Respuesta :
Answer : The correct option is, (B) bent, 2, and [tex]sp^3[/tex]
Explanation :
Formula used :
[tex]\text{Number of electron pair}=\frac{1}{2}[V+N-C+A][/tex]
where,
V = number of valence electrons present in central atom
N = number of monovalent atoms bonded to central atom
C = charge of cation
A = charge of anion
The given molecule is, [tex]NH_2^-[/tex]
[tex]\text{Number of electrons}=\frac{1}{2}\times [5+2+1]=4[/tex]
Number of bond pair = 2
Number of lone pair = 2
The total number of electron pair are 4 that means the hybridization will be [tex]sp^3[/tex] and the electronic geometry of the molecule will be tetrahedral.
But as there are 2 atoms around the central oxygen atom, the third and fourth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be bent or angular.
Hence, the correct option is, (B) bent, 2, and [tex]sp^3[/tex]
The shape , the number of lone pairs on nitrogen, and the hybridization at nitrogen are,
(B) bent, 2, and sp³
How to find Valence electrons?
The molecule is, [tex]NH_2^-[/tex]
Formula to be used : 1/2 * Valence electrons + monovalent atom + charge of cation - charge of anion
Thus,
Number of valance electrons in [tex]NH_2^-[/tex] = 1/2 * (5+2+1) = 4
Number of bond pair = 2
Number of lone pair = 2
Hybridization = sp³
Due to the repulsion between lone pair and bond pair of electrons is more and hence the molecular geometry will be bent or angular.
Hence, the correct option is, (B) bent, 2, and sp³.
Find more information Hybridization here:
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