Question

How many d-orbitals are involved in hybridization of $Cr{{O}_{4}}^{2-}$?
a.) 3
b.) 2
c.) 4
d.) 1

Answer 1

When one s, three p and one d-atomic orbitals mix, it forms five $s{{p}^{3}}d$ hybrid orbitals of equal energy called $s{{p}^{3}}d$ hybridization.
When one s, three p and two d orbitals mix, it forms six identical hybrid orbitals called $s{{p}^{3}}{{d}^{2}}$ hybridization.
Similarly, on mixing one s and three p orbitals, three identical hybrid orbitals are formed called $s{{p}^{3}}$ hybridization.

Complete step by step answer:
Let’s find the hybridization of $Cr{{O}_{4}}^{2-}$, we know the atomic number of chromium is 24 and its electronic configuration is [Ar] $4{{s}^{1}}3{{d}^{5}}$, it is half filled and more stable.
We know that the oxidation state of Oxygen is -2.
Let us assume the oxidation state of Chromium in this compound to be x.

& x+4\times (-2)=-2 \\\ & x=+6 \\\ \end{aligned}$$ So, the oxidation state of Chromium is +6. With the oxidation state of +6 the electronic configuration becomes [Ar] $$4{{s}^{0}}3{{d}^{0}}$$ Now, the $$C{{r}^{+6}}$$ ion contains 4s and 3d vacant orbitals. So, one 4s and three 3d orbitals of $$C{{r}^{+6}}$$ ion combine to give four $${{d}^{3}}s$$ hybridized orbitals. Thus, each $$C{{r}^{+6}}$$ ion in $$CrO_{4}^{2-}$$ ion is $${{d}^{3}}s$$ hybridized. While writing, the hybridization ‘d’ is written first when the inner d orbitals are involved in the hybridization. It has no unpaired electron, therefore, it is diamagnetic in nature. One of the four oxide ions, shares both the $$C{{r}^{+6}}$$ ions. In $$Cr{{O}_{4}}^{2-}$$, only three of the five molecular orbitals are used for hybridization and are used for bonding, which means that the four oxygen atoms contribute or donate electrons to 3 orbitals only. **So, the correct answer is “Option A”.** **Note:** Students often get confused that the hybridization of Cr in $$Cr{{O}_{4}}^{2-}$$ is $$s{{p}^{3}}$$,But the correct hybridization for $$Cr{{O}_{4}}^{2-}$$ is $${{d}^{3}}s$$. It is formally a Chromium (VI) compound. Hence, there are no 3d electrons.