Chapter 5, Problem 131QRT

(a)

Interpretation Introduction

Interpretation:

The effective nuclear charge on two p electrons in ${\text{O}}^{\text{2-}}{\text{ , F}}^{-}{\text{ , Na}}^{\text{+}}{\text{,and Mg}}^{\text{2+}}$ has to be calculated.

Concept Introduction:

Calculation of effective nuclear charge:

The effective nuclear charge equation $\text{Z* = Z - σ}$.  The electron configuration for an atom has to be written and grouped the subshell in this way $\text{(1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f) (5s, 5p) }\text{. }\text{. }\text{.}$  An electron for which effective nuclear charge to be calculated has chosen.  Sum these contributions to σ for all other electrons: Electrons in groups to the right of the selected electron contribute zero. Electrons in the same group as the selected electron contribute $0.35$ per electron (except for ($\text{1s}$) where the contribution is $0.30$). If the selected electron is in an ($\text{ns, np}$) group, electrons in the $\text{n - 1}$ shell contribute $\text{0}\text{.85}$ and electrons in the shells with lower n contribute $1.0$. If the selected electron is in an $\left(\text{nd}\right)\text{ or }\left(\text{nf}\right)$ group, all electrons in groups to the left count $1.0$.

Explanation of Solution

These ${\text{O}}^{\text{2-}}{\text{ , F}}^{-}{\text{ , Na}}^{\text{+}}{\text{,and Mg}}^{\text{2+}}$ electrons are all isoelectronic, with a grouped electron configuration: $\left({\text{1s}}^{\text{2}}\right)\left({\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}\right)$ so the shielding constant will be the same for each ion.

    $\begin{array}{l}\text{ σ}\text{=7x (0}\text{.35)+2}\times \text{ (0}\text{.85) = 4}\text{.15}\\ \\ {}_{\text{8}}{\text{O}}^{2-}\text{Z* = 8-4}\text{.15}\text{=3}\text{.85}\\ \\ {}_{\text{9}}{\text{F}}^{-}\text{ Z* = 9-4}\text{.15}\text{=4}\text{.85}\\ \\ {}_{\text{10}}{\text{Na}}^{\text{+}}\text{Z* = 10-4}\text{.15}\text{=4}\text{.85}\\ \\ {}_{\text{11}}{\text{Mg}}^{\text{2+}}\text{Z* = 11-4}\text{.15}\text{=6}\text{.85}\text{.}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The calculated effective nuclear charge consistent with relative size of ${\text{O}}^{\text{2-}}{\text{ , F}}^{-}{\text{ , Na}}^{\text{+}}{\text{,and Mg}}^{\text{2+}}$ ions or not has to be explained.

Concept Introduction:

Refer to part a.

Explanation of Solution

These ${\text{O}}^{\text{2-}}{\text{ , F}}^{-}{\text{ , Na}}^{\text{+}}{\text{,and Mg}}^{\text{2+}}$ electrons are all isoelectronic, with a grouped electron configuration: $\left({\text{1s}}^{\text{2}}\right)\left({\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}\right)$ so the shielding constant will be the same for each ion.

    $\begin{array}{l}\text{ σ}\text{=7x (0}\text{.35)+2}\times \text{ (0}\text{.85) = 4}\text{.15}\\ \\ {}_{\text{8}}{\text{O}}^{2-}\text{Z* = 8-4}\text{.15}\text{=3}\text{.85}\\ \\ {}_{\text{9}}{\text{F}}^{-}\text{ Z* = 9-4}\text{.15}\text{=4}\text{.85}\\ \\ {}_{\text{10}}{\text{Na}}^{\text{+}}\text{Z* = 10-4}\text{.15}\text{=4}\text{.85}\\ \\ {}_{\text{11}}{\text{Mg}}^{\text{2+}}\text{Z* = 11-4}\text{.15}\text{=6}\text{.85}\text{.}\end{array}$

The calculated effective nuclear charges are larger for more positive ions in an isoelectronic series. A greater effective nuclear charge has the capacity to pull the electrons closer to the nucleus, making the ions smaller. This does explain the reduction in relative sizes for more positive ions.

(c)

Interpretation Introduction

Interpretation:

The effective nuclear charge on four d electrons in a tin atom has to be calculated.

Concept Introduction:

Refer to part a.

Explanation of Solution

The electron configuration of tin is given by

    ${\text{(1s}}^{\text{2}}{\text{)(2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{)(3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{)(3d}}^{\text{10}}{\text{)(4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{)(4d}}^{\text{10}}{\text{)(5s}}^{\text{2}}{\text{5p}}^{\text{2}}\text{)}$

There are four electrons in a group to the right of $\text{4d}$; they contribute zero. Nine electrons are in the same group as the selected electron; they contribute $0.35$ per electron. Since the selected electron is in a ($\text{4d}$) group, all $\text{36}$ electrons in groups to the left of the ($\text{4d}$) group contribute $1.0$

    $\begin{array}{l}\text{ σ}\text{=}\text{9}\text{×}\text{(0}\text{.35)+(36)}\text{(1}\text{.00) = 39}\text{.15}\\ \\ \text{Z* = Z - σ}\\ \\ \text{=}\text{50 -39}\text{.15}\text{=}\text{10}\text{.85}\text{.}\end{array}$