Question

The number of protons in an ion from which $(Z-1)$ electrons are removed having its ionization energy $11800kJ$ $mo{l}^{-1}$ is :

• A. three
• B. five
• C. four
• D. insufficient information

Answer 1

The correct option is A three

The expression for the ionization energy is $I.E=\frac{Z^2}{n^2}\times 1312.4kJ/mol$
But the given ionization energy is 11800 kJ/mol.
Hence
$\frac{Z^2}{n^2}\times 1312.4=11800$
$\frac{Z^2}{n^2}=9$
$\frac{Z}{n}=3$
When $n=1$, $Z=3$
Since Z-1 (or 3-1 =2 ) electrons are removed, only 1 electron is remaining and hence $n=2$
Hence, the number of protons in the ion is 3.