Question

Question: The ground state ionization energy for the one electron ion X^m+ is $\mathbf{4}\mathbf{.}\mathbf{72}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}$ kJ/mol. Identify X and m .

Short answer

Answer

The element X is carbon and it is present in +5 oxidation state. Therefore,

$X^{m+}= {\text{C}}^{\text{+ 5}}$

Step-by-step solution

Ionization Energy:

Ionization Energy is the minimum energy required to remove an electron from outermost subshell / orbital of a gaseous isolated atom.

Step 2:

Given ion = X^m+

Ground state I.E = $4.72 \times {10}^4 KJ/mol$

From Bohr’s model:

$$\begin{gathered} E=-2.18\times {10}^{-18}\frac{z^2}{n^2}forGroundstaten=1 \\ -4.72\times {10}^4\times {10}^3\text{J}/\text{mol}=-2.18\times {10}^{-18}\frac{z^2}{{\left(1\right)}^2} \\ {z}^2=\frac{4.72\times {10}^{4}3\times {10}^3\text{J}/\text{mol}}{2.18\times {10}^{-18}7\times 6.023\times {10}^{23}\text{fal}}\text{atom} \\ {z}^2=\frac{2.16\times {10}^{25}}{6.023\times {10}^{23}} \\ {z}^2=0.358\times {10}^2 \\ z=\sqrt{35.8} \\ z\approx 6 \end{gathered}$$

It means atomic number of the molecule or atom is

Hence, X = Carbon

step :3

The Rydberg formula for the change in energy is

$\Delta E=-R_H{z}^2\left(\frac{1}{n_f^2}-\frac{1}{{n_i}^2}\right)$

To calculate the ionization energy from the ground state we get $n_f=1$ and $n_f=\infty$ for Carbon

$\Delta E=-R_H{z}^2=2.180\times {10}^{-18}\text{J}\times (6{)}^2\left(\frac{1}{\infty }-\frac{1}{1^2}\right)=7.848\times {10}^{-17}\text{J}$

On a molar basis the ionization energy is

$$\begin{gathered} (IE{)}_1=\left(I{E}_1\right)\times n \\ n=\frac{4.72\times {10}^{-1}}{7.848\times {10}^{-17}} \\ n=6 \\ \left(I{E}_1\right)\stackrel{∆}{\to }(IE{)}_1 \end{gathered}$$

Therefore ion is in +5 oxidation state. Therefore, the ion is C⁺⁵.