Arrange $L \dot{I}, \mathrm{Be}, \mathrm{B}$ and $\mathrm{C}$ in decreasing order of their first ionisation…
Arrange $L \dot{I}, \mathrm{Be}, \mathrm{B}$ and $\mathrm{C}$ in decreasing order of their first ionisation enthalpies.
Li $>$ B $>$ Be $>$ C
C $>$ Li $>$ Be $>$ B
$\mathrm{C}>\mathrm{Be}>\mathrm{B}>\mathrm{Li}$
$\mathrm{C}>\mathrm{B}>\mathrm{Be}>\mathrm{Li}$
Solution
As we go from left to right in period the ionisation potential increases but $\mathrm{Be}$ and $\mathrm{B}$ are exception as the Be has $1 s^2, 2 s^2$ and boron has $1 s^2, 2 s^2, 2 p^1$ so it's easy to remove electron from boron than berylium decreasing order of their first ionisation enthalpies.
$
\mathrm{C}>\mathrm{Be}>\mathrm{B}>\mathrm{Li}
$
So, option (3) is correct