Chapter 9, Problem 9.65P

(a)

Interpretation Introduction

Interpretation:

The magnetic behavior of $\text{CO}$ molecule has to be given.

Concept Introduction:

Bond order of a molecule or ion is calculates by the following formula,

Bond order $\text{=}\frac{\left({\text{N}}_{\text{b}}\right)\text{-}\left({\text{N}}_{\text{a}}\right)}{\text{2}}$

Where, ${\text{N}}_{\text{b}}$ is number of electrons in bonding orbital.

${\text{N}}_{\text{a}}$ is number of electrons in antibonding orbital.

Paramagnetic and diamagnetic compounds:

Paramagnetic compounds have unpaired electrons and are attracted to magnetic field whereas diamagnetic compounds have paired spins and are repelled by the magnetic field

Explanation of Solution

The molecular configuration of carbon monoxide is,

  ${\left(\text{σ1s}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}\text{1s}\right)}^{\text{2}}{\left(\text{σ2s}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}\text{2s}\right)}^{\text{2}}{\left({\text{π2p}}_{\text{x}}\right)}^{\text{2}}{\left({\text{π2p}}_{\text{y}}\right)}^{\text{2}}{\left({\text{σ2p}}_{\text{z}}\right)}^{\text{2}}$

There is no unpaired electron in the carbon monoxide.  Hence it is not a paramagnetic compound.

(b)

Interpretation Introduction

Interpretation:

The magnetic behavior of ${\text{F}}_{\text{2}}$ molecule has to be given.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The molecular electronic configuration of fluorine molecule is.

${\text{F}}_{\text{2}}$=${\left(\text{σ1s}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}\text{1s}\right)}^{\text{2}}{\left(\text{σ2s}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}\text{2s}\right)}^{\text{2}}{\left({\text{σ2p}}_{\text{z}}\right)}^{\text{2}}{\left({\text{π2p}}_{\text{x}}\right)}^{\text{2}}{\left({\text{π2p}}_{\text{y}}\right)}^{\text{2}}{\left({\text{π}}^{\text{*}}{\text{2p}}_{\text{x}}\right)}^{\text{2}}{\left({\text{π}}^{\text{*}}{\text{2p}}_{\text{y}}\right)}^{\text{2}}$

There is no unpaired electron in the ${\text{F}}_{\text{2}}$ molecule.  Hence it is not a paramagnetic compound.

(c)

Interpretation Introduction

Interpretation:

The magnetic behavior of ${\text{O}}_{\text{2}}$ molecule has to be given.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The molecular configuration of ${\text{O}}_{\text{2}}$ is,

  ${\left(\text{σ1s}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}\text{1s}\right)}^{\text{2}}{\left(\text{σ2s}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}\text{2s}\right)}^{\text{2}}{\left({\text{σ2p}}_{\text{z}}\right)}^{\text{2}}{\left({\text{π2p}}_{\text{x}}\right)}^{\text{2}}{\left({\text{π2p}}_{\text{y}}\right)}^{\text{2}}{\left({\text{π}}^{\text{*}}{\text{2p}}_{\text{x}}\right)}^{\text{1}}{\left({\text{π}}^{\text{*}}{\text{2p}}_{\text{y}}\right)}^{\text{1}}.$

Oxygen molecule has two unpaired electrons so it can act like magnet. ${\text{O}}_{\text{2}}$ is paramagnetic.

(d)

Interpretation Introduction

Interpretation:

The magnetic behavior of $\text{NO}$ molecule has to be given

Concept Introduction:

Refer to part (a).

Explanation of Solution

The molecular configuration of $\text{NO}$ molecule is,

  $\text{NO}\text{=}{\left(\text{σ1s}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}\text{1s}\right)}^{\text{2}}{\left(\text{σ2s}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}\text{2s}\right)}^{\text{2}}{\left({\text{σ2p}}_{\text{z}}\right)}^{\text{2}}{\left({\text{π2p}}_{\text{x}}\right)}^{\text{2}}{\left({\text{π2p}}_{\text{y}}\right)}^{\text{2}}{\left({\text{π}}^{\text{*}}{\text{2p}}_{\text{x}}\right)}^{\text{1}}$

$\text{NO}$ has one unpaired electron so it act like magnet. $\text{NO}$ is paramagnetic.

Since oxygen molecule has two unpaired electrons it shows strong mass shift in magnetic susceptibility balance.