Question

The type of hybrid orbital used by the oxygen atom in $C{l_2}O$ molecule is
$A)Sp$
$B)S{p^3}$
$C)S{p^2}$
$D)None$

Answer 1

The electronic configuration of oxygen is $1{s^2},2{s^2},2{p^4}$ . The hybridization concept is used to explain the concept. The oxygen atom is bound with chlorine and forms $2$ bond with chlorine and has $2$ lone pairs.

Complete answer:
Hybridization is an interaction of numerically joining at least two nuclear orbitals from a similar molecule to shape a totally new orbital not the same as its parts and thus being called a hybrid orbital.
The first nuclear orbitals are comparative in energy, yet not the same. The subsequent mixture orbitals are identical in energy to each other and are arranged so they can frame bonds with different ions.
The electronic configuration of oxygen is
$= 1{s^2},2{s^2},2P{x^2},2P{y^1},2P{z^1}$
It structures two bonds with chlorine and has two lone pairs of electrons present. Thus, a sum of four hybrid orbitals are needed.
So, the hybridization of oxygen atoms is $s{p^3}$ .
The correct answer is $B)$ .

Additional Information:
From all conceivable half and half orbitals of s and p the tetrahedral orbitals are the most appropriate for framing solid bonds. Even in deviated particles, the bond points around a tetrahedral carbon are normally somewhere in the range of $106$ and $113$ degrees. To examine bond points that vary enormously from the tetrahedral values, like cyclopropane, and note that the strain in these bonds makes them less steady, and hence simpler to break.

Note:
The resultant shape is $V$ -Type . We realize that in sp3 structure bond point is $109.47$ however a highly electro-negative oxygen molecule pulls in $Cl - O$ fortified electrons toward oxygen making repulsion.