SECTION - A [CHEMISTY - MCQ] - JEE STD 11 Science Questions - Vidyadip

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SECTION - A [CHEMISTY - MCQ]

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20 questions · 17 auto-graded MCQ + 3 self-marked written.

MCQ 14 Marks

Given below are two statements :
Statement I : $\mathrm{H}_{2} \mathrm{Se}$ is more acidic than $\mathrm{H}_{2} \mathrm{Te}$.
Statement II : $\mathrm{H}_{2} \mathrm{Se}$ has higher bond enthalpy for dissociation than $\mathrm{H}_{2} \mathrm{Te}$.
In the light of the above statements, choose the correct answer from the options given below.

A
Both statement I and Statement II are false.
B
Both statement I and Statement II are true.
C
Statement I is true but Statement II is false.
✓
Statement I is false but Statement II is true.

Answer

Correct option: D.

Statement I is false but Statement II is true.

(D) Statement I is false but Statement II is true.
Acidic strength : $\mathrm{H}_{2} \mathrm{Se}<\mathrm{H}_{2} \mathrm{Te}$
$\begin{array}{rlll}\Delta_{\text {dis }} H & : H _2 Se \quad>\quad H _2 Te \\ & {[276 KJ / mol ] \quad[238 KJ / mol ]}\end{array}$

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MCQ 24 Marks

Match the LIST-I with LIST-II

LIST-I (Complex/Species)		LIST-II (Shape & magnetic moment)
A.	$\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$	I.	Tetrahedral, 2.8 BM
B.	$\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$	II.	Square planar, 0 BM
C.	$\left[\mathrm{NiCl}_{4}\right]^{2-}$	III.	Tetrahedral, 0 BM
D.	$\left[\mathrm{MnBr}_{4}\right]^{2-}$	IV.	Tetrahedral, 5.9 BM

Choose the correct answer from the options given below :

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

Answer

(C) A-III, B-II, C-I, D-IV
(A) $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right], \mathrm{Ni}^{0}:[\mathrm{Ar}] 3 \mathrm{~d}^{8} 4 \mathrm{~s}^{2}$
Valence orbitals of $\mathrm{Ni}^{0}$ in pre-hybridisation state :

Tetrahedral, Diamagnetic, $\mu=0$ B.M.
(B) $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}, \mathrm{Ni}^{+2}:[\mathrm{Ar}] 3 \mathrm{~d}^{8} 4 \mathrm{~s}^{0}$
Valence orbitals of $\mathrm{Ni}^{+2}$ in pre-hybridisation state :

Square planar, Diamagnetic, $\mu=0$ B.M.
(C) $\left[\mathrm{NiCl}_{4}\right]^{2-}, \mathrm{Ni}^{+2}:[\mathrm{Ar}] 3 \mathrm{~d}^{8} 4 \mathrm{~s}^{0}$
Valence orbitals of $\mathrm{Ni}^{+2}$ in ground state

:Tetrahedral, paramagnetic, $\mu=\sqrt{8}=2.8$ B.M.
(D) $\left[\mathrm{MnBr}_{4}\right]^{2-}, \mathrm{Mn}^{+2}:[\mathrm{Ar}] 3 \mathrm{~d}^{5}$
Valence orbitals of $\mathrm{Mn}^{+2}$ in ground state :

Tetrahedral, paramagnetic, $\mu=\sqrt{35}=$ 5.9 B.M.

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MCQ 34 Marks

When

undergoes intramolecular aldol condensation, the major product formed is :

✓
B
C
D

Answer

Correct option: A.

(A)

Aldol condensation reaction

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MCQ 44 Marks

Match the LIST-I with LIST-II

LIST-I (Reagent)		LIST-II (Functional Group detected)
A.	Sodium bicarbonate solution	I.	double bond/unsaturation
B.	Neutral ferric chloride	II.	carboxylic acid
C.	ceric ammonium nitrate	III.	phenolic – OH
D.	alkaline $\mathrm{KMnO}_{4}$	IV.	alcoholic - OH

Choose the correct answer from the options given below :

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

Answer

Correct option: A.

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

(A) A-II, B-III, C-IV, D-I
(1) Carboxylic acid gives efferve scence with sodium bicarbonate solution
(2) Phenolic-OH gives voilet coloured complex with Neutral $\mathrm{FeCl}_{3}$.
(3) Alcoholic-OH gives Red colour with cerric ammonium Nitrate.
(4) When alkaline $\mathrm{KMnO}_{4}$ reacts with an unsaturated compound (Alkene or alkyne) the purple colour of $\mathrm{KMnO}_{4}$ solution disappears, indicating positive test for unsaturation.

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MCQ 54 Marks

The number of species from the following that are involved in $\mathrm{sp}^{3} \mathrm{~d}^{2}$ hybridization is $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}, \mathrm{SF}_{6},\left[\mathrm{CrF}_{6}\right]^{3-},\left[\mathrm{CoF}_{6}\right]^{3-},\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}$ and $\left[\mathrm{MnCl}_{6}\right]^{3-}$

A
5
B
6
✓
4
D
3

Answer

Correct option: C.

4

(C) 4
In $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right]_{6}\right]^{3+}, \mathrm{Co}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^{6}, \mathrm{NH}_{3}$ is S.F.L
Hybridisation state of $\mathrm{Co}^{3+}$ is $\mathrm{d}^{2} \mathrm{sp}^{3}$
In $\mathrm{SF}_{6}$, Hybridisation state of sulphur is $\mathrm{sp}^{3} \mathrm{~d}^{2}$
In $\left[\mathrm{CrF}_{6}\right]^{3-}, \mathrm{Cr}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^{3}$
Hybridisation state of $\mathrm{Cr}^{3+}$ is $\mathrm{d}^{2} \mathrm{sp}^{3}$
$\left[\mathrm{CoF}_{6}\right]^{3-}, \mathrm{Co}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^{6} \mathrm{~F}^{-}$is W.F.L
Hybridisation state of $\mathrm{Co}^{3+}$ is $\mathrm{sp}^{3} \mathrm{~d}^{2}$
$\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}, \mathrm{Mn}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^{4} \mathrm{CN}^{-}$is S.F.L
Hybridisation state of $\mathrm{Mn}^{3+}$ is $\mathrm{d}^{2} \mathrm{sp}^{3}$
$\left[\mathrm{MnCl}_{6}\right]^{3-}, \mathrm{Mn}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^{4} \mathrm{Cl}^{-}$is W.F.L
Hybridisation state of $\mathrm{Cl}^{-}$is $\mathrm{sp}^{3} \mathrm{~d}^{2}$
Total number of $\mathrm{sp}^{3} \mathrm{~d}^{2}$ hybridized molecules is 3

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MCQ 64 Marks

Correct statements for an element with atomic number 9 are
A. There can be 5 electrons for which $\mathrm{m}_{\mathrm{s}}=+\frac{1}{2}$ and 4 electrons for which $\mathrm{m}_{\mathrm{s}}=-\frac{1}{2}$
B. There is only one electron in $p_{z}$ orbital
C. The last electron goes to orbital with $\mathrm{n}=2$ and $l=1$
4. The sum of angular nodes of all the atomic orbitals is 1 .
Choose the correct answer from the options given below:

A
C and D Only
B
A and C Only
C
A, C and D Only
D
A and B Only

Answer

(B) A and C Only
Element with atomic number 9 is Fluorine

(A) 5 electrons can be up-spin $\left[\mathrm{m}_{\mathrm{s}}=+\frac{1}{2}\right]$ and 4 electrons can be down spin $\left[\mathrm{m}_{\mathrm{s}}=-\frac{1}{2}\right]$
(B) Unpaired electron can be in anyone of $p_{x}, p_{y}$ or $p_{z}$ orbital
(C) Last electron is in 2p subshell with $\mathrm{n}=2, \ell=1$
(D) Angular node for s-orbital $=0$ while of each p-orbital $=1$
Sum of all angular node $=3$

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MCQ 74 Marks

Choose the correct option for structures of A and B, respectively.

✓
B
C
D

Answer

Correct option: A.

(A)

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MCQ 84 Marks

On combustion 0.210 g of an orgainc compound containing $\mathrm{C}, \mathrm{H}$ and O gave $0.127 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}$ and 0.307 $\mathrm{g} \mathrm{CO}_{2}$. The percentages of hydrogen and oxygen in the given organic compound respectively are:

A
$53.41,39.6$
✓
$6.72,53.41$
C
$7.55,43.85$
D
$6.72,39.87$

Answer

Correct option: B.

$6.72,53.41$

(B) $6.72,53.41$
In the combustion of organic compound, all "C" in $\mathrm{CO}_{2}$ and all "H" in $\mathrm{H}_{2} \mathrm{O}$ comes from organic compound
$\underset{\substack{\\ 0.210 \mathrm{gm}}}{\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}}\mathrm{O}_{\mathrm{z}}}+\mathrm{O}_{2} \longrightarrow \underset{0.307 \mathrm{gm}}{\mathrm{CO}_{2}}+\underset{0.127 \mathrm{gm}}{\mathrm{H}_{2} \mathrm{O}}$
Weight of "C" in $\mathrm{CO}_{2}=\frac{12}{44} \times 0.307$
$=0.0837 \mathrm{gm}$
Weight of "H" in $\mathrm{H}_{2} \mathrm{O}=\frac{2}{18} \times 0.127=0.0141 \mathrm{~g}$
$\% ' H '$ in compound $=\frac{0.0141}{0.21} \times 100=6.719 \%$
$=6.72 \%$
Weight of "O" in compound
$\begin{array}{l}=0.210-(0.0837+0.0141) \\ =0.1122\end{array}$
$\%$ of "O" in compound $=\frac{0.1122}{0.21} \times 100$
$=53.41 \%$

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MCQ 94 Marks

Which one of the following reactions will not lead to the desired ether formation in major proportion?
(iso-$\mathrm{Bu} \Rightarrow$ isobutyl, sec-$\mathrm{Bu} \Rightarrow$ sec-butyl, $\mathrm{nPr} \Rightarrow$ n-propyl, ${ }^{\mathrm{t}} \mathrm{Bu} \Rightarrow$ tert-butyl, $\mathrm{Et} \Rightarrow$ ethyl)

A
${ }^{\mathrm{t}} \mathrm{Bu} \stackrel{\ominus}{\mathrm{O}} \stackrel{\oplus}{\mathrm{Na}}+\mathrm{EtBr} \longrightarrow{ }^{\mathrm{t}} \mathrm{Bu}-\mathrm{O}-\mathrm{Et}$
B
C
D
iso-$\mathrm{Bu} \stackrel{\ominus}{\mathrm{O}} \stackrel{\oplus}{\mathrm{Na}}+\mathrm{sec}-\mathrm{BuBr} \longrightarrow \mathrm{Sec}-\mathrm{Bu}-\mathrm{O}-$ iso-Bu

Answer

(D) iso-$\mathrm{Bu} \stackrel{\ominus}{\mathrm{O}} \stackrel{\oplus}{\mathrm{Na}}+\mathrm{sec}-\mathrm{BuBr} \longrightarrow \mathrm{Sec}-\mathrm{Bu}-\mathrm{O}-$ iso-Bu

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MCQ 104 Marks

The correct decreasing order of spin only magnetic moment values $(\mathrm{BM})$ of $\mathrm{Cu}^{+}, \mathrm{Cu}^{2+}, \mathrm{Cr}^{2+}$ and $\mathrm{Cr}^{3+}$ ions is :

A
$\mathrm{Cu}^{+}>\mathrm{Cu}^{2+}>\mathrm{Cr}^{3+}>\mathrm{Cr}^{2+}$
B
$\mathrm{Cu}^{2+}>\mathrm{Cu}^{+}>\mathrm{Cr}^{2+}>\mathrm{Cr}^{3+}$
✓
$\mathrm{Cr}^{2+}>\mathrm{Cr}^{3+}>\mathrm{Cu}^{2+}>\mathrm{Cu}^{+}$
D
$\mathrm{Cr}^{3+}>\mathrm{Cr}^{2+}>\mathrm{Cu}^{+}>\mathrm{Cu}^{2+}$

Answer

Correct option: C.

$\mathrm{Cr}^{2+}>\mathrm{Cr}^{3+}>\mathrm{Cu}^{2+}>\mathrm{Cu}^{+}$

(C) $\mathrm{Cr}^{2+}>\mathrm{Cr}^{3+}>\mathrm{Cu}^{2+}>\mathrm{Cu}^{+}$
$\mathrm{Cu}^{+}:[\mathrm{Ar}] 3 \mathrm{~d}^{10}$, Spin only magnetic moment $=0$ B.M.
$\mathrm{Cu}^{+2}:[\mathrm{Ar}] 3 \mathrm{~d}^{9}$, Spin only magnetic moment $=\sqrt{3}$ B.M.
$\mathrm{Cr}^{+2}:[\mathrm{Ar}] 3 \mathrm{~d}^{4}$, Spin only magnetic moment $=\sqrt{24}$ B.M.
$\mathrm{Cr}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^{3}$, Spin only magnetic moment $=\sqrt{15}$ B.M.
Order of $\mu: \mathrm{Cr}^{+2}>\mathrm{Cr}^{+3}>\mathrm{Cu}^{+2}>\mathrm{Cu}^{+}$

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MCQ 114 Marks

What is the correct IUPAC name of

A
4-Ethyl-1-hydroxycyclopent-2-ene
B
1-Ethyl-3-hydroxycyclopent-2-ene
C
1-Ethylcyclopent-2-en-3-ol
✓
4-Ethylcyclopent-2-en-1-ol

Answer

Correct option: D.

4-Ethylcyclopent-2-en-1-ol

(D) 4-Ethylcyclopent-2-en-1-ol

4-Ethylclopent-2-en-1-ol

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MCQ 124 Marks

$\mathrm{HA}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{A}^{-}(\mathrm{aq})$
The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is $0.20^{\circ} \mathrm{C}$.
The dissociation constant for the acid is
Given :
$\mathrm{K}_{\mathrm{f}}\left(\mathrm{H}_{2} \mathrm{O}\right)=1.8 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, molality $\equiv$ molarity

✓
$1.38 \times 10^{-3}$
B
$1.1 \times 10^{-2}$
C
$1.90 \times 10^{-3}$
D
$1.89 \times 10^{-1}$

Answer

Correct option: A.

$1.38 \times 10^{-3}$

(A) $1.38 \times 10^{-3}$
$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{ik}_{\mathrm{f}} \mathrm{m}$
$0.2=\mathrm{i} \times 1.8 \times 0.1$
$\mathrm{i}=\frac{20}{18}=\frac{10}{9}$
For $\quad \mathrm{HA}_{(\mathrm{aq})} \rightleftharpoons \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{A}_{(\mathrm{aq})}^{-}$
$\mathrm{t}=0 \quad 1$
$t=t_{\text {eq }} 1-\alpha \quad \alpha \quad \alpha$
$\mathrm{i}=1+\alpha$
$\frac{10}{9}=1+\alpha$
$\alpha=\frac{1}{9}$
$\mathrm{K}_{\mathrm{eq}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}=\frac{\mathrm{Ca}^{2}}{1-\alpha}$
$=\frac{0.1\left(\frac{1}{9}\right)^{2}}{1-\frac{1}{9}}=\frac{1}{720}$
$\mathrm{K}_{\text {eq.}}=1.38 \times 10^{-3}$

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MCQ 134 Marks

Which of the following binary mixture does not show the behaviour of minimum boiling azeotropes?

A
$\mathrm{H}_{2} \mathrm{O}+\mathrm{CH}_{3} \mathrm{COC}_{2} \mathrm{H}_{5}$
✓
$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}$
C
$\mathrm{CS}_{2}+\mathrm{CH}_{3} \mathrm{COCH}_{3}$
D
$\mathrm{CH}_{3} \mathrm{OH}+\mathrm{CHCl}_{3}$

Answer

Correct option: B.

$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}$

(B) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}$
Binary mixture of $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$ and $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}$ shows negative deviation from Raoult's law
So vapour pressure of solution is less than V.P of pure $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH} \& \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}$
So B.P. of solution is greater than boiling point of pure $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH} \& \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}$
So shows maximum Boiling azeotrope

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MCQ 144 Marks

The atomic number of the element from the following with lowest $1^{\text {st }}$ ionisation enthalpy is :

A
32
B
35
✓
87
D
19

Answer

Correct option: C.

87

(C) 87
Atomic no. $32 \Rightarrow \mathrm{Ge}$
Atomic no. $35 \Rightarrow \mathrm{Br}$
Atomic no. $87 \Rightarrow \mathrm{Fr}$
Atomic no. $19 \Rightarrow \mathrm{~K}$
Lowest first I.E. among the given element will be of Fr [87].
$\mathrm{Fr}-[\mathrm{Rn}] 7 \mathrm{~s}^{1}$

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MCQ 154 Marks

Given below are two statements :
Statement I : A homolepitc octahedral complex, formed using monodentate ligands, will not show stereoisomerism.
Statement II : cis- and trans- platin are heteroleptic complexes of Pd .
In the light of the above statements, choose the correct answer from the options given below.

A
Both statement I and Statement II are false.
B
Statement I is false but Statement II is true.
C
Both statement I and Statement II are true.
✓
Statement I is true but Statement II is false.

Answer

Correct option: D.

Statement I is true but Statement II is false.

(D) Statement I is true but Statement II is false.
Homoleptic complex of type $\left[\mathrm{Ma}_{6}\right]$ (Where $\mathrm{a} \Rightarrow$ monodentate ligand) cannot show geometrical as well as optical isomerism.
Cis-platin and trans-platin has formula $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]$ which is a heteroleptic complex of platinum.

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MCQ 164 Marks

A
✓
C
D

Answer

Correct option: B.

(B)

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MCQ 174 Marks

Choose the correct set of reagents for the following conversion.

✓
$\mathrm{Br}_{2}/\mathrm{Fe}; \mathrm{Cl}_{2}, \Delta$; alc. KOH
B
$\mathrm{Cl}_{2}/\mathrm{Fe}; \mathrm{Br}_{2} /$anhy.$\mathrm{AlCl}_{3}$; aq. KOH
C
$\mathrm{Br}_{2} /$anhy.$\mathrm{AlCl}_{3}; \mathrm{Cl}_{2}, \Delta$; aq. KOH
D
$\mathrm{Cl}_{2} /$anhy.$\mathrm{AlCl}_{3}; \mathrm{Br}_{2} / \mathrm{Fe}$; alc. KOH

Answer

Correct option: A.

$\mathrm{Br}_{2}/\mathrm{Fe}; \mathrm{Cl}_{2}, \Delta$; alc. KOH

(A) $\mathrm{Br}_{2} / \mathrm{Fe} ; \mathrm{Cl}_{2}, \Delta$; alc. KOH

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MCQ 184 Marks

$\mathrm{A} \xrightarrow[\text { (ii) } \mathrm{H}_{3} \mathrm{O}^{+}]{(\mathrm{i}) \mathrm{NaOH}} \mathrm{B} \xrightarrow[\text { (ii) } \mathrm{H}_{2} \mathrm{SO}_{4}, \Delta]{\text { (i) } \mathrm{EtOH}} \mathrm{C}$
'A' shows positive Lassaign's test for N and its molar mass is 121.
'B' gives effervescence with aq. $\mathrm{NaHCO}_{3}$.
'C' gives fruity smell.
Identify A, B and C from the following.

✓
B
C
D

Answer

Correct option: A.

(A)

Molar mass $=121$
A $\rightarrow$ Benzamide Shows positive Lassaigh's test.
$B \rightarrow$ Benzoic acid gives effervescence with aq. $\mathrm{NaHCO}_{3}$.
$\mathrm{C} \rightarrow$ Ester gives fruity smell.

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MCQ 194 Marks

Match the LIST-I with LIST-II

	LIST - I		LIST - II
A.	Carbocation	I.	Species that can supply a pair of electrons.
B.	C-Free radical	II.	Species that can receive a pair of electrons.
C.	Nucleophile	III.	$sp ^2$ hybridized carbon with empty p-orbital.
D.	Electrophile	IV.	$sp ^2 / sp ^3$ hybridized carbon with one unpaired electron.

Choose the correct answer from the options given below :

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

Answer

Correct option: D.

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

(D) A-III, B-IV, C-I, D-II
(A) Carbocation $\rightarrow \mathrm{sp}^{2}$ hybridised carbon with empty P-orbital

(B) Carbon free radical $\rightarrow \mathrm{sp}^{2} / \mathrm{sp}^{3}$ hybridised carbon with one unpaired electron.

(C) Nuecleophile $\rightarrow$ species of that can supply a pair of electron.
(D) Electrophile $\rightarrow$ species that can receive a pair of electron.

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MCQ 204 Marks

In a first order decomposition reaction, the time taken for the decomposition of reactant to one fourth and one eighth of its initial concentration are $t_{1}$ and $t_{2}(s)$, respectively. The ratio $t_{1} / t_{2}$ will :

A
$\frac{4}{3}$
B
$\frac{3}{2}$
C
$\frac{3}{4}$
✓
$\frac{2}{3}$

Answer

Correct option: D.

$\frac{2}{3}$

(D) $\frac{2}{3}$
For I ${ }^{\text {st }}$ order reaction
When $\mathrm{C}_{\mathrm{t}}=\mathrm{Co} / 4$
$\mathrm{t}_{1}=2 \mathrm{t}_{50 \%}$.
when $\mathrm{C}_{\mathrm{t}}=\mathrm{Co} / 8$
$\mathrm{t}_{2}=3 \mathrm{t}_{50 \%}$
so $\frac{t_{1}}{t_{2}}=\frac{2}{3}$

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