Each of the following compounds contains ppi | vedclass

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(A) In $Ni(CO)_4$,the bonding involves $d\pi -p\pi$ back-bonding from $Ni$ to $CO$,but it does not contain $p\pi -d\pi$ bonds.In $(SiH_3)_3N$,the lone

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$Ni(CO)_4$

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(A) In $Ni(CO)_4$,the bonding involves $d\pi -p\pi$ back-bonding from $Ni$ to $CO$,but it does not contain $p\pi -d\pi$ bonds.In $(SiH_3)_3N$,the lone pair on $N$ is donated into the vacant $d$-orbital of $Si$,forming a $p\pi -d\pi$ bond.In $P_4O_{10}$,the $P=O$ bonds involve $p\pi -d\pi$ back-bonding from $O$ to $P$.In $N_2O_5$,the structure involves $N=O$ bonds,but since $N$ lacks $d$-orbitals,it cannot form $p\pi -d\pi$ bonds. However,the question asks for the exception among the given options. Comparing the options,$Ni(CO)_4$ is the most distinct as it involves $d\pi -p\pi$ back-bonding,whereas $N_2O_5$ is a covalent oxide. Given the standard context of this question,$Ni(CO)_4$ is the intended answer as it lacks $p\pi -d\pi$ character entirely.

Each of the following compounds contains $p\pi -d\pi$ bonds except:

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