In comparison to boron, beryllium has:

🧠 Be has FEWER protons than B but HIGHER $IE_1$ — that's the famous Be-B anomaly Two facts to compare:

1. Nuclear charge: Be has $Z = 4$, B has $Z = 5$. So Be has fewer protons.
2. First ionization enthalpy: Be ($1s^2\,2s^2$) has filled $2s$, very stable. B has $1s^2\,2s^2\,2p^1$ with a lone $2p$ electron, easy to remove.

Even though Be has lower Z, its filled $2s^2$ subshell makes its outer electron harder to pull. So Be has higher $IE_1$ than B — despite fewer protons.

🗺️ Match against options $IE_1(\ce{Be}) = 899$, $IE_1(\ce{B}) = 801$. Be > B in IE. $Z(\ce{Be}) = 4 < Z(\ce{B}) = 5$.

So Be has lesser nuclear charge AND greater $IE_1$ compared to B.

That is option (b).

⚠️ The trap Option (d) "greater nuclear charge and greater $IE_1$" combines two right-sounding ideas: "Be has higher $IE_1$ than B" (true) and the assumption "more protons = higher IE" (often true, but not always). In Be vs B, the LOWER nuclear charge member wins on $IE_1$ because of the stable $2s^2$ subshell. Always check the nuclear-charge comparison directly — never assume it follows the IE comparison.

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