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CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313752 Match the species in Column I with the bond order in Column II and choose the correct option from the codes given below:
Column I
Column II
A
$N O$
P
1.5
B
$C O$
Q
2.0
C
$O_{2}^{-}$
R
2.5
D
$O_{2}$
S
3.0

1 A - Q, B - P, C - S, D - R

2 A - P, B - Q, C - R, D - S

3 A - Q, B - S, C - P, D - Q

4 A - R, B - S, C - P, D - Q

Explanation:

A-Q, B-S, C-P, D-Q
A. $N O (7 + 8 = 15) = σ 1 s^{2}, σ^{*} 1 s^{2}; σ 2 s^{2}; σ^{*} 2 s^{2}$ ,
${σ 2 p}_{z}^{2}, {π 2 p}_{x}^{2} \approx {π 2 p}_{y}^{2}, π^{*} {2 p}_{x}^{2}$
$B o n d o r d e r = \frac{1}{2} (N_{b} - N_{a}) = \frac{10 - 5}{2} = 2 .5$
B. $C O (6 + 8 = 14) = σ 1 s^{2}, σ^{*} 1 s^{2}, σ 2 s^{2}, σ^{*} 2 s^{2}$
${σ 2 p}_{z}^{2}, {π 2 p}_{x}^{2} \approx π^{*} {2 p}_{y}^{2}$
$B o n d o r d e r = \frac{10 - 4}{2} = 3 .0$
C. $O_{2}^{-} (8 + 8 + 1 = 17) = σ 1 s^{2}, σ^{*} 1 s^{2}, σ 2 s^{2}, σ^{*} 2 s^{2}$
${σ 2 p}_{z}^{2}, {π 2 p}_{x}^{2} = {π 2 p}_{y}^{2}, π^{*} {2 p}_{x}^{2} = π^{*} {2 p}_{x}^{1}$
$B o n d o r d e r = \frac{10 - 7}{2} = 1 .5$
D. $O_{2} (8 + 8 = 16) = σ 1 s^{2}, σ^{*} 1 s^{2}, σ 2 s^{2}, σ^{*} 2 s^{2} {, σ 2 p}_{z}^{2}$
${π 2 p}_{y}^{2}, π^{*} {2 p}_{x}^{1} ≃ π^{*} {2 p}_{y}^{1}$
$B o n d o r d e r = \frac{10 - 6}{2} = 2 .0$
So, the option (4) is correct.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313753 Which of the following is paramagnetic?

1

N O^{+}

2

C O

3

O_{2}^{2 -}

4

B_{2}

Explanation:

Total electrons
$N O^{+}$ 14
$C O$ 14
$O_{2}^{2 -}$ 18
$B_{2}$ 10

JEE - 2017

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313754 How many of the following species have bond order of 2.5 ?
$\begin{array}{l} (i) {N_{2}}^{\oplus} (i i) {N_{2}}^{⊖} (i i i) {O_{2}}^{\oplus} {(i v) O}_{2}^{⊖} \\ (v) N O (v i) C N \end{array}$

1 6

2 5

3 4

4 3

Explanation:

For (i) $N_{2}^{\oplus}$ and (ii) $N_{2}^{⊖}$
For (iii) $O_{2}^{\oplus}$ and (iv) $O_{2}$
For (v) NO and (vi) CN
Hence, $N_{2}^{\oplus}, N_{2}^{⊖}, O_{2}^{\oplus}$ , NO and CN have bond order $=$ 2.5.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313755 Which of the following statement is not correct from the view point molecular orbital theory?

1

{Be}_{2}

is not a stable molecule

2

{He}_{2}

is not stable but

{He}_{2}^{+}

is expected to exist

3 Bond strength of

N_{2}

is maximum amongst the homonuclear diatomic molecules belonging to the second period

4 The order of energies of molecular orbitals in

N_{2}

molecule is

Explanation:

The correct order of energies of molecular orbitals in $N_{2}$ molecule is
$σ 1 s < σ^{*} 1 s < σ 2 s < σ^{*} 2 s < (π 2 p_{x} \approx π 2 py)$
$< σ 2 p_{z} < π^{*} 2 p_{x} = π^{*} 2 p_{y} < σ^{*} 2 p_{z}$ .

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313752 Match the species in Column I with the bond order in Column II and choose the correct option from the codes given below:
Column I
Column II
A
$N O$
P
1.5
B
$C O$
Q
2.0
C
$O_{2}^{-}$
R
2.5
D
$O_{2}$
S
3.0

1 A - Q, B - P, C - S, D - R

2 A - P, B - Q, C - R, D - S

3 A - Q, B - S, C - P, D - Q

4 A - R, B - S, C - P, D - Q

Explanation:

A-Q, B-S, C-P, D-Q
A. $N O (7 + 8 = 15) = σ 1 s^{2}, σ^{*} 1 s^{2}; σ 2 s^{2}; σ^{*} 2 s^{2}$ ,
${σ 2 p}_{z}^{2}, {π 2 p}_{x}^{2} \approx {π 2 p}_{y}^{2}, π^{*} {2 p}_{x}^{2}$
$B o n d o r d e r = \frac{1}{2} (N_{b} - N_{a}) = \frac{10 - 5}{2} = 2 .5$
B. $C O (6 + 8 = 14) = σ 1 s^{2}, σ^{*} 1 s^{2}, σ 2 s^{2}, σ^{*} 2 s^{2}$
${σ 2 p}_{z}^{2}, {π 2 p}_{x}^{2} \approx π^{*} {2 p}_{y}^{2}$
$B o n d o r d e r = \frac{10 - 4}{2} = 3 .0$
C. $O_{2}^{-} (8 + 8 + 1 = 17) = σ 1 s^{2}, σ^{*} 1 s^{2}, σ 2 s^{2}, σ^{*} 2 s^{2}$
${σ 2 p}_{z}^{2}, {π 2 p}_{x}^{2} = {π 2 p}_{y}^{2}, π^{*} {2 p}_{x}^{2} = π^{*} {2 p}_{x}^{1}$
$B o n d o r d e r = \frac{10 - 7}{2} = 1 .5$
D. $O_{2} (8 + 8 = 16) = σ 1 s^{2}, σ^{*} 1 s^{2}, σ 2 s^{2}, σ^{*} 2 s^{2} {, σ 2 p}_{z}^{2}$
${π 2 p}_{y}^{2}, π^{*} {2 p}_{x}^{1} ≃ π^{*} {2 p}_{y}^{1}$
$B o n d o r d e r = \frac{10 - 6}{2} = 2 .0$
So, the option (4) is correct.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313753 Which of the following is paramagnetic?

1

N O^{+}

2

C O

3

O_{2}^{2 -}

4

B_{2}

Explanation:

Total electrons
$N O^{+}$ 14
$C O$ 14
$O_{2}^{2 -}$ 18
$B_{2}$ 10

JEE - 2017

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313754 How many of the following species have bond order of 2.5 ?
$\begin{array}{l} (i) {N_{2}}^{\oplus} (i i) {N_{2}}^{⊖} (i i i) {O_{2}}^{\oplus} {(i v) O}_{2}^{⊖} \\ (v) N O (v i) C N \end{array}$

1 6

2 5

3 4

4 3

Explanation:

For (i) $N_{2}^{\oplus}$ and (ii) $N_{2}^{⊖}$
For (iii) $O_{2}^{\oplus}$ and (iv) $O_{2}$
For (v) NO and (vi) CN
Hence, $N_{2}^{\oplus}, N_{2}^{⊖}, O_{2}^{\oplus}$ , NO and CN have bond order $=$ 2.5.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313755 Which of the following statement is not correct from the view point molecular orbital theory?

1

{Be}_{2}

is not a stable molecule

2

{He}_{2}

is not stable but

{He}_{2}^{+}

is expected to exist

3 Bond strength of

N_{2}

is maximum amongst the homonuclear diatomic molecules belonging to the second period

4 The order of energies of molecular orbitals in

N_{2}

molecule is

Explanation:

The correct order of energies of molecular orbitals in $N_{2}$ molecule is
$σ 1 s < σ^{*} 1 s < σ 2 s < σ^{*} 2 s < (π 2 p_{x} \approx π 2 py)$
$< σ 2 p_{z} < π^{*} 2 p_{x} = π^{*} 2 p_{y} < σ^{*} 2 p_{z}$ .

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313752 Match the species in Column I with the bond order in Column II and choose the correct option from the codes given below:
Column I
Column II
A
$N O$
P
1.5
B
$C O$
Q
2.0
C
$O_{2}^{-}$
R
2.5
D
$O_{2}$
S
3.0

1 A - Q, B - P, C - S, D - R

2 A - P, B - Q, C - R, D - S

3 A - Q, B - S, C - P, D - Q

4 A - R, B - S, C - P, D - Q

Explanation:

A-Q, B-S, C-P, D-Q
A. $N O (7 + 8 = 15) = σ 1 s^{2}, σ^{*} 1 s^{2}; σ 2 s^{2}; σ^{*} 2 s^{2}$ ,
${σ 2 p}_{z}^{2}, {π 2 p}_{x}^{2} \approx {π 2 p}_{y}^{2}, π^{*} {2 p}_{x}^{2}$
$B o n d o r d e r = \frac{1}{2} (N_{b} - N_{a}) = \frac{10 - 5}{2} = 2 .5$
B. $C O (6 + 8 = 14) = σ 1 s^{2}, σ^{*} 1 s^{2}, σ 2 s^{2}, σ^{*} 2 s^{2}$
${σ 2 p}_{z}^{2}, {π 2 p}_{x}^{2} \approx π^{*} {2 p}_{y}^{2}$
$B o n d o r d e r = \frac{10 - 4}{2} = 3 .0$
C. $O_{2}^{-} (8 + 8 + 1 = 17) = σ 1 s^{2}, σ^{*} 1 s^{2}, σ 2 s^{2}, σ^{*} 2 s^{2}$
${σ 2 p}_{z}^{2}, {π 2 p}_{x}^{2} = {π 2 p}_{y}^{2}, π^{*} {2 p}_{x}^{2} = π^{*} {2 p}_{x}^{1}$
$B o n d o r d e r = \frac{10 - 7}{2} = 1 .5$
D. $O_{2} (8 + 8 = 16) = σ 1 s^{2}, σ^{*} 1 s^{2}, σ 2 s^{2}, σ^{*} 2 s^{2} {, σ 2 p}_{z}^{2}$
${π 2 p}_{y}^{2}, π^{*} {2 p}_{x}^{1} ≃ π^{*} {2 p}_{y}^{1}$
$B o n d o r d e r = \frac{10 - 6}{2} = 2 .0$
So, the option (4) is correct.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313753 Which of the following is paramagnetic?

1

N O^{+}

2

C O

3

O_{2}^{2 -}

4

B_{2}

Explanation:

Total electrons
$N O^{+}$ 14
$C O$ 14
$O_{2}^{2 -}$ 18
$B_{2}$ 10

JEE - 2017

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313754 How many of the following species have bond order of 2.5 ?
$\begin{array}{l} (i) {N_{2}}^{\oplus} (i i) {N_{2}}^{⊖} (i i i) {O_{2}}^{\oplus} {(i v) O}_{2}^{⊖} \\ (v) N O (v i) C N \end{array}$

1 6

2 5

3 4

4 3

Explanation:

For (i) $N_{2}^{\oplus}$ and (ii) $N_{2}^{⊖}$
For (iii) $O_{2}^{\oplus}$ and (iv) $O_{2}$
For (v) NO and (vi) CN
Hence, $N_{2}^{\oplus}, N_{2}^{⊖}, O_{2}^{\oplus}$ , NO and CN have bond order $=$ 2.5.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313755 Which of the following statement is not correct from the view point molecular orbital theory?

1

{Be}_{2}

is not a stable molecule

2

{He}_{2}

is not stable but

{He}_{2}^{+}

is expected to exist

3 Bond strength of

N_{2}

is maximum amongst the homonuclear diatomic molecules belonging to the second period

4 The order of energies of molecular orbitals in

N_{2}

molecule is

Explanation:

The correct order of energies of molecular orbitals in $N_{2}$ molecule is
$σ 1 s < σ^{*} 1 s < σ 2 s < σ^{*} 2 s < (π 2 p_{x} \approx π 2 py)$
$< σ 2 p_{z} < π^{*} 2 p_{x} = π^{*} 2 p_{y} < σ^{*} 2 p_{z}$ .

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313752 Match the species in Column I with the bond order in Column II and choose the correct option from the codes given below:
Column I
Column II
A
$N O$
P
1.5
B
$C O$
Q
2.0
C
$O_{2}^{-}$
R
2.5
D
$O_{2}$
S
3.0

1 A - Q, B - P, C - S, D - R

2 A - P, B - Q, C - R, D - S

3 A - Q, B - S, C - P, D - Q

4 A - R, B - S, C - P, D - Q

Explanation:

A-Q, B-S, C-P, D-Q
A. $N O (7 + 8 = 15) = σ 1 s^{2}, σ^{*} 1 s^{2}; σ 2 s^{2}; σ^{*} 2 s^{2}$ ,
${σ 2 p}_{z}^{2}, {π 2 p}_{x}^{2} \approx {π 2 p}_{y}^{2}, π^{*} {2 p}_{x}^{2}$
$B o n d o r d e r = \frac{1}{2} (N_{b} - N_{a}) = \frac{10 - 5}{2} = 2 .5$
B. $C O (6 + 8 = 14) = σ 1 s^{2}, σ^{*} 1 s^{2}, σ 2 s^{2}, σ^{*} 2 s^{2}$
${σ 2 p}_{z}^{2}, {π 2 p}_{x}^{2} \approx π^{*} {2 p}_{y}^{2}$
$B o n d o r d e r = \frac{10 - 4}{2} = 3 .0$
C. $O_{2}^{-} (8 + 8 + 1 = 17) = σ 1 s^{2}, σ^{*} 1 s^{2}, σ 2 s^{2}, σ^{*} 2 s^{2}$
${σ 2 p}_{z}^{2}, {π 2 p}_{x}^{2} = {π 2 p}_{y}^{2}, π^{*} {2 p}_{x}^{2} = π^{*} {2 p}_{x}^{1}$
$B o n d o r d e r = \frac{10 - 7}{2} = 1 .5$
D. $O_{2} (8 + 8 = 16) = σ 1 s^{2}, σ^{*} 1 s^{2}, σ 2 s^{2}, σ^{*} 2 s^{2} {, σ 2 p}_{z}^{2}$
${π 2 p}_{y}^{2}, π^{*} {2 p}_{x}^{1} ≃ π^{*} {2 p}_{y}^{1}$
$B o n d o r d e r = \frac{10 - 6}{2} = 2 .0$
So, the option (4) is correct.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313753 Which of the following is paramagnetic?

1

N O^{+}

2

C O

3

O_{2}^{2 -}

4

B_{2}

Explanation:

Total electrons
$N O^{+}$ 14
$C O$ 14
$O_{2}^{2 -}$ 18
$B_{2}$ 10

JEE - 2017

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313754 How many of the following species have bond order of 2.5 ?
$\begin{array}{l} (i) {N_{2}}^{\oplus} (i i) {N_{2}}^{⊖} (i i i) {O_{2}}^{\oplus} {(i v) O}_{2}^{⊖} \\ (v) N O (v i) C N \end{array}$

1 6

2 5

3 4

4 3

Explanation:

For (i) $N_{2}^{\oplus}$ and (ii) $N_{2}^{⊖}$
For (iii) $O_{2}^{\oplus}$ and (iv) $O_{2}$
For (v) NO and (vi) CN
Hence, $N_{2}^{\oplus}, N_{2}^{⊖}, O_{2}^{\oplus}$ , NO and CN have bond order $=$ 2.5.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313755 Which of the following statement is not correct from the view point molecular orbital theory?

1

{Be}_{2}

is not a stable molecule

2

{He}_{2}

is not stable but

{He}_{2}^{+}

is expected to exist

3 Bond strength of

N_{2}

is maximum amongst the homonuclear diatomic molecules belonging to the second period

4 The order of energies of molecular orbitals in

N_{2}

molecule is

Explanation:

The correct order of energies of molecular orbitals in $N_{2}$ molecule is
$σ 1 s < σ^{*} 1 s < σ 2 s < σ^{*} 2 s < (π 2 p_{x} \approx π 2 py)$
$< σ 2 p_{z} < π^{*} 2 p_{x} = π^{*} 2 p_{y} < σ^{*} 2 p_{z}$ .