The correct statement regarding terminal/bridging CO groups in solid $\mathrm{Co_4(CO)_{12}}$ and $\mathrm{Ir_4(CO)_{12}}$ is
(a)both have equal number of bridging CO groups
(b)number of bridging CO groups in $\mathrm{Co}_{4}(\mathrm{CO})_{12}$ is 4
(c)the number of terminal CO groups in $\mathrm{Co}_{4}(\mathrm{CO})_{12}$ is 8
(d)the number of bridging CO groups in $\operatorname{Ir}_{4}(\mathrm{CO})_{12}$ is zero.
Answer
Answer (as printed): D
Explanation
Due to increase in size of Ir, compared to Co, the Metal-Metal bond length also increases, thus the ability to form bridge carbonyl decreases with increase in M-M bond.No. of bridging CO = 3No. of bridging CO = 0No. of Terminal CO = 9No. of terminal CO = 12