Question

Which of the following pairs are isostructural 1.so42-and bf4-
2.nh3 and nh4+
3.co2 and co32-
4.ch4 and bf3

Answer 1

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Answer 2

Answer : The correct option is, (1) $SO_4^{2-}\text{ and }BF_4^-$

Explanation :

Isostructural : It is defined as the species that have the same shape and hybridization.

Now we have to determine the hybridization of the following molecules.

Formula used  :

$\text{Number of electron pair}=\frac{1}{2}[V+N-C+A]$

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

(1) The given molecules are, $SO_4^{2-}\text{ and }BF_4^-$

$\text{Number of electron pair in }SO_4^{2-}=\frac{1}{2}\times [6+2]=4$

The number of electron pair are 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral.

$\text{Number of electron pair in }BF_4^-=\frac{1}{2}\times [3+4+1]=4$

The number of electron pair are 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral.

(2) The given molecules are, $NH_3\text{ and }NH_4^+$

$\text{Number of electron pair in }NH_3=\frac{1}{2}\times [5+3]=4$

The number of electron pair are 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral.

But as there are 3 atoms around the central nitrogen atom, the 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 trigonal planer.

$\text{Number of electron pair in }NH_4^+=\frac{1}{2}\times [5+4-1]=4$

The number of electron pair are 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetraheadral.

(3) The given molecules are, $CO_2\text{ and }CO_3^{2-}$

$\text{Number of electron pair in }CO_2=\frac{1}{2}\times [4]=2$

The number of electron pair are 2 that means the hybridization will be $sp$ and the electronic geometry of the molecule will be linear.

$\text{Number of electron pair in }CO_3^{2-}=\frac{1}{2}\times [4+2]=3$

The number of electron pair are 3 that means the hybridization will be $sp^2$ and the electronic geometry of the molecule will be trigonal planer.

(3) The given molecules are, $CH_4\text{ and }BF_3$

$\text{Number of electron pair in }CH_4=\frac{1}{2}\times [4+4]=4$

The number of electron pair are 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral.

$\text{Number of electron pair in }BF_3=\frac{1}{2}\times [3+3]=3$

The number of electron pair are 3 that means the hybridization will be $sp^2$ and the electronic geometry of the molecule will be trigonal planer.

From this we conclude that, $SO_4^{2-}\text{ and }BF_4^-$  pairs have the same shape and hybridization.  So, these pairs are isostructural.

Hence, the correct option is, (1) $SO_4^{2-}\text{ and }BF_4^-$