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$N_2$ forms $Ncl_3$ whereas P can form both $Pcl_3$ and $PCl_5$ why ?

$(a)\;\text{P has low lying 3d orbitals, which can be used for bonding but N does not have low lying 3d orditals.} \\ (b)\;\text{N atom is larger than P in size.} \\(c)\;\text{P is more reactive towards Cl than} \\(d)\;\text{none of these} $
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P has low lying 3d orbitals, which can be used for bonding , whereas N does not have low lying 3d orbital, So it cannot expand its octet.
Hence a is the correct answer.
answered Apr 2, 2014 by meena.p
 

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