Chapter 5, Problem 130QRT

Calculate the effective nuclear charge, Z*, on these electrons in a tin atom (a) 5s; (b) 5p; and (c) 4d.

Interpretation Introduction

Interpretation:

The effective nuclear charge on five s electrons in a tin atom 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

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 is no electron group to the right of the $\text{5s}$ group. Three electrons are in the same group as the selected electron; they contribute $\text{0}\text{.35}$ per electron. Since the selected electron is in a ($\text{5s, 5p}$) group, electrons in the $\text{n = 4}$ shell contribute $\text{0}\text{.85}$ and electrons in the $\text{3, 2, and 1}$ shells contribute $\text{1}\text{.0}$

    $\begin{array}{l}\text{ σ}\text{=}\text{(0}\text{.35)+(18) × (0}\text{.85) 28}\text{× (1}\text{.00) = 46}\text{.65}\\ \\ \text{Z* = Z - σ}\text{=}\text{46}\text{.65=3}\text{.35}\end{array}$

Interpretation Introduction

Interpretation:

The effective nuclear charge on five p 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

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

The electron is in the same group as $\text{5s}$, so the shielding and effective nuclear charge will be the same as $\text{5s}$.

    $\begin{array}{l}\text{ σ}\text{=}\text{(0}\text{.35)+(18) × (0}\text{.85) 28}\text{× (1}\text{.00) = 46}\text{.65}\\ \\ \text{Z* = Z - σ}\text{=}\text{46}\text{.65=3}\text{.35}\text{.}\end{array}$

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 $1.0$

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