Heating a sample of $[(eta^5 ext{-}mathrm{C_5H_5})mathrm{Mo(CO)_3}]_2$ results in the formation of $[(eta^5 ext{-}mathrm{C_5H_5})mathrm{Mo(CO)_2}]$ with elimination of 2 equivalents of CO. The Mo-Mo bond order in this reaction changes from
(a)2 to 3
(b)1 to 2
(c)1 to 3
(d)2 to 4
Answer
Answer (as printed): C
Explanation
$mathrm{TVE}=10+12+12=34=A$$B=(n imes 18)-A$$B=(2 imes 18)-34=36-34=2$Number of metal–metal bond $=2/2=1$On heating: $mathrm{TVE}=10+12+8=30=A$$B=(n imes 18)-A=36-30=6$Number of metal–metal bond $=B/2=6/2=3$