Question

What is the correct order of increasing second ionisation energies for the elements $Be,B,C,N$ and $F?$

• A. $Be<C<B<N<F$
• B. $F<C<B<N<Be$
• C. $Be<B<C<N<F$
• D. $Be<C<N<B<F$

Answer 1

The correct option is A $Be<C<B<N<F$

The second ionization energy $I{E}_2$ is the energy required to remove an electron from a 1+ cation in the gaseous state.

$X_{\left(g\right)}^{+}\to X^{2+}\left(g\right)+e^{-}$

Just like the first ionization energy, $I{E}_2$ is affected by size, effective nuclear charge, and electronic configuration. We would expect second ionization energies to increase from left to right as the ionic size decreases.

Electronic configuration of monocationic species:
$B{e}^{+}=1s^{2}2s^1$
$B^{+}=1s^{2}2s^2$
$C^{+}=1s^{2}2s^{2}2p^1$
$N^{+}=1s^{2}2s^{2}2p^2$
$F^{+}=1s^{2}2s^{2}2p^4$

$F^{+}$ will have the highest ionization energy. $B{e}^{+}$ will attain noble gas configuration and hence it will have the lowest ionization energy.
$B^{+}$ have higher ionization energy than $C^{+}$ since 2s orbital has more penetration power than 2p.
So, the correct order is $Be<C<B<N<F$