Question

How many atomic orbitals of each type mix to form hybrid orbitals of the central atom in cl2o?

Answer 1

One s and three p atomic orbitals mix to form hybrid orbitals of oxygen in ${\mathbf{C}}{{\mathbf{l}}_{\mathbf{2}}}{\mathbf{O}}$.

Further explanation:

Prediction of hybridization:

Hybridization is the process in which atomic orbitals of the same atom combine to form an equal number of hybrid orbitals with same energy and same shape.

It can be predicted by the calculation of the number of hybrid orbitals (X) formed by the atom. The formula to calculate X is as follows:

${\text{X = }}\frac{1}{2}\left[ {{\text{VE}} + {\text{MA}} - c + a} \right]$                              ...... (1)

Here,

VE is the total number of valence electrons of the central atom.

MA is the number of monovalent atoms/groups that surround the central atom.

c is the charge on the cation if the given species is a polyatomic cation.

a is the charge on the anion if the given species is a polyatomic anion.

In MA only monovalent species should be considered and for divalent atoms or groups, MA is equal to zero.

Usually, the least electronegative atom is considered as the central atom. The calculation of hybridization is,

1. If the value of X comes out to be 2, it means two hybrid orbitals are to be formed and therefore the hybridization is sp.

2. If the value of X comes out to be 3, it means three hybrid orbitals are to be formed and therefore the hybridization is$s{p^2}$.

3. If the value of X comes out to be 4, it means four hybrid orbitals are to be formed and therefore the hybridization is$s{p^3}$.

4. If the value of X comes out to be 5, it means five hybrid orbitals are to be formed and therefore the hybridization is$s{p^3}d$.

5. If the value of X comes out to be 6, it means six hybrid orbitals are to be formed and therefore the hybridization is$s{p^3}{d^2}$.

In ${\text{C}}{{\text{l}}_{\text{2}}}{\text{O}}$ , oxygen is the central atom and its ground state electronic configuration is $1{s^2}2{s^2}2{p^4}$. So it has 6 valence electrons in it.

Since chlorine is a monovalent atom thus the total number of monovalent atoms surrounding the central atom (MA) is 2.

As the molecule $\left( {{\text{C}}{{\text{l}}_2}{\text{O}}} \right)$ is a neutral species, therefore, the value of a and c is 0.

Substitute these values in equation (1).

$\begin{aligned}&{X = }}\frac{1}{2}\left[ {{\text{6}} + {\text{2}} - 0 + 0} \right] \\ &= \frac{1}{2}\left[ 8 \right] \\&= 4 \\\end{aligned}$

The value of X comes out to be 4 so four hybrid orbitals are formed and therefore the hybridization of ${\text{C}}{{\text{l}}_2}{\text{O}}$ is $s{p^3}$. It indicates that 4 atomic orbitals are utilized in the formation of 4 hybrid orbitals. So one s orbital and three p orbitals are used to make four hybrid orbitals of the oxygen atom.

Learn more:

1. Molecular shape around the central atoms in the amino acid glycine: brainly.com/question/4341225

2. What causes water molecules to have a bent shape? brainly.com/question/1979769

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Covalent bonding and molecular structure

Keywords: Hybridization, oxygen, sp3, s, p, four, X, 4, hybrid orbitals, atomic orbitals, one s orbital, three p orbital, Cl2O, 6, 2, 0, 1s22s22p4.

Answer 2

In here Oxygen is the central atom. It makes two bonds with Cl and has two lone pairs. Since, the shape is bent and the hybridization is sp3. Molecular geometry is a bit dissimilar from hybridization. Hybridization is reliant on the number of bonds and lone pairs. Since O has two bonds with Cl, its hybridization is sp3. It is like is this: 1 lone pair/bond = s. 2 lone pairs/bond = sp 3 lone pairs/bonds = sp2, etc. molecular geometry, you count the number of bonds and lone pairs. This has two bonds and 2 lone pairs so if they were all bonds, the molecule would be tetrahedral.