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CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313743 How many orbitals are singly occupied in $O_{2}$ molecule?

1 1

2 2

3 3

4 4

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313744 $O - O$ bond lengths in $O_{2}, O_{2} [A s F_{6}] a n d K O_{2}$ increasing order are

1

O_{2} (A s F_{6}) < O_{2} < K O_{2}

2

K O_{2} < O_{2} (A s F_{6}) < O_{2}

3

O_{2} < K O_{2} < O_{2} (A s F_{6})

4

O_{2} < K O_{2} = O_{2} (A s F_{6})

Explanation:

In $O_{2} (A s F_{6}), O_{2}^{+}$ has bond order 2.5.
$O_{2}$ has bond order 2.0
In $K O_{2}, O_{2}^{-}$ has bond order 1.5.
Higher the bond order, smaller is the bond length.
Hence, the bond order is $O_{2} (A s F_{6}) < O_{2} < K O_{2} .$

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313745 Four diatomic species are listed below in different sequences. The correct order of their bond orders is

1

C_{2}^{2 -} {< H e}_{2}^{+} {< N O < O}_{2}^{-}

2

{H e}_{2}^{+} {< O}_{2}^{-} {< N O < C}_{2}^{2 -}

3

O_{2}^{-} {< N O < C}_{2}^{2 -} {< H e}_{2}^{+}

4

{N O < C}_{2}^{2 -} {< O}_{2}^{-} {< H e}_{2}^{+}

Explanation:

Bond orders are : ${H e}_{2}^{+} = 0 {.5; O}_{2}^{-} = 1 .5;$
$N O = 2 {.5; C}_{2}^{2 -} = 3 .0$

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313746 The formatting issue $H_{2}^{+} a n d H_{2}^{-}$ is

1 Both

H_{2}^{+} a n d H_{2}^{-}

do not exist

2

H_{2}^{-}

is more stable than

H_{2}^{+}

3

H_{2}^{+}

is more stable than

H_{2}^{-}

4 Both

H_{2}^{+} a n d H_{2}^{-}

are equally stable

Explanation:

$H_{2}^{+} : (σ 1 s^{1})$
$∴ B . O . = \frac{1}{2} (1 - 0) = \frac{1}{2}$
$H_{2}^{-} : (σ 1 s^{2}) (σ^{*} 1 s^{1})$
$∴ B . O . = \frac{1}{2} (2 - 1) = \frac{1}{2}$
Even though the bond order of $H_{2}^{+} a n d H_{2}^{-}$ are equal but $H_{2}^{+}$ is more stable than $H_{2}^{-}$ as in the latter, an electron is present in the higher energy antibonding $(σ^{*} 1 s)$ orbital.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313747 Which of the following cannot be formed?

1

H e^{2 +}

2

H e^{+}

3

H e

4

H e_{2}

Explanation:

$H e_{2}$ cannot be formed as bond order is zero.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313743 How many orbitals are singly occupied in $O_{2}$ molecule?

1 1

2 2

3 3

4 4

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313744 $O - O$ bond lengths in $O_{2}, O_{2} [A s F_{6}] a n d K O_{2}$ increasing order are

1

O_{2} (A s F_{6}) < O_{2} < K O_{2}

2

K O_{2} < O_{2} (A s F_{6}) < O_{2}

3

O_{2} < K O_{2} < O_{2} (A s F_{6})

4

O_{2} < K O_{2} = O_{2} (A s F_{6})

Explanation:

In $O_{2} (A s F_{6}), O_{2}^{+}$ has bond order 2.5.
$O_{2}$ has bond order 2.0
In $K O_{2}, O_{2}^{-}$ has bond order 1.5.
Higher the bond order, smaller is the bond length.
Hence, the bond order is $O_{2} (A s F_{6}) < O_{2} < K O_{2} .$

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313745 Four diatomic species are listed below in different sequences. The correct order of their bond orders is

1

C_{2}^{2 -} {< H e}_{2}^{+} {< N O < O}_{2}^{-}

2

{H e}_{2}^{+} {< O}_{2}^{-} {< N O < C}_{2}^{2 -}

3

O_{2}^{-} {< N O < C}_{2}^{2 -} {< H e}_{2}^{+}

4

{N O < C}_{2}^{2 -} {< O}_{2}^{-} {< H e}_{2}^{+}

Explanation:

Bond orders are : ${H e}_{2}^{+} = 0 {.5; O}_{2}^{-} = 1 .5;$
$N O = 2 {.5; C}_{2}^{2 -} = 3 .0$

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313746 The formatting issue $H_{2}^{+} a n d H_{2}^{-}$ is

1 Both

H_{2}^{+} a n d H_{2}^{-}

do not exist

2

H_{2}^{-}

is more stable than

H_{2}^{+}

3

H_{2}^{+}

is more stable than

H_{2}^{-}

4 Both

H_{2}^{+} a n d H_{2}^{-}

are equally stable

Explanation:

$H_{2}^{+} : (σ 1 s^{1})$
$∴ B . O . = \frac{1}{2} (1 - 0) = \frac{1}{2}$
$H_{2}^{-} : (σ 1 s^{2}) (σ^{*} 1 s^{1})$
$∴ B . O . = \frac{1}{2} (2 - 1) = \frac{1}{2}$
Even though the bond order of $H_{2}^{+} a n d H_{2}^{-}$ are equal but $H_{2}^{+}$ is more stable than $H_{2}^{-}$ as in the latter, an electron is present in the higher energy antibonding $(σ^{*} 1 s)$ orbital.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313747 Which of the following cannot be formed?

1

H e^{2 +}

2

H e^{+}

3

H e

4

H e_{2}

Explanation:

$H e_{2}$ cannot be formed as bond order is zero.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313743 How many orbitals are singly occupied in $O_{2}$ molecule?

1 1

2 2

3 3

4 4

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313744 $O - O$ bond lengths in $O_{2}, O_{2} [A s F_{6}] a n d K O_{2}$ increasing order are

1

O_{2} (A s F_{6}) < O_{2} < K O_{2}

2

K O_{2} < O_{2} (A s F_{6}) < O_{2}

3

O_{2} < K O_{2} < O_{2} (A s F_{6})

4

O_{2} < K O_{2} = O_{2} (A s F_{6})

Explanation:

In $O_{2} (A s F_{6}), O_{2}^{+}$ has bond order 2.5.
$O_{2}$ has bond order 2.0
In $K O_{2}, O_{2}^{-}$ has bond order 1.5.
Higher the bond order, smaller is the bond length.
Hence, the bond order is $O_{2} (A s F_{6}) < O_{2} < K O_{2} .$

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313745 Four diatomic species are listed below in different sequences. The correct order of their bond orders is

1

C_{2}^{2 -} {< H e}_{2}^{+} {< N O < O}_{2}^{-}

2

{H e}_{2}^{+} {< O}_{2}^{-} {< N O < C}_{2}^{2 -}

3

O_{2}^{-} {< N O < C}_{2}^{2 -} {< H e}_{2}^{+}

4

{N O < C}_{2}^{2 -} {< O}_{2}^{-} {< H e}_{2}^{+}

Explanation:

Bond orders are : ${H e}_{2}^{+} = 0 {.5; O}_{2}^{-} = 1 .5;$
$N O = 2 {.5; C}_{2}^{2 -} = 3 .0$

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313746 The formatting issue $H_{2}^{+} a n d H_{2}^{-}$ is

1 Both

H_{2}^{+} a n d H_{2}^{-}

do not exist

2

H_{2}^{-}

is more stable than

H_{2}^{+}

3

H_{2}^{+}

is more stable than

H_{2}^{-}

4 Both

H_{2}^{+} a n d H_{2}^{-}

are equally stable

Explanation:

$H_{2}^{+} : (σ 1 s^{1})$
$∴ B . O . = \frac{1}{2} (1 - 0) = \frac{1}{2}$
$H_{2}^{-} : (σ 1 s^{2}) (σ^{*} 1 s^{1})$
$∴ B . O . = \frac{1}{2} (2 - 1) = \frac{1}{2}$
Even though the bond order of $H_{2}^{+} a n d H_{2}^{-}$ are equal but $H_{2}^{+}$ is more stable than $H_{2}^{-}$ as in the latter, an electron is present in the higher energy antibonding $(σ^{*} 1 s)$ orbital.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313747 Which of the following cannot be formed?

1

H e^{2 +}

2

H e^{+}

3

H e

4

H e_{2}

Explanation:

$H e_{2}$ cannot be formed as bond order is zero.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313743 How many orbitals are singly occupied in $O_{2}$ molecule?

1 1

2 2

3 3

4 4

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313744 $O - O$ bond lengths in $O_{2}, O_{2} [A s F_{6}] a n d K O_{2}$ increasing order are

1

O_{2} (A s F_{6}) < O_{2} < K O_{2}

2

K O_{2} < O_{2} (A s F_{6}) < O_{2}

3

O_{2} < K O_{2} < O_{2} (A s F_{6})

4

O_{2} < K O_{2} = O_{2} (A s F_{6})

Explanation:

In $O_{2} (A s F_{6}), O_{2}^{+}$ has bond order 2.5.
$O_{2}$ has bond order 2.0
In $K O_{2}, O_{2}^{-}$ has bond order 1.5.
Higher the bond order, smaller is the bond length.
Hence, the bond order is $O_{2} (A s F_{6}) < O_{2} < K O_{2} .$

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313745 Four diatomic species are listed below in different sequences. The correct order of their bond orders is

1

C_{2}^{2 -} {< H e}_{2}^{+} {< N O < O}_{2}^{-}

2

{H e}_{2}^{+} {< O}_{2}^{-} {< N O < C}_{2}^{2 -}

3

O_{2}^{-} {< N O < C}_{2}^{2 -} {< H e}_{2}^{+}

4

{N O < C}_{2}^{2 -} {< O}_{2}^{-} {< H e}_{2}^{+}

Explanation:

Bond orders are : ${H e}_{2}^{+} = 0 {.5; O}_{2}^{-} = 1 .5;$
$N O = 2 {.5; C}_{2}^{2 -} = 3 .0$

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313746 The formatting issue $H_{2}^{+} a n d H_{2}^{-}$ is

1 Both

H_{2}^{+} a n d H_{2}^{-}

do not exist

2

H_{2}^{-}

is more stable than

H_{2}^{+}

3

H_{2}^{+}

is more stable than

H_{2}^{-}

4 Both

H_{2}^{+} a n d H_{2}^{-}

are equally stable

Explanation:

$H_{2}^{+} : (σ 1 s^{1})$
$∴ B . O . = \frac{1}{2} (1 - 0) = \frac{1}{2}$
$H_{2}^{-} : (σ 1 s^{2}) (σ^{*} 1 s^{1})$
$∴ B . O . = \frac{1}{2} (2 - 1) = \frac{1}{2}$
Even though the bond order of $H_{2}^{+} a n d H_{2}^{-}$ are equal but $H_{2}^{+}$ is more stable than $H_{2}^{-}$ as in the latter, an electron is present in the higher energy antibonding $(σ^{*} 1 s)$ orbital.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313747 Which of the following cannot be formed?

1

H e^{2 +}

2

H e^{+}

3

H e

4

H e_{2}

Explanation:

$H e_{2}$ cannot be formed as bond order is zero.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313743 How many orbitals are singly occupied in $O_{2}$ molecule?

1 1

2 2

3 3

4 4

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313744 $O - O$ bond lengths in $O_{2}, O_{2} [A s F_{6}] a n d K O_{2}$ increasing order are

1

O_{2} (A s F_{6}) < O_{2} < K O_{2}

2

K O_{2} < O_{2} (A s F_{6}) < O_{2}

3

O_{2} < K O_{2} < O_{2} (A s F_{6})

4

O_{2} < K O_{2} = O_{2} (A s F_{6})

Explanation:

In $O_{2} (A s F_{6}), O_{2}^{+}$ has bond order 2.5.
$O_{2}$ has bond order 2.0
In $K O_{2}, O_{2}^{-}$ has bond order 1.5.
Higher the bond order, smaller is the bond length.
Hence, the bond order is $O_{2} (A s F_{6}) < O_{2} < K O_{2} .$

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313745 Four diatomic species are listed below in different sequences. The correct order of their bond orders is

1

C_{2}^{2 -} {< H e}_{2}^{+} {< N O < O}_{2}^{-}

2

{H e}_{2}^{+} {< O}_{2}^{-} {< N O < C}_{2}^{2 -}

3

O_{2}^{-} {< N O < C}_{2}^{2 -} {< H e}_{2}^{+}

4

{N O < C}_{2}^{2 -} {< O}_{2}^{-} {< H e}_{2}^{+}

Explanation:

Bond orders are : ${H e}_{2}^{+} = 0 {.5; O}_{2}^{-} = 1 .5;$
$N O = 2 {.5; C}_{2}^{2 -} = 3 .0$

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313746 The formatting issue $H_{2}^{+} a n d H_{2}^{-}$ is

1 Both

H_{2}^{+} a n d H_{2}^{-}

do not exist

2

H_{2}^{-}

is more stable than

H_{2}^{+}

3

H_{2}^{+}

is more stable than

H_{2}^{-}

4 Both

H_{2}^{+} a n d H_{2}^{-}

are equally stable

Explanation:

$H_{2}^{+} : (σ 1 s^{1})$
$∴ B . O . = \frac{1}{2} (1 - 0) = \frac{1}{2}$
$H_{2}^{-} : (σ 1 s^{2}) (σ^{*} 1 s^{1})$
$∴ B . O . = \frac{1}{2} (2 - 1) = \frac{1}{2}$
Even though the bond order of $H_{2}^{+} a n d H_{2}^{-}$ are equal but $H_{2}^{+}$ is more stable than $H_{2}^{-}$ as in the latter, an electron is present in the higher energy antibonding $(σ^{*} 1 s)$ orbital.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313747 Which of the following cannot be formed?

1

H e^{2 +}

2

H e^{+}

3

H e

4

H e_{2}

Explanation:

$H e_{2}$ cannot be formed as bond order is zero.

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