In which of the following species the interatomic bond angle is \[109^\circ 28\prime \;\].
A.\[N{H_3},{\rm{ }}{\left( {B{F_4}} \right)^{ - 1}}\;\]
B.\[{\left( {N{H_4}} \right)^ + },{\rm{ }}B{F_3}\]
C.\[N{H_3},{\rm{ }}B{F_4}\]
D.\[{\left( {N{H_2}} \right)^{ - 1}},{\rm{ }}B{F_3}\]

Hint: Bond angle 109°28′
is demonstrated by a regular tetrahedral molecular geometry.
When the central metal atom having a tetrahedral shape has 4 bond pairs and no lone pairs, it will demonstrate this bond angle.

Complete step by step solution:The angle between two bonds arising from the same atom in a covalent species is comprehended as the bond angle.
The given bond angle is in species with regular tetrahedral geometry and they carry a hybridization of\[s{p^3}\].
For this, the steric number is 4. The steric number is the no.of atoms, groups, or lone pairs for the central metal atom.
For the given bond angle, there are generally 4 bonding pairs of electrons and no lone pairs.
So currently we have to study the provided molecules and attempt to find out the number of bond pairs in them.
Out of the given options, \[N{H_3},{\rm{ }}{\left( {B{F_4}} \right)^{ - 1}}\;\]both have interatomic bond angles of \[109^\circ 28\prime \;\].
\[N{H_3}\;\]
There is an N atom which is the central atom. It has 5 valence electrons.
Three electrons undergo bond formation with each of the three hydrogen atoms and the rest lone pair of electrons is there.
So, there are three bond pairs and one lone pair.
So, it has a pyramidal structure due to the repulsion between bond pairs and the lone pair.
So, the bond angle is \[109^\circ 28\prime \;\].
\[{\left( {B{F_4}} \right)^{ - 1}}\;\]
The shape is tetrahedral and, in the molecule, there are four bond pairs and zero lone pairs.
There are four B-F bonds.
So, the bond angle is \[109^\circ 28\prime \;\].


Image: Structure of \[N{H_3},{\rm{ }}{\left( {B{F_4}} \right)^{ - 1}}\;\]


So, option A is correct.

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