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Q. Assertion (A): Boron has a smaller first ionization enthalpy than beryllium
Reason (R): The penetration of a $2s$ electron to the nucleus is more than the $2p$ electron hence $2p$ electron is more shielded by the inner core of electrons than the $2s$ electrons.

AP EAMCETAP EAMCET 2020

Solution:

Assertion $( A )$ Electronic configuration of boron $(Z=5)$ is $[ He ] 2 s^{2} 2 p^{1} .$ So, in first ionisation $\left(1 I E_{1}\right.$ or $\left.\Delta_{i} H_{1}\right)$ removal will take place from unpaired $p^{1}$-electron. Whereas that of Be will be from paired $2 s^{2}$-electrons which requires more energy.
Electronic configuration of $Be(Z=4):[H e] 2 s^{2}$
So, the Assertion is a correct statement.
Reason $( R ) s$-orbital is being symmetrical in shape (spherical), it shields nuclear force (nuclear charge) strongly. So, $2 p^{1}$-electron of $B$ is experiences lesser nuclear attractive force for ionisation. As a result, $I E_{1}$ or $\Delta_{i} H_{1}: B<$ Be
So, the Reason is correct explanation for Assertion.