B has a smaller first ionization enthalpy than Be. The correct statements are: (I) It is easier to remove 2p electron…

🧠 B < Be in $IE_1$ because B's $2p^1$ electron is higher in energy AND better shielded — but B's atomic radius is SMALLER than Be (statement IV is the false one) Boron has one more proton than beryllium. In the same period, more protons means a tighter atom — so B is smaller than Be. The lower IE of B has nothing to do with size; it comes from orbital energy differences.

🗺️ Evaluate each statement (I) "It is easier to remove the $2p$ electron than the $2s$ electron": $2p$ is at higher energy than $2s$ in a multi-electron atom. So yes, $2p$ is easier. True.

(II) "B's $2p$ electron is more shielded than Be's $2s$ electron": B's $2p$ is shielded by both the $1s^2$ core AND the $2s^2$ subshell sitting beneath it. Be's $2s$ is shielded by only $1s^2$. So B's outer electron experiences more shielding. True.

(III) "$2s$ has more penetration power than $2p$": $2s$ orbitals have a small inner lobe near the nucleus that 2p orbitals don't. So $2s$ penetrates more. True.

(IV) "B has bigger atomic radius than Be": B = 88 pm, Be = 111 pm. Wait — actually Be > B in radius. So the statement is FALSE.

Correct: I, II, III.

⚠️ The trap Statement (IV) is the killer. Students see "B has more electrons than Be" and think "more electrons = bigger atom". But in the same period, the dominant factor is increasing nuclear charge, which shrinks the atom. Be → B → C → ... → F all shrink left to right. So Be is bigger than B, not smaller.

$\boxed{\text{Answer: (c)}}$