Question

Which set hybridization is correct for the following compounds
A) \[N{O_2} - sp\] , \[S{F_4} - s{p^2}\] , \[P{F_6} - s{p^3}\]
B) \[N{O_2} - sp\] , \[S{F_4} - s{p^3}d\] , \[P{F_6} - s{p^3}{d^2}\]
C) \[N{O_2} - s{p^2}\] , \[S{F_4} - s{p^3}\] , \[P{F_6} - {d^2}s{p^3}\]
D) \[N{O_2} - s{p^3}\] , \[S{F_4} - s{p^3}{d^2}\] , \[P{F_6} - s{p^3}{d^2}\]

Answer 1

Hint: Here, we have to check the hybridization of the given compounds. For that, we have to use a formula of hybridization. There is another way of identifying hybridization of a compound by counting the number of groups that surround the central atom.

Formula Used:The formula of hybridization is,
\[H = \dfrac{{V + X - C + A}}{2}\]
Here, V stands for the count of valence electrons, X stands for count of monovalent atoms of the compound, C stands for charge of cation and A is for the charge of anion.

Complete step by step solution:Let’s first check the hybridization of \[{\rm{N}}{{\rm{O}}_{\rm{2}}}^ + \] .We know, the count of valence electrons of Nitrogen atom is 5 and the monovalent atoms in the compound is zero and cationic charge is 1.
\[H = \dfrac{{5 - 1}}{2} = 2\]
Therefore, \[{\rm{N}}{{\rm{O}}_{\rm{2}}}^ + \]molecule is \[sp\] hybridized
Let’s check the hybridization of \[S{F_4}\] . We know, the count of valence electrons of the Sulphur atom is 6 and the monovalent atoms in the compound is four.
\[H = \dfrac{{6 + 4}}{2} = 5\]
If H value is 5, the molecule is \[s{p^3}d\] hybridized.
Let’s check the hybridization of \[P{F_6}^ - \] . We know, the count of valence electrons of the Phosphorus atom is 5 and the monovalent atoms in the compound is six and the anionic charge is 1.
\[H = \dfrac{{5 + 6 + 1}}{2} = 6\]
If H value is 6, the molecule is \[s{p^3}{d^2}\]hybridized .

Therefore, option B is right.

Note: Always remember that sp hybridization indicates the molecular geometry of linear. If the hybridization is \[s{p^3}d\], the electron geometry is trigonal pyramidal. And the \[s{p^3}{d^2}\]hybridization is for octahedral geometry.