Question

Determine the set of isoelectronic species.
A. ${\text{C}} _2^{2 - },{\text{N}}{{\text{O}}^ + },{\text{C}}{{\text{N}}^ - },{\text{O}}_2^{2 + }$
B. ${\text {CO}}, \,{\text{NO}}\,,{{\text{O}}_2},\,{\text{CN}}$
C. ${\text{C}}{{\text{O}}_2},\,{\text{N}}{{\text{O}}_2},{{\text{O}}_2},{{\text{N}}_2}{{\text{O}}_3}$
D. ${\text {CO}}, \,{\text{C}}{{\text{O}}_2},\,{\text{NO}}\,,\,{\text{N}}{{\text{O}}_2}$

Answer 1

Hint: The species having the same number of electrons are known as isoelectronic species. The total count of the electrons of a species is the sum of atomic number and negative charge and subtraction of positive charge.

Complete step by step answer:
Determine the total electrons in each of the given species as follows:
For a neutral molecule sum of the atomic number gives the total electron count.
For a positive charged molecule, subtract the positive charge from the atomic number to get the total electron count.
For a negatively charged molecule, add the negative charge to the atomic number to get the total electron count.
The atomic number of carbon is $6$, nitrogen is $7$ and oxygen is $8$.

SET-A

Total electrons in ${\text{C}}_2^{2 - }$,
${\text{C}}_2^{2 - }\, = \,6 + 6 + 2$
${\text{C}}_2^{2 - }\, = \,14$

Total electrons in ${\text{N}}{{\text{O}}^ + }$,
${\text{N}}{{\text{O}}^ + }\, = \,7 + 8 - 1$
${\text{N}}{{\text{O}}^ + }\, = \,14$

Total electrons in ${\text{C}}{{\text{N}}^ - }$,
${\text{C}}{{\text{N}}^ - }\, = \,6 + 7 + 1$
${\text{C}}{{\text{N}}^ - }\, = \,14$

Total electrons in ${\text{O}}_2^{2 + }$,
${\text{O}}_2^{2 + }\, = \,8 + 8 - 2$
${\text{O}}_2^{2 + }\, = \,14$

All the molecules of SET-A have an electron count 14. So, the SET-A represents the isoelectronic species.

SET-B

Total electrons in ${\text{CO}}$,
${\text{CO}}\, = \,6 + 8$
${\text{CO}}\, = \,14$

Total electrons in ${\text{NO}}$,
${\text{NO}} = \,7 + 8$
${\text{NO}}\, = \,15$

Total electrons in ${{\text{O}}_{\text{2}}}$,
${{\text{O}}_2} = \,8 + 8$
${{\text{O}}_2}\, = \,16$

Total electrons in ${\text{CN}}$,
${\text{CN}} = \,6 + 7$
${\text{CN}} = 13$

All the molecules of SET-B have a different electron count. So, the SET-B does not represent the isoelectronic species.

SET-C

Total electrons in ${\text{C}}{{\text{O}}_2}$,
${\text{C}}{{\text{O}}_2}\, = \,6 + 8\, + 8$
${\text{C}}{{\text{O}}_2}\, = \,22$

Total electrons in ${\text{N}}{{\text{O}}_2}$,
${\text{N}}{{\text{O}}_2} = \,7 + 8\, + 8$
${\text{N}}{{\text{O}}_2} = 23$

Total electrons in ${{\text{O}}_{\text{2}}}$,
${{\text{O}}_2} = \,8 + 8$
${{\text{O}}_2}\, = \,16$

Total electrons in ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{3}}}$,
${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{3}}} = \,7 + 7 + 8 + 8 + 8$
${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{3}}} = \,38$

All the molecules of SET-C have a different electron count. So, the SET-C does not represent the isoelectronic species.

SET-D

Total electrons in ${\text{CO}}$,
${\text{CO}}\, = \,6 + 8$
${\text{CO}}\, = \,14$

Total electrons in ${\text{C}}{{\text{O}}_2}$,
${\text{C}}{{\text{O}}_2}\, = \,6 + 8\, + 8$
${\text{C}}{{\text{O}}_2}\, = \,22$

Total electrons in ${\text{NO}}$,
${\text{NO}} = \,7 + 8$
${\text{NO}}\, = \,15$

Total electrons in${\text{N}}{{\text{O}}_2}$,
${\text{N}}{{\text{O}}_2} = \,7 + 8\, + 8$
${\text{N}}{{\text{O}}_2} = 23$

All the molecules of SET-D have a different electron count. So, the SET-D does not represent the isoelectronic species.

Thus, the correct option is A.


Note:
The subscript represents the number of atoms, so the atomic number should be added as many times as denoted in the subscript. Superscript represents the charge that will be added in case of negative charge and subtracted in case of a positive charge.