Q. Nitrogen forms $N_{2}$ , but phosphorus is converted into $P_{4}$ from P, the reason is

NTA AbhyasNTA Abhyas 2020The p-Block Elements Report Error

A

Triple bond is present between phosphorus atom

B

$\mathrm{p} \pi-\mathrm{p} \pi$ bonding is strong in nitrogen

C

$\mathrm{p} \pi-\mathrm{p} \pi$ bonding is weak in nitrogen

D

Multiple bond is formed easily

Solution:

Nitrogen form $\mathrm{N}_2$ (i.e. $\mathrm{N} \equiv \mathrm{N}$ ) but phosphorus form $\mathrm{P}_4$, because in $\mathrm{P}_2$, $\mathrm{p} \pi-\mathrm{p} \pi$ bonding is present which is a weaker bonding. Nitrogen is smaller in size hence it forms multiple bond with other nitrogen atom $(\mathrm{N} \equiv \mathrm{N})$ Nitrogen has unique ability to form $\mathrm{p} \pi-\mathrm{p} \pi$ multiple bonds with itself and with other elements having small size and high electron negativity. However heavier elements of this group do not form $\mathrm{p} \pi-\mathrm{p} \pi$ bonds as their atomic orbitals are so large and diffuse that they cannot have effective overlapping. Thus nitrogen exist as diatomic molecule with a triple bonds. The single P - P bond is stronger and has higher catenation tendency

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