MCQ
${N_2}$ forms $NC{l_3}$, whereas $P$ can form both $PC{l_3}$ and $PC{l_5}$. Why
  • $P$ has low lying $3d$ orbitals, which can be used for bonding but ${N_2}$ does not have low lying $3d$ orbital
  • B
    ${N_2}$ atom is larger than $P$ in size
  • C
    $P$ is more reactive towards $Cl$ than ${N_2}$
  • D
    None of these

Answer

Correct option: A.
$P$ has low lying $3d$ orbitals, which can be used for bonding but ${N_2}$ does not have low lying $3d$ orbital
a
(a) $P$ has low lying $3d$ orbitals, which can be used for bonding, where as ${N_2}$ does not have low lying $3d$ orbitals.

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