Question

Ground-state ionization energies of some one-electron species are

Answer 1

The ionization energy of $B^{4+}$ is $3.28199 \cdot 10^{7} \mathrm{~J} / \mathrm{mol}$.

What is ionization energy?

It is the energy needed to remove one electron from a neutral atom, which results in the formation of an ion.

The measurement is based on an isolated atom in its gaseous phase and is often expressed in kJ/mol.

b) Atom is ionized when the electron is completely removed from its electron cloud (so it being moved from first, $n_{\text {initial }}=1$ to the infinity's shell and $n_{\text {final }}=\infty$ ), and now the equation can be written as

$\Delta E=-2.18 \cdot 10^{-18} J\left(\frac{1}{n_{\text {final }}^{2}}-\frac{1}{n_{\text {initial }}^{2}}\right)\\=-2.18 \cdot 10^{-18} J\left(\frac{1}{\infty^{2}}-\frac{1}{1}\right)$

So the ionization energy is affected by the charge of the nucleus and the general formula can be represented as:

$\Delta E_{I E}=-2.18 \cdot 10^{-18} J\left(\frac{1}{\infty^{2}}-\frac{1}{1}\right) \cdot Z^{2} \cdot\left(6.022 \cdot 10^{23}\right molecules/mol)$ and as $\frac{1}{\infty}=0$

$\begin{gathered}\Delta E_{I E}=-2.18 \cdot 10^{-18} \mathrm{~J} \cdot Z^{2} \cdot\left(6.022 \cdot 10^{23} \text { molecules } / \mathrm{mol}\right) \\\Delta E_{I E}=1.312796 \cdot 10^{6} \mathrm{~J} / \mathrm{mol} \cdot Z^{2}\end{gathered}$

The $B^{4+}$ has an atomic number Z=5, therefore using the formula when $n_{\text {initial }}$=1, we get

$\Delta E_{I E}=1.312796 \cdot \mathrm{J} / \mathrm{mol} \cdot 5^{2}\\=3.28199 \cdot 10^{7} \mathrm{~J} / \mathrm{mol}$

The ionization energy of $B^{4+}$ is $3.28199 \cdot 10^{7} \mathrm{~J} / \mathrm{mol}$.

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