Question

Among the following series of transition of metal ions, the one where all metal ions have $3{{d}^{2}}$ electronic configuration is:
A. $T{{i}^{+}},{{V}^{4+}},C{{r}^{6+}},M{{n}^{7+}}$
B. $T{{i}^{4+}},{{V}^{3+}},C{{r}^{2+}},M{{n}^{3+}}$
C. $T{{i}^{2+}},{{V}^{3+}},C{{r}^{4+}},M{{n}^{5+}}$
D. $T{{i}^{3+}},{{V}^{2+}},C{{r}^{3+}},M{{n}^{4+}}$

Answer 1

Hint: d- block elements are also called transition metals. In transition elements the valence electrons are present in d-orbital. We have to know the electronic configuration of the given metal atoms. The positive charge on the transition metal ion indicates the number of electrons lost by the respective transition metal atom.

Complete step by step answer:
- In the question it is asked to find the transition metal ions which have an electronic configuration of $3{{d}^{2}}$ .
- We have to check the charge on the given metal atoms and have to match with the electronic configuration of $3{{d}^{2}}$ .
- After losing the electrons we have to count the number electrons present in the transition metal atoms.
- Coming to Titanium (Ti), the atomic number of Titanium (Ti) is 22.
- The electronic configuration of Titanium is $[Ar]3{{d}^{2}}4{{s}^{2}}$ . Means to get the electronic configuration of $3{{d}^{2}}$ , titanium has to lose two electrons.
- After losing two electrons from 4s orbital Titanium will get $3{{d}^{2}}$ electronic configuration. So, $T{{i}^{2+}}$ has $3{{d}^{2}}$ electronic configuration.
- Coming to Vanadium (V), the atomic number of Vanadium is 23.
- The electronic configuration of Vanadium is $[Ar]3{{d}^{3}}4{{s}^{2}}$ . Means to get the electronic configuration of $3{{d}^{2}}$ , vanadium has to lose three electrons.
- After losing two electrons from 4s orbital and one electron from 3d orbital Vanadium will get $3{{d}^{2}}$ electronic configuration. So, ${{V}^{3+}}$ has $3{{d}^{2}}$ electronic configuration.
- Coming to Chromium (Cr) the atomic number of Chromium is 24.
- The electronic configuration of Chromium is $[Ar]3{{d}^{5}}4{{s}^{1}}$ . Means to get the electronic configuration of $3{{d}^{2}}$ , Chromium has to lose four electrons.
- After losing three electrons from 3d orbital and one electron from 4s orbital Chromium will get $3{{d}^{2}}$ electronic configuration. So, $C{{r}^{4+}}$ has $3{{d}^{2}}$ electronic configuration.
- Coming to Manganese (Mn) the atomic number of Chromium is 25.
- The electronic configuration of Manganese is $[Ar]3{{d}^{5}}4{{s}^{2}}$ . Means to get the electronic configuration of $3{{d}^{2}}$ , Manganese has to lose five electrons.
- After losing three electrons from 3d orbital and two electrons from 4s orbital Manganese will get $3{{d}^{2}}$ electronic configuration. So, $M{{n}^{5+}}$ has $3{{d}^{2}}$ electronic configuration.

So, the correct answer is “Option C”.

Note: Transition metal atoms lose or donate some electrons and form bonds with other ligands to form coordination complexes. The ligands accept the electrons donated by transition metal atoms and coordinate with them.