The metal complex that exhibits a triplet as well as doublet in its ${}^{31}\mathrm{P}$ NMR spectrum is
(a)mer-$[\mathrm{IrCl_3(PPh_3)_3}]$
(b)trans-$[\mathrm{IrCI(Co)(PPh_3)_2}]$
(c)fac-$[\mathrm{IrCl_3(PPh_3)_3}]$
(d)$[\mathrm{Ir(PPh_3)_4}]^{+}$
Answer
Answer (as printed): A
Explanation
Triplet as well as doublet indicate 3 phosphorus atoms in the molecule, in two sets: 2P = doublet, 1P = triplet. That is possible in mer-$[\mathrm{IrCl_3(PPh_3)_3}]$. fac-$[\mathrm{IrCl_3(PPh_3)_3}]$ will give a singlet of 3P. trans-$[\mathrm{IrCI(Co)(PPh_3)_2}]$ will give a singlet of 2P. $[\mathrm{Ir(PPh_3)_4}]^{+}$ will give a singlet of 4P.