
Which pair represents isostructural species?
A.$C{H_3}^ + $ and $C{H_3}^ - $
B.$N{H_4}^ + $ and $N{H_3}$
C.$S{O_4}^{2 - }$ and $B{F_4}^ - $
D.$N{H_2}^ - $ and $Be{F_2}$
Answer
582.3k+ views
Hint: Isostructural species: Those species which have the same shape and hybridization, are known as isostructural species. For example: $Xe{F_2}$ and $I{F_2}^ - $ they are isostructural species as they both are $s{p^3}d$ hybridized molecules.
Complete step by step answer:
Let us first talk on isostructural species: Those species which have the same shape and hybridization, are known as isostructural species. For example: $Xe{F_2}$ and $I{F_2}^ - $ they are isostructural species as they both are $s{p^3}d$ hybridized molecules.
Hybridisation: It is defined as the process of combining two or more atomic orbitals from the same atom to form a new orbital having different components.
In $s{p^3}$ hybridisation the mixing of one $2s$ orbital with three $2p$ orbitals takes place to form four hybrid orbitals.
In $s{p^3}{d^3}$ hybridisation the mixing of one $s$ orbital with three $p$ orbitals and three $d$ orbitals takes place to form seven hybrid orbitals.
In ${d^3}s{p^3}$ hybridisation the mixing of three $d$ orbitals with one $s$ orbital and three $p$ orbitals takes place to form seven hybrid orbitals.
In $ds{p^2}$ hybridisation the mixing of one $d$ orbital with one $s$ orbital and two $p$ orbitals takes place to form four hybrid orbitals.
$C{H_3}^ - $ has hybridization $s{p^3}$ in which three bonds one lone pair is there so the shape of this molecule is trigonal pyramidal.
$C{H_3}^ + $ has hybridization $s{p^2}$ in which only three bond pairs are there so the shape of the molecule is trigonal planar.
Thus, $C{H_3}^ - $ and $C{H_3}^ + $ are not isostructural species.
$N{H_4}^ + $ has hybridization $s{p^3}$ in which four bond pairs are present in the molecule so the shape of the molecule is tetrahedral.
$N{H_3}$ has hybridization $s{p^2}$ in which three bond pairs are present in the molecule so the shape of the molecule is trigonal pyramidal.
Thus, $N{H_4}^ + $ and $N{H_3}$ are not isostructural species.
$S{O_4}^{2 - }$ has hybridisation $s{p^3}$ in which four bond pairs are present in the molecule so the shape of the molecule is tetrahedral.
$B{F_4}^ - $ has hybridization $s{p^3}$ in which four bond pairs are present in the molecule so the shape of the molecule is tetrahedral.
Thus, $S{O_4}^{2 - }$ and $B{F_4}^ - $ are isostructural species.
$N{H_2}^ - $ and $Be{F_2}$ are not isostructural because $N{H_2}^ - $ has $s{p^2}$ hybridisation and $Be{F_2}$ has $sp$ hybridization.
Hence the correct option is (C) $S{O_4}^{2 - }$ and $B{F_4}^ - $.
Note:
For the species to be isostructural hybridization should be the same i.e. shape of molecules should be same.
Isoelectronic species: The species which have the same number of valence electrons, are known as isoelectronic species.
Complete step by step answer:
Let us first talk on isostructural species: Those species which have the same shape and hybridization, are known as isostructural species. For example: $Xe{F_2}$ and $I{F_2}^ - $ they are isostructural species as they both are $s{p^3}d$ hybridized molecules.
Hybridisation: It is defined as the process of combining two or more atomic orbitals from the same atom to form a new orbital having different components.
In $s{p^3}$ hybridisation the mixing of one $2s$ orbital with three $2p$ orbitals takes place to form four hybrid orbitals.
In $s{p^3}{d^3}$ hybridisation the mixing of one $s$ orbital with three $p$ orbitals and three $d$ orbitals takes place to form seven hybrid orbitals.
In ${d^3}s{p^3}$ hybridisation the mixing of three $d$ orbitals with one $s$ orbital and three $p$ orbitals takes place to form seven hybrid orbitals.
In $ds{p^2}$ hybridisation the mixing of one $d$ orbital with one $s$ orbital and two $p$ orbitals takes place to form four hybrid orbitals.
$C{H_3}^ - $ has hybridization $s{p^3}$ in which three bonds one lone pair is there so the shape of this molecule is trigonal pyramidal.
$C{H_3}^ + $ has hybridization $s{p^2}$ in which only three bond pairs are there so the shape of the molecule is trigonal planar.
Thus, $C{H_3}^ - $ and $C{H_3}^ + $ are not isostructural species.
$N{H_4}^ + $ has hybridization $s{p^3}$ in which four bond pairs are present in the molecule so the shape of the molecule is tetrahedral.
$N{H_3}$ has hybridization $s{p^2}$ in which three bond pairs are present in the molecule so the shape of the molecule is trigonal pyramidal.
Thus, $N{H_4}^ + $ and $N{H_3}$ are not isostructural species.
$S{O_4}^{2 - }$ has hybridisation $s{p^3}$ in which four bond pairs are present in the molecule so the shape of the molecule is tetrahedral.
$B{F_4}^ - $ has hybridization $s{p^3}$ in which four bond pairs are present in the molecule so the shape of the molecule is tetrahedral.
Thus, $S{O_4}^{2 - }$ and $B{F_4}^ - $ are isostructural species.
$N{H_2}^ - $ and $Be{F_2}$ are not isostructural because $N{H_2}^ - $ has $s{p^2}$ hybridisation and $Be{F_2}$ has $sp$ hybridization.
Hence the correct option is (C) $S{O_4}^{2 - }$ and $B{F_4}^ - $.
Note:
For the species to be isostructural hybridization should be the same i.e. shape of molecules should be same.
Isoelectronic species: The species which have the same number of valence electrons, are known as isoelectronic species.
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