Chemical Bonding & Molecular Structure — Page 12 | JEE Chemistry

Q111

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) :

\mathrm{NH}_3

and

\mathrm{NF}_3

molecule have pyramidal shape with a lone pair of electrons on nitrogen atom. The resultant dipole moment of

\mathrm{NH}_3

is greater than that of

\mathrm{NF}_3

. Reason (R) : In

\mathrm{NH}_3

, the orbital dipole due to lone pair is in the same direction as the resultant dipole moment of the

\mathrm{N}-\mathrm{H}

bonds.

\mathrm{F}

is the most electronegative element. In the light of the above statements, choose the correct answer from the options given below :

A (A) is false but (R) is true

B (A) is true but (R) is false

C Both (A) and (R) are true but (R) is NOT the correct explanation of (A)

D Both (A) and (R) are true and (R) is the correct explanation of (A)

Correct Answer

Option D

Solution

Dipole moment of

\mathrm{NH}_3

is higher than

\mathrm{NF}_3

as $\mu$ of lone-pair is in same direction as the resultant dipole moment of

\mathrm{N}-\mathrm{H}

bonds. So, both (A) & (R) are true, & (R) is the correct explanation of (A)

Q112

The molecules having square pyramidal geometry are

A

\mathrm{BrF}_5 ~\&~ \mathrm{PCl}_5

B

\mathrm{SbF}_5 ~\&~ \mathrm{PCl}_5

C

\mathrm{BrF}_5 ~\&~ \mathrm{XeOF}_4

D

\mathrm{SbF}_5 ~\&~ \mathrm{XeOF}_4

Correct Answer

Option C

Solution

$\mathrm{BrF}_5$ : Square pyramedal $\mathrm{XeOF}_4$ : Square pyramedal $\mathrm{SbF}_5$ : Trigonal bipyramidal $\mathrm{PCl}_5$ : Trigonal bipyramidal

Q113

Arrange the following compounds in increasing order of their dipole moment :

\mathrm{HBr}, \mathrm{H}_2 \mathrm{~S}, \mathrm{NF}_3

and

\mathrm{CHCl}_3

A

\mathrm{HBr}<\mathrm{H}_2 \mathrm{~S}<\mathrm{NF}_3<\mathrm{CHCl}_3

B

\mathrm{H}_2 \mathrm{~S}<\mathrm{HBr}<\mathrm{NF}_3<\mathrm{CHCl}_3

C

\mathrm{NF}_3<\mathrm{HBr}<\mathrm{H}_2 \mathrm{~S}<\mathrm{CHCl}_3

D

\mathrm{CHCl}_3<\mathrm{NF}_3<\mathrm{HBr}<\mathrm{H}_2 \mathrm{S}

Correct Answer

Option C

Solution

We need to compare the experimental (or well‐established) dipole moments of the given molecules: $\mathrm{HBr}, \mathrm{H_2S}, \mathrm{NF_3},$ and $\mathrm{CHCl_3}$ .

1.

$\mathbf{NF_3}$ Structure: Trigonal pyramidal (like $\mathrm{NH_3}$ ), but each bond is more polar toward fluorine.

Net dipole moment: Quite small, about $0.23\text{–}0.24\,D$ (the bond dipoles partly oppose the lone‐pair contribution).

2.

$\mathbf{HBr}$ Structure: Simple diatomic.

Dipole moment: About $0.78\,D$ .

3.

$\mathbf{H_2S}$ Structure: Bent (like $\mathrm{H_2O}$ ) but S is less electronegative than O, and the S–H bond angle is wider than the O–H angle in water.

Dipole moment: About $0.95\,D$ .

4. $\mathbf{CHCl_3}$ (chloroform) Structure: Tetrahedral around carbon with three Cl and one H.

Dipole moment: About $1.01\,D$ .

Ordering from smallest to largest dipole moment $\boxed{\mathrm{NF_3} \; Hence, the correct choice is:$ \boxed{\text{Option (C)}\quad NF_3

Q114

According to molecular orbital theory, which of the following is true with respect to Li2+ and Li2

-

?

A

Li_2^ +

is unstable and

Li_2^ -

is stable

B

Li_2^ +

is stable and

Li_2^ -

unstable

C Both are stable

D Both are unstable

Correct Answer

Option C

Solution

Li_2^ +

(5 electrons) =

{\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^1}}}

Li_2^ -

(7 electrons) =

{\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}

\,\sigma _{2{s^1}}^ * \,

$\therefore$ Bond order of

Li_2^ +

=

{{3 - 2} \over 2}

= 0.5 Bond order of

Li_2^ -

=

{{4 - 3} \over 2}

= 0.5 As both

Li_2^ +

and

Li_2^ -

has non-zero bond order, so both are stable.

Q115

Given below are two statements : Statement (I) : Experimentally determined oxygen-oxygen bond lengths in the

\mathrm{O}_3

are found to be same and the bond length is greater than that of a

\mathrm{O}=\mathrm{O}

(double bond) but less than that of a single

(\mathrm{O}-\mathrm{O})

bond. Statement (II) : The strong lone pair-lone pair repulsion between oxygen atoms is solely responsible for the fact that the bond length in ozone is smaller than that of a double bond

(\mathrm{O}=\mathrm{O})

but more than that of a single bond

(\mathrm{O}-\mathrm{O})

. In the light of the above statements, choose the correct answer from the options given below :

A Statement I is false but Statement II is true

B Both Statement I and Statement II are true

C Statement I is true but Statement II is false

D Both Statement I and Statement II are false

Correct Answer

Option C

Solution

Due to resonance bond length is identical in ozone. Therefore statement I is true and statement II is false.

Q116

A molecule with the formula

\mathrm{AX}_4 \mathrm{Y}

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

\mathrm{A}, \mathrm{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 Pentagonal planar

B Square pyramidal

C Trigonal bipyramidal

D Octahedral

Correct Answer

Option B

Solution

Given A is rarest, monoatomic, non-radioactive p-block element and form $\mathrm{AX}_4 \mathrm{Y}$ type of molecule.

$\therefore$ 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 $\mathrm{X} \& \mathrm{Y}$ is highest and second highest respectively among all element. $\therefore \mathrm{X}$ & Y are F & O $\therefore$ Compound is consider as $\mathrm{XeOF}_4$ with square pyramidal shape.

Q117

Among

\mathrm{SO}_2, \mathrm{NF}_3, \mathrm{NH}_3, \mathrm{XeF}_2, \mathrm{ClF}_3

and

\mathrm{SF}_4

, the hybridization of the molecule with nonzero dipole moment and highest number of lone-pairs of electrons on the central atom is

A

\mathrm{sp}^3

B

\mathrm{dsp}^2

C

\mathrm{sp}^3 \mathrm{~d}^2

D

\mathrm{sp}^3 \mathrm{~d}

Correct Answer

Option D

Solution

Molecules with non-zero dipole moment SO₂ (bent) NF₃ (pyramidal) NH₃ (pyramidal) ClF₃ (T-shaped) SF₄ (seesaw) (XeF₂ is linear → zero dipole) Lone-pair count on central atom SO₂: 1 NF₃: 1 NH₃: 1 SF₄: 1 ClF₃: 2 → ClF₃ has the highest number (2) of lone pairs among those with nonzero dipole.

Hybridization of Cl in ClF₃ • Total electron domains = 3 bonding pairs + 2 lone pairs = 5 • For 5 domains → $sp^3d$ hybridization Answer: Option D ( $sp^3d$ ).

Q118

Given below are two statements: Statement I : Wet cotton clothes made of cellulose based carbohydrate takes comparatively longer time to get dried than wet nylon polymer based clothes. Statement II : Intermolecular hydrogen bonding with water molecule is more in nylon-based clothes than in the case of cotton clothes. In the light of 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 false but Statement II is true

D Statement I is true but Statement II is false

Correct Answer

Option D

Solution

The correct choice is Option D: Statement I is true but Statement II is false.

Statement I (True) Cotton is made from cellulose, which is a kind of carbohydrate.

Cellulose has many –OH groups (hydroxyl groups) which can make strong hydrogen bonds with water.

Because of this, cotton holds more water and takes more time to dry than nylon.

Statement II (False) Nylon is made of polymer chains with only a few special groups (amide –NH and C=O) that can form hydrogen bonds with water.

Nylon forms fewer hydrogen bonds with water compared to cotton.

So, cotton actually makes more hydrogen bonds with water than nylon does.

Q119

Which of the following pair of species have the same bond order?

A CN- and NO+

B CN- and CN+

C

O_2^-

and CN-

D NO+ and CN+

Correct Answer

Option A

Solution

Number of electron in NO+ = number of electron in CN– = 14 electrons.

As both have same number of electrons so their bond order is equal.

Moleculer orbital configuration of NO+ (14 electrons) is =

{\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^0}^ * = \,\,\pi _{2p_y^0}^ *

$\therefore$ B.O =

{1 \over 2}\left[ {10 - 4} \right]

= 3 Moleculer orbital configuration of CN- (14 electrons) is =

{\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}

$\therefore$ B.O =

{1 \over 2}\left[ {10 - 4} \right]

= 3

Q120

Match List I with List II. .tg .tg LIST I LIST II A.

\mathrm{ICl}

I. T - shape B.

\mathrm{ICl}_3

II. Square pyramidal C.

\mathrm{ClF}_5

III. Pentagonal bipyramidal D.

\mathrm{IF}_7

IV. Linear Choose the correct answer from the options given below :

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

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

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

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

Correct Answer

Option B

Solution

\begin{aligned} & \mathrm{ICl} \rightarrow s p^3 \rightarrow 1 \mathrm{bp}+3 \mathrm{lp} \rightarrow \text{ Linear } \\ & \mathrm{ICl}_3 \rightarrow s p^3 d \rightarrow 3 \mathrm{bp}+2 \mathrm{lp} \rightarrow \mathrm{T} \text{-shape } \\ & \mathrm{CIF}_5 \rightarrow s p^3 d^2 \rightarrow 5 \mathrm{bp}+1 \mathrm{lp} \rightarrow \text{ Square Pyramidal } \\ & \mathrm{IF}_7 \rightarrow s p^3 d^3 \rightarrow 7 \text{ bp only } \rightarrow \text{ Pentagonal bipyramidal } \\ & \text{ (A)-(IV), (B)-(I), (C)-(II), D-(III) } \end{aligned}