Question

Hybridization of $P$ in phosphate ion $(PO_4^{3 - })$ is the same as in:
(A) $I\,in\,ICl_4^ -$
(B) $S\,in\,S{O_3}$
(C) $N\,in\,NO_3^ -$
(D) $S\,in\,SO_3^{2 - }$

Answer 1

The hybridization central atom can be determined by finding out the total number of electrons involved in the chemical bonding. Here in this question, the hybridization of phosphorus can be found out by using phosphate ion $(PO_4^{3 - })$ , and one can relate this configuration with the given options and find out the correct option. One can use the formula to calculate the hybridization.
Complete step by step answer:

1. First of all we will see the formula to calculate the total number orbitals participating in the hybridization as follows,
   Hybridization $H = \dfrac{1}{2}(V + M - C + A)$
   Where, $V$ is the number of electrons in the valence shell of the central atom, $M$ is the number of monovalent atoms around the central atom, $C$ is the value of cationic charge and $A$ is the value of the anionic charge.
2. Now let's calculate the hybridization of $PO_4^{3 - }$ the molecule,
   Total valence electrons on central atom P$= 5$, Charge $= - 3$, Number of monovalent atoms attached to the central atom $= 0$. Let’s put the obtained values in the following formula,
   Hybridization of $PO_4^{3 - }$ molecule $= \dfrac{{5 + 3}}{2} = 4$
   The value $4$ stands for the four bonds which shows the hybridization is $s{p^3}$
3. Now the hybridization of $ICl_4^ -$ the molecule,
   Total valence electrons on central atom $= 7$, Charge $= - 1$, Number of monovalent atoms attached to the central atom $= 4$ . Let’s put the obtained values in the following formula,
   Hybridization of $ICl_4^ -$ molecule $= \dfrac{{7 + 4 + 1}}{2} = 6$
   The value $6$ stands for the six bonds which shows the hybridization is $s{p^3}{d^2}$ .
4. For hybridization of $S{O_3}$ molecule,
   Total valence electrons on central atom $= 6$, Charge $= 0$, number of monovalent atoms attached to the central atom $= 0$. Let’s put the obtained values in the following formula,
   Hybridization of $S{O_3}$ molecule $= \dfrac{6}{2} = 3$
   The value $3$ stands for the three bonds which shows the hybridization is $s{p^2}$ .
5. For the hybridization of $NO_3^ -$ molecule,
   Total valence electrons on central atom $= 5$, Charge $= - 1$, number of monovalent atoms attached to the central atom $= 0$ . Let’s put the obtained values in the following formula,
   hybridization of $NO_3^ -$ molecule $= \dfrac{6}{2} = 3$
   The value $3$ stands for the three bonds which shows the hybridization is $s{p^2}$ .
6. For hybridization of $SO_3^{2 - }$ molecule,
   Total valence electrons on central atom $= 6$, Charge $= - 2$, number of monovalent atoms attached to the central atom $= 0$. Let’s put the obtained values in the following formula,
   hybridization of $SO_3^{2 - }$ molecule $= \dfrac{{6 + 2}}{2} = 4$
   The value $4$ stands for the three bonds which shows the hybridization is $s{p^3}$ .
7. After analyzing all the structure hybridization both $SO_3^{2 - }\,and\,PO_4^{3 - }$ have $s{p^3}$ hybridization which shows option ‘D’ is correct.
   Hence option D is correct.

Note:
The orbital hybridization can be elaborated as mixing of atomic orbitals into new hybrid orbitals suitable for the pairing of electrons to form chemical bonds in valence bond theory. Only the number of monovalent atoms surrounding the central atom should be taken into account. Examples of monovalent atoms include Cl, F, I, Na, K, etc.