Question

Which of the following should be the correct order of the second ionisation potential of Li, Be, B and C?

• A. Li $<$ Be $>$ B $<$ C
• B. Be $<$ B $>$ C $<$ Li
• C. Be $<$ C $<$ B $<$ Li
• D. Li $<$ C $<$ B $<$ Be

Answer 1

The correct option is C Be $<$ C $<$ B $<$ Li

The $I{E}_2$ of $Be$ is the lowest then $C$. For $I{E}_2$, $e^{-}s$ are removed from $2s$ orbital of $Be$ and in $e$ it is removed from $2p$ and then $B$, removal of $e^{-}$ from a filled shell is quite tough. $e^{-}s$ of $Li$ are also present in $1s$ after the first removal of $e^{-}$.

Hence, $Be<C<B<Li$