Chapter 9, Problem 99AP

Interpretation Introduction

Interpretation:

The electronic configuration of the cyanide ion, and a stable molecule isoelectronic to the cyanide ion, are to be determined.

Concept introduction:

The electronic configuration of a molecule is the arrangement of electrons in the orbitals of a molecule.

Two atomic orbitals combine to form a bonding and an antibonding molecular orbital. Orbitals that lie on the internuclear axis combine to form sigma $\sigma$

molecular orbital, and orbitals parallel to each other combine to form $\pi$

molecular orbitals.

The molecular orbital formed by the combination of the $\text{1s}$ orbital forms the bonding molecular orbital, designated as ${\sigma }_{1s}$, and the antibonding molecular orbital, ${\sigma }^{\ast }{}_{1s}$. The $\text{2s}$

orbital forms the corresponding molecular orbitals.

The molecular orbital formed by the combination of the ${\text{2p}}_{\text{x}}$ orbital forms a bonding molecular orbital, designated as ${\sigma }_{2{\text{p}}_{\text{x}}}$, and an antibonding molecular orbital, designated as ${\sigma }^{\ast }{}_{2{\text{p}}_{\text{x}}}$.

The molecular orbitals formed by combining the ${\text{2p}}_{\text{y}}$ and ${\text{2p}}_{\text{z}}$ orbitals form bonding molecular orbitals, designated as ${\text{π}}_{{\text{2p}}_{\text{y}}}$ and ${\text{π}}_{{\text{2p}}_{\text{z}}}$, and antibonding molecular orbitals, designated as ${\text{π}}^{\ast }{}_{{\text{2p}}_{\text{y}}}$

and ${\text{π}}^{\ast }{}_{{\text{2p}}_{\text{z}}}$.

Isoelectronic molecules contain an equal number of electrons.

Answer to Problem 99AP

Solution: ${\left({\sigma }_{1s}\right)}^2{\left({\sigma }^{\ast }{}_{1s}\right)}^2{\left({\sigma }_{2s}\right)}^2{\left({\sigma }^{\ast }{}_{2s}\right)}^2{\left({\text{π}}_{{\text{2p}}_{\text{y}}}\right)}^2{\left({\text{π}}_{{\text{2p}}_{\text{z}}}\right)}^2{\left({\sigma }_{{\text{2p}}_{\text{x}}}\right)}^2$; $\text{CO}$

Explanation of Solution

The electronic configuration of a carbon atom is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{2}}$ and that of a nitrogen atom is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{3}}$.

In a ${\text{CN}}^{-}$

ion, one carbon and one nitrogen atom, and a negative charge, are present. Thus, there are 14 electrons, in total, to be accommodated in the molecular orbitals of ${\text{CN}}^{-}$.

The electron configuration for ${\text{CN}}^{-}$ is as follows:

${\left({\sigma }_{1s}\right)}^2{\left({\sigma }^{\ast }{}_{1s}\right)}^2{\left({\sigma }_{2s}\right)}^2{\left({\sigma }^{\ast }{}_{2s}\right)}^2{\left({\text{π}}_{{\text{2p}}_{\text{y}}}\right)}^2{\left({\text{π}}_{{\text{2p}}_{\text{z}}}\right)}^2{\left({\sigma }_{{\text{2p}}_{\text{x}}}\right)}^2$

Compounds having same number of electrons are said to be isoelectronic. The ${\text{CN}}^{-}$ ion contains 14 electrons. So, a molecule isoelectronic to ${\text{CN}}^{-}$ would have 14 electrons. Carbon contains six electrons and oxygen contains eight electrons. Thus, $\text{CO}$ contains 14 electrons. Hence, the $\text{CO}$

molecule is isoelectronic to the ${\text{CN}}^{-}$ ion.

Conclusion

$\text{CO}$ molecule is a stable molecule isoelectronic to the cyanide ion and the electronic configuration of the cyanide ion is as follows:

${\left({\sigma }_{1s}\right)}^2{\left({\sigma }^{\ast }{}_{1s}\right)}^2{\left({\sigma }_{2s}\right)}^2{\left({\sigma }^{\ast }{}_{2s}\right)}^2{\left({\text{π}}_{{\text{2p}}_{\text{y}}}\right)}^2{\left({\text{π}}_{{\text{2p}}_{\text{z}}}\right)}^2{\left({\sigma }_{{\text{2p}}_{\text{x}}}\right)}^2$