Question

The order of size of $sp$ , $s{p_2}$ , $s{p_3}$ is -
A.$s{p_3} < s{p_2} < sp$
B.$sp < s{p_2} < s{p_3}$
C.$s{p_2} < sp < s{p_3}$
D.$s{p_2} < s{p_3} < sp$

Answer 1

Hint: It is known that an orbital is a three dimensional description of the most likely location of an electron around an atom. They are just combined when bonds are formed in between atoms in a molecule. It basically describes properties characteristic of no more than two electrons in the vicinity of an atomic nucleus or of a system of nuclei as in a molecule.

Complete answer:
We know that the s orbital is basically a spherically symmetric around the nucleus of the atom, like a hollow ball which is made of rather fluffy material with the nucleus at its centre. And, as the energy levels increase, the electrons which are located further from the nucleus, so the orbitals get bigger.
Now, not all electrons are going to inhabit the s orbitals. In the first energy level, the only orbital which is available to electrons is the 1s orbital. But however, at the second level, there are also orbitals which are called 2p orbitals in addition to the 2s orbital.
And unlike the s orbital, a p orbital points in a particular direction.
Now, looking at the question -
As the p character increases, it is noticed that the size of the hybrid orbital also increases. So, the increasing order of the size of the hybrid orbital is $sp < s{p_2} < s{p_3}$ .

Therefore , the correct option is (b). $sp < s{p_2} < s{p_3}$.

Note:
Orbitals, it basically describes properties characteristic of no more than two electrons in the vicinity of an atomic nucleus or of a system of nuclei as in a molecule.