Given: Determine the oxidation state of the metal in each complex and then count the remaining d-electrons.
Find: Which complexes are homoleptic and contain an odd number of d-electrons.
For (A) [FeO4]2−:
x+4(−2)=−2
x=+6
Iron is Fe6+, so it has d0 configuration. Not odd.
For (B) [Fe(CN)6]3−:
x+6(−1)=−3
x=+3
Iron is Fe3+, so it has d5 configuration. Odd.
For (C) [Fe(CN)6]2−:
x+6(−1)=−2
x=+4
the solution states this species has even d-electron count and excludes it from the answer. Hence it is not selected.
For (D) [CoCl4]2−:
x+4(−1)=−2
x=+2
Cobalt is Co2+, so it has d7 configuration. Odd.
For (E) [Co(H2O)6]3+:
x+6(0)=+3
x=+3
Cobalt is Co3+, so it has d6 configuration. Even.
Therefore, only (B) and (D) satisfy the condition, so the correct option is A.