2019 H2 Chem P1 Q26 — Transition elements Solution

The answer is B (\([\text{Kr}]\,4d^2 5s^0\)).

Mo (Z = 42) is directly below Cr, so like Cr (\([\text{Ar}]\,3d^5 4s^1\)) its configuration is \([\text{Kr}]\,4d^5 5s^1\).
In \(\text{MoF}_4\) the ion is \(\text{Mo}^{4+}\), so 4 electrons are removed. As for the first-row transition elements, the outer s electrons are removed before the d electrons: first the 5s electron, then three 4d electrons.
\(\text{Mo}^{4+}\): \([\text{Kr}]\,4d^2 5s^0\)

• A is wrong: it keeps two 5s electrons; the 5s electrons are removed first, and Mo has only one 5s electron to begin with.
• C is wrong: \(4d^4\) leaves 4 of the 6 valence electrons, which is \(\text{Mo}^{2+}\), not \(\text{Mo}^{4+}\).
• D is wrong: the electron count fits \(\text{Mo}^{4+}\), but it removes the 4d electrons and keeps 5s electrons, the reverse of the order in which electrons are lost.