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In the complex $K_3[Cr(C_2O_4)_3]$, the coordination number and oxidation state of Cr respectively are:
A3 and +3
B6 and +3
C6 and +6
D3 and +6
Answer & 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.
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