Question

Consider the following graph.

What could be the possible reason for a dip at boron $(B)$ in the above plot?

• A. Size of boron is smaller than $Be$
• B. $2p$ electrons are more shielded as compared to $2s$ electrons.
• C. The penetration of a $2s$-electron is less than that of a $2p$-electron
• D. Both (a) and (b)

Answer 1

Answer: (B)

In boron $(B)$, the electron is removed from $p$-orbital. The $2p$-electrons of boron is more shielded from the nucleus by inner core of electrons than the $2s$-electrons. Therefore, it is easier to remove the $2p$-electrons from boron compared to the removal of a $2s$-electron from beryllium. Also, removal of one $2p$-electron from boron leads to a comparatively stable configuration of $2s^2$, where the two electrons are paired.
Hence, the first ionisation energy of boron $(B)$ is lower than both $Be$ and $C$.