Chemical Bonding & Molecular Structure — Page 13 | JEE Chemistry

Q121

If the magnetic moment of a dioxygen species is 1.73 B.M, it may be :

A

O_2^ -

or

O_2^ +

B O2,

O_2^ -

or

O_2^ +

C O2 or

O_2^ +

D O2 or

O_2^ -

Correct Answer

Option A

Solution

Magnetic moment = 1.73 BM $\therefore$ Unpaired electron = 1 (1)

O_2

has 16 electrons. Moleculer orbital configuration of

O_2

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^1}^ * = \,\,\pi _{2p_y^1}^ *

Here 2 unpaired electron present. (2)

O_2^{+}

has 15 electrons. Moleculer orbital configuration of

O_2^{+}

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^1}^ * = \,\,\pi _{2p_y^0}^ *

Here 1 unpaired electron present. (3)

O_2^{-}

has 17 electrons. Moleculer orbital configuration of

O_2^{-}

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^2}^ * = \,\,\pi _{2p_y^1}^ *

Here 1 unpaired electron present. Hence

O_2^{-}

&

O_2^{+}

have one unpaired electron.

Q122

Which among the following species has unequal bond lengths?

A XeF4

B BF

_4^ -

C SiF4

D SF4

Correct Answer

Option D

Solution

(a) XeF4 is sp3d2 hybridised with 4 bond pairs and 1 lone pair and structure is square planar.

Here all the bond lengths are equal. (b)

\,\,\,\,

BF

_4^ -

, 4 bond pair present so angle is 109o 28' and sp3 hybridised.

So structure is regular tetrahedral.

Here all the bond lengths are equal. (c) SiF4 is sp3 hybridisation and regular tetrahedral geometry.

Here all the bond lengths are equal. (d) SF4 is sp3d hybridised with 4 bond pairs and 1 lone pair and its expected trigonal bipyramidal geometry gets distorted due to presence of a lone pair of electrons and it becomes distorted tetrahedral or see-saw with the bond angles equal to < 120o and 179o instead of the expected angles of 120o and 180o respectively.

Here axial and equitorial both bonds are presents.

And we know axial bonds are longer and weaker.

Q123

Match the type of interaction in column A with the distance dependence of their interaction energy in column B $$

List - I		List - II
(b)	${1 \over {{r^2}}}$	(ii)	dipole-dipole
(c)	${1 \over {{r^3}}}$	(iii)	London dispersion
(d)	${1 \over {{r^6}}}$	()

A (I)-(a), (II)-(b), (III)-(d)

B (I)-(b), (II)-(d), (III)-(c)

C (I)-(a), (II)-(b), (III)-(c)

D (I)-(a), (II)-(c), (III)-(d)

Correct Answer

Option D

Solution

Ion-ion interaction energy $\propto$

{1 \over r}

Dipole-dipole interaction energy $\propto$

{1 \over {{r^3}}}

London dispersion $\propto$

{1 \over {{r^6}}}

Q124

Match with : .tg .tg (molecule) (hybridization ; shape)

List - I		List - II
(A)	XeO $_3$	(I)	sp $^3$ d ; linear
(B)	XeF $_2$	(II)	sp $^3$ ; pyramidal
(C)	XeOF $_4$	(III)	sp $^3$ d $^3$ ; distorted octahedral
(D)	XeF $_6$	(IV)	sp $^3$ d $^2$ ; square pyramidal

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

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

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

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

Correct Answer

Option A

Solution

\mathrm{XeO}_{3}-s p^{3}

, Pyramidal

\mathrm{XeF}_{2}-s p^{3} d

, linear

\mathrm{XeOF}_{4}-s p^{3} d^{2}

, Square Pyramidal

\mathrm{XeF}_{6}-s p^{3} d^{3}

, distorted octahedral

Q125

Match with : $$

List - I		List - II
(C)	O $_3$	(II)	square pyramidal
(D)	PCl $_5$	(III)	trigonal bipyramidal
()		(IV)	octahedral

A

(\mathrm{A})-(\mathrm{I}),(\mathrm{B})-(\mathrm{II}),(\mathrm{C})-(\mathrm{III}),(\mathrm{D})-(\mathrm{IV})

B

(\mathrm{A})-(\mathrm{IV}),(\mathrm{B})-(\mathrm{III}),(\mathrm{C})-(\mathrm{II}),(\mathrm{D})-(\mathrm{I})

C

(\mathrm{A})-(\mathrm{II}),(\mathrm{B})-(\mathrm{IV}),(\mathrm{C})-(\mathrm{I}),(\mathrm{D})-(\mathrm{III})

D

(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{IV}),(\mathrm{C})-(\mathrm{II}),(\mathrm{D})-(\mathrm{I})

Correct Answer

Option C

Solution

(A)

\mathrm{BrF}_{5}-

square pyramidal (B)

\left[\mathrm{CrF}_{6}\right]^{3-}-

octahedral (C)

\mathrm{O}_{3}-

bent (D)

\mathrm{PCl}_{5}

- trigonal bipyramidal

Q126

Match with . }=\psi_{\mathrm{A}}-\psi_{\mathrm{B}}$$

List - I		List - II
(B)	$\mu=Q \times r$	(I)	Dipole moment
(C)	$\dfrac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}$	(II)	Bonding molecular orbital
(D)	$\psi_{\mathrm{MO}}=\psi_{\mathrm{A}}+\psi_{\mathrm{B}}$	(III)	Anti-bonding molecular orbital
()		(IV)	Bond order

A

(\mathrm{A})-(\mathrm{II}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{IV}),(\mathrm{D})-(\mathrm{III})

B

(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{IV}),(\mathrm{C})-(\mathrm{I}),(\mathrm{D})-(\mathrm{II})

C

(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{IV}),(\mathrm{D})-(\mathrm{II})

D

(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{IV}),(\mathrm{C})-(\mathrm{II}),(\mathrm{D})-(\mathrm{I})

Correct Answer

Option C

Solution

$\Psi_{A}-\Psi_{B}=\Psi_{M O}$ is anti-boding molecular orbital

\begin{aligned} &\mu=Q \times r \text{ is dipole moment } \\\\ &\frac{N_{b}-N_{a}}{2}=\text{ bond order } \\\\ &\Psi_{\mathrm{A}}+\Psi_{\mathrm{B}}=\Psi_{\mathrm{MO}} \text{ is boding molecular orbital. } \end{aligned}

Q127

The states of hybridization of boron and oxygen atoms in boric acid (H3BO3) are respectively

A sp2 and sp2

B sp3 and sp3

C sp3 and sp2

D sp2 and sp3

Correct Answer

Option D

Solution

Structure of H3BO3 is Here Boron has 3 sigma bond.

So it is sp2 hybridised.

And oxygen has two sigma bond and two lone pair.

So it sp3 hybridised.

Q128

Match List I with List II .tg .tg List I (molecules/ions) List II (No. of lone pairs of e

^-

on central atom) A.

\mathrm{IF_7}

I. Three B.

\mathrm{ICl}

_4^ -

II. One C.

\mathrm{XeF_6}

III. Two D.

\mathrm{XeF_2}

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

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

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

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

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

Correct Answer

Option C

Solution

(A) IF

_7

$-$ 0 lone pairs (B) ICI

_4^ -

$-$ 2 lone pairs (C) XeF

_6

$-$ 1 lone pair (D) XeF

_2

$-$ 3 lone pairs

Q129

Match List - I with List - II: .tg .tg List - I Species List - II Geometry/Shape A.

\mathrm{H_3O^+}

I. Tetrahedral B. Acetylide anion II. Linear C.

\mathrm{NH_4^+}

III. Pyramidal D.

\mathrm{ClO_2^-}

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

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

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

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

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

Correct Answer

Option B

Solution

Let's consider each species in List I and determine their geometries/shapes.

A.

$\mathrm{H_3O^+}$ : This ion is formed by the addition of a proton to a water molecule.

The central atom (O) is surrounded by three Hydrogen atoms and one lone pair, making its shape pyramidal.

B.

Acetylide anion: The acetylide ion ( $\mathrm{C_2H^-}$ or $\mathrm{C_2^{2-}}$ ) has a linear geometry as the three atoms are in a straight line.

C.

$\mathrm{NH_4^+}$ : The ammonium ion has a central nitrogen atom surrounded by four hydrogen atoms, giving it a tetrahedral geometry.

D.

$\mathrm{ClO_2^-}$ : The chlorite ion has a central chlorine atom surrounded by two oxygen atoms and one lone pair, which gives it a bent or V-shaped geometry.

Matching these to List II, we get: A.

$\mathrm{H_3O^+}$ - III.

Pyramidal B.

Acetylide anion - II.

Linear C.

$\mathrm{NH_4^+}$ - I.

Tetrahedral D.

$\mathrm{ClO_2^-}$ - IV.

Bent

Q130

Match List I with List II .tg .tg LIST I (Molecule) LIST II (Shape) A.

\mathrm{NH_3}

I. Square pyramid B.

\mathrm{BrF_5}

II. Tetrahedral C.

\mathrm{PCl_5}

III. Trigonal pyramidal D.

\mathrm{CH_4}

IV. Trigonal bipyramidal Choose the correct answer from the options given below:

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

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

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

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

Correct Answer

Option B

Solution

To match the molecules in List I with their corresponding shapes in List II, we need to consider the electronic geometry and the VSEPR (Valence Shell Electron Pair Repulsion) theory.

Here’s the detailed analysis:

\mathrm{NH_3}

(Ammonia): Ammonia has a central nitrogen atom bonded to three hydrogen atoms and has one lone pair of electrons.

This leads to a trigonal pyramidal shape.

\mathrm{BrF_5}

(Bromine pentafluoride): Bromine pentafluoride has a central bromine atom bonded to five fluorine atoms and has one lone pair of electrons.

This leads to a square pyramidal shape.

\mathrm{PCl_5}

(Phosphorus pentachloride): Phosphorus pentachloride has a central phosphorus atom bonded to five chlorine atoms and no lone pairs.

This results in a trigonal bipyramidal shape.

\mathrm{CH_4}

(Methane): Methane has a central carbon atom bonded to four hydrogen atoms with no lone pairs, forming a tetrahedral structure.

Using the above analyses, we can match the molecules and shapes as follows: A (

\mathrm{NH_3}

) - III (Trigonal pyramidal) B (

\mathrm{BrF_5}

) - I (Square pyramid) C (

\mathrm{PCl_5}

) - IV (Trigonal bipyramidal) D (

\mathrm{CH_4}

) - II (Tetrahedral) Therefore, the correct matching is option B: Option B: A-III, B-I, C-IV, D-II