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Chemical Bonding & Molecular Structure

JEE Chemistry · 150 questions · Page 8 of 15 · Click an option or "Show Solution" to reveal answer

Q71
A central atom in a molecule has two lone pairs of electrons and forms three single bonds. The shape of this molecule is :
A trigonal pyramidal
B T-shaped
C see-saw
D planar triangular
Correct Answer
Option B
Solution

sp3d hybridised T-shaped

Q72
In the following the correct bond order sequence is :
A O22−>O2+>O2−>O2O_2^{2 - } > O_2^ + > O_2^ - > {O_2}
B O2+>O2−>O22−>O2O_2^ + > O_2^ - > O_2^{2 - } > {O_2}
C O2+>O2>O2−>O22−O_2^ + > {O_2} > O_2^ - > O_2^{2 - }
D O2>O2−>O22−>O2+{O_2} > O_2^ - > O_2^{2 - } > O_2^ +
Correct Answer
Option C
Solution

Note : (1)

    \,\,\,\,

Bond strength ∝\propto Bond order (2)

    \,\,\,\,

Bond length ∝\propto

1Bond  order{1 \over {Bond\,\,order}}

(3)

 \,

Bond order

=12= {1 \over 2}

[Nb −- Na] Nb = No of electrons in bonding molecular orbital Na == No of electrons in anti bonding molecular orbital (4)

    \,\,\,\,

upto 14 electrons, molecular orbital configuration is Here Na = Anti bonding electron == 4 and Nb = 10 (5)

    \,\,\,\,

After 14 electrons to 20 electrons molecular orbital configuration is - - - Here Na = 10 and Nb = 10 In O atom 8 electrons present, so in O2, 8 ×\times 2 = 16 electrons present.

Then in

O2+O_2^ +

no of electrons = 15 in

O2−O_2^ -

no of electrons = 17 in

O22−O_2^{2 - }

no of electrons = 18

∴    \therefore\,\,\,\,

Molecular orbital configuration of O2 (16 electrons) is

σ1s2  σ1s2∗ {\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,
σ2s2  σ2s2∗ {\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,
σ2pz2  π2px2=π2py2  π2px1∗  =π2py1∗{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ *
∴    \therefore\,\,\,\,

Na = 6 Nb = 10

∴    \therefore\,\,\,\,

BO =

12[10−6]=2{1 \over 2}\left[ {10 - 6} \right] = 2

Molecular orbital configuration of O

2+_2^ +

(15 electrons) is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 π2px1∗ = π2pyo∗{\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^o}^ *
∴    \therefore\,\,\,\,

Nb = 10 Na = 5

∴    \therefore\,\,\,\,

BO =

12[10−5]{1 \over 2}\left[ {10 - 5} \right]

= 2.5 Molecular orbital configuration of

O2−O_2^ -

(17 electrons) is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 π2px2∗ = π2py1∗{\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}^ *
∴    \therefore\,\,\,\,

Nb = 10 Na = 7

∴    \therefore\,\,\,\,

BO =

12[10−7]{1 \over 2}\left[ {10 - 7} \right]

= 1.5 Molecular orbital configuration of O

22−_2^{2 - }

(18 electrons) is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 π2px2∗ = π2py2∗{\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^2}^ *
∴    \therefore\,\,\,\,

Nb = 10 Na = 8

∴    \therefore\,\,\,\,

BO =

12{1 \over 2}

[ 10 −- 8] = 1 As Bond strength ∝\propto Bond order so, correct order is

O22−<O2−<O2<O2+O_2^{2 - } < O_2^ - < {O_2} < O_2^ +
Q73
Number of paramagnetic oxides among the following given oxides is ____________. Li2O, CaO, Na2O2, KO2, MgO and K2O
A 1
B 2
C 3
D 0
Correct Answer
Option A
Solution

Li2O ⇒\Rightarrow 2Li+ O2−- CaO ⇒\Rightarrow Ca2+ O2−- Na2O2 ⇒\Rightarrow 2Na+ O

22−_2^{2 - }

KO2 ⇒\Rightarrow O

2−_2^{- }

MgO ⇒\Rightarrow Mg2+ O2−- K2O ⇒\Rightarrow 2K+ O2−- O2−- ⇒\Rightarrow Complete octet, diamagnetic

O22−O_2^{2−}

has 18 electrons. Moleculer orbital configuration of

O22−O_2^{2−}

is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 π2px2∗ = π2py2∗{\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^2}^ *

Here is no unpaired electron so it is diamagnetic.

O2−O_2^{−}

has 17 electrons. Moleculer orbital configuration of

O2−O_2^{−}

is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 π2px2∗ = π2py1∗{\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 is 1 unpaired electron so it is paramagnetic.

Q74
Consider the species CH4, NH4+_4^ + and BH4−_4^ - . Choose the correct option with respect to the these species.
A They are isoelectronic and only two have tetrahedral structures.
B They are isoelectronic and all have tetrahedral structures.
C Only two are isoelectronic and all have tetrahedral structures.
D Only two are isoelectronic and only two have tetrahedral structures.
Correct Answer
Option B
Solution

All are tetrahedral and each have 10 electrons.

Q75
Number of lone pair(s) of electrons on central atom and the shape BrF3 molecule respectively, are
A 0, triangular planar.
B 1, pyramidal.
C 2, bent T-shape.
D 1, bent T-shape.
Correct Answer
Option C
Solution

Steric no. = 5 (sp3d), lone pair = 2 Bent T shape.

Q76
In the structure of SF4, the lone pair of electrons on S is in.
A equatorial position and there are two lone pair - bond pair repulsions at 90∘^\circ.
B equatorial position and there are three lone pair - bond pair repulsions at 90∘^\circ.
C axial position and there are three lone pair - bond pair repulsion at 90∘^\circ.
D axial position and there are two lone pair - bond pair repulsion at 90∘^\circ.
Correct Answer
Option A
Solution

SF4 → sp3d hybridisation.

The lone pair of electrons on S is in an equatorial position and there are two lone pair-bond pair repulsions at 90°.

Q77
Consider the ions/molecule O2+_2^ + , O2, O2−_2^ - , O22−_2^ {2-} For increasing bond order the correct option is :
A O22−_2^ {2-} 2 < O2+_2^ +
B O2−_2^ - 2 < O2+_2^ +
C O2−_2^ - 2
D O2−_2^ - 2
Correct Answer
Option A
Solution

Note : (1)

    \,\,\,\,

Bond strength ∝\propto Bond order (2)

    \,\,\,\,

Bond length ∝\propto

1Bond  order{1 \over {Bond\,\,order}}

(3)

 \,

Bond order

=12= {1 \over 2}

[Nb −- Na] Nb = Number of electrons in bonding molecular orbital Na == Number of electrons in anti bonding molecular orbital (4)

    \,\,\,\,

upto 14 electrons, molecular orbital configuration is Here Na = Anti bonding electron == 4 and Nb = 10 (5)

    \,\,\,\,

After 14 electrons to 20 electrons molecular orbital configuration is - - - Here Na = 10 and Nb = 10 In O atom 8 electrons present, so in O2, 8 ×\times 2 = 16 electrons present.

Then in

O2+O_2^ +

no of electrons = 15 in

O2−O_2^ -

no of electrons = 17 in

O22−O_2^{2 - }

no of electrons = 18

∴    \therefore\,\,\,\,

Molecular orbital configuration of O2 (16 electrons) is

σ1s2  σ1s2∗ {\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,
σ2s2  σ2s2∗ {\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,
σ2pz2  π2px2=π2py2  π2px1∗  =π2py1∗{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ *
∴    \therefore\,\,\,\,

Na = 6 Nb = 10

∴    \therefore\,\,\,\,

BO =

12[10−6]=2{1 \over 2}\left[ {10 - 6} \right] = 2

Molecular orbital configuration of O

2+_2^ +

(15 electrons) is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 π2px1∗ = π2pyo∗{\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^o}^ *
∴    \therefore\,\,\,\,

Nb = 10 Na = 5

∴    \therefore\,\,\,\,

BO =

12[10−5]{1 \over 2}\left[ {10 - 5} \right]

= 2.5 Molecular orbital configuration of

O2−O_2^ -

(17 electrons) is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 π2px2∗ = π2py1∗{\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}^ *
∴    \therefore\,\,\,\,

Nb = 10 Na = 7

∴    \therefore\,\,\,\,

BO =

12[10−7]{1 \over 2}\left[ {10 - 7} \right]

= 1.5 Molecular orbital configuration of O

22−_2^{2 - }

(18 electrons) is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 π2px2∗ = π2py2∗{\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^2}^ *
∴    \therefore\,\,\,\,

Nb = 10 Na = 8

∴    \therefore\,\,\,\,

BO =

12{1 \over 2}

[ 10 −- 8] = 1 So, correct order of Bond order is

O22−<O2−<O2<O2+O_2^{2 - } < O_2^ - < {O_2} < O_2^ +
Q78
Bonding in which of the following diatomic molecule(s) become(s) stronger, on the basis of MO Theory, by removal of an electron? (A) NO (B) N2 (C) O2 (D) C2 (E) B2 Choose the most appropriate answer from the options given below :
A (A), (B), (C) only
B (B), (C), (E) only
C (A), (C) only
D (D) only
Correct Answer
Option C
Solution

If an electron is removed from the anti-bonding orbital, then it will tend to increase the bond order.

The HOMO in NO and O2 is antibonding molecular orbital .

Hence, in NO and O2 bond order will increase on loss of electron.

Q79
Number of electron deficient molecules among the following PH3, B2H6, CCl4, NH3, LiH and BCl3 is
A 0
B 1
C 2
D 3
Correct Answer
Option C
Solution

Electron deficient species have less than 8 electrons (or two electrons for

H\mathrm{H}

) in their valence (incomplete octet).

B2H6,BCl3\mathrm{B}_{2} \mathrm{H}_{6}, \mathrm{BCl}_{3}

have incomplete octet.

Q80
The correct order of bond orders of C22−{C_2}^{2 - }, N22−{N_2}^{2 - } and O22−{O_2}^{2 - } is, respectively
A C22−{C_2}^{2 - } < N22−{N_2}^{2 - } < O22−{O_2}^{2 - }
B O22−{O_2}^{2 - } < N22−{N_2}^{2 - } < C22−{C_2}^{2 - }
C C22−{C_2}^{2 - } < O22−{O_2}^{2 - } < N22−{N_2}^{2 - }
D N22−{N_2}^{2 - } < C22−{C_2}^{2 - } < O22−{O_2}^{2 - }
Correct Answer
Option B
Solution

Note : (1)

 \,

Bond order

=12= {1 \over 2}

[Nb −- Na] Nb = No of electrons in bonding molecular orbital Na == No of electrons in anti bonding molecular orbital (2)

    \,\,\,\,

upto 14 electrons, molecular orbital configuration is Here Na = Anti bonding electron == 4 and Nb = 10 (3)

    \,\,\,\,

After 14 electrons to 20 electrons molecular orbital configuration is - - - Here Na = 10 and Nb = 10 In O atom 8 electrons present, so in O2, 8 ×\times 2 = 16 electrons present. in

O22−O_2^{2 - }

no of electrons = 18 (A) Molecular orbital configuration of O

22−_2^{2 - }

(18 electrons) is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 π2px2∗ = π2py2∗{\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^2}^ *
∴    \therefore\,\,\,\,

Nb = 10 Na = 8

∴    \therefore\,\,\,\,

BO =

12{1 \over 2}

[ 10 −- 8] = 1 (B)

C22−C_2^{2 - }

has 14 electrons. Moleculer orbital configuration of

C22−C_2^{2 - }

is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ π2px2 = π2py2 σ2pz2{\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\,\,\,\,

Na = 4 Nb = 10

∴    \therefore\,\,\,\,

BO =

12[10−4]=3{1 \over 2}\left[ {10 - 4} \right] = 3

(C)

N22−N_2^{2 - }

has 16 electrons. Moleculer orbital configuration of

N22−N_2^{2 - }

is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ π2px2 = π2py2 σ2pz2 π2px1∗ = π2py1∗{\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}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^1}^ *
∴    \therefore\,\,\,\,

Na = 6 Nb = 10

∴    \therefore\,\,\,\,

BO =

12[10−6]=2{1 \over 2}\left[ {10 - 6} \right] = 2

The correct order of bond orders of

C22−{C_2}^{2 - }

,

N22−{N_2}^{2 - }

and

O22−{O_2}^{2 - }
O22−{O_2}^{2 - }

<

N22−{N_2}^{2 - }

<

C22−{C_2}^{2 - }
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