2018 H2 Chem P1 Q20 — Transition elements Solution

The answer is D (3 only).
V is \([\text{Ar}]3\text{d}^3 4\text{s}^2\) and Ca is \([\text{Ar}]4\text{s}^2\). V has 3 more protons, but the extra electrons go into the inner 3d subshell, which shields the 4s electrons poorly. So the 4s electrons of V experience a greater effective nuclear charge, are more strongly attracted to the nucleus, and V has a smaller atomic radius (Data Booklet: V 0.122 nm, Ca 0.197 nm). Smaller atoms of greater mass (\(A_r\): V 50.9, Ca 40.1) pack more mass into each unit volume, so V is denser.
• Statement 1 is false: the outer 4s electrons of V are less shielded, because the diffuse 3d electrons shield poorly.
• Statement 2 does not explain it: V and Ca both have two electrons in the outer (n = 4) shell, and in any case electrons in the same shell do not shield one another effectively, so the number of outer shell electrons does not account for the higher effective nuclear charge.
• Statement 3 is correct: the greater attraction of the nucleus for the outer electrons gives V the smaller atomic radius and hence the higher density.