Let [ . ] denotes greatest integer function. Then the value of ∫_0^3 ( e^x+e^-x [x]! )dx is :

(1)/(2) (e^2+e^3-(1)/(e^2)-(1)/(e^3) )

JEE Main 2026 April 02, Shift 1 Question Paper with Solutions

All 75 questions from the JEE Main 2026 (April 02, Shift 1) shift — Physics (25), Chemistry (25) and Mathematics (25) — with the correct answer and a step-by-step solution for every question.

Physics PYQs 2026 Chemistry PYQs 2026 Mathematics PYQs 2026

Chemistry25 questions

Q51Single correctSome Basic Concepts in Chemistry

The mass of iron converted into

\text{Fe}_3\text{O}_4

by the action of 18g of steam is (Given :Molar mass of

H_2O

and Fe are 1,16 and 56 g

\text{mol}^{-1}

respectively) Assume iron is present in excess.

(A)

2.1 gm

(B)

4.2 gm

(C)

21 gm

(D)

42 gm

Q52Single correctAtomic Structure

What is the energy (in J ato

m^{-1}

) required for the following process?

\text{Li}_{(g)}^{2+} \rightarrow \text{Li}_{(g)}^{3+} + e^{-}

(Take the ionization energy for the H atom in the ground state as

2.18\times 10^{-18}

J ato

m^{-1}

)

(A)

\displaystyle 8.72\times 10^{-18}

(B)

\displaystyle 1.962\times 10^{-18}

(C)

\displaystyle 1.962\times 10^{-17}

(D)

\displaystyle 6.542\times 10^{-17}

Q53Single correctChemical Bonding and Molecular Structure

Given below are two statements

Statement-I : The correct sequence of bond length in the following species is

O_2^{\oplus} < O_2 < O_2^{-} < O_2^{2-}

Statement-II : The correct sequence of number of unpaired electrons in the following species is

O_2 > O_2^{-} > O_2^{-} > O_2^{2-}

In the light of the above statements choose the correct answers from the options given below

(A)

Both Statement-I and Statement-II are true

(B)

Both Statement-I and Statement-II are false

(C)

Statement-I is true but Statement-II is false

(D)

Statement-I is false but Statement-II is true

SolutionAnswer: Option 3Statement-I is true but Statement-II is false

Approach:

Use molecular-orbital theory to obtain bond orders and unpaired-electron counts for

\text{O}_2^{+},\text{O}_2,\text{O}_2^{-},\text{O}_2^{2-}

, then evaluate both statements.

Step 1:Given data and target.

\text{species: }\text{O}_2^{+},\text{O}_2,\text{O}_2^{-},\text{O}_2^{2-}

Step 2:List MO electrons (total) and antibonding

\pi^{*}

occupancy.

\text{O}_2^{+}:15e^{-},\;\text{O}_2:16e^{-},\;\text{O}_2^{-}:17e^{-},\;\text{O}_2^{2-}:18e^{-}

Step 3:Compute bond orders.

\text{BO}(\text{O}_2^{+})=2.5,\;\text{BO}(\text{O}_2)=2,\;\text{BO}(\text{O}_2^{-})=1.5,\;\text{BO}(\text{O}_2^{2-})=1

Step 4:Invert to get bond lengths (smaller BO

\Rightarrow

longer bond).

\text{O}_2^{+}<\text{O}_2<\text{O}_2^{-}<\text{O}_2^{2-}

Step 5:List unpaired electrons (from

\pi^{*}_{2p}

occupancy).

\text{O}_2^{+}:1,\;\text{O}_2:2,\;\text{O}_2^{-}:1,\;\text{O}_2^{2-}:0

Step 6:Examine Statement II — the correct decreasing chain is

\text{O}_2(2)>\text{O}_2^{+}(1)=\text{O}_2^{-}(1)>\text{O}_2^{2-}(0)

, with equality between

\text{O}_2^{+}

and

\text{O}_2^{-}

2>1=1>0

Step 7:Cross-check by combining both findings: I true and II false.

\text{I true},\;\text{II false}

Step 8:Result.

\text{Statement-I true, Statement-II false}

Final answer: Statement-I is true but Statement-II is false

Q54Single correctChemical Thermodynamics

Consider the following data i)

2Al_{(s)}+6\text{HCl}_{(aq)}\rightarrow Al_2Cl_{6(aq)}+3H_{2(g)}+1200\,kJ/\text{mol}

ii)

H_{2(g)}+Cl_{2(g)}\rightarrow 2\text{HCl}_{(g)}+164\,kJ/\text{mol}

iii)

\text{HCl}_{(g)}+aq\rightarrow \text{HCl}_{(aq)}+83\,kJ/\text{mol}

iv)

Al_2Cl_{6(s)}+aq\rightarrow Al_2Cl_{6(aq)}+663\,kJ/\text{mol}

The enthalpy of formation of anhydrous solid

Al_2Cl_6

is :

(A)

\displaystyle -648\,kJ\ \text{mol}^{-1}

(B)

\displaystyle -1350\,kJ\ \text{mol}^{-1}

(C)

\displaystyle -2002\,kJ\ \text{mol}^{-1}

(D)

\displaystyle -1527\,kJ\ \text{mol}^{-1}

SolutionAnswer: Option 4

-1527\,kJ\ \text{mol}^{-1}

Approach:

Combine the four given thermochemical equations via Hess's law to obtain

2\text{Al}(s)+3\text{Cl}_2(g)\rightarrow \text{Al}_2\text{Cl}_6(s)

and sum the corresponding enthalpy changes.

Step 1:Given data and target reaction.

2\text{Al}(s)+3\text{Cl}_2(g)\rightarrow \text{Al}_2\text{Cl}_6(s),\;\Delta H_f=?

Step 2:Read the four enthalpies (heat released

\Rightarrow

negative

\Delta H

\Delta H_1=-1200,\;\Delta H_2=-164,\;\Delta H_3=-83,\;\Delta H_4=-663\,\text{kJ/mol}

Step 3:Take combination

(\text{i})+3(\text{ii})+6(\text{iii})-(\text{iv})

and verify cancellation of HCl(g), HCl(aq),

H_2

, A

l_2Cl_6

(aq), and aq.

2\text{Al}+6\text{HCl}(aq)+3\text{H}_2+3\text{Cl}_2+6\text{HCl}(g)+aq+\text{Al}_2\text{Cl}_6(aq)\rightarrow \text{Al}_2\text{Cl}_6(aq)+3\text{H}_2+6\text{HCl}(g)+6\text{HCl}(aq)+\text{Al}_2\text{Cl}_6(s)+aq

Step 4:Substitute the enthalpy values.

\Delta H_f=-1200+3(-164)+6(-83)-(-663)

Step 5:Arithmetic.

\Delta H_f=-1200-492-498+663=-1527\,\text{kJ mol}^{-1}

Step 6:Cross-check by recomputing partial sums.

-1200-492=-1692;\;-1692-498=-2190;\;-2190+663=-1527

Step 7:Result.

\Delta H_f(\text{Al}_2\text{Cl}_6,s)=-1527\,\text{kJ mol}^{-1}

Final answer:

-1527\,kJ\ \text{mol}^{-1}

Q55Single correctSolutions

19.5 g of fluoro acetic acid (molar mass

=78

\text{mol}^{-1}

) is dissolved in 500g of water at 298K. The depression in the freezing point of water was

1^{0}C

. What is

k_a

of fluoro acetic acid? (For water,

K_f =1.86

K kg

\text{mol}^{-1}

). Assume molarity and molality to have same values.

(A)

\displaystyle 10^{-6}

(B)

\displaystyle 4\times 10^{-4}

(C)

\displaystyle 3\times 10^{-5}

(D)

\displaystyle 3\times 10^{-3}

Q56Single correctEquilibrium

The solubility product constants of

Ag_2\text{CrO}_4

and AgBr are

32x

and

4y

respectively at 298K. The value of

\left(\dfrac{molarity\ of\ Ag_2CrO_4}{molarity\ of\ AgBr}\right)

can be expressed as :

(A)

\displaystyle \dfrac{2\sqrt[3]{x}}{y}

(B)

Option 2 expression rendered as a small image in the PDF; appears to contain a graphical/mathematical figure rather than plain text

(C)

\displaystyle \sqrt{\dfrac{x}{y}}

(D)

\displaystyle \dfrac{\sqrt[3]{x}}{\sqrt{y}}

Q57Single correctRedox Reactions and Electrochemistry

An electrochemical cell is constructed using half cells in the direction of spontaneous change

Fe(OH)_2(s)+2e^{-}\rightarrow Fe(s)+2OH^{-}(aq)\quad E^{0}=-0.88V

\text{AgBr}(s)+e^{-}\rightarrow Ag(s)+Br^{-}(aq)\quad E^{0}=0.07V

Which of the following option is correct

(A)

Overall reaction

\displaystyle Fe(s)+2OH^{-}(aq)+2\text{AgBr}(s)\rightleftharpoons Fe(OH)_2(s)+2Ag(s)+2Br^{-}(aq)

(B)

\displaystyle E^{0}_{cell} = -0.95\,V

(C)

Fe is reduced in the electrochemical cell

(D)

\displaystyle E^{0}_{cell}

is an extensive property

SolutionAnswer: Option 1Overall reaction

Fe(s)+2OH^{-}(aq)+2\text{AgBr}(s)\rightleftharpoons Fe(OH)_2(s)+2Ag(s)+2Br^{-}(aq)

Approach:

Identify the cathode (higher

E^{0}

) and anode (lower

E^{0}

), build the spontaneous overall cell reaction, compute

E^{0}_{\text{cell}}

, and evaluate each option.

Step 1:Given data and target.

E^{0}(\text{Fe(OH)}_2/\text{Fe})=-0.88\,V,\;E^{0}(\text{AgBr}/\text{Ag})=+0.07\,V

Step 2:The half-cell with the higher reduction potential is the cathode.

+0.07>-0.88\Rightarrow \text{AgBr/Ag is cathode, Fe(OH)}_2/\text{Fe is anode}

Step 3:Write the cathode (reduction) half-reaction.

2\text{AgBr}(s)+2e^{-}\rightarrow 2\text{Ag}(s)+2\text{Br}^{-}(aq)

Step 4:Write the anode (oxidation) half-reaction (reverse of Fe(OH

)_2

reduction).

\text{Fe}(s)+2\text{OH}^{-}(aq)\rightarrow \text{Fe(OH)}_2(s)+2e^{-}

Step 5:Add the half-reactions to obtain the overall spontaneous reaction.

\text{Fe}(s)+2\text{OH}^{-}(aq)+2\text{AgBr}(s)\rightleftharpoons \text{Fe(OH)}_2(s)+2\text{Ag}(s)+2\text{Br}^{-}(aq)

Step 6:Compute

E^{0}_{\text{cell}}

E^{0}_{\text{cell}}=0.07-(-0.88)=+0.95\,V

Step 7:In the cell Fe is at the anode and is oxidized, so option 3 is incorrect;

E^{0}_{\text{cell}}

is intensive (independent of amount), so option 4 is incorrect.

\text{Fe oxidized};\;E^{0}_{\text{cell}}\text{ intensive}

Step 8:Cross-check by checking sign and direction:

E^{0}_{\text{cell}}>0

confirms the written direction is spontaneous.

E^{0}_{\text{cell}}=+0.95\,V>0

Step 9:Result.

\text{Overall: }\text{Fe}(s)+2\text{OH}^{-}(aq)+2\text{AgBr}(s)\rightleftharpoons \text{Fe(OH)}_2(s)+2\text{Ag}(s)+2\text{Br}^{-}(aq)

Final answer: Overall reaction

Fe(s)+2OH^{-}(aq)+2\text{AgBr}(s)\rightleftharpoons Fe(OH)_2(s)+2Ag(s)+2Br^{-}(aq)

Q58Single correctChemical Kinetics

t_{100\%}

is the time required for the 100% completion of the reaction while

t_{1/2}

is the time required for the 50% of the reaction to be completed which of the following option correctly represents relation between

t_{100\%}

and

t_{1/2}

for zero and first order reactions respectively?

(A)

\displaystyle t_{100\%}=(t_{1/2})^2

and

\displaystyle t_{100\%}=(t_{1/2})^{-\infty}

(B)

\displaystyle t_{100\%}=2\times t_{1/2}: t_{100\%}=(t_{1/2})^{\infty}

(C)

\displaystyle t_{100\%}=2\times t_{1/2}: t_{100\%}=(2t_{1/2})^2

(D)

\displaystyle t_{100\%}=(t_{1/2})^{\infty}: t_{100\%}=2t_{1/2}

SolutionAnswer: Option 2

t_{100\%}=2\times t_{1/2}: t_{100\%}=(t_{1/2})^{\infty}

Approach:

Use the integrated rate laws for zero- and first-order reactions to express

t_{100\%}

and

t_{1/2}

, then compare.

Step 1:Given data and target.

\text{compare } t_{100\%}\text{ vs }t_{1/2}\text{ for zero and first order}

Step 2:(zero order): set

[A]=0

[A]_0-kt=0

to find completion time.

t_{100\%}=\dfrac{[A]_0}{k}

Step 3:(zero order): take ratio with

t_{1/2}=[A]_0/(2k)

\dfrac{t_{100\%}}{t_{1/2}}=\dfrac{[A]_0/k}{[A]_0/(2k)}=2\Rightarrow t_{100\%}=2\,t_{1/2}

Step 4:(first order): set

[A]=0

[A]_0e^{-kt}=0

e^{-kt}=0\Rightarrow kt\rightarrow \infty \Rightarrow t_{100\%}\rightarrow \infty

Step 5:(first order): since

t_{1/2}=\ln 2/k

is finite,

t_{100\%}

is infinitely many half-lives, symbolically written

t_{100\%}=(t_{1/2})^{\infty}

t_{100\%}=(t_{1/2})^{\infty}

Step 6:Cross-check by inspecting integrated forms: zero-order concentration reaches zero in finite time, first-order only asymptotically.

\text{zero: linear decay};\;\text{first: exponential decay}

Step 7:Result.

t_{100\%}=2\,t_{1/2}\;(\text{zero}),\;t_{100\%}=(t_{1/2})^{\infty}\;(\text{first})

Final answer:

t_{100\%}=2\times t_{1/2}: t_{100\%}=(t_{1/2})^{\infty}

Q59Single correctClassification of Elements and Periodicity

Given below are two statements

Statement – I : The first Ionization enthalpy of the elements

\text{Na}, \text{Mg}, \text{Cl}

and

\text{Ar}

follows the order

\text{Na} > \text{Mg} > \text{Cl} > \text{Ar}

Statement – II : Among

\text{Ca}, \text{Al}, \text{Fe}

and

\text{B}

, the third ionisation enthalpy is very high for

\text{Ca}

.

In the light of the above statements choose the correct answers from the options given below

(A)

Both Statement-I and Statement-II are true

(B)

Both Statement-I and Statement-II are false

(C)

Statement-I is true but Statement-II is false

(D)

Statement-I is false and Statement-II is true

SolutionAnswer: Option 4Statement-I is false and Statement-II is true

Approach:

Evaluate Statement-I against the standard first ionisation enthalpy trend across Na, Mg, Cl, Ar and Statement-II by comparing successive (third) ionisation enthalpies for Ca, Al, Fe, B based on the stability of the resulting electron configurations.

Step 1:The targets: validate (i) the order of first ionisation enthalpy among

\text{Na}, \text{Mg}, \text{Cl}, \text{Ar}

and (ii) which element among

\text{Ca}, \text{Al}, \text{Fe}, \text{B}

has the largest third ionisation enthalpy.

\text{IE}_1(\text{Na,Mg,Cl,Ar}) \;\text{and}\; \text{IE}_3(\text{Ca,Al,Fe,B})

Step 2:The first ionisation enthalpy values (kJ mo

l^{-1}

\text{Na}=496

\text{Mg}=738

\text{Cl}=1251

\text{Ar}=1521

\text{IE}_1(\text{Na})=496,\;\text{IE}_1(\text{Mg})=738,\;\text{IE}_1(\text{Cl})=1251,\;\text{IE}_1(\text{Ar})=1521

Step 3:Statement-I claims

\text{Na} > \text{Mg} > \text{Cl} > \text{Ar}

which is the exact reverse of the true period trend; hence Statement-I is FALSE.

\text{Statement-I order} = \text{reverse of actual}

Step 4:Third ionisation enthalpies (kJ mo

l^{-1}

\text{Ca}^{2+}([\text{Ar}]) \to \text{Ca}^{3+}

breaks a noble-gas core,

\text{IE}_3(\text{Ca}) = 4912

;

\text{IE}_3(\text{B}) = 3660

(removal from

2s^1

);

\text{IE}_3(\text{Fe}) = 2957

;

\text{IE}_3(\text{Al}) = 2745

\text{IE}_3: \text{Ca}(4912) > \text{B}(3660) > \text{Fe}(2957) > \text{Al}(2745)

Step 5:The reasoning for

\text{Ca}

: after losing two valence

4s

electrons,

\text{Ca}^{2+}

has the stable

[\text{Ar}]

noble-gas configuration; removing a 3rd electron requires very high energy.

\text{Ca}^{2+}: [\text{Ar}] \xrightarrow{\text{IE}_3} \text{Ca}^{3+}: [\text{Ne}]3s^2 3p^5

Step 6:Cross-check by comparing tabulated successive ionisation enthalpies —

\text{Ca}

shows the characteristic large jump between

\text{IE}_2

and

\text{IE}_3

confirming that

\text{IE}_3(\text{Ca})

is the largest of the four.

\text{IE}_3(\text{Ca}) \gg \text{IE}_3(\text{Al}, \text{Fe}, \text{B})

Step 7:Result: Statement-I is FALSE and Statement-II is TRUE, which corresponds to option (4).

\text{Statement-I: FALSE},\;\text{Statement-II: TRUE}

Final answer: Statement-I is false and Statement-II is true

Q60Single correctp-Block Elements

Given below are two statements :

Statement I : Oxidising power of halogens decreases in the order

F_2 > Cl_2 > Br_2 > I_2

which is the basis of “Layer test”

Statement II : “Layer test” to identify

Br_2

and

I_2

in aqueous solution involves the oxidation of bromide or iodide into

Br_2

I_2

respectively with

Cl_2

, which is a type of displacement redox reaction.

In the light of the above statements choose the correct answers from the options given below

(A)

Both Statement-I and Statement-II are true

(B)

Both Statement-I and Statement-II are false

(C)

Statement-I is true but Statement-II is false

(D)

Statement-I is false but Statement-II is true

SolutionAnswer: Option 1Both Statement-I and Statement-II are true

Approach:

Cross-check: the oxidising-power trend of the halogens using standard reduction potentials and confirm that the layer test displacement of

\text{Br}^-

\text{I}^-

\text{Cl}_2

is a redox reaction.

Step 1:Given data: Statement-I claims the oxidising power decreases as

F_2 > Cl_2 > Br_2 > I_2

and Statement-II describes the layer test as a displacement redox process.

\text{Statements about oxidising power and layer test}

Step 2:The standard reduction potentials

E^{\circ}(X_2/X^-)

F_2/F^- = +2.87\,\text{V}

Cl_2/Cl^- = +1.36\,\text{V}

Br_2/Br^- = +1.09\,\text{V}

I_2/I^- = +0.54\,\text{V}

E^{\circ}: F_2(+2.87) > Cl_2(+1.36) > Br_2(+1.09) > I_2(+0.54)\,\text{V}

Step 3:The chemistry of the layer test: chlorine water added to a mixture containing

\text{Br}^-/\text{I}^-

oxidises them to

\text{Br}_2/\text{I}_2

; carbon tetrachloride or

\text{CS}_2

extracts the liberated halogen into the organic (lower) layer.

Cl_2 + 2Br^- \to 2Cl^- + Br_2;\; Cl_2 + 2I^- \to 2Cl^- + I_2

Step 4:Colour observations:

\text{Br}_2

gives an orange/brown organic layer,

\text{I}_2

gives a violet/pink organic layer — the very basis of qualitative identification.

\text{Br}_2: \text{orange};\; \text{I}_2: \text{violet}

Step 5:Cross-check by oxidation-state book-keeping:

\text{Cl}

goes from

0 \to -1

(reduced),

\text{Br}/\text{I}

goes from

-1 \to 0

(oxidised); a balanced electron transfer — a genuine redox.

\Delta(\text{ox. state}): Cl: 0 \to -1,\; X: -1 \to 0

Step 6:Result: both Statement-I and Statement-II are true, matching option (1).

\text{Both statements TRUE}

Final answer: Both Statement-I and Statement-II are true

Q61Single correctd- and f- Block Elements

Which of the following sets includes all the species that will change the orange colour of

K_2Cr_2O_7

in acidic medium?

(A)

\displaystyle Fe^{2+}, Sn^{2+}, I^-, S^{2-}

(B)

\displaystyle S^{2-}, Fe^{3+}, I^-, C_2O_4^{2-}

(C)

\displaystyle Fe^{2+}, NO_2^-, SO_2, Sn^{4+}

(D)

\displaystyle Fe^{3+}, SO_4^{2-}, S^{2-}, Sn^{4+}

SolutionAnswer: Option 1

Fe^{2+}, Sn^{2+}, I^-, S^{2-}

Approach:

Acidified

K_2Cr_2O_7

is a strong oxidising agent (

Cr^{+6} \to Cr^{+3}

). Identify which option contains only species capable of being oxidised (in lower oxidation states) so that they reduce dichromate and discharge its orange colour.

Step 1:The target: only reducing species (i.e. species in lower oxidation states relative to dichromate) can discharge the orange colour by being oxidised.

\text{Reducing species} + Cr_2O_7^{2-}/H^+ \to \text{oxidised products} + Cr^{3+}(\text{green})

Step 2:Option (1) species — each is in its lower stable oxidation state and is readily oxidised:

Fe^{2+}\to Fe^{3+}

Sn^{2+}\to Sn^{4+}

I^-\to I_2

S^{2-}\to S

Fe^{2+}, Sn^{2+}, I^-, S^{2-}

Step 3:Balanced equation for iron:

6Fe^{2+} + Cr_2O_7^{2-} + 14H^+ \to 6Fe^{3+} + 2Cr^{3+} + 7H_2O

6Fe^{2+} + Cr_2O_7^{2-} + 14H^+ \to 6Fe^{3+} + 2Cr^{3+} + 7H_2O

Step 4:For tin:

3Sn^{2+} + Cr_2O_7^{2-} + 14H^+ \to 3Sn^{4+} + 2Cr^{3+} + 7H_2O

3Sn^{2+} + Cr_2O_7^{2-} + 14H^+ \to 3Sn^{4+} + 2Cr^{3+} + 7H_2O

Step 5:For iodide and sulphide:

6I^- + Cr_2O_7^{2-} + 14H^+ \to 3I_2 + 2Cr^{3+} + 7H_2O

and

3S^{2-} + Cr_2O_7^{2-} + 14H^+ \to 3S + 2Cr^{3+} + 7H_2O

6I^- + Cr_2O_7^{2-} + 14H^+ \to 3I_2 + 2Cr^{3+} + 7H_2O

Step 6:Cross-check: other options fail: option (2) contains

Fe^{3+}

(highest stable state, cannot be further oxidised by

Cr_2O_7^{2-}

); option (3) contains

Sn^{4+}

(cannot be oxidised); option (4) contains

Fe^{3+}, SO_4^{2-}, Sn^{4+}

(all highest oxidation state, non-reducing).

Fe^{3+}, Sn^{4+}, SO_4^{2-}

cannot reduce

Cr_2O_7^{2-}/H^+

Step 7:Result: only option (1) contains a complete set of reducing species.

\text{Set} = \{Fe^{2+}, Sn^{2+}, I^-, S^{2-}\}

Final answer:

Fe^{2+}, Sn^{2+}, I^-, S^{2-}

Q62Single correctCoordination Compounds

Match List – I with List – II

List– I (Chromium (III) Complexes, en = ethylene diamine)	List– II $\Delta_0\,(cm^{-1})$
A) $[Cr(CN)_6]^{3-}$	I) 15060
B) $[\text{CrF}_6]^{3-}$	II) 17400
C) $[Cr(H_2O)_6]^{3+}$	III) 22300
D) $[Cr(en)_3]^{3+}$	IV) 26600

Choose the correct answer from the options given below:

(A)

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

(B)

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

(C)

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

(D)

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

SolutionAnswer: Option 4

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

Approach:

Crystal-field splitting energy

\Delta_0

for a given metal centre in the same oxidation state increases with ligand-field strength as ordered in the spectrochemical series. Rank the four Cr(III) complexes by ligand strength and assign the four

\Delta_0

values accordingly.

Step 1:The four ligands and their roles: all four complexes are octahedral Cr(III) (

d^3

) species, so

\Delta_0

depends only on the ligand-field strength.

\text{Common metal centre}: Cr^{3+},\; d^3,\; \text{octahedral}

Step 2:The spectrochemical position of each ligand:

CN^-

is a strong-field

\pi

-acceptor (largest

\Delta_0

); ethylenediamine (en) is a strong-field neutral

\sigma

-donor and chelating ligand;

H_2O

is intermediate;

F^-

is a weak-field ligand (smallest

\Delta_0

CN^- > en > H_2O > F^-

Step 3:Assignment of

\Delta_0

values from List-II in decreasing order to the complexes in decreasing ligand strength: A

[Cr(CN)_6]^{3-} \to 26600\,\text{cm}^{-1}

(IV); D

[Cr(en)_3]^{3+} \to 22300\,\text{cm}^{-1}

(III).

A \to IV,\; D \to \text{III}

Step 4:Remaining assignments: C

[Cr(H_2O)_6]^{3+} \to 17400\,\text{cm}^{-1}

(II); B

[CrF_6]^{3-} \to 15060\,\text{cm}^{-1}

(I).

C \to II,\; B \to I

Step 5:Cross-check: monotonicity:

\Delta_0

decreases as the ligand becomes weaker —

26600 > 22300 > 17400 > 15060\,\text{cm}^{-1}

exactly tracks

CN^- > en > H_2O > F^-

26600 > 22300 > 17400 > 15060\,\text{cm}^{-1}

Step 6:Result.

A-IV,\; B-I,\; C-II,\; D-\text{III}

Final answer:

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

Q63Single correctPurification and Characterisation of Organic Compounds

Given below are two statements :

Statement-I : 1, 2, 3-trihydroxy propane can be separated from water by simple distillation.

Statement-II : An azeotropic mixture cannot be separated by fractional Distillation

In the light of the above statements choose the correct answers from the options given below

(A)

Both Statement-I and Statement-II are true

(B)

Both Statement-I and Statement-II are false

(C)

Statement-I is true but Statement-II is false

(D)

Statement-I is false but Statement-II is true

SolutionAnswer: Option 4Statement-I is false but Statement-II is true

Approach:

Identify the appropriate distillation technique for glycerol (1,2,3-trihydroxypropane), which has a high boiling point and decomposes on heating at atmospheric pressure, and recall the property of an azeotrope that prevents fractional-distillation separation.

Step 1:The target: judge Statement-I (glycerol separable from water by simple distillation) and Statement-II (azeotrope inseparable by fractional distillation).

\text{Statements about glycerol and azeotrope separation}

Step 2:Glycerol behaviour: glycerol boils at

\approx 290\,^{\circ}\text{C}

and undergoes thermal dehydration to acrolein at temperatures approaching its boiling point.

\text{Glycerol} \xrightarrow{\Delta,\;\text{1 atm}} \text{acrolein} + 2H_2O

Step 3:The technique used: glycerol is separated by distillation under reduced pressure (vacuum distillation), which lowers the boiling point below the decomposition temperature.

\text{Reduced-pressure distillation}: P \downarrow \Rightarrow T_b \downarrow

Step 4:Azeotrope theory: at the azeotropic composition the vapour has the same composition as the liquid, so no enrichment occurs across stages of fractional distillation.

y_i = x_i \Rightarrow \text{no fractionation possible}

Step 5:Consequence: separation of an azeotrope requires alternative methods (azeotropic agent, pressure swing, membrane separation), confirming Statement-II.

\text{Alternative methods needed for azeotropes}

Step 6:Cross-check by considering the standard glycerol–water industrial separation: vacuum distillation is the accepted technique, not simple atmospheric distillation.

\text{Industrial glycerol purification} = \text{vacuum distillation}

Step 7:Result: Statement-I FALSE, Statement-II TRUE corresponds to option (4).

\text{I: FALSE, II: TRUE}

Final answer: Statement-I is false but Statement-II is true

Q64Single correctOrganic Compounds Containing Halogens

Given below are two statements

Statement-1 : Benzyl chloride reacts faster in

S_N^{1}

mechanism than ethyl chloride

Statement-II : Ethyl carbocation intermediate is less stabilized by hyperconjugation than benzyl carbocation by resonance.

In the light of the above statements choose the correct answers from the options given below

(A)

Both Statement-I and Statement-II are true

(B)

Both Statement-I and Statement-II are false

(C)

Statement-I is true but Statement-II is false

(D)

Statement-I is false but Statement-II is true

SolutionAnswer: Option 1Both Statement-I and Statement-II are true

Approach:

S_N1

rate is governed by the stability of the carbocation intermediate. Compare benzyl cation (resonance-delocalised onto the aromatic ring) with the ethyl cation (only hyperconjugation), then judge both statements.

Step 1:The target: relate the

S_N1

rate of benzyl chloride versus ethyl chloride to the stability of their carbocation intermediates.

\text{PhCH}_2\text{Cl}\;\text{vs}\;\text{CH}_3\text{CH}_2\text{Cl}

Step 2:The carbocation from benzyl chloride: ionisation gives

\text{PhCH}_2^+

whose positive charge is delocalised onto the ortho and para positions of the phenyl ring through three additional resonance structures.

\text{PhCH}_2^+ \leftrightarrow \text{ring resonance contributors}

Step 3:The carbocation from ethyl chloride: ionisation gives

\text{CH}_3\text{CH}_2^+

which is a primary cation stabilised only by hyperconjugation with three C–H bonds of the methyl group.

\text{CH}_3\text{CH}_2^+: \text{three hyperconjugative C–H bonds}

Step 4:The stability comparison: resonance delocalisation (energy lowering

\sim 25–35\,\text{kcal/mol}

) is much larger than hyperconjugation (a few kcal/mol per C–H), so

\text{PhCH}_2^+

is far more stable than

\text{CH}_3\text{CH}_2^+

E_{\text{stab}}(\text{benzyl})_{\text{resonance}} \gg E_{\text{stab}}(\text{ethyl})_{\text{hyperconj.}}

Step 5:The link to kinetics: a more stable carbocation leads to a lower activation energy for ionisation, hence a faster

S_N1

rate for benzyl chloride.

\Delta G^{\ddagger}(\text{PhCH}_2\text{Cl}) < \Delta G^{\ddagger}(\text{CH}_3\text{CH}_2\text{Cl})

Step 6:Cross-check: Statement-II: the reason given ("ethyl cation less stabilised by hyperconjugation than benzyl cation by resonance") is the standard physical-organic chemistry argument and is itself correct.

\text{Stabilisation: resonance} > \text{hyperconjugation}

Step 7:Result: both statements are true and Statement-II is the correct explanation of Statement-I; this matches option (1).

\text{Both statements TRUE}

Final answer: Both Statement-I and Statement-II are true

Q65Single correctSome Basic Principles of Organic Chemistry

In IUPAC nomenclature, the correct order of decreasing priority of functional group is :

(A)

\displaystyle -\text{CONH}_2, \;\rangle C=O, -\text{CHO}, -NH_2, -C\equiv C-

(B)

\displaystyle -\text{CONH}_2, -\text{COOCH}_3, -\text{CHO}, -NH_2, -OH

(C)

\displaystyle -\text{CONH}_2, -\text{CHO}, \;\rangle C=O, -NH_2, -C\equiv C-

(D)

\displaystyle -\text{CONH}_2, -\text{CHO}, -CN, -NH_2, -C\equiv C-

SolutionAnswer: Option 3

-\text{CONH}_2, -\text{CHO}, \;\rangle C=O, -NH_2, -C\equiv C-

Approach:

Use the IUPAC seniority order of principal characteristic groups (carboxylic acid > acid derivatives in the order anhydride > ester > acid halide > amide > nitrile > aldehyde > ketone > alcohol > amine > alkyne > alkene) and pick the option whose listed sequence respects this order.

Step 1:The target: among the four options, select the one whose listed functional groups are arranged in strictly decreasing order of IUPAC priority.

\text{Pick} \;\text{decreasing-priority sequence}

Step 2:The relative ranks within the option set: amide (

-\text{CONH}_2

) outranks aldehyde (

-\text{CHO}

); aldehyde outranks ketone (

>C=O

); ketone outranks amine (

-NH_2

); amine outranks alkyne (

-C\equiv C-

-\text{CONH}_2 > -\text{CHO} > >\!C=O > -NH_2 > -C\equiv C-

Step 3:Option (1) check:

-\text{CONH}_2, >\!C=O, -\text{CHO}, -NH_2, -C\equiv C-

— places ketone above aldehyde, which violates the priority order.

>\!C=O > -CHO\;\text{(false)}

Step 4:Option (2) check:

-\text{CONH}_2, -\text{COOCH}_3, -\text{CHO}, -NH_2, -OH

— places amide above ester (

-\text{COOR}

is senior to

-\text{CONH}_2

in IUPAC), so order is reversed.

-COOR > -CONH_2\;\text{(reverse needed)}

Step 5:Option (3) check:

-\text{CONH}_2, -\text{CHO}, >\!C=O, -NH_2, -C\equiv C-

— matches the expected decreasing-priority order term-by-term.

-\text{CONH}_2 > -\text{CHO} > >\!C=O > -NH_2 > -C\equiv C-

Step 6:Option (4) check:

-\text{CONH}_2, -\text{CHO}, -CN, -NH_2, -C\equiv C-

— places aldehyde above nitrile, but

-C\equiv N

is senior to

-\text{CHO}

in IUPAC priority, so order is wrong.

-C\equiv N > -CHO\;\text{(reverse needed)}

Step 7:Cross-check: option (3) against the seniority list: each adjacent pair satisfies the standard IUPAC ranking; no inversions.

\text{All adjacent pairs ordered correctly}

Step 8:Result.

-\text{CONH}_2, -\text{CHO}, >\!C=O, -NH_2, -C\equiv C-

Final answer:

-\text{CONH}_2, -\text{CHO}, \;\rangle C=O, -NH_2, -C\equiv C-

Q66Single correctOrganic Compounds Containing Nitrogen

For the given molecule, "

x

", the preferred site for the attack of the electrophile is :

N-phenylbenzamide skeletal structure labelled x=. Two benzene rings linked by a -C(=O)-N(H)- amide group. Left (benzoyl) ring carries position labels r (meta) and p (ortho to the carbonyl). Right (anilide) ring carries position labels s (ortho to the N-H) and u (para to the N-H).

(A)

Predominantly at “r”

(B)

“r” and “u”

(C)

“p” and “s”

(D)

Predominantly at “u”

SolutionAnswer: Option 4Predominantly at “u”

Approach:

Classify the two substituents in N-phenylbenzamide (

\text{Ph–CO–NH–Ph}

): the

-\text{NHCOR}

group (on the aniline ring carrying positions s/u) is a

+M

donor, ortho/para directing and activating; the

-\text{CONHR}

group (on the benzoyl ring carrying positions r/p) is a

-M

acceptor, meta directing and deactivating. Then choose between the activated ortho (s) and para (u) sites by steric considerations.

Step 1:The substrate: N-phenylbenzamide

\text{Ph–CO–NH–Ph}

has two phenyl rings — the benzoyl ring carries the carbonyl side (

-\text{CONHR}

, positions r ortho and p para) while the anilide ring carries the nitrogen side (

-\text{NHCOR}

, positions s ortho and u para).

\text{Ph}_{\text{benzoyl}}-CO-NH-\text{Ph}_{\text{anilide}}

Step 2:The resonance behaviour on the anilide ring: the nitrogen lone pair is donated into the ring (

+M

effect of

-NHCOR

), creating partial negative charge at ortho and para positions s and u.

Ar - \overset{..}{N}H - \text{COR} \leftrightarrow Ar^{(-)} = \overset{+}{N}H - \text{COR}

Step 3:The behaviour on the benzoyl ring: the carbonyl group of

-CONHR

withdraws electron density via

-M

, making the ring electron-poor; positions r (ortho) and p (para) are deactivated and the group is meta-directing.

Ar - C(=O) - \text{NHR}\; (-M)

Step 4:The selection between s and u: both are activated by

-NHCOR

, but ortho (s) is sterically hindered by the bulky

-NHCOPh

substituent, whereas para (u) is free from such steric crowding.

\text{Steric strain at s} > \text{Steric strain at u}

Step 5:The comparison: the resonance activation at u (para to N) is also strong; therefore u is the predominant site of electrophilic attack.

E^+ \to \text{position u}

Step 6:Cross-check by ruling out alternatives: r (ortho on deactivated ring) and p (para on deactivated ring) are slow; s is slowed by steric hindrance; only u combines activation with low steric demand.

\text{Rates}: u \gg s > p, r

Step 7:Result.

\text{Predominant site} = u

Final answer: Predominantly at “u”

Q67Single correctPrinciples Related to Practical Chemistry

Match List - I with List - II

List-I (Mixture of compounds)	List-II (Reagent used to distinguish)
A. Diethyl amine + Ethyl amine	I. Bromine water
B. Acetaldehyde + Acetone	II. $\text{CHCl}_3 + \text{KOH}, \Delta$
C. Ethanol + Phenol	III. Neutral $\text{FeCl}_3$
D. Benzoic acid + Cinnamic acid	IV. Ammonical silver nitrate

Choose the correct answer from the options given below :

(A)

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

(B)

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

(C)

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

(D)

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

SolutionAnswer: Option 4A - II, B - IV, C - III, D - I

Approach:

Pair each two-component mixture with the reagent that gives a positive test for exactly one component:

\text{CHCl}_3/\text{KOH}

(carbylamine) selects

1^\circ

amine; Tollens distinguishes aldehyde from ketone; neutral

\text{FeCl}_3

identifies phenol; bromine water adds to the

\text{C=C}

of cinnamic acid.

Step 1:Pick a reagent for each mixture that reacts with only one of the two compounds.

\text{Targets: A, B, C, D} \leftrightarrow \text{Reagents: I, II, III, IV}

Step 2:A: ethyl amine is

1^\circ

and gives the carbylamine isocyanide odour with

\text{CHCl}_3/\text{KOH}/\Delta

; diethyl amine is

2^\circ

and is unreactive.

\text{C}_2\text{H}_5\text{NH}_2 + \text{CHCl}_3 + 3\,\text{KOH} \xrightarrow{\Delta} \text{C}_2\text{H}_5\text{NC} + 3\,\text{KCl} + 3\,\text{H}_2\text{O}

Step 3:B: acetaldehyde reduces

[\text{Ag(NH}_3)_2]^+

to a silver mirror; acetone (ketone) does not.

\text{CH}_3\text{CHO} + 2[\text{Ag(NH}_3)_2]^+ + 3\,\text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + 2\,\text{Ag}\!\downarrow + 4\,\text{NH}_3 + 2\,\text{H}_2\text{O}

Step 4:C: phenol forms violet

[\text{Fe(OAr)}_6]^{3-}

with neutral

\text{FeCl}_3

; ethanol gives no colour.

6\,\text{C}_6\text{H}_5\text{OH} + \text{FeCl}_3 \rightarrow [\text{Fe(OC}_6\text{H}_5)_6]^{3-} + 3\,\text{HCl} + 3\,\text{H}^+

Step 5:D: cinnamic acid has a conjugated

\text{C=C}

that decolourises red bromine water; benzoic acid is unreactive toward

\text{Br}_2/\text{H}_2\text{O}

\text{C}_6\text{H}_5\text{CH=CHCOOH} + \text{Br}_2 \rightarrow \text{C}_6\text{H}_5\text{CHBr-CHBrCOOH}

Step 6:Cross-check by ensuring every reagent is used exactly once and uniquely identifies one compound in each pair.

\{A,B,C,D\} \leftrightarrow \{II, IV, III, I\}

(bijection)

Step 7:Result.

A\text{-}II,\; B\text{-}IV,\; C\text{-}III,\; D\text{-}I

Final answer: A - II, B - IV, C - III, D - I

Q68Single correctOrganic Compounds Containing Nitrogen

Consider the three aromatic molecules (P, Q and R) whose structures have been given below:

The correct order regarding the reactivity of these compounds with

\text{Ph-N}^{+} \equiv \text{NCl}^{-}

under optimum but slightly acidic medium is :

Three substituted N,N-dimethylaniline structures arranged on the page. P: unsubstituted N,N-dimethylaniline (phenyl ring bonded to N with two methyl groups; lone pair shown on N). Q: 2-methyl-N,N-dimethylaniline (ortho-methyl on the ring; N still bears two methyl groups). R: 2,6-dimethyl-N,N-dimethylaniline (two ortho-methyl substituents flanking the N; N still bears two methyl groups). Lone pair on N shown in each structure.

(A)

\displaystyle P > Q > R

(B)

\displaystyle R > P > Q

(C)

\displaystyle R > Q > P

(D)

\displaystyle P > R > Q

SolutionAnswer: Option 1

P > Q > R

Approach:

Diazo coupling with benzenediazonium chloride is an electrophilic aromatic substitution whose rate is proportional to ring nucleophilicity. Ring nucleophilicity from the

-\text{NMe}_2

group depends on its ability to donate its lone pair into the ring by resonance (

+M

); ortho methyl groups twist

-\text{NMe}_2

out of the ring plane and switch off this

+M

donation (steric inhibition of resonance).

Step 1:The electrophile and the structural variable:

\text{PhN}_2^+

couples para to

-\text{NMe}_2

; the only difference between P, Q, R is the number of ortho methyl groups on the ring (0, 1, 2).

P:\,0\,o\text{-Me};\; Q:\,1\,o\text{-Me};\; R:\,2\,o\text{-Me}

Step 2:With no ortho substituent (P),

-\text{NMe}_2

is coplanar with the ring and the N lone pair is fully conjugated, giving maximum

+M

and the highest ring electron density.

P:\;-\text{NMe}_2\;\text{coplanar} \Rightarrow \text{full }+M

Step 3:One ortho methyl (Q) sterically pushes

-\text{NMe}_2

partially out of the ring plane, reducing p-overlap and weakening

+M

donation.

Q:\;\text{partial twist} \Rightarrow \text{reduced }+M

Step 4:Two ortho methyls (R) push

-\text{NMe}_2

nearly perpendicular to the ring; the N lone pair can no longer overlap with the

\pi

-system, so

+M

is essentially lost.

R:\;-\text{NMe}_2 \perp\;\text{ring} \Rightarrow \text{minimal }+M

Step 5:Combine: ring nucleophilicity, and therefore rate of diazo coupling, follows

P > Q > R

\text{Rate: } P > Q > R

Step 6:Cross-check by independence on inductive donation: all three molecules share the same

-\text{NMe}_2

and ortho methyls are weak

+I

donors, so the dominant effect that varies among P, Q, R is steric inhibition of resonance, which monotonically lowers nucleophilicity as ortho methyls increase.

+I(\text{Me}) \ll +M(\text{NMe}_2)

Step 7:Result: and map to the option.

P > Q > R

Final answer:

P > Q > R

Q69Single correctBiomolecules

Match List - I with List - II

List-I (Vitamin)	List-II (Name)
A. Vitamin $\text{B}_1$	I. Pyridoxine
B. Vitamin $\text{B}_2$	II. Ascorbic acid
C. Vitamin $\text{B}_6$	III. Thiamine
D. Vitamin C	IV. Riboflavin

Choose the correct answer from the options given below:

(A)

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

(B)

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

(C)

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

(D)

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

Q70Single correctPrinciples Related to Practical Chemistry

A salt with few drops of conc. HCl gives apple green colour in flame test. The group precipitate of the salt is dissolved in acetic acid and treated with

\text{K}_2\text{CrO}_4

To give yellow precipitate. When the sodium carbonate extract of the salt solution is heated with conc.

\text{HNO}_3

and ammonium molybdate it resulted a canary yellow precipitate. The cation and anion present in the salt are respectively.

(A)

\displaystyle \text{Ca}^{+2}

and

\displaystyle \text{SO}_4^{2-}

(B)

\displaystyle \text{Ba}^{+2}

and

\displaystyle \text{PO}_4^{3-}

(C)

\displaystyle \text{Mn}^{2+}

and

\displaystyle \text{PO}_4^{3-}

(D)

\displaystyle \text{Ba}^{2+}

and

\displaystyle \text{SO}_4^{2-}

SolutionAnswer: Option 2

\text{Ba}^{+2}

and

\text{PO}_4^{3-}

Approach:

Identify the cation from the flame colour and confirm with the chromate test, then identify the anion from the ammonium molybdate test in conc.

\text{HNO}_3

Step 1:Given observations: apple-green flame, yellow ppt with

\text{K}_2\text{CrO}_4

in acetic-acid medium, canary-yellow ppt with ammonium molybdate in conc.

\text{HNO}_3

\text{Flame: apple green;\; Group ppt + AcOH + K}_2\text{CrO}_4: \text{yellow;\; Na}_2\text{CO}_3\text{ extract + HNO}_3 + (\text{NH}_4)_2\text{MoO}_4: \text{canary yellow}

Step 2:Observation 1: among the listed cations,

\text{Ba}^{2+}

gives an apple-green flame;

\text{Ca}^{2+}

gives brick-red,

\text{Sr}^{2+}

crimson-red, and

\text{Mn}^{2+}

no characteristic colour.

\text{Ba}^{2+} \xrightarrow{\text{flame}} \text{apple green}

Step 3:Observation 2:

\text{Ba}^{2+}

in dilute acetic acid precipitates yellow

\text{BaCrO}_4

with

\text{K}_2\text{CrO}_4

, confirming barium.

\text{Ba}^{2+} + \text{CrO}_4^{2-} \rightarrow \text{BaCrO}_4\!\downarrow

Step 4:Observation 3:

\text{PO}_4^{3-}

heated with ammonium molybdate in conc.

\text{HNO}_3

gives the canary-yellow ammonium phosphomolybdate

(\text{NH}_4)_3[\text{P(Mo}_3\text{O}_{10})_4]

\text{PO}_4^{3-} + (\text{NH}_4)_2\text{MoO}_4 + \text{HNO}_3 \rightarrow (\text{NH}_4)_3[\text{P(Mo}_3\text{O}_{10})_4]\!\downarrow\;\text{(canary yellow)}

Step 5:Cross-check: only option 2 lists both

\text{Ba}^{2+}

and

\text{PO}_4^{3-}

; options 1, 3, 4 fail at least one of the three observations.

\text{Ba}^{2+},\;\text{PO}_4^{3-}

Step 6:Result: and select the option.

\text{Cation: Ba}^{2+},\;\text{Anion: PO}_4^{3-}

Final answer:

\text{Ba}^{+2}

and

\text{PO}_4^{3-}

Q71NumericalCoordination Compounds

5.33

g of

\text{CrCl}_3 \cdot 6\text{H}_2\text{O}

which is a 1:3 electrolyte, is dissolved in water and is passed through a cation exchanger. The Chloride ions in the eluted solution on treatment with

\text{AgNO}_3

results in

8.61

g of

\text{AgCl}

. The ratio of moles of complex reacted and moles of

\text{AgCl}

formed is ________

\times 10^{-2}

. (Nearest integer)

[Molar mass in g

\text{mol}^{-1}

Cr : 52, Ag : 108, Cl : 35.5, H : 1, O : 16]

SolutionAnswer: 33

Approach:

Since

\text{CrCl}_3 \cdot 6\text{H}_2\text{O}

behaves as a 1:3 electrolyte, the complex is

[\text{Cr}(\text{H}_2\text{O})_6]\text{Cl}_3

which releases three ionizable

\text{Cl}^-

after passage through a cation exchanger. Compute moles of complex from its mass and moles of

\text{AgCl}

from its mass, take the ratio, and express it as the integer multiplied by

10^{-2}

Step 1:Given data:

m_{\text{complex}} = 5.33

m_{\text{AgCl}} = 8.61

g, atomic masses

\text{Cr}=52,\,\text{Ag}=108,\,\text{Cl}=35.5,\,\text{H}=1,\,\text{O}=16

. Target:

\dfrac{n_{\text{complex}}}{n_{\text{AgCl}}}\times 100

m_{\text{complex}}=5.33\,\text{g},\;m_{\text{AgCl}}=8.61\,\text{g}

Step 2:The molar mass of

[\text{Cr}(\text{H}_2\text{O})_6]\text{Cl}_3

M = 52 + 6(2\!\cdot\!1 + 16) + 3(35.5) = 52 + 108 + 106.5 = 266.5\,\text{g mol}^{-1}

M_{[\text{Cr}(\text{H}_2\text{O})_6]\text{Cl}_3} = 52 + 108 + 106.5 = 266.5\,\text{g mol}^{-1}

Step 3:Moles of complex.

n_{\text{complex}} = \dfrac{5.33}{266.5} = 0.02\,\text{mol}

Step 4:Moles of

\text{AgCl}

using

M_{\text{AgCl}} = 108 + 35.5 = 143.5\,\text{g mol}^{-1}

n_{\text{AgCl}} = \dfrac{8.61}{143.5} = 0.06\,\text{mol}

Step 5:Form the required ratio and scale by 100.

\dfrac{n_{\text{complex}}}{n_{\text{AgCl}}}\times 100 = \dfrac{0.02}{0.06}\times 100 = \dfrac{1}{3}\times 100 = 33.33

Step 6:Cross-check by stoichiometry: from

[\text{Cr}(\text{H}_2\text{O})_6]\text{Cl}_3 \rightarrow 3\,\text{AgCl}

, the theoretical mole ratio is exactly

1:3

, so the percentage is

\tfrac{1}{3}\times 100 = 33.33

1:3 \Rightarrow \tfrac{1}{3}\times 100 = 33.33

Step 7:Result.

33.33 \;\xrightarrow{\text{nearest integer}}\; 33

Final answer: 33

Q72NumericalHydrocarbons

Consider the isomers of hydrocarbon with molecular formula

\text{C}_5\text{H}_{10}

. These isomers do not decolorise

\text{KMnO}_4

solutions. These isomers are subjected to chlorination with chlorine in presence of light to give monochloro compounds. The total number of monochloro compounds (structural isomers only) formed is ________

SolutionAnswer: 14

Approach:

Isomers of

\text{C}_5\text{H}_{10}

that do not decolourise

\text{KMnO}_4

have no

\pi

-bond, so they are the saturated cycloalkanes only. Enumerate the five cycloalkane structural isomers and, for each, count the number of distinct H-environments; each environment yields one monochloro structural isomer on photochlorination.

Step 1:

\text{C}_5\text{H}_{10}

has DoU

= 1

. "Not decolorising

\text{KMnO}_4

" rules out alkenes, leaving only the cycloalkanes.

\text{DoU} = 1, \;\text{no C=C} \Rightarrow \text{ring only}

Step 2:The cycloalkane structural isomers of

\text{C}_5\text{H}_{10}

: (i) cyclopentane, (ii) methylcyclobutane, (iii) ethylcyclopropane, (iv) 1,1-dimethylcyclopropane, (v) 1,2-dimethylcyclopropane.

\{\text{C}_5\text{H}_{10}\}_{\text{sat}} = \{\text{cyclopentane, MeC}_4\text{H}_7,\;\text{EtC}_3\text{H}_5,\;1{,}1\text{-Me}_2\text{C}_3\text{H}_4,\;1{,}2\text{-Me}_2\text{C}_3\text{H}_4\}

Step 3:Cyclopentane: all 10 H atoms are equivalent (single environment), giving 1 monochloro product.

\text{Cyclopentane}:\; 1\;\text{H environment} \Rightarrow 1

Step 4:Methylcyclobutane: the four ring carbons differ as C1 (CH-Me), C2 and C4 (equivalent

\alpha

to C1), C3 (

\beta

); plus the C

H_3

group. This gives 4 distinct H-environments and thus 4 monochloro products.

\text{Methylcyclobutane}:\; \{\text{C1-H},\;\text{C2/4-H},\;\text{C3-H},\;\text{CH}_3\} \Rightarrow 4

Step 5:Ethylcyclopropane: ring C1 (bearing Et), ring C2 and C3 (equivalent), and the ethyl group

\text{-CH}_2\text{-CH}_3

contributes two more (the

\text{CH}_2

and the

\text{CH}_3

). Four distinct H-environments, hence 4 monochloro products.

\text{Ethylcyclopropane}:\; \{\text{C1-H},\;\text{C2/3-H},\;\text{CH}_2,\;\text{CH}_3\} \Rightarrow 4

Step 6:1,1-dimethylcyclopropane: the two methyls on C1 are equivalent; ring C

H_2

groups (C2, C3) are equivalent. Two distinct H-environments give 2 monochloro products.

1{,}1\text{-Dimethylcyclopropane}:\; \{\text{CH}_3 (\text{on C1}),\;\text{CH}_2 (\text{C2/3})\} \Rightarrow 2

Step 7:1,2-dimethylcyclopropane: equivalent ring methyls; equivalent ring C-H (on C1 and C2); the two H atoms on C3 are diastereotopic (cis and trans to the methyls) and give two different monochloro structural products. Total = 3 monochloro structural isomers.

1{,}2\text{-Dimethylcyclopropane}:\; \{\text{CH}_3,\;\text{C1/2-H},\;\text{C3-H}_{\text{cis}}\;\text{vs}\;\text{C3-H}_{\text{trans}}\} \Rightarrow 3

Step 8:Sum the monochloro structural isomers across all five parent cycloalkanes.

1 + 4 + 4 + 2 + 3 = 14

Step 9:Cross-check by independent counting (replacement-of-each-distinct-H method) on each parent; the totals 1, 4, 4, 2, 3 are reproduced, summing to 14.

\sum n_i = 14

Step 10:Result.

\text{Total monochloro structural isomers} = 14

Final answer: 14

Q73NumericalHydrocarbons

One mole of an alkane (x) requires 8 moles oxygen for complete combustion. Sum of number of carbon and hydrogen atoms in the alkane (x) is ________

Q74NumericalChemical Kinetics

For reaction

A \rightarrow P

, rate constant

k = 1.5 \times 10^3 \, s^{-1}

27^\circ C

. If activation energy for the above reaction is

60

kJ mo

l^{-1}

, then the temperature

(in\, ^\circ C)

at which rate constant,

k = 4.5 \times 10^3 \, s^{-1}

is ________. (Nearest integer)

Given :

\log 2 = 0.30, \log 3 = 0.48, R = 8.3 \, JK^{-1} \text{mol}^{-1}, \ln 10 = 2.3

SolutionAnswer: 41

Approach:

Apply the two-temperature form of the Arrhenius equation,

\ln(k_2/k_1) = (E_a/R)\,(T_2 - T_1)/(T_1 T_2)

, with

k_2/k_1 = 3

and

T_1 = 300\,\text{K}

; solve for

T_2

in Kelvin and convert to

^\circ

Step 1:Given data:

k_1 = 1.5\times 10^3\,\text{s}^{-1}

T_1 = 27\,^\circ\text{C} = 300\,\text{K}

;

k_2 = 4.5\times 10^3\,\text{s}^{-1}

;

E_a = 60\,\text{kJ mol}^{-1} = 60\,000\,\text{J mol}^{-1}

;

R = 8.3\,\text{J K}^{-1}\text{mol}^{-1}

;

\log 3 = 0.48

\ln 10 = 2.3

k_2/k_1 = 3,\; T_1=300\,\text{K},\;E_a=60\,000\,\text{J/mol}

Step 2:LHS:

\ln 3 = 2.303 \cdot \log 3 = 2.3 \times 0.48 = 1.104

\ln(k_2/k_1) = \ln 3 = 2.3 \times 0.48 = 1.104

Step 3:Plug into the Arrhenius two-temperature form.

1.104 = \dfrac{60\,000}{8.3}\!\left[\dfrac{T_2 - 300}{300\,T_2}\right]

Step 4:Evaluate

E_a/R = 60\,000/8.3 = 7228.9\,\text{K}

E_a/R = 7228.9\,\text{K}

Step 5:Isolate

(T_2 - 300)/T_2 = 1 - 300/T_2

\dfrac{T_2 - 300}{T_2} = \dfrac{1.104 \times 300}{7228.9} = \dfrac{331.2}{7228.9} = 0.04582

Step 6:Solve for

T_2

\dfrac{300}{T_2} = 1 - 0.04582 = 0.95418 \Rightarrow T_2 = \dfrac{300}{0.95418} = 314.4\,\text{K}

Step 7:Convert to Celsius:

t = T_2 - 273 = 314.4 - 273 = 41.4\,^\circ\text{C}

t = 41.4\,^\circ\text{C}

Step 8:Cross-check by back-substitution:

\ln(k_2/k_1) = (60\,000/8.3)\,(314.4 - 300)/(300 \times 314.4) = 7228.9 \times 14.4/94320 = 1.104

, which matches

\ln 3

7228.9 \times \dfrac{14.4}{94320} = 1.104 = \ln 3

Step 9:Result.

t \approx 41\,^\circ\text{C}

Final answer: 41

Q75NumericalChemical Thermodynamics

At the transition temperature T,

A = B

and

\Delta G^0 = 105 - 35 \log T

where A and B are two states of substance X. The transition temperature in

^\circ C

when pressure is 1 atm is ________ (Nearest integer)

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