Chapter 6, Problem 86A

Interpretation Introduction

Interpretation:

The electronic configuration of sodium when it forms 1+ ion and fluorine when it forms $1-$ ions needs to be written. The reason for the two elements not forming 2+ and $2-$ ions respectively needs to be explained.

Concept introduction:

The electron configuration of an element defines the distribution of electrons in its atomic orbitals. Electron configurations of atoms obey a standard representation in which all the atomic sub-shells are arranged in a sequence.

Answer to Problem 86A

Electronic configuration of ${\text{Na}}^{+}$ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}$ and ${\text{F}}^{-}$ is

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}$.

Explanation of Solution

The electronic configuration of sodium is as follows:

Atomic number of $\text{Na=11}$

${\text{Na=1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{1}}$

To form 1+ ion, it loses one electron from 3s sub-shell. Hence, the electron configuration become,

${\text{Na}}^{+}{\text{=1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}$

The electronic configuration of fluorine is as follows:

Atomic number of $\text{F=9}$

${\text{F=1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^5$

To form $1-$ ion, it gains one electron from 2p sub-shell. Hence, the electron configuration for ${\text{F}}^{-}$ become,

${\text{F}}^{-}{\text{=1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}$

In case of sodium or fluorine 1+ ion and $1-$ ions forms, they satisfy the octet rule. If adding or withdrawing of an additional electron takes place, they become unstable. Therefore, Na does not form 2+ ion and F does not form $2-$ ions.