The correct option (b) [Fe(CO)5]
Explanation:
(a) Mn+= 3d5 ,4s1 .
In the presence of CO effective configuration
= 3d6 4s0,
Three lone pair for back bonding with vacant orbital of C in CO.
(b) Fe0 = 3d6, 4s2. ln presence of CO effective configuration = 3d8,
Four lone pair for back bonding with CO.
(c) Cr0 = 3d5 ,4s1,effective configuration = 3d6 Three lone pair for back bonding with CO.
(d) V- = 3d4 , 4s2, effective configurat ion = 3d6 three lone pair for back bonding with CO.
Maximum back bonding in Fe(CO)5, therefore CO bond order is lowest here.