Question

An ion has electronic configuration $\left[Kr\right]4d^4$ placed in group 6 in the modern periodic table. The ion is:

• A. $C{r}^{2+}$
• B. $M{n}^{3+}$
• C. $M{o}^{2+}$
• D. $T{c}^{3+}$

Answer 1

The correct option is C $M{o}^{2+}$

Since the metal belongs to 6th group so it has 6 valence electrons in its ground state. As the given configuration has 2 valence electrons less i.e. element is in (II) oxidation state.
The ground state electronic configuration would be: $\left[Kr\right]4d^{5}5s^1$ and the element is molybdenum $\left(Mo\right)$.