1.

{:("COLUMN-I","COLUMAN-II"),((A)[Ni(CO)_(4)],(p)"Tetrahedral"),((B)[Fe(CO)_(2)(NO)_(2)],(q)pi-"back bonding"),((C)[Ni(PF_(3))_(4)],(r)"diamagnetic"),((D)[Ni(PPh_(3))_(2)Br_(2)],(s)"one of the ligand is " 3e^(-)"donar"):}

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Solution :CO and `PF_(3)` are both `pi`- acceptors (`pi` ACIDS). They form `pi`-back bonding high EN and small at SIZE of F makes `PF_(3)-pi` acceptor.
.NO. donates 3 ELECTRONS to the METAL atom and BECOMES `NO^(+)`.


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