Question

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

Answer 1

Hint: 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.
\[\begin{align}
  & x+4\times (-2)=-2 \\
 & x=+6 \\
\end{align}\]
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.