Solution

The electronic configuration of $$B$$ and $$Be$$ are:

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

(I) $$2s$$ - orbital experienced more $$Z_{eff}$$ than $$2p$$ thefefore $$2p$$ electron could be easily removed than $$2s$$.

(II) In case of $$B$$ there are total 4 inner core electrons but in $$Be$$ only $$2e^-$$ (in $$1s^2$$) therefore $$2p$$ more shielded in $$B$$ than $$2s$$ of $$Be$$.

(III) $$s$$-orbital has spherical shape through that $$Z_{eff}$$ work $$10\%$$ but p-orbital is dumbell shape therefore its $$Z_{eff}$$ less and penetrating power less than $$2s$$.

(IV) Atomic radius of $$B$$ less than $$Be$$ due more $$Z_{eff}$$.

According to the above explanation only (I), (II) and (III) are correct,

Hence, option $$C$$ is correct.