Question

$\mathrm{B}$ has a smaller first ionization enthalpy than $\mathrm{Be}$. Consider the following statements:

(i) It is easier to remove an electron from $2\mathrm{p}$ than $2\mathrm{s}$.

(ii) $2\mathrm{p}$ electron of $\mathrm{B}$ is more shielded from the nucleus by the inner core of electrons than the $2\mathrm{s}$ electron of $\mathrm{Be}$

(iii) $2\mathrm{s}$ electron has more penetration power than $2\mathrm{p}$electron.

(iv)The atomic radius of Boron $\left(\mathrm{B}\right)$ is more than$\mathrm{Be}$.

• A. $\left(\mathrm{i}\right),\left(\mathrm{ii}\right),\mathrm{and}\left(\mathrm{iii}\right)$
• B. $\left(\mathrm{i}\right),\left(\mathrm{iii}\right),\mathrm{and}\left(\mathrm{iv}\right)$
• C. $\left(\mathrm{ii}\right),\left(\mathrm{ii}\right),\mathrm{and}\left(\mathrm{iii}\right)$
• D. $\left(\mathrm{i}\right),\left(\mathrm{ii}\right),\mathrm{and}\left(\mathrm{iv}\right)$

Answer 1

The correct option is A

$\left(\mathrm{i}\right),\left(\mathrm{ii}\right),\mathrm{and}\left(\mathrm{iii}\right)$

The explanation for the correct option:

$\mathrm{Be}$: $1{\mathrm{s}}^{2}2{\mathrm{s}}^2$

$\mathrm{B}$: $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^1$

The penetrating power order: $\mathrm{s}>\mathrm{p}>\mathrm{d}>\mathrm{f}$ .

The shielding power order: $\mathrm{s}>\mathrm{p}>\mathrm{d}>\mathrm{f}$.

Hence $\mathrm{s}$ have high shielding power compared to the $\mathrm{p}$.thus electron can easily be removed from $\mathrm{p}$ orbital. It is easier to remove $2\mathrm{p}$ an electron than $2\mathrm{s}$ an electron.

The explanation for the incorrect options:

As we move along the period, the size decreases, as ${\mathrm{Z}}_{\mathrm{eff}}$ increases. Hence the radius of B is smaller than the radius of Be.

Thus, option (A) is correct.