Give reasons for the following:
$R_{3} P = O$ exists but $R_{3} N = O$ does not, $R$ is an alkyl group?

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Solution

Nitrogen cannot have covalency more than 4. This is the restriction which comes in nitrogen(N) to expand its coordination number beyond four. Hence, $R_{3} N = O$ does not exist. But P can show covalency of 5. Hence, $R_{3} P = O$ exists.

Nitrogen cannot form $d π - p π$ bonds but phosphorous can form $d π - p π$ bonds.

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Why does R₃P=O exist but R₃N=O does not (R = alkyl group)?

R_{3} P = O

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R_{3} N = O

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