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

Answer 1

Answer: (B)

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