(IE) $_{1}$ of Be is greater than that of B but (IE) $_{2}$ of Be is less than that of B.
If true enter 1, else enter 0.

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Solution

The electronic configuratopn of Be is $([H e] 2 s^{2})$ while that of B is $([H e] 2 s^{2} 2 p^{1})$ .
As we can see that the loss of electron from B will be easier as compared to Be which has completely filled s orbital. Besides, loss of electron from p orbital is easier than s orbital.
However, loss of second electron from B is difficult because now, electron has to be removed from a stale configuration unlike in Be where the loss of second electron is easier as this loss will result in a stable configuration.
Hence, the statement is true.

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