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CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313765 The bond order and magnetic property of acetylide ion are same as that of

1

O_{2}^{+}

2

N O^{+}

3

O_{2}^{-}

4

N_{2}^{+}

Explanation:

$C_{2}^{- 2} - 14 e^{-} - B . O = 3$ ; Diamagnetic
$N_{2}^{+} - 13 e^{-} -$ B.O. $= 2.5$ ; Paramagnetic
${NO}^{+} - 14 e^{-} -$ B.O. $= 3$ ; Diamagnetic
$O_{2}^{-} - 17 e^{-}$ – B.O. $= 1.5$ ; Paramagnetic
$O_{2}^{+} - 15 e^{-} -$ B.O. $= 2.5$ ; Paramagnetic
So, correct option is (2).

JEE - 2023

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313766 Assertion :
$B_{2}$ molecule is diamagnetic
Reason :
The highest occupied molecular orbital is of $σ$ type.

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 Both Assertion and Reason are incorrect

Explanation:

The total number of electrons in $B_{2} = 10$
Molecular orbital configuration:
$σ 1 s^{2}, \overset{*}{σ} 1 s^{2}, σ 2 s^{2}, \overset{*}{σ} 2 s^{2}, π 2 p_{x}^{1} \approx π 2 p_{y}^{1}$
It has two unpaired electrons.
$∴$ It is paramagnetic and the highest occupied molecular orbital is of $π$ type.
Hence, both $A$ and $R$ are incorrect

AIIMS - 2005

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313768 In which of the following ionisation process the bond energy has increased and also the magentic behaviour has changed from paramagnetic to diamagnetic?

1

N O \to N O^{+}

2

N_{2} \to N_{2}^{+}

3

C_{2} \to C_{2}^{+}

4

O_{2} \to O_{2}^{+}

Explanation:

For NO
Total no. of electrons = 15
B.O = 2.5
Magnetic Behaviour = Paramagnetic
For $N O^{+}$
Total no. of electrons = 14
B.O = 3
Magnetic Behaviour = Diamagnetic

JEE - 2013

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313769 According to molecular orbital theory, which of the following will not be a viable molecule?

1

{H e}_{2}^{+}

2

H_{2}^{-}

3

H_{2}^{2 -}

4

{H e}_{2}^{2 +}

Explanation:

The electronic configuration of $H_{2}^{2 -}$
$\Rightarrow (σ 1 s^{2}), {(σ^{*} 1 s)}^{2}$
Bond order of $H_{2}^{2 -} = \frac{N_{b} - N_{a}}{2} = \frac{2 - 2}{2} = 0$
Hence $H_{2}^{2 -}$ does not exist

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313765 The bond order and magnetic property of acetylide ion are same as that of

1

O_{2}^{+}

2

N O^{+}

3

O_{2}^{-}

4

N_{2}^{+}

Explanation:

$C_{2}^{- 2} - 14 e^{-} - B . O = 3$ ; Diamagnetic
$N_{2}^{+} - 13 e^{-} -$ B.O. $= 2.5$ ; Paramagnetic
${NO}^{+} - 14 e^{-} -$ B.O. $= 3$ ; Diamagnetic
$O_{2}^{-} - 17 e^{-}$ – B.O. $= 1.5$ ; Paramagnetic
$O_{2}^{+} - 15 e^{-} -$ B.O. $= 2.5$ ; Paramagnetic
So, correct option is (2).

JEE - 2023

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313766 Assertion :
$B_{2}$ molecule is diamagnetic
Reason :
The highest occupied molecular orbital is of $σ$ type.

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 Both Assertion and Reason are incorrect

Explanation:

The total number of electrons in $B_{2} = 10$
Molecular orbital configuration:
$σ 1 s^{2}, \overset{*}{σ} 1 s^{2}, σ 2 s^{2}, \overset{*}{σ} 2 s^{2}, π 2 p_{x}^{1} \approx π 2 p_{y}^{1}$
It has two unpaired electrons.
$∴$ It is paramagnetic and the highest occupied molecular orbital is of $π$ type.
Hence, both $A$ and $R$ are incorrect

AIIMS - 2005

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313768 In which of the following ionisation process the bond energy has increased and also the magentic behaviour has changed from paramagnetic to diamagnetic?

1

N O \to N O^{+}

2

N_{2} \to N_{2}^{+}

3

C_{2} \to C_{2}^{+}

4

O_{2} \to O_{2}^{+}

Explanation:

For NO
Total no. of electrons = 15
B.O = 2.5
Magnetic Behaviour = Paramagnetic
For $N O^{+}$
Total no. of electrons = 14
B.O = 3
Magnetic Behaviour = Diamagnetic

JEE - 2013

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313769 According to molecular orbital theory, which of the following will not be a viable molecule?

1

{H e}_{2}^{+}

2

H_{2}^{-}

3

H_{2}^{2 -}

4

{H e}_{2}^{2 +}

Explanation:

The electronic configuration of $H_{2}^{2 -}$
$\Rightarrow (σ 1 s^{2}), {(σ^{*} 1 s)}^{2}$
Bond order of $H_{2}^{2 -} = \frac{N_{b} - N_{a}}{2} = \frac{2 - 2}{2} = 0$
Hence $H_{2}^{2 -}$ does not exist

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313765 The bond order and magnetic property of acetylide ion are same as that of

1

O_{2}^{+}

2

N O^{+}

3

O_{2}^{-}

4

N_{2}^{+}

Explanation:

$C_{2}^{- 2} - 14 e^{-} - B . O = 3$ ; Diamagnetic
$N_{2}^{+} - 13 e^{-} -$ B.O. $= 2.5$ ; Paramagnetic
${NO}^{+} - 14 e^{-} -$ B.O. $= 3$ ; Diamagnetic
$O_{2}^{-} - 17 e^{-}$ – B.O. $= 1.5$ ; Paramagnetic
$O_{2}^{+} - 15 e^{-} -$ B.O. $= 2.5$ ; Paramagnetic
So, correct option is (2).

JEE - 2023

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313766 Assertion :
$B_{2}$ molecule is diamagnetic
Reason :
The highest occupied molecular orbital is of $σ$ type.

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 Both Assertion and Reason are incorrect

Explanation:

The total number of electrons in $B_{2} = 10$
Molecular orbital configuration:
$σ 1 s^{2}, \overset{*}{σ} 1 s^{2}, σ 2 s^{2}, \overset{*}{σ} 2 s^{2}, π 2 p_{x}^{1} \approx π 2 p_{y}^{1}$
It has two unpaired electrons.
$∴$ It is paramagnetic and the highest occupied molecular orbital is of $π$ type.
Hence, both $A$ and $R$ are incorrect

AIIMS - 2005

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313768 In which of the following ionisation process the bond energy has increased and also the magentic behaviour has changed from paramagnetic to diamagnetic?

1

N O \to N O^{+}

2

N_{2} \to N_{2}^{+}

3

C_{2} \to C_{2}^{+}

4

O_{2} \to O_{2}^{+}

Explanation:

For NO
Total no. of electrons = 15
B.O = 2.5
Magnetic Behaviour = Paramagnetic
For $N O^{+}$
Total no. of electrons = 14
B.O = 3
Magnetic Behaviour = Diamagnetic

JEE - 2013

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313769 According to molecular orbital theory, which of the following will not be a viable molecule?

1

{H e}_{2}^{+}

2

H_{2}^{-}

3

H_{2}^{2 -}

4

{H e}_{2}^{2 +}

Explanation:

The electronic configuration of $H_{2}^{2 -}$
$\Rightarrow (σ 1 s^{2}), {(σ^{*} 1 s)}^{2}$
Bond order of $H_{2}^{2 -} = \frac{N_{b} - N_{a}}{2} = \frac{2 - 2}{2} = 0$
Hence $H_{2}^{2 -}$ does not exist

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313765 The bond order and magnetic property of acetylide ion are same as that of

1

O_{2}^{+}

2

N O^{+}

3

O_{2}^{-}

4

N_{2}^{+}

Explanation:

$C_{2}^{- 2} - 14 e^{-} - B . O = 3$ ; Diamagnetic
$N_{2}^{+} - 13 e^{-} -$ B.O. $= 2.5$ ; Paramagnetic
${NO}^{+} - 14 e^{-} -$ B.O. $= 3$ ; Diamagnetic
$O_{2}^{-} - 17 e^{-}$ – B.O. $= 1.5$ ; Paramagnetic
$O_{2}^{+} - 15 e^{-} -$ B.O. $= 2.5$ ; Paramagnetic
So, correct option is (2).

JEE - 2023

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313766 Assertion :
$B_{2}$ molecule is diamagnetic
Reason :
The highest occupied molecular orbital is of $σ$ type.

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 Both Assertion and Reason are incorrect

Explanation:

The total number of electrons in $B_{2} = 10$
Molecular orbital configuration:
$σ 1 s^{2}, \overset{*}{σ} 1 s^{2}, σ 2 s^{2}, \overset{*}{σ} 2 s^{2}, π 2 p_{x}^{1} \approx π 2 p_{y}^{1}$
It has two unpaired electrons.
$∴$ It is paramagnetic and the highest occupied molecular orbital is of $π$ type.
Hence, both $A$ and $R$ are incorrect

AIIMS - 2005

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313768 In which of the following ionisation process the bond energy has increased and also the magentic behaviour has changed from paramagnetic to diamagnetic?

1

N O \to N O^{+}

2

N_{2} \to N_{2}^{+}

3

C_{2} \to C_{2}^{+}

4

O_{2} \to O_{2}^{+}

Explanation:

For NO
Total no. of electrons = 15
B.O = 2.5
Magnetic Behaviour = Paramagnetic
For $N O^{+}$
Total no. of electrons = 14
B.O = 3
Magnetic Behaviour = Diamagnetic

JEE - 2013

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313769 According to molecular orbital theory, which of the following will not be a viable molecule?

1

{H e}_{2}^{+}

2

H_{2}^{-}

3

H_{2}^{2 -}

4

{H e}_{2}^{2 +}

Explanation:

The electronic configuration of $H_{2}^{2 -}$
$\Rightarrow (σ 1 s^{2}), {(σ^{*} 1 s)}^{2}$
Bond order of $H_{2}^{2 -} = \frac{N_{b} - N_{a}}{2} = \frac{2 - 2}{2} = 0$
Hence $H_{2}^{2 -}$ does not exist

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