Question

Which of the following is isoelectronic as well as has same structure as that of \[{{\rm{N}}_{\rm{2}}}{\rm{O}}\]
A) \[{{\rm{N}}_{\rm{3}}}{\rm{H}}\]
B) \[{{\rm{H}}_{\rm{2}}}{\rm{O}}\]
C) \[{\rm{N}}{{\rm{O}}_{\rm{2}}}\]
D) \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\]

Answer 1

Hint: Isoelectronic species are those chemical species that have an equal number of electrons. To check if a species is isoelectronic or not, we have to count the total electrons of each atom of the compound and also add the charge that the species carry.

Complete step by step solution:Let’s first find out the electrons and shape of \[{{\rm{N}}_{\rm{2}}}{\rm{O}}\].
Electrons of \[{{\rm{N}}_{\rm{2}}}{\rm{O = 2}} \times {\rm{7 + 8 = 22}}\]
The hybridization of \[{{\rm{N}}_{\rm{2}}}{\rm{O}}\]has the linear shape.
Let’s count electrons of all the given compounds.
Option A is \[{{\rm{N}}_{\rm{3}}}{\rm{H}}\]. Electrons in \[{{\rm{N}}_{\rm{3}}}{\rm{H}} = 3 \times 7 + 1 = 22\]and its shape is trigonal planar.
Option B is \[{{\rm{H}}_{\rm{2}}}{\rm{O}}\]. Electrons in \[{{\rm{H}}_{\rm{2}}}{\rm{O}} = {\rm{2 + 8 = 10}}\]. So, B \[{{\rm{H}}_{\rm{2}}}{\rm{O}}\]is not isoelectronic with\[{{\rm{N}}_{\rm{2}}}{\rm{O}}\]. Also the shape of water molecule is bent.
Option C is \[{\rm{N}}{{\rm{O}}_{\rm{2}}}\] . The count of electrons in \[{\rm{N}}{{\rm{O}}_{\rm{2}}} = 7 + 8 \times 2 = 23\]. So, \[{\rm{N}}{{\rm{O}}_{\rm{2}}}\] is also not isoelectronic with \[{{\rm{N}}_{\rm{2}}}{\rm{O}}\]. And the shape of \[{\rm{N}}{{\rm{O}}_{\rm{2}}}\]is bent.
Option D is \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\].The count of electrons in \[{\rm{C}}{{\rm{O}}_{\rm{2}}} = 6 + 2 \times 8 = 22\]electrons. So, carbon dioxide is isoelectronic with \[{{\rm{N}}_{\rm{2}}}{\rm{O}}\]. Now, we have to identify the shape of the carbon dioxide molecule. In the \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\]molecule, two atoms of oxygen forms bond with the central atom,i.e, carbon. And no lone pair is present on C atom. So, \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\] is \[sp\] hybridized. Therefore, it is linear in shape.
Therefore, \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\]is isoelectronic and has similar shape with \[{{\rm{N}}_{\rm{2}}}{\rm{O}}\].

Hence, option D is right.

Note: Chemical species such as \[{\rm{Ne,}}{{\rm{O}}^{2 - }},{{\rm{F}}^ - },{\rm{M}}{{\rm{g}}^{2 + }},{\rm{N}}{{\rm{a}}^ + }{\rm{,}}\] are isoelectronic (10 electrons)but their ionic sizes are different. The anions with greater charge, are bigger in size than the neutral atom. And the cation with higher charge has lower atomic size than the neutral atom. So, the decreasing order of their ionic radius is, \[{{\rm{O}}^{2 - }}{\rm{ > }}{{\rm{F}}^ - }{\rm{ > Ne > N}}{{\rm{a}}^ + }{\rm{ > M}}{{\rm{g}}^{2 + }}\].