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

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SECTION - A [CHEMISTY - MCQ]

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20 questions · 19 auto-graded MCQ + 1 self-marked written.

MCQ 14 Marks

A molecule with the formula AX₄Y has all it's elements from p-block. Element A is rarest, monoatomic, non-radioactive from its group and has the lowest ionization enthalpy value among A, X and Y. Elements X and Y have first and second highest electronegativity values respectively among all the known elements. The shape of the molecule is :

A
Square pyramidal
B
Octahedral
C
Pentagonal planar
D
Trigonal bipyramidal

Answer

(A) Square pyramidal
Explanation: Given A is rarest, monoatomic, non-radioactive p-block element and form AXY type of molecule.
∴ It is concluded that it is Xe
It is given the electronegativity of A is less than X & Y
It is given the electronegativity of X & Y is highest and second highest respectively among all element.
∴ X & Y are F & O
∴ Compound is consider as $XeOF _4$ with square pyramidal shape.

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MCQ 24 Marks

Consider the following molecules :

The correct order of rate of hydrolysis is :

A
r > q > p > s
B
q > p > r > s
C
p > r > q > s
✓
p > q > r > s

Answer

Correct option: D.

p > q > r > s

(D) p > q > r > s
Explanation: Rate of hydrolysis $\propto$ Leaving group ability

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MCQ 34 Marks

Consider the following compound (X)

The most stable and least stable carbon radicals, respectively, produced by homolytic cleavage of corresponding C - H bond are :

A
II, IV
B
III, II
C
I, IV
✓
II, I

Answer

Correct option: D.

II, I

(D) II, I
Explanation:

II most stable carbon radical due to resonance stablise
I least stable carbon radical due to no stabilising factor.

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MCQ 44 Marks

On complete combustion 1.0 g of an organic compound (X) gave 1.46 g of $CO _2$ and 0.567 g of $H _2 O$. The empirical formula mass of compound ( X ) is __________ g.
(Given molar mass in $g mol ^{-1} C : 12, H : 1, O : 16$ )

✓
30
B
45
C
60
D
15

Answer

Correct option: A.

30

(A) 30
Explanation: Moles of ' C ' $= n _{ CO _2}=\frac{1.46}{44}=0.033$
Moles of ' C ' $= W _{ C }=0.033 \times 12$
Moles of 'H' $=2 \times n _{ H _2 O }=2 \times \frac{0.567}{18}=0.063$
Mass of 'H' $=0.0063$
Mass of Oxygen $( O )=1-\left( W _{ C }+ W _{ H }\right)$
$=1-(0.033 \times 12+0.063 \times 1)=0.541 gm$
Moles of ' O ' $=\frac{0.541}{16}=0.033$
Empirical formula $= CH _2 O$
Empirical formula mass $=30$.

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MCQ 54 Marks

Choose the correct tests with respective observations.
(A) $CuSO _4$ (acidified with acetic acid) + $K _4\left[ Fe ( CN )_6\right] \rightarrow$ Chocolate brown precipitate.
(B) $FeCl _3+ K _4\left[ Fe ( CN )_6\right] \rightarrow$ Prussian blue precipitate.
(C) $ZnCl _2+ K _4\left[ Fe ( CN )_6\right]$, neutralised with $NH _4 OH$ $\rightarrow$ White or bluish white precipitate.
(D) $MgCl _2+ K _4\left[ Fe ( CN )_6\right] \rightarrow$ Blue precipitate.
(E) $BaCl _2+ K _4\left[ Fe ( CN )_6\right]$, neutralised with NaOH $\rightarrow$ White precipitate.
Choose the correct answer from the options given below:

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

Answer

Correct option: C.

A, B and C only

(C) A, B and C only
Explanation:

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MCQ 64 Marks

Given below are two statements :
Statement (I): In octahedral complexes, when $\Delta_{ o }$ < P high spin complexes are formed. When $\Delta_{ o }$ > P low spin complexes are formed.
Statement (II): In tetrahedral complexes because of $\Delta_{ t }$ <P, low spin complexes are rarely formed. 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 incorrect
C
Statement I is incorrect but Statement II is correct
✓
Both Statement I and Statement II are correct

Answer

Correct option: D.

Both Statement I and Statement II are correct

(D) Both Statement I and Statement II are correct
Explanation: In octahedral complex (CN = 6)
If $\Delta_0$ <P.E., then high spin complexes are formed
If $\Delta_0$ > P.E., then low spin complexes are formed
But in tetrahedral complex (CN = 4)
$\Delta_t$ < P.E., then mainly high spin complexes are formed and rarely low spin complexes are formed.

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MCQ 74 Marks

Given below are two statements :
Statement (I): The metallic radius of Al is less than that of Ga.
Statement (II) : The ionic radius of $Al ^{3+}$ is less than that of $Ga ^{3+}$.
In the light of the above statements, choose the most appropriate answer from the options given below :

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

Answer

Correct option: B.

Statement I is incorrect but Statement II is correct

(B) Statement I is incorrect but Statement II is correct
Explanation: ⇒ The metallic radius order of Al & Ga is
$B <\underbrace{ Ga < Al }_{\downarrow}< In < T \ell$
(due to poor shielding of d-subshell electrons)
$\Rightarrow$ The ionic radius order of $Al ^{+3} \& Ga ^{+3}$ is
$B ^{+3}< Al ^{+3}< Ga ^{+3}< In ^{+3}< T \ell^{+3}$

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MCQ 84 Marks

The correct order of basic nature on aqueous solution for the bases $NH _3, H _2 N- NH _2, CH _3 CH _2 NH _2$, $\left( CH _3 CH _2\right)_2 NH$ and $\left( CH _3 CH _2\right)_3 N$ is :

A
$NH _3< H _2 N- NH _2<\left( CH _3 CH _2\right)_3 N< CH _3 CH _2 NH _2<$ $\left( CH _3 CH _2\right)_2 NH$
B
$NH _3< H _2 N- NH _2< CH _3 CH _2 NH _2<\left( CH _3 CH _2\right)_2 NH <$ $\left( CH _3 CH _2\right)_3 N$
C
$H _2 N- NH _2< NH _3<\left( CH _3 CH _2\right)_3 N< CH _3 CH _2 NH _2<$ $\left( CH _3 CH _2\right)_2 NH$
✓
$NH _2- NH _2< NH _3< CH _3 CH _2 NH _2<\left( CH _3 CH _2\right)_3 N<$ $\left( CH _3 CH _2\right)_2 NH$

Answer

Correct option: D.

$NH _2- NH _2< NH _3< CH _3 CH _2 NH _2<\left( CH _3 CH _2\right)_3 N<$ $\left( CH _3 CH _2\right)_2 NH$

(D) $NH _2- NH _2< NH _3< CH _3 CH _2 NH _2<\left( CH _3 CH _2\right)_3 N<$ $\left( CH _3 CH _2\right)_2 NH$
Explanation: Basic strength of amine depends on hydrogen bonding and electronic inductive effect.
$NH ( Et )_2> N ( Et )_3> NH _2 Et >\ddot{ N }_3> NH _2- NH _2$

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MCQ 94 Marks

Identify the correct statement among the following:

A
All naturally occurring amino acids except glycine contain one chiral centre.
B
All naturally occurring amino acids are optically active.
C
Glutamic acid is the only amino acid that contains a -COOH group at the side chain.
✓
Amino acid, cysteine easily undergo dimerization due to the presence of free SH group.

Answer

Correct option: D.

Amino acid, cysteine easily undergo dimerization due to the presence of free SH group.

(D) Amino acid, cysteine easily undergo dimerization due to the presence of free SH group.
Explanation: * Isoleucine has 2 chiral centre
* Glycine is optically inactive
* Aspartic acid also contain COOH group at the side chain.
* Cysteine easily dimmerise due to free SH group

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MCQ 104 Marks

Given below are two statements:
Statement (I):

will react with NaOH and also with Tollen's reagent.
Statement (II):

will undergo self aldol condensation very easily.
In the light of the above statements, choose the most appropriate answer from the options given below:

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

Answer

Correct option: B.

Statement I is correct but Statement II is incorrect

(B) Statement I is correct but Statement II is incorrect
Explanation:

Vanillin does not give self-aldol reaction due to lack of acidic H for condensation.

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MCQ 114 Marks

Among $SO _2, NF _3, NH _3, XeF _2, ClF _3$ and $SF _4$, the hybridization of the molecule with non-zero dipole moment and highest number of lone-pairs of electrons on the central atom is

A
$sp ^3$
B
$dsp ^2$
C
$sp ^3 d^2$
✓
$sp ^3 d$

Answer

Correct option: D.

$sp ^3 d$

(D) sp³d
Explanation:

Molecule	Hybridisation	Dipole Moment	Lone pair on the central atom
$SO _2$	sp²	Non-zero	1
$NF _3$	sp³	Non-zero	1
$NH _3$	sp³	Non-zero	1
$XeF _2$	sp³d	zero	3
$C \ell F _3$	sp³d	Non-zero	2
$SF _4$	sp³d	Non-zero	1

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MCQ 124 Marks

If equal volumes of AB₂ and XY (both are salts) aqueous solutions are mixed, which of the following combination will give a precipitate of AY₂ at 300 K?
(Given K_sp (at 300 K) for AY₂ = $5.2 \times 10^{-7}$)

A
$3.6 \times 10^{-3} M AB _2, 5.0 \times 10^{-4} M XY$
B
$2.0 \times 10^{-4} M AB _2, 0.8 \times 10^{-3} M XY$
✓
$2.0 \times 10^{-2} M AB _2, 2.0 \times 10^{-2} M XY$
D
$1.5 \times 10^{-4} M AB _2, 1.5 \times 10^{-3} M XY$

Answer

Correct option: C.

$2.0 \times 10^{-2} M AB _2, 2.0 \times 10^{-2} M XY$

(C) $2.0 \times 10^{-2} M AB _2, 2.0 \times 10^{-2} M XY$
Explanation: When equal volumes are mixed molarity reduce to half.
For precipitation $Q _{ SP }=\left[ A ^{+2}\right][ Y ]^2> K _{ SP }$
(1) $Q _{ SP }=\left(1.8 \times 10^{-3}\right)\left(\frac{5}{2} \times 10^{-4}\right)^2< K _{ SP }$
(2) $Q _{ SP }=\left(10^{-4}\right)\left(0.4 \times 10^{-3}\right)^2< K _{ sP }$
(3) $Q _{ SP }=\left(10^{-2}\right)\left(10^{-2}\right)^2> K _{ SP }$
(4) $Q _{ SP }=\left(\frac{1.5}{2} \times 10^{-4}\right)\left(\frac{1.5}{2} \times 10^{-3}\right)^2< K _{ SP }$

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MCQ 134 Marks

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) H₂O(l)
Consider the above reaction, what mass of CaCl₂ will be formed if 250 mL of 0.76 M HCl reacts with 1000 g of CaCO₃?
(Given: Molar mass of Ca, C, O, H and Cl are 40, 12, 16, 1 and 35.5 g mol^-1, respectively)

A
3.908 g
B
2.636 g
✓
10.545 g
D
5.272 g

Answer

Correct option: C.

10.545 g

(C) 10.545 g
Explanation:
$\begin{array}{l} CaCO _3+2 HCl \rightarrow CaCl _2+ CO _2+ H _2 O \\
\text { Moles of } CaCO _3=\frac{1000}{100}=10 \\
\text { Moles of } HCl =0.76 \times \frac{250}{1000}=0.19 \text { (L.R.) } \\
\text { Moles of } CaCl _2 \text { formed }=\frac{0.19}{2} \\
\text { Mass of } CaCl _2=\frac{0.19}{2} \times 111=10.545 gm \end{array}$

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MCQ 144 Marks

A solution is made by mixing one mole of volatile liquid A with 3 moles of volatile liquid B. The vapour pressure of pure A is 200 mm Hg and that of the solution is 500 mm Hg. The vapour pressure of pure B and the least volatile component of the solution, respectively, are:

A
1400 mm Hg, A
B
1400 mm Hg, B
C
600 mm Hg, B
✓
600 mm Hg, A

Answer

Correct option: D.

600 mm Hg, A

(D) 600 mm Hg, A
Explanation: $P _{ S }= P _{ A }^{ o } \cdot X _{ A }+ P _{ B }^0 \cdot X _{ B }$
$500=200 \times \frac{1}{4}+ P _{ B }^{ o } \cdot \frac{3}{4}$
$P _{ B }^0=$ 600 mm Hg
As $P _{ A }^{ o }< P _{ B }^{ o } \Rightarrow A$ is least volatile.

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MCQ 154 Marks

Two vessels A and B are connected via stopcock. The vessel A is filled with a gas at a certain pressure. The entire assembly is immersed in water and is allowed to come to thermal equilibrium with water. After opening the stopcock the gas from vessel A expands into vessel B and no change in temperature is observed in the thermometer. Which of the following statement is true?

A
$d w \neq 0$
B
$dq \neq 0$
C
$dU \neq 0$
✓
The pressure in the vessel B before opening the stopcock is zero.

Answer

Correct option: D.

The pressure in the vessel B before opening the stopcock is zero.

(D) The pressure in the vessel B before opening the stopcock is zero.
Explanation: It is free expansion of gas $\Rightarrow P_{\text {ext }}=0$
Where $w =0, q =0$ and $\Delta U =0$

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MCQ 164 Marks

According to Bohr's model of hydrogen atom, which of the following statement is incorrect?

A
Radius of 3^rdorbit is nine times larger than that of 1^st orbit.
B
Radius of 8^th orbit is four times larger than that of 4^thorbit.
✓
Radius of 6^th orbit is three time larger than that of 4^th orbit.
D
Radius of 4^th orbit is four times larger than that of 2^ndorbit.

Answer

Correct option: C.

Radius of 6^th orbit is three time larger than that of 4^th orbit.

(C) Radius of 6^th orbit is three time larger than that of 4^th orbit.
Explanation: $r \propto n ^2$
(1) $\frac{ r _3}{ r _1}=\frac{9}{1}$
(2) $\frac{ r _8}{ r _4}=\frac{64}{16}=4$
(3) $\frac{r_6}{r_4}=\left(\frac{6}{4}\right)^2=\frac{9}{4}$
(4) $\frac{r_4}{r_2}=\left(\frac{4}{2}\right)^2=4$

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MCQ 174 Marks

Which of the following graph correctly represents the plots of K_H at 1 bar gases in water versus H temperature?

A
B
C
✓

Answer

Correct option: D.

(D)

Explanation: As temperature increases solubility first decrease then increase hence $K _{ H }$ first increase than decrease also at moderate temperature $K _{ H }$ value $He > N _2>$ $CH _4$.

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MCQ 184 Marks

The property/properties that show irregularity in first four elements of group-17 is/are:
(A) Covalent radius
(B) Electron affinity
(C) Ionic radius
(D) First ionization energy
Choose the correct answer from the options given below:

A
B and D only
B
A and C only
✓
B only
D
A, B, C and D

Answer

Correct option: C.

B only

(C) B only
Explanation: The order of first four elements of group-17 are as follows.
F < Cl < Br < I (Covalent radius)
Cl > F > Br > I (Electron affinity)
F^- < Cl^- < Br^- < I^- (Ionic radius)
F > Cl > Br > I (I^st ionization energy)
Electron affinity order is irregular.

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MCQ 194 Marks

An optically active alkyl halide $C _4 H _9 Br [ A ]$ reacts with hot KOH dissolved in ethanol and forms alkene [B] as major product which reacts with bromine to give dibromide [C]. The compound [C] is converted into a gas [D] upon reacting with alcoholic $NaNH _2$. During hydration 18 gram of water is added to 1 mole of gas [D] on warming with mercuric sulphate and dilute acid at 333 K to form compound [E]. The IUPAC name of compound $[ E ]$ is :

A
But-2-yne
B
Butan-2-ol
✓
Butan-2-one
D
Butan-1-al

Answer

Correct option: C.

Butan-2-one

(C) Butan-2-one
Explanation:

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MCQ 204 Marks

Designate whether each of the following compounds is aromatic or not aromatic.

A
e, g aromatic and a, b, c, d, f, h not aromatic
B
b, e, f, g aromatic and a, c, d, h not aromatic
C
a, b, c, d aromatic and e, f, g, h not aromatic
✓
a, c, d, e, h aromatic and b, f, g not aromatic

Answer

Correct option: D.

a, c, d, e, h aromatic and b, f, g not aromatic

(D) a, c, d, e, h aromatic and b, f, g not aromatic
Explanation: Aromatic compounds

a, c, d, e, h follow Huckel's rule
(Not Aromatic)

b, f, g, are not aromatic, these compounds do not
follow Huckel's rule.

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