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CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313694 About the species $O_{2}^{+}, O_{2}^{2 -}, O_{2}, O_{2}^{-}$ which one of the following statements is correct?

1

O_{2}^{+}

has the highest bond order and largest bond length

2

O_{2}

has the highest bond order.

3 The bond order of

O_{2}

equals the average bond orders of

O_{2}^{+}

and

O_{2}^{-}

.

4 The sum of bond orders of these species is 1 .

Explanation:

$\begin{matrix} O_{2} O_{2}^{+} O_{2}^{-} \\ Bond order : 2.0 2.5 1.5 \end{matrix}$
Average bond order of $O_{2}^{+}$ and $O_{2}^{-}$ = B.O of $O_{2}$ $\frac{2.5 + 1.5}{2} = 2$

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313695 Stability of the species $L i_{2} {, L i}_{2}^{-} a n d {L i}_{2}^{+}$ increases in the order of :

1

L i_{2} {< L i}_{2}^{+} {< L i}_{2}^{-}

2

{L i}_{2}^{-} {< L i}_{2}^{+} < L i_{2}

3

L i_{2} {< L i}_{2}^{-} {< L i}_{2}^{+}

4

{L i}_{2}^{-} < L i_{2} {< L i}_{2}^{+}

Explanation:

$L i_{2} = σ 1 s^{2} σ^{*} 1 s^{2} σ 2 s^{2}$
$∴ B o n d o r d e r = \frac{1}{2} (4 - 2) = 1$
${L i}_{2}^{+} = σ 1 s^{2} σ^{*} 1 s^{2} σ 2 s^{1}$
$B . O . \frac{1}{2} (3 - 2) 0 .5$
${L i}_{2}^{-} = σ 1 s^{2} σ^{*} 1 s^{2} σ 2 s^{2} σ^{*} 2 s^{1}$
$B . O . = \frac{1}{2} (4 - 3) 0 .5$
The bond order of ${L i}_{2}^{+}$ and ${L i}_{2}^{-}$ is same but ${L i}_{2}^{+}$ is more stable than ${L i}_{2}^{-}$ because ${L i}_{2}^{+}$ is smaller in size and has 2 electrons in antibonding orbitals whereas ${L i}_{2}^{-}$ has 3 electrons in antibonding orbitals. Hence ${L i}_{2}^{+}$ is more stable than ${L i}_{2}^{-}$ .

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313696 Assertion :
Fluorine molecule has bond order one.
Reason :
The number of electrons in the antibonding molecular orbitals is two less than that in bonding molecular orbitals.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

The electronic configuration of the $F_{2}$ molecule can be represented as follows:
$F_{2} : σ 1 S^{2}, σ^{*} 1 S^{2}, σ 2 S^{2}, σ^{*} 2 S^{2}, σ 2 P_{z}^{2}$ , $(π 2 P_{x}^{2} = π 2 P_{y}^{2}), π^{*} 2 P^{2} = π^{*} 2 P_{y}^{2}$
Bond Order $(BO) = ($ No. of electrons in bonding MOs - No. of electrons in antibonding MOs) / 2
No. of electrons in bonding MOs $= 10$ electrons No. of electrons in antibonding MOs $= 8$ electrons
$BO = (10 - 8) / 2 = 2 / 2 = 1$
Thus, the bond order of the $F_{2}$ molecule is 1. So the option (1) is correct.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313697 Which of the following is more stable?

1

N_{2}

2

O_{2}

3

{Li}_{2}

4 All of these

Explanation:

Bond order:
$1 2 3$
$\begin{array}{l} L i_{2} & O_{2} & N_{2} \end{array}$
As bond order of nitrogen is more as compared to other molecules. Hence, nitrogen is more stable.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313694 About the species $O_{2}^{+}, O_{2}^{2 -}, O_{2}, O_{2}^{-}$ which one of the following statements is correct?

1

O_{2}^{+}

has the highest bond order and largest bond length

2

O_{2}

has the highest bond order.

3 The bond order of

O_{2}

equals the average bond orders of

O_{2}^{+}

and

O_{2}^{-}

.

4 The sum of bond orders of these species is 1 .

Explanation:

$\begin{matrix} O_{2} O_{2}^{+} O_{2}^{-} \\ Bond order : 2.0 2.5 1.5 \end{matrix}$
Average bond order of $O_{2}^{+}$ and $O_{2}^{-}$ = B.O of $O_{2}$ $\frac{2.5 + 1.5}{2} = 2$

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313695 Stability of the species $L i_{2} {, L i}_{2}^{-} a n d {L i}_{2}^{+}$ increases in the order of :

1

L i_{2} {< L i}_{2}^{+} {< L i}_{2}^{-}

2

{L i}_{2}^{-} {< L i}_{2}^{+} < L i_{2}

3

L i_{2} {< L i}_{2}^{-} {< L i}_{2}^{+}

4

{L i}_{2}^{-} < L i_{2} {< L i}_{2}^{+}

Explanation:

$L i_{2} = σ 1 s^{2} σ^{*} 1 s^{2} σ 2 s^{2}$
$∴ B o n d o r d e r = \frac{1}{2} (4 - 2) = 1$
${L i}_{2}^{+} = σ 1 s^{2} σ^{*} 1 s^{2} σ 2 s^{1}$
$B . O . \frac{1}{2} (3 - 2) 0 .5$
${L i}_{2}^{-} = σ 1 s^{2} σ^{*} 1 s^{2} σ 2 s^{2} σ^{*} 2 s^{1}$
$B . O . = \frac{1}{2} (4 - 3) 0 .5$
The bond order of ${L i}_{2}^{+}$ and ${L i}_{2}^{-}$ is same but ${L i}_{2}^{+}$ is more stable than ${L i}_{2}^{-}$ because ${L i}_{2}^{+}$ is smaller in size and has 2 electrons in antibonding orbitals whereas ${L i}_{2}^{-}$ has 3 electrons in antibonding orbitals. Hence ${L i}_{2}^{+}$ is more stable than ${L i}_{2}^{-}$ .

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313696 Assertion :
Fluorine molecule has bond order one.
Reason :
The number of electrons in the antibonding molecular orbitals is two less than that in bonding molecular orbitals.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

The electronic configuration of the $F_{2}$ molecule can be represented as follows:
$F_{2} : σ 1 S^{2}, σ^{*} 1 S^{2}, σ 2 S^{2}, σ^{*} 2 S^{2}, σ 2 P_{z}^{2}$ , $(π 2 P_{x}^{2} = π 2 P_{y}^{2}), π^{*} 2 P^{2} = π^{*} 2 P_{y}^{2}$
Bond Order $(BO) = ($ No. of electrons in bonding MOs - No. of electrons in antibonding MOs) / 2
No. of electrons in bonding MOs $= 10$ electrons No. of electrons in antibonding MOs $= 8$ electrons
$BO = (10 - 8) / 2 = 2 / 2 = 1$
Thus, the bond order of the $F_{2}$ molecule is 1. So the option (1) is correct.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313697 Which of the following is more stable?

1

N_{2}

2

O_{2}

3

{Li}_{2}

4 All of these

Explanation:

Bond order:
$1 2 3$
$\begin{array}{l} L i_{2} & O_{2} & N_{2} \end{array}$
As bond order of nitrogen is more as compared to other molecules. Hence, nitrogen is more stable.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313694 About the species $O_{2}^{+}, O_{2}^{2 -}, O_{2}, O_{2}^{-}$ which one of the following statements is correct?

1

O_{2}^{+}

has the highest bond order and largest bond length

2

O_{2}

has the highest bond order.

3 The bond order of

O_{2}

equals the average bond orders of

O_{2}^{+}

and

O_{2}^{-}

.

4 The sum of bond orders of these species is 1 .

Explanation:

$\begin{matrix} O_{2} O_{2}^{+} O_{2}^{-} \\ Bond order : 2.0 2.5 1.5 \end{matrix}$
Average bond order of $O_{2}^{+}$ and $O_{2}^{-}$ = B.O of $O_{2}$ $\frac{2.5 + 1.5}{2} = 2$

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313695 Stability of the species $L i_{2} {, L i}_{2}^{-} a n d {L i}_{2}^{+}$ increases in the order of :

1

L i_{2} {< L i}_{2}^{+} {< L i}_{2}^{-}

2

{L i}_{2}^{-} {< L i}_{2}^{+} < L i_{2}

3

L i_{2} {< L i}_{2}^{-} {< L i}_{2}^{+}

4

{L i}_{2}^{-} < L i_{2} {< L i}_{2}^{+}

Explanation:

$L i_{2} = σ 1 s^{2} σ^{*} 1 s^{2} σ 2 s^{2}$
$∴ B o n d o r d e r = \frac{1}{2} (4 - 2) = 1$
${L i}_{2}^{+} = σ 1 s^{2} σ^{*} 1 s^{2} σ 2 s^{1}$
$B . O . \frac{1}{2} (3 - 2) 0 .5$
${L i}_{2}^{-} = σ 1 s^{2} σ^{*} 1 s^{2} σ 2 s^{2} σ^{*} 2 s^{1}$
$B . O . = \frac{1}{2} (4 - 3) 0 .5$
The bond order of ${L i}_{2}^{+}$ and ${L i}_{2}^{-}$ is same but ${L i}_{2}^{+}$ is more stable than ${L i}_{2}^{-}$ because ${L i}_{2}^{+}$ is smaller in size and has 2 electrons in antibonding orbitals whereas ${L i}_{2}^{-}$ has 3 electrons in antibonding orbitals. Hence ${L i}_{2}^{+}$ is more stable than ${L i}_{2}^{-}$ .

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313696 Assertion :
Fluorine molecule has bond order one.
Reason :
The number of electrons in the antibonding molecular orbitals is two less than that in bonding molecular orbitals.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

The electronic configuration of the $F_{2}$ molecule can be represented as follows:
$F_{2} : σ 1 S^{2}, σ^{*} 1 S^{2}, σ 2 S^{2}, σ^{*} 2 S^{2}, σ 2 P_{z}^{2}$ , $(π 2 P_{x}^{2} = π 2 P_{y}^{2}), π^{*} 2 P^{2} = π^{*} 2 P_{y}^{2}$
Bond Order $(BO) = ($ No. of electrons in bonding MOs - No. of electrons in antibonding MOs) / 2
No. of electrons in bonding MOs $= 10$ electrons No. of electrons in antibonding MOs $= 8$ electrons
$BO = (10 - 8) / 2 = 2 / 2 = 1$
Thus, the bond order of the $F_{2}$ molecule is 1. So the option (1) is correct.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313697 Which of the following is more stable?

1

N_{2}

2

O_{2}

3

{Li}_{2}

4 All of these

Explanation:

Bond order:
$1 2 3$
$\begin{array}{l} L i_{2} & O_{2} & N_{2} \end{array}$
As bond order of nitrogen is more as compared to other molecules. Hence, nitrogen is more stable.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313694 About the species $O_{2}^{+}, O_{2}^{2 -}, O_{2}, O_{2}^{-}$ which one of the following statements is correct?

1

O_{2}^{+}

has the highest bond order and largest bond length

2

O_{2}

has the highest bond order.

3 The bond order of

O_{2}

equals the average bond orders of

O_{2}^{+}

and

O_{2}^{-}

.

4 The sum of bond orders of these species is 1 .

Explanation:

$\begin{matrix} O_{2} O_{2}^{+} O_{2}^{-} \\ Bond order : 2.0 2.5 1.5 \end{matrix}$
Average bond order of $O_{2}^{+}$ and $O_{2}^{-}$ = B.O of $O_{2}$ $\frac{2.5 + 1.5}{2} = 2$

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313695 Stability of the species $L i_{2} {, L i}_{2}^{-} a n d {L i}_{2}^{+}$ increases in the order of :

1

L i_{2} {< L i}_{2}^{+} {< L i}_{2}^{-}

2

{L i}_{2}^{-} {< L i}_{2}^{+} < L i_{2}

3

L i_{2} {< L i}_{2}^{-} {< L i}_{2}^{+}

4

{L i}_{2}^{-} < L i_{2} {< L i}_{2}^{+}

Explanation:

$L i_{2} = σ 1 s^{2} σ^{*} 1 s^{2} σ 2 s^{2}$
$∴ B o n d o r d e r = \frac{1}{2} (4 - 2) = 1$
${L i}_{2}^{+} = σ 1 s^{2} σ^{*} 1 s^{2} σ 2 s^{1}$
$B . O . \frac{1}{2} (3 - 2) 0 .5$
${L i}_{2}^{-} = σ 1 s^{2} σ^{*} 1 s^{2} σ 2 s^{2} σ^{*} 2 s^{1}$
$B . O . = \frac{1}{2} (4 - 3) 0 .5$
The bond order of ${L i}_{2}^{+}$ and ${L i}_{2}^{-}$ is same but ${L i}_{2}^{+}$ is more stable than ${L i}_{2}^{-}$ because ${L i}_{2}^{+}$ is smaller in size and has 2 electrons in antibonding orbitals whereas ${L i}_{2}^{-}$ has 3 electrons in antibonding orbitals. Hence ${L i}_{2}^{+}$ is more stable than ${L i}_{2}^{-}$ .

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313696 Assertion :
Fluorine molecule has bond order one.
Reason :
The number of electrons in the antibonding molecular orbitals is two less than that in bonding molecular orbitals.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

The electronic configuration of the $F_{2}$ molecule can be represented as follows:
$F_{2} : σ 1 S^{2}, σ^{*} 1 S^{2}, σ 2 S^{2}, σ^{*} 2 S^{2}, σ 2 P_{z}^{2}$ , $(π 2 P_{x}^{2} = π 2 P_{y}^{2}), π^{*} 2 P^{2} = π^{*} 2 P_{y}^{2}$
Bond Order $(BO) = ($ No. of electrons in bonding MOs - No. of electrons in antibonding MOs) / 2
No. of electrons in bonding MOs $= 10$ electrons No. of electrons in antibonding MOs $= 8$ electrons
$BO = (10 - 8) / 2 = 2 / 2 = 1$
Thus, the bond order of the $F_{2}$ molecule is 1. So the option (1) is correct.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313697 Which of the following is more stable?

1

N_{2}

2

O_{2}

3

{Li}_{2}

4 All of these

Explanation:

Bond order:
$1 2 3$
$\begin{array}{l} L i_{2} & O_{2} & N_{2} \end{array}$
As bond order of nitrogen is more as compared to other molecules. Hence, nitrogen is more stable.

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