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

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SECTION - A [CHEMISTY - MCQ]

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20 questions · self-marked practice — reveal the answer and mark yourself.

MCQ 14 Marks

In Dumas' method for estimation of nitrogen, 0.5 gram of an organic compound gave 60 mL of nitrogen collected at 300 K temperature and 715 mm Hg pressure. The percentage composition of nitrogen in the compound (Aqueous tension at $300 \mathrm{~K}=15 \mathrm{~mm} \mathrm{Hg}$ ) is

A
1.257
B
20.87
C
18.67
D
12.57

Answer

D. 12.57
Pressure of $\mathrm{N}_{2}$ gas $=(715-15)$
$=700 \mathrm{mmHg}$
$\mathrm{n}_{\mathrm{N}_{2}}=\frac{\mathrm{PV}}{\mathrm{RT}}$
$\mathrm{n}_{\mathrm{N}_{2}}=\frac{700 \times 60 \times 10^{-3}}{760 \times 0.0821 \times 300}$
$\begin{aligned}=2.24 & \times 10^{-3} mol \\ \text { Mass of } N _2 & =2.24 \times 10^3 \times 28 g \\ & =0.06272 g\end{aligned}$
$\% \mathrm{~N}_{2}=\frac{0.06272}{0.5} \times 100 \simeq 12.57$

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MCQ 24 Marks

A tetrapeptide "$x$" on complete hydrolysis produced glycine (Gly), alanine (Ala), valine (Val), leucine (Leu) in equimolar proportion each. The number of tetrapeptides (sequences) possible involving each of these amino acids is

A
16
B
32
C
8
D
24

Answer

D. 24
The number of tetrapeptides (sequences) possible involving each of these amino acids (glycine, alanine, valine, leucine) ; It has three (3) peptides linkage the number of permutations in which they can be arranged
$=4 \times 3 \times 2 \times 1$
$=24$

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MCQ 34 Marks

Which of the following graphs correctly represents the variation of thermodynamic properties of Haber's process ?

A
B
C
D

Answer

$\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}$
$\Delta \mathrm{H}^{\circ}=-\mathrm{ve}$
$\Delta \mathrm{S}^{\circ}=-\mathrm{ve}$
(As gaseous moles decreases).
(1) As temperature increases $\frac{-\Delta \mathrm{H}_{R}^{\circ}}{T}$, decreases
(2) $\Delta \mathrm{G}^{\circ}=-\mathrm{RT} \ln \mathrm{K}_{\mathrm{eq}}$
$\mathrm{R} \ln \mathrm{K}_{\mathrm{eq}}=-\frac{\Delta \mathrm{G}^{\mathrm{o}}}{\mathrm{T}}$
(on increasing temperature in exothermic reaction $\mathrm{K}_{\mathrm{eq}}$ decreases)
$\Delta \mathrm{H}^{\circ}$ and $\Delta \mathrm{S}^{\circ}$ are almost constant with temperature.

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MCQ 44 Marks

Consider the following chemical equilibrium of the gas phase reaction at a constant temperature : $\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})$
If $p$ being the total pressure, $K_{p}$ is the pressure equilibrium constant and $\alpha$ is the degree of dissociation, then which of the following is true at equilibrium?

A
If p value is extremely high compared to $\mathrm{K}_{\mathrm{p}}, \alpha \approx 1$
B
When $p$ increases $\alpha$ decreases
C
If $k_{p}$ value is extremely high compared to $p, \alpha$ becomes much less than unity
D
When $p$ increases $\alpha$ increases

Answer

B. When $p$ increases $\alpha$ decreases
$\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})$
$\begin{array}{cccc} t =0 & a & 0 & 0 \\ t = t & a (1-\alpha) & a \alpha & a \alpha\end{array}$
a moles of $\mathrm{A}(\mathrm{g})$ taken initially and at time
Now moles fraction of $\mathrm{A}(\mathrm{g}), \mathrm{B}(\mathrm{g})$ and $\mathrm{C}(\mathrm{g})$ are
$X_{A}=\frac{a-a \alpha}{a+a \alpha}=\frac{1-\alpha}{1+\alpha}$
$X_{B}=\frac{a \alpha}{a+a \alpha}=\frac{\alpha}{1+\alpha}$
$X_{C}=\frac{a \alpha}{a+a \alpha}=\frac{\alpha}{1+\alpha}$
Now if P is total pressure then partial pressure of $\mathrm{A}(\mathrm{g}), \mathrm{B}(\mathrm{g})$ and $\mathrm{C}(\mathrm{g})$ are
$\mathrm{P}_{\mathrm{A}}=\left(\frac{1-\alpha}{1+\alpha}\right) \mathrm{P}$
$\mathrm{P}_{\mathrm{B}}=\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P}$
$\mathrm{P}_{\mathrm{C}}=\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P}$
$K_{P}=\frac{\left(\frac{\alpha}{1+\alpha}\right) P\left(\frac{\alpha}{1+\alpha}\right) P}{\left(\frac{1-\alpha}{1+\alpha}\right) P}$
$K_{P}=\frac{\alpha^{2} P}{1-\alpha^{2}}$
As $K_{P}$ is only function of temperature.
So as $\mathrm{P} \uparrow \quad \alpha \downarrow$

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MCQ 54 Marks

Match List-I with List-II

Choose the correct answer from the options given below :

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

Answer

B. (A)-(III), (B)-(IV), (C)-(II), (D)-(I)

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MCQ 64 Marks

The nature of oxide $\left(\mathrm{TeO}_{2}\right)$ and hydride $\left(\mathrm{TeH}_{2}\right)$ formed by Te , respectively are :

A
Oxidising and acidic
B
Reducing and basic
C
Reducing and acidic
D
Oxidising and basic

Answer

A. Oxidising and acidic
$\mathrm{TeO}_{2}$ is oxidizing in nature because it can be reduced from +4 oxidation state to lower oxidation state.
$\mathrm{TeH}_{2}$ due to less bond dissociation energy easily breaks and hence acidic in nature.

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MCQ 74 Marks

Reactant A converts to product D through the given mechanism (with the net evolution of heat) :
$\mathrm{A} \rightarrow \mathrm{B}$ slow ; $\Delta \mathrm{H}=+\mathrm{ve}$
$\mathrm{B} \rightarrow \mathrm{C}$ fast; $\Delta \mathrm{H}=-\mathrm{ve}$
$\mathrm{C} \rightarrow \mathrm{D}$ fast; $\Delta \mathrm{H}=-\mathrm{ve}$
Which of the following represents the above reaction mechanism?

A
B
C
D

Answer

\begin{array}{ll}
A \rightarrow B & & Slow\\
\Delta H-+ve & E_{a_1} \rightarrow \text { High } \\
B \rightarrow C & & Fast\\
\Delta H=-ve & E_{a_2} \rightarrow \text { Low } \\
C \rightarrow D & \\
\Delta H=-ve & E_{a_3} \rightarrow \text { Low }
\end{array}

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MCQ 84 Marks

Arrange the following in order of magnitude of work done by the system / on the system at constant temperature :
(a) $\left|\omega_{\text {reversible }}\right|$ for expansion in infinite stage.
(b) $\left|w_{\text {irreversible }}\right|$ for expansion in single stage.
(c) $\left|\omega_{\text {reversible }}\right|$ for compression in infinite stage.
(d) $\left|w_{\text {irreversible }}\right|$ for compression in single stage.
Choose the correct answer from the options given below :

A
$a>b>c>d$
B
$d>c=a>b$
C
$c=a>d>b$
D
$a>c>b>d$

Answer

B. $d>c=a>b$
For isothermal process
$\left|\mathrm{W}_{\text {reversible }}\right|_{\text {expansion }}=\left|\mathrm{W}_{\text {reversible }}\right|_{\text {compression }}$
$=-n R T \ln \frac{\mathrm{~V}_{\mathrm{f}}}{\mathrm{V}_{\mathrm{i}}}$
$\left|\mathrm{W}_{\text {irreversible }}\right|_{\text {expansion }}<\left|\mathrm{W}_{\text {irreversible }}\right|_{\text {compression }}$
$\mathrm{d}>\mathrm{c}=\mathrm{a}>\mathrm{b}$
$\mid W_{\text {irreversiblele }}$ expansion $=-P_{\text {ext }}\left(V_{f}-V_{i}\right)$

(Single step irreversible)
Graphical representation
We can compare work by area of PV graph.

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MCQ 94 Marks

'x' g of NaCl is added to water in a beaker with a lid. The temperature of the system is raised from $1^{\circ} \mathrm{C}$ to $25^{\circ} \mathrm{C}$. Which out of the following plots, is best suited for the change in the molarity (M) of the solution with respect to temperature?
[Consider the solubility of NaCl remains unchanged over the temperature range]

A
B
C
D

Answer

$\underset{1^{\circ} C }{\text { Water }} \xrightarrow[\text { decreases }]{\text { Volume }} \underset{4^{\circ} C }{\text { Water }} \xrightarrow[\text { thermal expansion }]{\text { Volume increases }} \underset{25^{\circ} C }{\text { Water }}$
Molarity $=\frac{\mathrm{n}_{\text {solute }}}{(\text { Volume of solution })_{\ell}}$
Hence
$1^{\circ} \mathrm{C} \xrightarrow[\text { increases }]{\text { molarity }} 4^{\circ} \mathrm{C} \xrightarrow[\text { decreases }]{\text { molarity }} 25^{\circ} \mathrm{C}$

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MCQ 104 Marks

Match List-I with List-II

List-I (Purification technique)		List-II (Mixture of organic compounds)
(A)	Distillation (simple)	(I)	Diesel + Petrol
(B)	Fractional distillation	(II)	Aniline + Water
(C)	Distillation under reduced pressure	(III)	Chloroform + Aniline
(D)	Steam distillation	(IV)	Glycerol + Spent-lye

Choose the correct answer from the options given below :

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

Answer

D. (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

List-I (Purification technique)		List-II (Mixture of organic compounds)
(A)	Distillation (simple)	(III)	Chloroform + Aniline
(B)	Fractional distillation	(I)	Diesel + Petrol
(C)	Distillation under reduced pressure	(IV)	Glycerol + Spent-lye
(D)	Steam distillation	(II)	Aniline + Water

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MCQ 114 Marks

Electronic configuration of four elements $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and D are given below :
(A) $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{3}$
(B) $1 s^{2} 2 s^{2} 2 p^{4}$
(C) $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{5}$
(D) $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{2}$
Which of the following is the correct order of increasing electronegativity (Pauling's scale)?

A
A $<$ D $<$ B $<$ C
B
A $<$ C $<$ B $<$ D
C
A $<$ B $<$ C $<$ D
D
D $<$ A $<$ B $<$ C

Answer

D. D $<$ A $<$ B $<$ C
$\mathrm{N}:-1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{3}$ (Electronegativity $=3$ )
$\mathrm{O}:-1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{4}$ (Electronegativity $=3.5$ )
$\mathrm{F}:-1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{5}$ (Electronegativity $=4$ )
$\mathrm{C}:-1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{2}$ (Electronegativity $=2.55$ )
Correct order $=\mathrm{C}>\mathrm{B}>\mathrm{A}>\mathrm{D}$

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MCQ 124 Marks

Consider the following reactions. From these reactions which reaction will give carboxylic acid as a major product?
(A) $\mathrm{R}-\mathrm{C} \equiv \mathrm{N} \xrightarrow[\text { mild condition }]{\text { (i) } \mathrm{H}^{+} / \mathrm{H}_{2} \mathrm{O}}$
(B) $\mathrm{R}-\mathrm{MgX} \xrightarrow[\text { (ii) } \mathrm{H}_{3} \mathrm{O}^{+}]{\text {(i) } \mathrm{CO}_{2}}$
(C) $\mathrm{R}-\mathrm{C} \equiv \mathrm{N} \xrightarrow[\text { (ii) } \mathrm{H}_{3} \mathrm{O}^{+}]{\text {(i) } \mathrm{SnCl}_{2} / \mathrm{HCl}}$
(D) $\mathrm{R} \cdot \mathrm{CH}_{2} \cdot \mathrm{OH} \xrightarrow{\mathrm{PCC}}$
(E)

Choose the correct answer from the options given below :

A
A and D only
B
A, B and E only
C
B, C and E only
D
B and E only

Answer

D. B and E only
(A) $\mathrm{R}-\mathrm{C} \equiv \mathrm{N} \xrightarrow[\text { mild condition }]{\text { (i) } \mathrm{H}^{+} / \mathrm{H}_{2} \mathrm{O}} \mathrm{R}-\mathrm{CONH}_{2}$
Under mild condition amide is formed because this reaction is typically slow if further more heat will supplied then it gets convert in to -COOH .
(B) $\mathrm{R}-\mathrm{MgX} \xrightarrow[\text { (ii) } \mathrm{H}_{3} \mathrm{O}^{+}]{\text {(i) } \mathrm{CO}_{2}} \mathrm{R}-\mathrm{COOH}$
(C) $\mathrm{R}-\mathrm{C} \equiv \mathrm{N} \xrightarrow[\text { (ii) } \mathrm{H}_{3} \mathrm{O}^{+}]{\text {(i) } \mathrm{SnCl}_{2} / \mathrm{HCl}} \mathrm{R}-\mathrm{CHO}$
(D) $\mathrm{R} \cdot \mathrm{CH}_{2} \cdot \mathrm{OH} \xrightarrow{\mathrm{PCC}} \mathrm{R}-\mathrm{CHO}$
(E)

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MCQ 134 Marks

The type of hybridization and the magnetic property of $\left[\mathrm{MnCl}_{6}\right]^{3-}$ are :

A
$d^{2} s p^{3}$, paramagnetic with four unpaired electrons
B
$s p^{3} d^{2}$, paramagnetic with four unpaired electrons
C
$d ^2 sp ^3$, paramagnetic with two unpaired electrons
D
$\mathrm{sp}^{3} \mathrm{~d}^{2}$, paramagnetic with two unpaired electrons

Answer

B. $s p^{3} d^{2}$, paramagnetic with four unpaired electrons
$\left[\mathrm{MnCl}_{6}\right]^{3-}$ contains $\mathrm{Mn}^{+3}$
$\mathrm{Mn}^{+3}$ :- $[\mathrm{Ar}] 3 \mathrm{~d}^{4}$
Ligand $\Rightarrow \mathrm{Cl}^{-}$(WFL)

Hybridisation $=\mathrm{sp}^{3} \mathrm{~d}^{2}$
4 unpaired electrons

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MCQ 144 Marks

Which of the following statements are true ?
(A) The subsidiary quantum number $l$ describes the shape of the orbital occupied by the electron.

(C) The + and - signs in the wave function of the $2 p_{x}$ orbital refer to charge.
(D) The wave function of $2 p_{x}$ orbital is zero everywhere in the xy plane.

A
(B) and (D) only
B
(A), (B) and (C) only
C
(C) and (D) only
D
(A) and (B) only

Answer

D. (A) and (B) only
(A) Azimuthal quantum number ( $\ell$ ) indicates the shape of orbital occupied by the electron
(B)

(C) The + and - sign in the wave function of $2 p_{x}$ orbital refer to the sign (Phase) of the wave function, not the charge
(D) The wave function of $2 p_{x}$ orbital will be zero in yz plane (Nodal plane).

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MCQ 154 Marks

In 3, 3-dimethylhex-1-ene-4-yne, there are __________ $\mathrm{sp}^{3}$, __________ $\mathrm{sp}^{2}$ and __________ sp hybridised carbon atoms respectively :

A
$4,2,2$
B
$3,3,2$
C
$2,4,2$
D
$2,2,4$

Answer

$4-\mathrm{sp}^{3}, 2-\mathrm{sp}^{2}, 2-\mathrm{sp}$

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MCQ 164 Marks

Formation of $\mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{5} \mathrm{NOS}\right]$, a purple coloured complex formed by addition of sodium nitroprusside in sodium carbonate extract of salt indicates the presence of:

A
Sodium ion
B
Sulphate ion
C
Sulphide ion
D
Sulphite ion

Answer

C. Sulphide ion
$\underset{\substack{\text { Sodiun } \\ \text { sulphide }}}{\mathrm{Na}_{2} \mathrm{~S}}+\underset{\substack{\text { Sodium nitro } \\ \text { Prusside }}}{\mathrm{Na}_{2}}\left[\mathrm{Fe}(\mathrm{CN})_{5} \mathrm{NO}\right] \rightarrow \underset{4}{ } \rightarrow \underset{\text { Violet Colour }}{\mathrm{Na}_{4}}\left[\mathrm{Fe}(\mathrm{CN})_{5} \mathrm{NOS}\right]$

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MCQ 174 Marks

Which among the following molecules is (a) involved in $\mathrm{sp}^{3} \mathrm{~d}$ hybridization, (b) has different bond lengths and (c) has lone pair of electrons on the central atom ?

A
$\mathrm{PF}_{5}$
B
$\mathrm{XeF}_{4}$
C
$\mathrm{SF}_{4}$
D
$\mathrm{XeF}_{2}$

Answer

C. $\mathrm{SF}_{4}$

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MCQ 184 Marks

Given below are two statements :
Statement (I) : Neopentane forms only one monosubstituted derivative.
Statement (II) : Melting point of neopentane is higher than n-pentane
In the light of the above statements, choose the most appropriate answer from the options given below :

A
Statement I is correct but Statement II is incorrect
B
Both Statement I and Statement II are correct
C
Both Statement I and Statement II are incorrect
D
Statement I is incorrect but Statement II is correct

Answer

B. Both Statement I and Statement II are correct
Both Statement-I and Statement-II are correct.

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MCQ 194 Marks

The d-orbital electronic configuration of the complex among $\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}, \left[\mathrm{CoF}_{6}\right]^{3-}$, $\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ and $\left[\mathrm{Zn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ that has the highest CFSE is :

A
$t_{2 g}^{6} e_{g}^{0}$
B
$t_{2 g}^{6} e_{g}^{4}$
C
$t_{2 g}^{3} e_{g}^{2}$
D
$t_{2 g}^{4} e_{g}^{2}$

Answer

A. $t_{2 g}^{6} e_{g}^{0}$
In $\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{+3}$ S.F.L. is present and hence highest value of CFSE
In rest all complexes WFL is present hence CFSE will be low.
$\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{+3} \Rightarrow \mathrm{Co}^{+3}$ and en (SFL)
$\mathrm{Co}^{+3} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^{6}$

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MCQ 204 Marks

When a concentrated solution of sulphanilic acid and 1-naphthylamine is treated with nitrous acid ( 273 K ) and acidified with acetic acid, the mass (g) of 0.1 mole of product formed is :
(Given molar mass in $\mathrm{g} \mathrm{mol}^{-1} \mathrm{H}: 1, \mathrm{C}: 12, \mathrm{~N}: 14$, O : 16, S : 32)

A
343
B
330
C
33
D
66

Answer

0.1 mole of red-azo dye (Molar Mass $=327 \mathrm{gm} / \mathrm{mol})$ will have 32.7 gm mass. Nearly 33 gm .

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