In the complex $K_3[Cr(C_2O_4)_3]$, the coordination number and oxidation state of Cr respectively are:
- 3 and +3
- 6 and +3
- 6 and +6
- 3 and +6
Show answer and solution
Correct answer: B. 6 and +3
Each oxalate ($C_2O_4^{2-}$) is bidentate (2 donor O atoms).
Three oxalates → 6 donor atoms
→ CN = 6. Charge: Cr + 3(-2) = -3 (matching 3K⁺) ⇒ Cr = +3.