Chapter 7, Problem 123AP

One way to estimate the effective charge $\left({\text{Z}}_{\text{eff}}\right)$ of a many-electron atom is to use the equation $I{E}_{\text{1}}\text{= }\left(\text{1312 kJ/mol}\right)\left(Z^{\text{2}}{}_{\text{eff}}\text{/}{n}^{\text{2}}\right)$, where $I{E}_1$ is the first ionization energy and n is the principal quantum number of the shell in which the electron resides. Use this equation to calculate the effective nuclear charges of Li, Na, and K. Also calculate $Z_{\text{eff}}/n$ for each metal. Comment on your results.

Interpretation Introduction

Interpretation: The effective nuclear charge($Z_{eff}$) of lithium, sodium, and potassium are to be calculated. The value of $Z_{eff}/n$ for each given metal is to be calculated.

Concept introduction:

Ionization energy is the energy needed to remove a valence electron from a neutral atom in gaseous phase.

The ionization energy is expressed in $\text{kJ}/\text{mol}$.

Across the period, effective nuclear charge increases and down the group, effective nuclear charge decreases.

Answer to Problem 123AP

Solution: The effective nuclear charge of lithium, sodium, and potassium are1.26, 1.84, and 2.26, respectively. The value of $Z_{eff}/n$ for lithium, sodium, and potassium are 0.630, 0.613, and 0.565, respectively.

Explanation of Solution

Given information: $\mathit{\text{I}}{\mathit{\text{E}}}_1=\left(1312\text{kJ/mol}\right)\left({\mathit{\text{Z}}}_{eff}{}^2/{\mathit{\text{n}}}^2\right)$.

The equation for first ionization energy is given as follows:

$\mathit{\text{I}}{\mathit{\text{E}}}_1=\left(1312\text{kJ/mol}\right)\left({\mathit{\text{Z}}}_{eff}{}^2/{\mathit{\text{n}}}^2\right)$

Here, $n$ is the principle quantum number of shell in which the electron resides, $\mathit{\text{I}}{\mathit{\text{E}}}_1$ is first the ionization energy, and ${\mathit{\text{Z}}}_{eff}$ is the effective charge.

Rearrange the above equation as follows:

${\mathit{\text{Z}}}_{eff}=\mathit{\text{n}}\sqrt{\frac{\mathit{\text{I}}{\mathit{\text{E}}}_1}{1312\text{kJ/mol}}}$ …… (1)

For lithium, the principle quantum number is 2 and first ionization energy $\left(\mathit{\text{I}}{\mathit{\text{E}}}_1\right)$ is $520\text{ kJ/mol}$.

So, the effective nuclear charge for lithium is calculated as follows:

Substitute 2 for $n$ and $520\text{ kJ/mol}$ for $\mathit{\text{I}}{\mathit{\text{E}}}_1$ in equation (1) as follows:

$\begin{array}{c}{\mathit{\text{Z}}}_{eff}(Li)=2\times \sqrt{\frac{520\text{ kJ/mo}l}{1312\text{kJ/mol}}}\\ =1.26\end{array}$

For sodium, the principle quantum number is 3 and the first ionization energy $\left(\mathit{\text{I}}{\mathit{\text{E}}}_1\right)$ is $\text{495}\text{.9 kJ/mol}$.

So, the effective nuclear charge for lithium is calculated as follows:

Substitute 3 for $n$ and $\text{495}\text{.9 kJ/mol}$ for $\mathit{\text{I}}{\mathit{\text{E}}}_1$ in equation (1) as follows:

$\begin{array}{c}{\mathit{\text{Z}}}_{eff}(Na)=3\times \sqrt{\frac{495.9\text{ kJ/mo}l}{1312\text{kJ/mol}}}\\ =1.84\end{array}$

Similarly, for potassium, the principle quantum number is 4 and first ionization energy $\left(\mathit{\text{I}}{\mathit{\text{E}}}_1\right)$ is $\text{418}\text{.7 kJ/mol}$.

So, the effective nuclear charge for lithium is calculated as follows:

Substitute 4 for $n$ and $\text{418}\text{.7 kJ/mol}$ for $\mathit{\text{I}}{\mathit{\text{E}}}_1$ in equation (1) as follows:

$\begin{array}{c}{\mathit{\text{Z}}}_{eff}(K)=4\times \sqrt{\frac{418.7\text{ kJ/mo}l}{1312\text{kJ/mol}}}\\ =2.26\end{array}$

Down the group, the effective charge decreases because shells with large value of $n$ are less effective than the outer electrons from the effective nuclear charge at shielding.

For lithium, the value of $Z_{eff}/n$ is given as follows:

$\begin{array}{c}\frac{{\mathit{\text{Z}}}_{eff}}{\mathit{\text{n}}}=\frac{1.26}{2}\\ =0.630\end{array}$

For sodium, the value of $Z_{eff}/n$ is given as follows:

$\begin{array}{c}\frac{{\mathit{\text{Z}}}_{eff}}{\mathit{\text{n}}}=\frac{1.84}{3}\\ =0.613\end{array}$

For potassium, the value of $Z_{eff}/n$ is given as follows:

$\begin{array}{c}\frac{{\mathit{\text{Z}}}_{eff}}{\mathit{\text{n}}}=\frac{2.26}{4}\\ =0.565\end{array}$

Since, the $Z_{eff}$ values are relatively constant, therefore, the shielding per shell is nearly the same.