Question

Which of the following has $s{{p}^{2}}$ hybridisation?
A. $C{{O}_{2}}$
B. $S{{O}_{2}}$
C. ${{N}_{2}}O$
D. $CO$

Answer 1

Hint: We have to find out the hybridisation, which means when different atomic orbitals are combined having different energies to give the equivalent orbitals. Here we will see the combination of one s orbital and two p orbitals to give the $s{{p}^{2}}$ hybridisation.

Complete step by step answer:
- To find hybridisation we need to find the electron pair count, which is denoted by the symbol Z.
- Electron pair count can be found by the formula:
\[Z=\dfrac{1}{2}\times \left( V+M-C+A \right)\]
Where, Z= Electron pair count
V= Valence electron of central atom
M=Monovalent atom attached to central atom,
C= Cationic charge
A= Anionic charge
- Let’s first see about $C{{O}_{2}}$ :
The electronic configuration of carbon is- $1{{s}^{2}}2{{s}^{2}}2{{p}^{4}}$ , so we can see that there is four valence electrons present in it. In this compound there are zero monovalent atoms present. So, the hybridisation in this is:
\[\begin{align}
& Z=\dfrac{1}{2}\times \left( 4+0-0+0 \right) \\
& \implies \dfrac{1}{2}\times 4 \\
& \therefore 2 \\
\end{align}\]
Here, the hybridisation will be $sp$. And the shape will be linear
- Let’s see about $S{{O}_{2}}$ :
The electronic configuration of sulphur is- $\left[ Ne \right]3{{s}^{2}}3{{p}^{4}}$ , so we can see that there is six valence electrons present in it. In this compound there are zero monovalent atoms present. So, the hybridisation in this is:
\[\begin{align}
& Z=\dfrac{1}{2}\times \left( 6+0-0+0 \right) \\
& \implies \dfrac{1}{2}\times 6 \\
& \therefore 3 \\
\end{align}\]
Here, the hybridisation will be $s{{p}^{2}}$. And the shape will be trigonal planar
- Let’s see about CO:
The electronic configuration of carbon is-$1{{s}^{2}}2{{s}^{2}}2{{p}^{4}}$ , so we can see that there is four valence electrons present in it. In this compound there are zero monovalent atoms present. So, the hybridisation in this is:
\[\begin{align}
& Z=\dfrac{1}{2}\times \left( 4+0-0+0 \right) \\
& \implies \dfrac{1}{2}\times 4 \\
& \therefore 2 \\
\end{align}\]
Here, the hybridisation will be sp. And the shape will be linear
- Let’s see about ${{N}_{2}}O$:
We can see here that based on hybridisation theory, the hybridisation states of the atoms are sp, sp and $s{{p}^{3}}$,for the atoms N,N and O. The shape of the molecule will be linear.

Hence, we can say that the correct option is (B), that is $S{{O}_{2}}$ has $s{{p}^{2}}$ hybridisation.

Note: - Many of the properties of a substance are determined by the geometry of the compound.
- We should write the electronic configuration correctly in order to find the valence electron in the atom, and should see whether any cationic or anionic charge is present in the compound to find the hybridisation.