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CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313715 Which one of the following properties is not shown by NO?

1 It is diamagnetic in gaseous state

2 It is neutral oxide

3 It combines with oxygen to form nitrogen dioxide

4 It’s bond order is 2.5

Explanation:

Nitric oxide is paramagnetic. In the gaseous state because of the presence of one unpaired electron in its outermost shell. The electronic configuration of NO is ${(σ 1 s)}^{2} {(σ^{*} 2 s)}^{2} {(σ^{*} 2 s)}^{2} {(σ 2 p_{z})}^{2} ({π 2 p}_{x}^{2} {= π 2 p}_{y}^{2}) (π^{*} {2 p}_{x}^{1})$

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313716 Bond order normally gives idea of stability of a molecular species. All the molecules viz. $H_{2}, L i_{2} a n d B_{2}$ have the same bond order yet they are not equally stable. Their stability order is

1

H_{2} > L i_{2} > B_{2}

2

L i_{2} > H_{2} > B_{2}

3

L i_{2} > B_{2} > H_{2}

4

B_{2} > H_{2} > L i_{2}

Explanation:

The molecular orbital configuration of the given molecules is $H_{2} = σ 1 s^{2}$ (no electron anti-bonding)
$L i_{2} = σ 1 s^{2} σ^{*} 1 s^{2} σ 2 s^{2}$ (two anti-bonding electrons)
$B_{2} = σ 1 s^{2} σ^{*} 1 s^{2} σ 2 s^{2} σ^{*} 2 s^{2} {{π 2 p}_{y}^{1} {= π 2 p}_{z}^{1}}$
(4 anti-bonding electrons)
Though the bond order of all the species are same (B.O = 1) but stability is different. This is due to difference in the presence of no. of anti bonding electron.
Higher the no. of anti-bonding electron lower is the stabiltiy hence the correct order is $H_{2} > L i_{2} > B_{2}$

JEE - 2013

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313717 The common features among the species ${CN}^{-}, CO$ and ${NO}^{+}$ are

1 bond order three and isoelectronic

2 bond order three and weak field ligands

3 bond order two and

π

-acceptor

4 isoelectronic and weak field ligands

Explanation:

${CN}^{-}$ ion Total number of electrons
$6 + 7 + 1 = 14$
MO configuration is
$σ 1 s^{2}, \overset{*}{σ} 1 s^{2}, \overset{}{σ} 2 s^{2}, \overset{*}{σ} 2 s^{2}, π 2 p_{x}^{2}$
$\approx π 2 p_{y}^{2}, σ 2 p_{z}^{2}$
Bond order $= \frac{10 - 4}{2} = 3$
CO molecule Total number of valence electrons
$= 6 + 8 = 14$
MO configuration is
$σ 1 s^{2}, \overset{*}{σ} 1 s^{2}, σ 2 s^{2}, \overset{*}{σ} 2 s^{2}, π {2 p}_{x}^{2},$
$π 2 p_{y}^{2}, σ 2 p_{z}^{2}$
Bond order $= \frac{10 - 4}{2} = 3$
${NO}^{+}$ ion Total number of valence electrons $= 7 + 8 - 1 = 14$
MO configuration is
$σ 1 s^{2}, \overset{*}{σ} 1 s^{2}, σ 2 s^{2}, \overset{*}{σ} 2 s^{2}, π 2 p_{x}^{2}$ ,
$\approx π 2 p_{y}^{2}, σ 2 p_{z}^{2}$
Bond order $= \frac{10 - 4}{2} = 3$
All ${CN}^{-}, CO$ and ${NO}^{+}$ ion have bond order 3 and they are isoelectronic

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313718 Using MO theory, predict which of the following species has the shortest bond length?

1

O_{2}^{+}

2

O_{2}^{-}

3

O_{2}^{2 -}

4

O_{2}^{2 +}

Explanation:

Bond order
$= \frac{(\begin{array}{l} N o o f b o n d i n g e l e c t r o n s - \\ N o . o f a n t i - b o n d i n g e l e c t r o n s \end{array})}{2}$
Bond order in $O_{2}^{+} = \frac{10 - 5}{2} = 2 .5$
Bond order in $O_{2}^{-} = \frac{10 - 7}{2} = 1 .5$
Bond order in $O_{2}^{2 -} = \frac{10 - 8}{2} = 1$
Bond order in $O_{2}^{2 +} = \frac{10 - 4}{2} = 3$
Since Bond order $\propto \frac{1}{B o n d l e n g t h}$
$∴$ Bond length is shortest in $O_{2}^{2 +}$ .

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313715 Which one of the following properties is not shown by NO?

1 It is diamagnetic in gaseous state

2 It is neutral oxide

3 It combines with oxygen to form nitrogen dioxide

4 It’s bond order is 2.5

Explanation:

Nitric oxide is paramagnetic. In the gaseous state because of the presence of one unpaired electron in its outermost shell. The electronic configuration of NO is ${(σ 1 s)}^{2} {(σ^{*} 2 s)}^{2} {(σ^{*} 2 s)}^{2} {(σ 2 p_{z})}^{2} ({π 2 p}_{x}^{2} {= π 2 p}_{y}^{2}) (π^{*} {2 p}_{x}^{1})$

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313716 Bond order normally gives idea of stability of a molecular species. All the molecules viz. $H_{2}, L i_{2} a n d B_{2}$ have the same bond order yet they are not equally stable. Their stability order is

1

H_{2} > L i_{2} > B_{2}

2

L i_{2} > H_{2} > B_{2}

3

L i_{2} > B_{2} > H_{2}

4

B_{2} > H_{2} > L i_{2}

Explanation:

The molecular orbital configuration of the given molecules is $H_{2} = σ 1 s^{2}$ (no electron anti-bonding)
$L i_{2} = σ 1 s^{2} σ^{*} 1 s^{2} σ 2 s^{2}$ (two anti-bonding electrons)
$B_{2} = σ 1 s^{2} σ^{*} 1 s^{2} σ 2 s^{2} σ^{*} 2 s^{2} {{π 2 p}_{y}^{1} {= π 2 p}_{z}^{1}}$
(4 anti-bonding electrons)
Though the bond order of all the species are same (B.O = 1) but stability is different. This is due to difference in the presence of no. of anti bonding electron.
Higher the no. of anti-bonding electron lower is the stabiltiy hence the correct order is $H_{2} > L i_{2} > B_{2}$

JEE - 2013

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313717 The common features among the species ${CN}^{-}, CO$ and ${NO}^{+}$ are

1 bond order three and isoelectronic

2 bond order three and weak field ligands

3 bond order two and

π

-acceptor

4 isoelectronic and weak field ligands

Explanation:

${CN}^{-}$ ion Total number of electrons
$6 + 7 + 1 = 14$
MO configuration is
$σ 1 s^{2}, \overset{*}{σ} 1 s^{2}, \overset{}{σ} 2 s^{2}, \overset{*}{σ} 2 s^{2}, π 2 p_{x}^{2}$
$\approx π 2 p_{y}^{2}, σ 2 p_{z}^{2}$
Bond order $= \frac{10 - 4}{2} = 3$
CO molecule Total number of valence electrons
$= 6 + 8 = 14$
MO configuration is
$σ 1 s^{2}, \overset{*}{σ} 1 s^{2}, σ 2 s^{2}, \overset{*}{σ} 2 s^{2}, π {2 p}_{x}^{2},$
$π 2 p_{y}^{2}, σ 2 p_{z}^{2}$
Bond order $= \frac{10 - 4}{2} = 3$
${NO}^{+}$ ion Total number of valence electrons $= 7 + 8 - 1 = 14$
MO configuration is
$σ 1 s^{2}, \overset{*}{σ} 1 s^{2}, σ 2 s^{2}, \overset{*}{σ} 2 s^{2}, π 2 p_{x}^{2}$ ,
$\approx π 2 p_{y}^{2}, σ 2 p_{z}^{2}$
Bond order $= \frac{10 - 4}{2} = 3$
All ${CN}^{-}, CO$ and ${NO}^{+}$ ion have bond order 3 and they are isoelectronic

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313718 Using MO theory, predict which of the following species has the shortest bond length?

1

O_{2}^{+}

2

O_{2}^{-}

3

O_{2}^{2 -}

4

O_{2}^{2 +}

Explanation:

Bond order
$= \frac{(\begin{array}{l} N o o f b o n d i n g e l e c t r o n s - \\ N o . o f a n t i - b o n d i n g e l e c t r o n s \end{array})}{2}$
Bond order in $O_{2}^{+} = \frac{10 - 5}{2} = 2 .5$
Bond order in $O_{2}^{-} = \frac{10 - 7}{2} = 1 .5$
Bond order in $O_{2}^{2 -} = \frac{10 - 8}{2} = 1$
Bond order in $O_{2}^{2 +} = \frac{10 - 4}{2} = 3$
Since Bond order $\propto \frac{1}{B o n d l e n g t h}$
$∴$ Bond length is shortest in $O_{2}^{2 +}$ .

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313715 Which one of the following properties is not shown by NO?

1 It is diamagnetic in gaseous state

2 It is neutral oxide

3 It combines with oxygen to form nitrogen dioxide

4 It’s bond order is 2.5

Explanation:

Nitric oxide is paramagnetic. In the gaseous state because of the presence of one unpaired electron in its outermost shell. The electronic configuration of NO is ${(σ 1 s)}^{2} {(σ^{*} 2 s)}^{2} {(σ^{*} 2 s)}^{2} {(σ 2 p_{z})}^{2} ({π 2 p}_{x}^{2} {= π 2 p}_{y}^{2}) (π^{*} {2 p}_{x}^{1})$

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313716 Bond order normally gives idea of stability of a molecular species. All the molecules viz. $H_{2}, L i_{2} a n d B_{2}$ have the same bond order yet they are not equally stable. Their stability order is

1

H_{2} > L i_{2} > B_{2}

2

L i_{2} > H_{2} > B_{2}

3

L i_{2} > B_{2} > H_{2}

4

B_{2} > H_{2} > L i_{2}

Explanation:

The molecular orbital configuration of the given molecules is $H_{2} = σ 1 s^{2}$ (no electron anti-bonding)
$L i_{2} = σ 1 s^{2} σ^{*} 1 s^{2} σ 2 s^{2}$ (two anti-bonding electrons)
$B_{2} = σ 1 s^{2} σ^{*} 1 s^{2} σ 2 s^{2} σ^{*} 2 s^{2} {{π 2 p}_{y}^{1} {= π 2 p}_{z}^{1}}$
(4 anti-bonding electrons)
Though the bond order of all the species are same (B.O = 1) but stability is different. This is due to difference in the presence of no. of anti bonding electron.
Higher the no. of anti-bonding electron lower is the stabiltiy hence the correct order is $H_{2} > L i_{2} > B_{2}$

JEE - 2013

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313717 The common features among the species ${CN}^{-}, CO$ and ${NO}^{+}$ are

1 bond order three and isoelectronic

2 bond order three and weak field ligands

3 bond order two and

π

-acceptor

4 isoelectronic and weak field ligands

Explanation:

${CN}^{-}$ ion Total number of electrons
$6 + 7 + 1 = 14$
MO configuration is
$σ 1 s^{2}, \overset{*}{σ} 1 s^{2}, \overset{}{σ} 2 s^{2}, \overset{*}{σ} 2 s^{2}, π 2 p_{x}^{2}$
$\approx π 2 p_{y}^{2}, σ 2 p_{z}^{2}$
Bond order $= \frac{10 - 4}{2} = 3$
CO molecule Total number of valence electrons
$= 6 + 8 = 14$
MO configuration is
$σ 1 s^{2}, \overset{*}{σ} 1 s^{2}, σ 2 s^{2}, \overset{*}{σ} 2 s^{2}, π {2 p}_{x}^{2},$
$π 2 p_{y}^{2}, σ 2 p_{z}^{2}$
Bond order $= \frac{10 - 4}{2} = 3$
${NO}^{+}$ ion Total number of valence electrons $= 7 + 8 - 1 = 14$
MO configuration is
$σ 1 s^{2}, \overset{*}{σ} 1 s^{2}, σ 2 s^{2}, \overset{*}{σ} 2 s^{2}, π 2 p_{x}^{2}$ ,
$\approx π 2 p_{y}^{2}, σ 2 p_{z}^{2}$
Bond order $= \frac{10 - 4}{2} = 3$
All ${CN}^{-}, CO$ and ${NO}^{+}$ ion have bond order 3 and they are isoelectronic

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313718 Using MO theory, predict which of the following species has the shortest bond length?

1

O_{2}^{+}

2

O_{2}^{-}

3

O_{2}^{2 -}

4

O_{2}^{2 +}

Explanation:

Bond order
$= \frac{(\begin{array}{l} N o o f b o n d i n g e l e c t r o n s - \\ N o . o f a n t i - b o n d i n g e l e c t r o n s \end{array})}{2}$
Bond order in $O_{2}^{+} = \frac{10 - 5}{2} = 2 .5$
Bond order in $O_{2}^{-} = \frac{10 - 7}{2} = 1 .5$
Bond order in $O_{2}^{2 -} = \frac{10 - 8}{2} = 1$
Bond order in $O_{2}^{2 +} = \frac{10 - 4}{2} = 3$
Since Bond order $\propto \frac{1}{B o n d l e n g t h}$
$∴$ Bond length is shortest in $O_{2}^{2 +}$ .

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313715 Which one of the following properties is not shown by NO?

1 It is diamagnetic in gaseous state

2 It is neutral oxide

3 It combines with oxygen to form nitrogen dioxide

4 It’s bond order is 2.5

Explanation:

Nitric oxide is paramagnetic. In the gaseous state because of the presence of one unpaired electron in its outermost shell. The electronic configuration of NO is ${(σ 1 s)}^{2} {(σ^{*} 2 s)}^{2} {(σ^{*} 2 s)}^{2} {(σ 2 p_{z})}^{2} ({π 2 p}_{x}^{2} {= π 2 p}_{y}^{2}) (π^{*} {2 p}_{x}^{1})$

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313716 Bond order normally gives idea of stability of a molecular species. All the molecules viz. $H_{2}, L i_{2} a n d B_{2}$ have the same bond order yet they are not equally stable. Their stability order is

1

H_{2} > L i_{2} > B_{2}

2

L i_{2} > H_{2} > B_{2}

3

L i_{2} > B_{2} > H_{2}

4

B_{2} > H_{2} > L i_{2}

Explanation:

The molecular orbital configuration of the given molecules is $H_{2} = σ 1 s^{2}$ (no electron anti-bonding)
$L i_{2} = σ 1 s^{2} σ^{*} 1 s^{2} σ 2 s^{2}$ (two anti-bonding electrons)
$B_{2} = σ 1 s^{2} σ^{*} 1 s^{2} σ 2 s^{2} σ^{*} 2 s^{2} {{π 2 p}_{y}^{1} {= π 2 p}_{z}^{1}}$
(4 anti-bonding electrons)
Though the bond order of all the species are same (B.O = 1) but stability is different. This is due to difference in the presence of no. of anti bonding electron.
Higher the no. of anti-bonding electron lower is the stabiltiy hence the correct order is $H_{2} > L i_{2} > B_{2}$

JEE - 2013

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313717 The common features among the species ${CN}^{-}, CO$ and ${NO}^{+}$ are

1 bond order three and isoelectronic

2 bond order three and weak field ligands

3 bond order two and

π

-acceptor

4 isoelectronic and weak field ligands

Explanation:

${CN}^{-}$ ion Total number of electrons
$6 + 7 + 1 = 14$
MO configuration is
$σ 1 s^{2}, \overset{*}{σ} 1 s^{2}, \overset{}{σ} 2 s^{2}, \overset{*}{σ} 2 s^{2}, π 2 p_{x}^{2}$
$\approx π 2 p_{y}^{2}, σ 2 p_{z}^{2}$
Bond order $= \frac{10 - 4}{2} = 3$
CO molecule Total number of valence electrons
$= 6 + 8 = 14$
MO configuration is
$σ 1 s^{2}, \overset{*}{σ} 1 s^{2}, σ 2 s^{2}, \overset{*}{σ} 2 s^{2}, π {2 p}_{x}^{2},$
$π 2 p_{y}^{2}, σ 2 p_{z}^{2}$
Bond order $= \frac{10 - 4}{2} = 3$
${NO}^{+}$ ion Total number of valence electrons $= 7 + 8 - 1 = 14$
MO configuration is
$σ 1 s^{2}, \overset{*}{σ} 1 s^{2}, σ 2 s^{2}, \overset{*}{σ} 2 s^{2}, π 2 p_{x}^{2}$ ,
$\approx π 2 p_{y}^{2}, σ 2 p_{z}^{2}$
Bond order $= \frac{10 - 4}{2} = 3$
All ${CN}^{-}, CO$ and ${NO}^{+}$ ion have bond order 3 and they are isoelectronic

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313718 Using MO theory, predict which of the following species has the shortest bond length?

1

O_{2}^{+}

2

O_{2}^{-}

3

O_{2}^{2 -}

4

O_{2}^{2 +}

Explanation:

Bond order
$= \frac{(\begin{array}{l} N o o f b o n d i n g e l e c t r o n s - \\ N o . o f a n t i - b o n d i n g e l e c t r o n s \end{array})}{2}$
Bond order in $O_{2}^{+} = \frac{10 - 5}{2} = 2 .5$
Bond order in $O_{2}^{-} = \frac{10 - 7}{2} = 1 .5$
Bond order in $O_{2}^{2 -} = \frac{10 - 8}{2} = 1$
Bond order in $O_{2}^{2 +} = \frac{10 - 4}{2} = 3$
Since Bond order $\propto \frac{1}{B o n d l e n g t h}$
$∴$ Bond length is shortest in $O_{2}^{2 +}$ .

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