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CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313770 In forming (i) $N_{2} \to N_{2}^{+}$ and (ii) $O_{2} \to O_{2}^{+}$ ; the electrons respectively are removed from

1

(π^{*} 2 p_{y} o r π^{*} 2 p_{x}) a n d (π^{*} 2 p_{y} o r π^{*} 2 p_{x})

2

(π 2 p_{y} o r π 2 p_{x}) a n d (π 2 p_{y} o r π 2 p_{x})

3

(π 2 p_{y} o r π 2 p_{x}) a n d (π^{*} 2 p_{y} o r π^{*} 2 p_{x})

4

(π^{*} 2 p_{y} o r π^{*} 2 p_{x}) a n d (π 2 p_{y} o r π 2 p_{x})

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313771 The decreasing order of bond length for $N_{2}$ and its ions is correctly shown in

1

N_{2} {> N}_{2}^{-} {= N}_{2}^{+} {> N}_{2}^{2 -} {= N}_{2}^{2 +}

2

N_{2}^{2 -} {= N}_{2}^{2 -} {> N}_{2}^{-} {= N}_{2}^{+} > N_{2}

3

N_{2}^{2 -} {> N}_{2}^{1 +} {> N}_{2}^{-} {> N}_{2}^{2 +} > N_{2}

4

N_{2}^{2 -} {> N}_{2}^{-} > N_{2} {> N}_{2}^{+} {> N}_{2}^{2 +}

Explanation:

Bond order of $N_{2}^{2 -}$ and $N_{2}^{2 +}$ is 2
Bond order of $N_{2}^{-}$ and $N_{2}^{+}$ is 2.5
Bond order of $N_{2}$ is 3.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313772 What is the effect of the following processes on the bond order in $N_{2} a n d O_{2}$ ?
i) $N_{2} \to N_{2}^{+} + e^{-}$
ii) $O_{2} \to O_{2}^{+} + e^{-}$

1 Decreases in both (i) & (ii)

2 Increases in both (i) & (ii)

3 Increases in (i) but decreases in (ii)

4 Decreases in (i) but increases in (ii)

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313773 Consider the following statements
I. When $O_{2}$ is converted into $O_{2}^{2 +}$ bond order decreases.
II. $O_{2}$ molecule is paramagnetic because it contains two unpaired electrons in $π^{*} 2 p_{x}$ and $π^{*} 2 p_{y}$ . molecular orbitals.
III. The bond length in NO is greater than in $N O^{+}$ .
Choose the correct statements.

1 Only I

2 I and II

3 II and III

4 II and IV

Explanation:

Statements II and III are correct while statement I is incorrect. It’s correct form is as follows:
When $O_{2}$ is converted into $O_{2}^{2 +}$ bond order increases. So, the option (3) is correct.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313770 In forming (i) $N_{2} \to N_{2}^{+}$ and (ii) $O_{2} \to O_{2}^{+}$ ; the electrons respectively are removed from

1

(π^{*} 2 p_{y} o r π^{*} 2 p_{x}) a n d (π^{*} 2 p_{y} o r π^{*} 2 p_{x})

2

(π 2 p_{y} o r π 2 p_{x}) a n d (π 2 p_{y} o r π 2 p_{x})

3

(π 2 p_{y} o r π 2 p_{x}) a n d (π^{*} 2 p_{y} o r π^{*} 2 p_{x})

4

(π^{*} 2 p_{y} o r π^{*} 2 p_{x}) a n d (π 2 p_{y} o r π 2 p_{x})

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313771 The decreasing order of bond length for $N_{2}$ and its ions is correctly shown in

1

N_{2} {> N}_{2}^{-} {= N}_{2}^{+} {> N}_{2}^{2 -} {= N}_{2}^{2 +}

2

N_{2}^{2 -} {= N}_{2}^{2 -} {> N}_{2}^{-} {= N}_{2}^{+} > N_{2}

3

N_{2}^{2 -} {> N}_{2}^{1 +} {> N}_{2}^{-} {> N}_{2}^{2 +} > N_{2}

4

N_{2}^{2 -} {> N}_{2}^{-} > N_{2} {> N}_{2}^{+} {> N}_{2}^{2 +}

Explanation:

Bond order of $N_{2}^{2 -}$ and $N_{2}^{2 +}$ is 2
Bond order of $N_{2}^{-}$ and $N_{2}^{+}$ is 2.5
Bond order of $N_{2}$ is 3.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313772 What is the effect of the following processes on the bond order in $N_{2} a n d O_{2}$ ?
i) $N_{2} \to N_{2}^{+} + e^{-}$
ii) $O_{2} \to O_{2}^{+} + e^{-}$

1 Decreases in both (i) & (ii)

2 Increases in both (i) & (ii)

3 Increases in (i) but decreases in (ii)

4 Decreases in (i) but increases in (ii)

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313773 Consider the following statements
I. When $O_{2}$ is converted into $O_{2}^{2 +}$ bond order decreases.
II. $O_{2}$ molecule is paramagnetic because it contains two unpaired electrons in $π^{*} 2 p_{x}$ and $π^{*} 2 p_{y}$ . molecular orbitals.
III. The bond length in NO is greater than in $N O^{+}$ .
Choose the correct statements.

1 Only I

2 I and II

3 II and III

4 II and IV

Explanation:

Statements II and III are correct while statement I is incorrect. It’s correct form is as follows:
When $O_{2}$ is converted into $O_{2}^{2 +}$ bond order increases. So, the option (3) is correct.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313770 In forming (i) $N_{2} \to N_{2}^{+}$ and (ii) $O_{2} \to O_{2}^{+}$ ; the electrons respectively are removed from

1

(π^{*} 2 p_{y} o r π^{*} 2 p_{x}) a n d (π^{*} 2 p_{y} o r π^{*} 2 p_{x})

2

(π 2 p_{y} o r π 2 p_{x}) a n d (π 2 p_{y} o r π 2 p_{x})

3

(π 2 p_{y} o r π 2 p_{x}) a n d (π^{*} 2 p_{y} o r π^{*} 2 p_{x})

4

(π^{*} 2 p_{y} o r π^{*} 2 p_{x}) a n d (π 2 p_{y} o r π 2 p_{x})

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313771 The decreasing order of bond length for $N_{2}$ and its ions is correctly shown in

1

N_{2} {> N}_{2}^{-} {= N}_{2}^{+} {> N}_{2}^{2 -} {= N}_{2}^{2 +}

2

N_{2}^{2 -} {= N}_{2}^{2 -} {> N}_{2}^{-} {= N}_{2}^{+} > N_{2}

3

N_{2}^{2 -} {> N}_{2}^{1 +} {> N}_{2}^{-} {> N}_{2}^{2 +} > N_{2}

4

N_{2}^{2 -} {> N}_{2}^{-} > N_{2} {> N}_{2}^{+} {> N}_{2}^{2 +}

Explanation:

Bond order of $N_{2}^{2 -}$ and $N_{2}^{2 +}$ is 2
Bond order of $N_{2}^{-}$ and $N_{2}^{+}$ is 2.5
Bond order of $N_{2}$ is 3.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313772 What is the effect of the following processes on the bond order in $N_{2} a n d O_{2}$ ?
i) $N_{2} \to N_{2}^{+} + e^{-}$
ii) $O_{2} \to O_{2}^{+} + e^{-}$

1 Decreases in both (i) & (ii)

2 Increases in both (i) & (ii)

3 Increases in (i) but decreases in (ii)

4 Decreases in (i) but increases in (ii)

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313773 Consider the following statements
I. When $O_{2}$ is converted into $O_{2}^{2 +}$ bond order decreases.
II. $O_{2}$ molecule is paramagnetic because it contains two unpaired electrons in $π^{*} 2 p_{x}$ and $π^{*} 2 p_{y}$ . molecular orbitals.
III. The bond length in NO is greater than in $N O^{+}$ .
Choose the correct statements.

1 Only I

2 I and II

3 II and III

4 II and IV

Explanation:

Statements II and III are correct while statement I is incorrect. It’s correct form is as follows:
When $O_{2}$ is converted into $O_{2}^{2 +}$ bond order increases. So, the option (3) is correct.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313770 In forming (i) $N_{2} \to N_{2}^{+}$ and (ii) $O_{2} \to O_{2}^{+}$ ; the electrons respectively are removed from

1

(π^{*} 2 p_{y} o r π^{*} 2 p_{x}) a n d (π^{*} 2 p_{y} o r π^{*} 2 p_{x})

2

(π 2 p_{y} o r π 2 p_{x}) a n d (π 2 p_{y} o r π 2 p_{x})

3

(π 2 p_{y} o r π 2 p_{x}) a n d (π^{*} 2 p_{y} o r π^{*} 2 p_{x})

4

(π^{*} 2 p_{y} o r π^{*} 2 p_{x}) a n d (π 2 p_{y} o r π 2 p_{x})

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313771 The decreasing order of bond length for $N_{2}$ and its ions is correctly shown in

1

N_{2} {> N}_{2}^{-} {= N}_{2}^{+} {> N}_{2}^{2 -} {= N}_{2}^{2 +}

2

N_{2}^{2 -} {= N}_{2}^{2 -} {> N}_{2}^{-} {= N}_{2}^{+} > N_{2}

3

N_{2}^{2 -} {> N}_{2}^{1 +} {> N}_{2}^{-} {> N}_{2}^{2 +} > N_{2}

4

N_{2}^{2 -} {> N}_{2}^{-} > N_{2} {> N}_{2}^{+} {> N}_{2}^{2 +}

Explanation:

Bond order of $N_{2}^{2 -}$ and $N_{2}^{2 +}$ is 2
Bond order of $N_{2}^{-}$ and $N_{2}^{+}$ is 2.5
Bond order of $N_{2}$ is 3.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313772 What is the effect of the following processes on the bond order in $N_{2} a n d O_{2}$ ?
i) $N_{2} \to N_{2}^{+} + e^{-}$
ii) $O_{2} \to O_{2}^{+} + e^{-}$

1 Decreases in both (i) & (ii)

2 Increases in both (i) & (ii)

3 Increases in (i) but decreases in (ii)

4 Decreases in (i) but increases in (ii)

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313773 Consider the following statements
I. When $O_{2}$ is converted into $O_{2}^{2 +}$ bond order decreases.
II. $O_{2}$ molecule is paramagnetic because it contains two unpaired electrons in $π^{*} 2 p_{x}$ and $π^{*} 2 p_{y}$ . molecular orbitals.
III. The bond length in NO is greater than in $N O^{+}$ .
Choose the correct statements.

1 Only I

2 I and II

3 II and III

4 II and IV

Explanation:

Statements II and III are correct while statement I is incorrect. It’s correct form is as follows:
When $O_{2}$ is converted into $O_{2}^{2 +}$ bond order increases. So, the option (3) is correct.