Question

In which of the following the central atom does not use \[s{p^3}\] hybrid orbitals in its bonding?
A) \[{\rm{Be}}{{\rm{F}}_{\rm{3}}}^ - \]
B) \[{\rm{O}}{{\rm{H}}_{\rm{3}}}^ + \]
C) \[N{H_2}^ - \]
D) \[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{OH}}\]

Answer 1

Hint: Here, first we will calculate the hybridization of the central atom of all the compounds. If the value of hybridization of a molecule is 4, then the molecule is \[s{p^3}\] hybridised and the electron geometry of the molecule is tetrahedral.

Formula used: \[H = \dfrac{{V + X - C + A}}{2}\]; Here, H stands for hybridization, V is the count of valence electrons, X is the count of monovalent groups bonded to the central atom, C stands for cationic charge and A stands for the charge of anion.

Complete Step by Step Answer:
Option A is \[{\rm{Be}}{{\rm{F}}_{\rm{3}}}^ - \]. We know that valence electrons of Be atom=2, count of monovalent groups=3, anionic charge=1.
\[H = \dfrac{{2 + 3 + 1}}{2} = 3\]
As the value of hybridization is 3, the hybridization of the molecule is \[s{p^2}\] .
Option B is \[{\rm{O}}{{\rm{H}}_{\rm{3}}}^ + \], We know, valence electrons of Be atom=6, count of monovalent groups=3, cationic charge=1.
\[H = \dfrac{{6 + 3 - 1}}{2} = 4\]
As the value of hybridization is 4, the hybridization of the molecule is \[s{p^3}\].
Option C is \[N{H_2}^ - \], We know, valence electrons of Ne atom=5, count of monovalent groups=2, anionic charge=1.
\[H = \dfrac{{5 + 2 + 1}}{2} = 4\]
As the value of hybridization is 4, the hybridization of the molecule is \[s{p^3}\].
Option D is \[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{OH}}\]. Here, all the carbon atoms are \[s{p^3}\] hybridised.
Therefore, in\[{\rm{Be}}{{\rm{F}}_{\rm{3}}}^ - \], no \[s{p^3}\] hybridised central atom is present.
Hence, Option A is right.

Note: It is to be noted that, if the hybridization is \[s{p^3}\], the electron geometry of the molecule is tetrahedral. And if the molecule has\[s{p^2}\] hybridization, then the molecule's electron geometry is trigonal planar.