© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the ligands and determine their denticity
In the given complex, [E(en)2(C2O4)], there are two ligands:
Ethylene diamine ($en$), which is a bidentate ligand (each $en$ provides two donor atoms).
Oxalate ($C_2O_4^{2-}$), which is also a bidentate ligand (each oxalate provides two donor atoms).
Hence, each $en$ ligand coordinates through two sites, and the oxalate ligand also coordinates through two sites.
Step 2: Calculate the coordination number of E
Since $en$ is bidentate and there are two $en$ ligands, that contributes $2 \times 2 = 4$ coordination sites. The oxalate ligand $C_2O_4^{2-}$ contributes 2 more sites. Thus, the total coordination number (CN) is:
$ \text{CN} = 4 + 2 = 6. $
Step 3: Determine the oxidation state of E
Let the oxidation state of E be $x$. Ethylene diamine ($en$) is a neutral ligand, so it contributes no charge. Oxalate ($C_2O_4^{2-}$) has a charge of $-2$. The complex is associated with $NO_2^{-}$ outside the coordination sphere, which indicates the entire complex cation must have a $+1$ charge. Therefore, setting up the charge balance:
$ x + (-2) = +1 $
Solving for $x$:
$ x - 2 = 1 \quad \Rightarrow \quad x = +3.
$
Step 4: State the final answer
Thus, the coordination number of E is 6, and the oxidation state of E is $+3$.
