Question

${\left(IE\right)}_1$ of Be is greater than that of B but ${\left(IE\right)}_2$ of (B) is greater than that of Be.If true enter 1, else 0.

Answer 1

Unpaired electron in B is easier to remove than that paired electron in Be. Hence ${\left(IE\right)}_1$ of be > ${\left(IE\right)}_1$ of B, and by ${\left(IE\right)}_1$ ${Be}^{+}$ and $B^{+}$ are formed.
Now reserved is the case; removal of unpaired electron from ${Be}^{+}$ is easier than that of paired electron from $B^{+}$. Hence ${\left(IE\right)}_2$ of B > ${\left(IE\right)}_2$ of Be.