Question

Which among the group 15 elements does not exist as a tetratomic molecule?
A. Nitrogen
B. Phosphorous
C. Arsenic
D. antimony

Answer 1

Hint: Nitrogen has a small size and high electronegativity to forms \[p\pi - p\pi \;\] multiple bonds and nitrogen doesn’t form a \[p\pi - d\pi \;\] bond. If there is bonding between two atoms where one atom is having one vacant orbital and another is having one lone pair of electrons, then this electron pair is donated to that respective vacant orbital. Then this bonding is called p-p or p-d depending upon the orbital to which the pair is donated and from which the electron pair is donated.

Complete step by step answer:

Element	Atomic number	Electronic configuration	Group number	Period number
Nitrogen	7	$\left[ {{\text{He}}} \right]2{s^2}2{p^3}$	15	2
Phosphorous	15	$\left[ {{\text{Ne}}} \right]3{s^2}3{p^3}$	15	3
Arsenic	33	$\left[ {{\text{Ar}}} \right]3{d^{10}}4{s^2}4{p^3}$	15	4
Antimony	51	$\left[ {{\text{Kr}}} \right]4{d^{10}}5{s^2}5{p^3}$	15	5
Bismuth	83	$\left[ {{\text{Xe}}} \right]4{f^{14}}5{d^{10}}6{s^2}6{p^3}$	15	6

Nitrogen forms a diatomic structure because of its small size and high electronegativity forms \[p\pi - p\pi \]  multiple bonds
Therefore, it exists as a diatomic molecule, N≡N. Phosphorous, due to its large size and low electronegativity cannot form \[p\pi - p\pi \] multiple bonds with itself. 
The octet of both N atoms is complete. P has a large atomic size and little tendency to form triple bonds. It can complete its octet by sharing valence electrons with three other P atoms to form a tetra-atomic \[{P_4}\] molecule.
While arsenic and antimony form tetra-atomic structure due to the presence of \[p\pi - d\pi \;\] bonding
Hence, the correct option is (A), Nitrogen.

Note: Do not get confused at \[p\pi - p\pi \;\]and \[p\pi - d\pi \;\], make sure to remember that nitrogen doesn’t form \[p\pi - d\pi \;\]. Those elements cannot form \[p\pi - d\pi \;\]bond as they do not possess d - orbitals in their valence shell.