Question

Consider the complex ion ${\left[Cr{\left(N{H}_3\right)}_{2}C{l}_2\left(C_2{O}_4\right)\right]}^{-}$. What is the
( i ) oxidation number of the metal atom?
( ii ) coordination number of the metal atom?
( iii ) charge on the complex if all ligands were chloride ions?

• A. $\begin{array}{ccc}-1& 6& -3\end{array}$
• B. $\begin{array}{ccc}+3& 6& -3\end{array}$
• C. $\begin{array}{ccc}+3& 6& -1\end{array}$
• D. $\begin{array}{ccc}-1& 5& -3\end{array}$

Answer 1

Answer: (B)

$\begin{array}{ccc}+3& 6& -3\end{array}$

In ${\left[Cr{\left(N{H}_3\right)}_{2}C{l}_2\left(C_2{O}_4\right)\right]}^{-}$, ligands are 2 ammonia molecules with zero charge, 2 chloride ion with -1 charge, one oxalate ion with -2 charge. So coordination no. of metal atom is 6 & oxidation no. of central atom is +3. If we replace all ligands with chloride ions then charge on complex is $+3-6\times \left(1\right)=-3$.