Question

The species in which central atom uses $s{p^2}$- hybrid orbitals in its bonding is:
A.$P{H_3}$
B.$N{H_3}$
C.$C{H_3}^ +$
D.$\;Sb{H_3}$

Answer 1

Either find out the hybridization of each species and then reach the answer or it can also be told by looking at the structure and then number of bond pairs and lone pairs or any charge present. This could lead to the answer very easily.

Step by step explanation:
In the given question, the $s{p^2}$ hybrid orbitals of species are asked and this is the hybridization in which only 3 bond pairs are used which leads to formation of $s{p^2}$ hybrid orbital.
Now we will see the hybridization of each species and then conclude the answer.
The first species given is PH3 and the hybridization of this can be calculated by using the formula i.e. Hybridization number = $\dfrac{1}{2}(V + X - C + A)$ where V = valency, X = monovalent atom C = cationic charge A = anionic charge
So, calculating for each species, the hybridization number =
PH3 = $\dfrac{1}{2}(5 + 3) = 4 = s{p^3}$ where P valency = 5 and Hydrogen present = 3
NH3 = $\dfrac{1}{2}(5 + 3) = 4 = s{p^3}$
CH3+ = $\dfrac{1}{2}(4 + 3 - 1) = 3 = s{p^2}$
SbH3 = $\dfrac{1}{2}(5 + 3) = 4 = s{p^3}$
So, by finding the hybridization we can conclude that species containing $s{p^2}$ hybrid orbital is $C{H_3}^ +$.
We can also find out the hybridization by looking at the number of bond pairs and lone pairs present on the given species and if you see all the species except $C{H_3}^ +$ has lone pair on them whereas Carbon has positive charge on it.
Now three hydrogen atoms are attached to it means 3 hybrid orbitals are used and 1 lone pair is also present which is used by the orbital. So overall the hybridization becomes $s{p^3}$.
Whereas in $C{H_3}^ +$ there is no lone pair present instead there is positive charge present so only 3 hybrid orbitals are used which leads to ultimately the $s{p^2}$ hybridization.
Therefore, the correct answer is C i.e. $C{H_3}^ +$

Note: To answer similar types of questions, the hybridization concept should be well known which can be calculated by the formula as well as by looking at the structure and then number of bond pairs and lone pairs. So, any method could be used to find out the hybridization of any atom or species by this.