Chemical Bonding and Molecular Structure — DPP 1

20 NEET questions from Chemical Bonding and Molecular Structure, with answers and explanations. Attempt it as a timed test to get scoring, accuracy breakdown and mistake tracking.

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1
Among the following which one has the highest cation to anion size ratio?
  1. A CsI
  2. B CsF
  3. C LiF
  4. D NaF
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Correct answer: B

2
The dipole moment of the given molecules are such that:
  1. A $\mathrm{BF}_{3}>\mathrm{NF}_{3}>\mathrm{NH}_{3}$
  2. B $\mathrm{NF}_{3}>\mathrm{BF}_{3}>\mathrm{NH}_{3}$
  3. C $\mathrm{NH}_{3}>\mathrm{NF}_{3}>\mathrm{BF}_{3}$
  4. D $\mathrm{NH}_{3}>\mathrm{BF}_{3}>\mathrm{NF}_{3}$
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Correct answer: C

3
Which one of the following can not exist on the basis of molecular orbital theory?
  1. A $\mathrm{H}_{2}^{+}$
  2. B $\mathrm{He}_{2}^{+}$
  3. C $\mathrm{C}_{2}$
  4. D $\mathrm{He}_{2}$
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Correct answer: D

4
The ONO angle is maximum in:
  1. A $\mathrm { HNO } _ { 3 }$
  2. B $\mathrm { NO } _ { 2 } ^ { + }$
  3. C $\mathrm { HNO } _ { 2 }$
  4. D $\mathrm { NO } _ { 2 }$
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Correct answer: B

5
Which of the following compounds would have significant intermolecular hydrogen bonding?
$\mathrm{HF}, \mathrm{CH}_{3} \mathrm{OH}, \mathrm{N}_{2} \mathrm{O}_{4}, \mathrm{CH}_{4}$
  1. A $\mathrm{HF}, \mathrm{N}_{2} \mathrm{O}_{4}$
  2. B $\mathrm{HF}, \mathrm{CH}_{4}, \mathrm{CH}_{3} \mathrm{OH}$
  3. C $\mathrm{HF}, \mathrm{CH}_{3} \mathrm{OH}$
  4. D $\mathrm{CH}_{3} \mathrm{OH}, \mathrm{CH}_{4}$
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Correct answer: C

6
Phosphorus pentachloride, $\mathrm{PCl}_{5}$, dissociates in the vapour phase according to the equation:
$\mathrm{PCl}_{5(\mathrm{~g})} \rightleftharpoons \mathrm{PCl}_{4}^{+}+\mathrm{Cl}^{-}$
What is the hybridisation of P in $\mathrm{PCl}_{4}^{+}$ion?
  1. A $\mathrm{sp}^{3}$
  2. B $\mathrm{sp}^{3} \mathrm{~d}$
  3. C $\mathrm{sp}^{3} \mathrm{~d}^{2}$
  4. D $\mathrm{dsp}^{2}$
Show answer & explanation

Correct answer: A

$\mathrm{PCl}_{4}^{+}$is isostructrual with $\mathrm{NH}_{4}^{+}$so N and P both will have same hybridisation. Hence, the hybridisation of P in $\mathrm{PCl}_{4}^{+}$ is $\mathrm{sp}^{3}$.
Therefore, option (1) is the correct answer.
7
Choose the molecules in which hybridisation occurs in the ground state.
(A) BCl$_{3}$
(B) NH$_{3}$
(C) PCl$_{3}$
(D) BeF$_{2}$
The correct answer is
  1. A A, B, C
  2. B A, B
  3. C B, C
  4. D C, D
Show answer

Correct answer: C

8
The charge/size ratio of a cation determines its polarizing power. Which one of the following sequences represents the increasing order of the polarizing power of these cationic species, $\mathrm{K}^{+}, \mathrm{Ca}^{2+}, \mathrm{Mg}^{2+}, \mathrm{Be}^{2+}$ ?
  1. A $\mathrm{K}^{+}<\mathrm{Ca}^{2+}<\mathrm{Mg}^{2+}<\mathrm{Be}^{2+}$
  2. B $\mathrm{Ca}^{2+}<\mathrm{Mg}^{2+}<\mathrm{Be}^{2+}<\mathrm{K}^{+}$
  3. C $\mathrm{Mg}^{2+}<\mathrm{Be}^{2+}<\mathrm{K}^{+}<\mathrm{Ca}^{2+}$
  4. D $\mathrm{Be}^{2+}<\mathrm{K}^{+}<\mathrm{Ca}^{2+}<\mathrm{Mg}^{2+}$
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Correct answer: A

9
The minimum number of $180^{\circ}$ angles possible between X-M-X bond for compounds with $\mathrm{sp}^{3} \mathrm{~d}^{2}$ and $\mathrm{sp}^{3} \mathrm{~d}$ hybridisation respectively are:
  1. A 3 and 3
  2. B 3 and 1
  3. C 1 and 3
  4. D 3 and 0
Show answer & explanation

Correct answer: B

sp$^{3}$d is MX$_{5}$ kind of geometry in which number of X-M-X bond of 180° is (1 axial angle) whereas sp$^{3}$d$^{2}$ is octahedral geometry hence there would be 3 bonds with straight 180° angle.
Therefore, option (2) is the correct answer
10
Which of the following has fractional bond order?
  1. A $\mathrm{B}_{2}$
  2. B $\mathrm{O}_{2}{ }^{2-}$
  3. C $\mathrm{F}_{2}$
  4. D $\mathrm{H}_{2}^{-}$
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Correct answer: D

11
Which cannot be explained by VBT?
  1. A Overlapping
  2. B Bond formation
  3. C Paramagnetic nature of oxygen
  4. D Shapes of molecules
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Correct answer: C

12
Among the following compounds the one that is polar and has central atom with $\mathrm{sp}^{3}$ hybridisation is:
  1. A $\mathrm{H}_{2} \mathrm{CO}_{3}$
  2. B $\mathrm{SiF}_{4}$
  3. C $\mathrm{BF}_{3}$
  4. D $\mathrm{HClO}_{2}$
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Correct answer: D

13
The ion which is not tetrahedral in shape is:
  1. A $\mathrm{BF}_{4}^{-}$
  2. B $\mathrm{NH}_{4}^{+}$
  3. C $\mathrm{XeO}_{4}$
  4. D $\mathrm{ICl}_{4}^{-}$
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Correct answer: D

14
The correct order of increasing bond angle in $\mathrm{NO}_{2}, \mathrm{NO}_{2}^{+}$ and $\mathrm{NO}_{2}^{-}$is:
  1. A $\mathrm{NO}_{2}^{-}<\mathrm{NO}_{2}<\mathrm{NO}_{2}^{+}$
  2. B $\mathrm{NO}_{2}<\mathrm{NO}_{2}^{+}<\mathrm{NO}_{2}^{-}$
  3. C $\mathrm{NO}_{2}<\mathrm{NO}_{2}^{-}<\mathrm{NO}_{2}^{+}$
  4. D $\mathrm{NO}_{2}^{+}<\mathrm{NO}_{2}^{-}<\mathrm{NO}_{2}$
Show answer & explanation

Correct answer: A

In $\mathrm{NO}_{2}$, the bond angle is $134^{\circ}$ (experimentally determined) $\mathrm{NO}_{2}^{+}$is linear bond angle is $180^{\circ} . \mathrm{NO}_{2}^{-}$is bent with bond angle slightly less than $120^{\circ}$.
Therefore, option (1) is the correct answer.
15
In which of the following pair, both the species have the same bond order?
  1. A $\mathrm{N}_{2}, \mathrm{~N}_{2}^{+}$
  2. B $\mathrm{O}_{2}, \mathrm{~F}_{2}$
  3. C $\mathrm{C}_{2}, \mathrm{O}_{2}$
  4. D $\mathrm{C}_{2}, \mathrm{~N}_{2}$
Show answer & explanation

Correct answer: C

In the pair $\left(\mathrm{N}_{2}, \mathrm{~N}_{2}^{+}\right)$, bond order of $\mathrm{N}_{2}$ is 3 whereas $\mathrm{N}_{2}^{+}$has a bond order of 2.5. In ( $\mathrm{O}_{2}, \mathrm{~F}_{2}$ ) pair, bond order of $\mathrm{O}_{2}$ and $\mathrm{F}_{2}$ is 2 and 1 respectively whereas in the pair $\left(\mathrm{C}_{2}, \mathrm{~N}_{2}\right)$, it is 2 and 3 respectively.
Only in the pair ( $\mathrm{C}_{2}, \mathrm{O}_{2}$ ) both the species have the same bond order of 2 .
Therefore, option (3) is the correct answer.
16
XeF$_{4}$ is planar whereas CCl$_{4}$ is tetrahedral because:
  1. A In $\mathrm{XeF}_{4}, \mathrm{Xe}$ is $\mathrm{sp}^{2}$ hybridised and C is $\mathrm{sp}^{3}$ hybridised in $\mathrm{CCl}_{4}$.
  2. B Xe is $\mathrm{sp}^{3} \mathrm{~d}^{2}$ hybridised in $\mathrm{XeF}_{4}$ but due to the presence of two lone pairs of electrons, the shape is planar whereas in $\mathrm{CCl}_{4}, \mathrm{C}$ is $\mathrm{sp}^{3}$ hybridised.
  3. C In both $\mathrm{XeF}_{4}$ and $\mathrm{CCl}_{4}$, the central atom is $\mathrm{sp}^{3}$ hybridised.
  4. D Xe is a noble gas whereas C is a non-metal.
Show answer & explanation

Correct answer: B

$\mathrm{XeF}_{4}$ is planar because Xe is $\mathrm{sp}^{3} \mathrm{~d}^{2}$ hybridised. Due to the presence of two lone pairs of electrons, the shape is square planar. $\mathrm{CCl}_{4}$ is tetrahedral because C is $\mathrm{sp}^{3}$ hybridised. Therefore, option (2) is the correct answer.
17
Which of the following compounds have some bond angle as nearly $90^{\circ}$ ?
  1. A $\mathrm{CH}_{4}$
  2. B $\mathrm{CO}_{2}$
  3. C $\mathrm{H}_{2} \mathrm{O}$
  4. D $\mathrm{SF}_{6}$
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Correct answer: D

18
In which of the following molecules number of lone pairs and bond pairs on central atom are not equal?
  1. A H₂O
  2. B I$_{3}$$^{–}$
  3. C O₂F$_{2}$
  4. D SCl$_{2}$
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Correct answer: B

19
The number and type of bond between two carbon atoms in CaC$_{2}$ are:
  1. A one sigma and one pi bond.
  2. B one sigma and two pi bonds
  3. C two pi bonds only
  4. D one sigma bond only
Show answer & explanation

Correct answer: B

Calcium carbide ($\mathrm{CaC}_{2}$) is an ionic compound having anion $\mathrm{C}_{2}^{2-}$. The calcium atom is bonded to the two carbon atoms through ionic bonding. The two carbon atoms have one sigma and two pi bonds $(:\mathrm{C} \equiv \mathrm{C}:)^{2-}$.
Therefore, option (2) is the correct answer.
20
Carbonate ion $\left(\mathrm{CO}_{3}^{2-}\right)$ and nitrate ion $\left(\mathrm{NO}_{3}^{-}\right)$are isoelectronic. If $\mathrm{NO}_{3}^{-}$has a trigonal planar geometry, then $\mathrm{CO}_{3}^{2-}$ will have a $\_\_\_\_$ geometry.
  1. A bent
  2. B tetrahedral
  3. C trigonal pyramidal
  4. D trigonal planar
Show answer & explanation

Correct answer: D

Isoelectronic species have same number of electrons and geometry. Therefore, $\mathrm{CO}_{3}^{2-}$ will have the trigonal planar geometry.
Therefore, option (4) is the correct answer.

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