i) CH _{3} MgBr } A \xrightarrow[\Delta]{\text{ Conc} H _{2} SO _{4}} B \xrightarrow[(ii)H_2O,OH^-]{(i) B _{2} H _{6}} CA$ and $C$ are

Oxidation is addition of oxygen or electronegative element;
Oxidation is removal of hydrogen or electropositive element.

Q15. Which is the most suitable reagent for the following conversion?
$CH_3 - CH = CH - CH_2 -\overset{\underset{||}{O}}{C}-CH_3 \to CH_3 - CH = CH- CH_2 - \overset{\underset{||}{O}}{C}-OH$

$I_2$ and $NaOH$ solution with subsequent acidification

$Sn$ and $NaOH$ solution

Q16. $C_6H_5CH_2Cl \ce{->[\text{alc}.NH_3]} A \ce{->[2CH_3Cl]}B$ . The product $B$ is

Reduction of nitrobenzene with $H_2 / Pd$ is ethanol

$MCl + H _{2} SO _{4} \longrightarrow MHSO _{4}+ HCl _{( g )}$

$12 : 5$ Membered ring
$20 : 6$ Membered ring

Q25. In Chrysoberyl, a compound containing Beryllium, Aluminium and oxygen, oxide ions form cubic close packed structure. Aluminium ions occupy $1/4^\text{ th}$ of tetrahedral voids and Beryllium ions occupy $1/4^\text{ th}$ of octahedral voids. The formula of the compound is

$BeAlO _{4}$

$BeAl _{2} O _{4}$

$Be _{2} AlO _{2}$

$BeAlO _{2}$

$CCP = N =4$
Octahedral void $= N =1 \times 4=4 \times \frac{1}{4}=1: Be$
Tetrahedral void $=2 N =2 \times 4=2 \times \frac{1}{4}=2: A l$

Q27. A metal crystallises in $BCC$ lattice with unit cell edge length of $300\, pm$ and density $6.15\, g\, cm^{-3}$ . The molar mass of the metal is

$50\, g\, mol ^{-1}$

$60\, g\, mol ^{-1}$

$40\, g\, mol ^{-1}$

$70\, g\, mol ^{-1}$

$d =\frac{ ZM }{ a ^{3} N _{ A }}$
$6.15=\frac{2 \times M }{\left(3 \times 10^{-8}\right)^{3} \times 6.02 \times 10^{23}}$
$M =\frac{6.15 \times 27 \times 10^{-24} \times 6.02 \times 10^{23}}{2}$
$M =50\, g / mol$