Question

What is the hybridization of the central atom in $NOBr$ ?

Answer 1

Hint: We need to know that the redistribution of the energy of orbitals of individual atoms to give orbitals of equivalent energy happens when two atomic orbitals combine to form a hybrid orbital in a molecule. This process is called hybridization. The new orbitals thus formed are known as hybrid orbitals.

Complete answer:

As we know, the $NOBr$ has a chemical name which is nitrosyl bromide. The structure of nitrosyl bromide can be represented as below:

In nitrosyl bromide structure, there are three atoms; nitrogen, oxygen and bromine. Out of which the central atom is nitrogen as you can see in the given structure.

So we need to calculate the hybridization of the central atom in nitrosyl bromide which is Nitrogen in this case: The formula for hybridization can be calculated as shown below:

Hybridization=\[ = \dfrac{1}{2}\left( {V + M - C + A} \right)\]

 \[V\]= number of valence electrons,

\[M\] = monovalent

\[C\]= positive charge

\[A\]= negative charge

So in case of nitrosyl bromide hybridization will be

\[ = \dfrac{1}{2}\left( {5 + 1} \right)\]

\[ = \dfrac{1}{2} \times 6 = 3\]

Thus, on getting value $3$ in the hybridization formula for nitrosyl bromide we can say that it has a hybridization $s{p^2}$. The geometry involved in this hybridization is trigonal bipyramidal. A mixture of s and p orbital forms in trigonal symmetry is maintained at $120^\circ $ .

Note:

In case of $s{p^2}$ hybridization. All the three hybrid orbitals remain in one plane and make an angle of $120^\circ $ with one another. Each of the hybrid orbitals formed has $33.33\% $ s character and $66.66\% $‘p’ character. The molecules in which the central atom is linked to $3$ atoms and is $s{p^2}$ hybridized have a triangular planar shape.