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Why is phosphorus $ {{P}_{4}} $ ?

Last updated date: 24th Jul 2024
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Hint :Phosphorus (atomic no. $ 15 $ ) belongs to the nitrogen family i.e Group $ 15 $ . Phosphorus having an electronic configuration $ 1{{s}^{2}},2{{s}^{2}},2{{p}^{6}},3{{s}^{2}},3{{p}^{3}} $ since it has a deficiency of $ 3 $ electrons to attain noble gas configuration therefore its valency is $ 3 $ .

Complete Step By Step Answer:
In group $ 15 $ except nitrogen which is a diatomic molecule all other elements in group $ 15 $ exist in polyatomic form in nature.
Nitrogen being a diatomic have $ 1 $ sigma and $ 2 $ pi bond whereas in case of phosphorus $ 4 $ phosphorus atom are connected with the help of sigma bond there is no pi bonds present in case of $ {{P}_{4}} $ .
Nitrogen being small in size has an ability to form a $ p\pi -p\pi $ multiple bond. This is due to effective overlapping of orbital takes place, more overlapping stronger the bond formed whereas in phosphorus effective overlapping of $ p\pi -p\pi $ multiple bonds does not occur due to large size of phosphorus as compared to nitrogen . Due to increase in size of an atom the $ p- $ orbital becomes taller and steeper which decreases the effectiveness of overlapping. As a result phosphorus forms polyatomic molecules.
Phosphorus ( $ {{P}_{4}} $ ) exists in many allotropic forms but common are White, Red and Black Phosphorus.
The stability of allotropes depends on the structure in which an allotrope exists. White phosphorus is most reactive and less stable than other allotropes because the structure has high angular strain.
Red Phosphorus is more stable than white phosphorus but less stable than black phosphorus. As it comprises polymeric structure which reduces the angular strain compared to white phosphorus.
Black Phosphorus is the most stable allotrope of phosphorus as it has a layered structure just like the structure of graphite ( an allotrope of carbon). Which makes it most stable.

Note :
 $ P{{H}_{3}} $ ( Phosphine gas) has a rotten fish-like smell. The concept of hybridization is not applicable for $ P{{H}_{3}} $ as hybridization occur between orbital having comparable energy but energy gap in case of $ P{{H}_{3}} $ is high .