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 $2s$ electrons 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. $\left(i\right),\left(iii\right)$ and $\left(iv\right)$
• B. $\left(i\right),\left(ii\right)$ and $\left(iii\right)$
• C. $\left(i\right),\left(ii\right)$ and $\left(iv\right)$
• D. $\left(ii\right),\left(iii\right)$ and $\left(iv\right)$

Answer 1

The correct option is B $\left(i\right),\left(ii\right)$ and $\left(iii\right)$

$Be\Rightarrow 1s^{2}2s^2$
$B\Rightarrow 1s^{2}2s^{2}2p^1$

It is easier to remove $2p$ electron than $2s$ electron due to less penetration effect of $2p$ than $2s.$
$2p$ electron of Boron is more shielded from the nucleus by the inner core of electron than the $2s$ electron of $Be$
s electrons have more penetration than p electrons.
$B$ has a smaller size than $Be$
Correct option is (b).