Question

Set of molecules which are isoelectronic is:
a) ${{\text{N}}_{\text{2}}}$ and ${\text{CO}}$
b) ${\text{C}}{{\text{O}}_{\text{2}}}$ and laughing gas $\left( {{{\text{N}}_{\text{2}}}{\text{O}}} \right)$
c) ${\text{CaO}}$ and ${\text{MgS}}$
d) Benzene and Borazine $\left( {{{\text{B}}_{\text{3}}}{{\text{N}}_{\text{3}}}{{\text{H}}_{\text{6}}}} \right)$
(A)- a, b, d
(B)- b, c, d
(C)- c, d
(D)- a, b, c, d

Answer 1

Hint: Iso’ delivers the meaning of the same & ‘electronic’ delivers the meaning of electrons, so isoelectronic species are that species which are having the same no. of electrons in that.

Complete Solution :
In the given question some molecules are given & we have to choose isoelectronic species among them, so for that we have to add all the electrons present in each atom in the molecule.

-In option (a) ${{\text{N}}_{\text{2}}}$ and ${\text{CO}}$ is given, so we have to add all the electrons present in each atom of given molecule.
No. of electrons in ${{\text{N}}_{\text{2}}}$= $7 + 7 = 14$
No. of electrons in ${\text{CO}}$ = $6 + 8 = 14$
Thus, ${{\text{N}}_{\text{2}}}$ and ${\text{CO}}$ molecules are isoelectronic in nature.

-In option (b) ${\text{C}}{{\text{O}}_{\text{2}}}$ and ${{\text{N}}_{\text{2}}}{\text{O}}$ is given, so we have to add all the electrons present in each atom of given molecule.
No. of electrons in ${\text{C}}{{\text{O}}_{\text{2}}}$= $6 + 8 + 8 = 22$
No. of electrons in ${{\text{N}}_{\text{2}}}{\text{O}}$ = $7 + 7 + 8 = 22$
Thus, ${\text{C}}{{\text{O}}_{\text{2}}}$ and ${{\text{N}}_{\text{2}}}{\text{O}}$ molecules are isoelectronic in nature.

-In option (c) ${\text{CaO}}$ and ${\text{MgS}}$ is given, so we have to add all the electrons present in each atom of a given molecule.
No. of electrons in ${\text{CaO}}$= $20 + 8 = 28$
No. of electrons in ${\text{MgS}}$ = $12 + 16 = 28$
Thus, ${\text{CaO}}$ and ${\text{MgS}}$ molecules are isoelectronic in nature.

-In option (d) Benzene $\left( {{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}} \right)$ and Borazine $\left( {{{\text{B}}_{\text{3}}}{{\text{N}}_{\text{3}}}{{\text{H}}_{\text{6}}}} \right)$ is given, so we have to add all the electrons present in each atom of given molecule.
No. of electrons in ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}$= $\left( {6 \times 6} \right) + \left( {6 \times 1} \right) = 36 + 6 = 42$
No. of electrons in ${{\text{B}}_{\text{3}}}{{\text{N}}_{\text{3}}}{{\text{H}}_{\text{6}}}$ = $\left( {3 \times 5} \right) + \left( {3 \times 7} \right) + \left( {6 \times 1} \right) = 15 + 21 + 6 = 42$
Thus, Benzene $\left( {{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}} \right)$ and Borazine $\left( {{{\text{B}}_{\text{3}}}{{\text{N}}_{\text{3}}}{{\text{H}}_{\text{6}}}} \right)$ molecules are isoelectronic in nature.
So, the correct answer is “Option D”.

Note: In this question some of you may do wrong calculation if you are only taking outermost shell or valence shell no. of electrons because all atoms which are present in a molecule may have different valencies. So, always take the whole no. of electrons present in each atom then add them.