Question

$B$ has a smaller first ionization enthalpy than $Be$. Consider the following statements:
(i) It is easier to remove $2p$ electron than $2s$ electron
(ii) $2p$ electron of $B$ is more shielded from the nucleus by the inner core of electrons than the $2s$ electron of $Be$
(iii) $2s$ electron has more penetration power than $2p$ electron
(iv) Atomic radius of $B$ is more than $Be$ (Atomic number $B$=5, $Be$=4) The correct statements are:

• A. (i), (ii), and (iii)
• B. (i), (iii) and (iv)
• C. (ii), (iii) and (iv)
• D. (i), (ii), (iv)

Answer 1

The correct option is A (i), (ii), and (iii)

$Be\left(4\right):1s^{2}2s^2$
$B\left(5\right):1s^{2}2s^{2}2p^1$
The electron in $2p^1$ can easily be extracted.
The penetrating power is of the order: $s>p>d>f$
The shielding power order: s>p>d>f
As we move along the period, the size decreases, as $Z_{eff}$ increases. Hence the radius of $B$ is smaller than the radius of $Be$.