Question

The hybridization involved in complex [ZnBr4]2-

1.sp3
2.sp2
3.dsp2
4.sp3d2​

Answer 1

To find:

Hybridisation of complex [Zn(Br)4]2- ?

Solution:

Oxidation state of Zn be x:

$\therefore \: x + 4( - 1) = - 2$

$\implies \: x + (- 4)= - 2$

$\implies \: x = - 2 + 4$

$\implies \: x = + 2$

So, Zn is in +2 oxidation state.

Now, Zn has atomic number = 30.

$\therefore \: Zn \rightarrow \: 1 {s}^{2} \: 2 {s}^{2} \: 2 {p}^{6} \: 3{s}^{2} \: 3 {p}^{6} \: 4 {s}^{2} \: 3{d}^{10}$

$\implies \: {Zn}^{ + 2} \rightarrow \: 1 {s}^{2} \: 2 {s}^{2} \: 2 {p}^{6} \: 3{s}^{2} \: 3 {p}^{6} \: \: 3{d}^{10}$

So, d¹⁰ orbital in Zn+2 means that the d orbital is full filled.

So four Br can donate electrons in the 4s and 4p orbitals.

Hence, the required hybridisation will be:

$\boxed{ \bold{hybridisation = s {p}^{3} }}$

• We can also understand that this coordinate complex will be a TETRAHEDRAL complex as it's hybridisation is sp³.

Hope It Helps.