MCQ
The coordination number and the oxidation state of the element $'E'$ in the complex
$[E (en)_2 (C_2O_4)]NO_2$ (where $(en)$ is ethylene diamine) are, respectively,
- A$6$ and $2$
- B$4$ and $2$
- C$4$ and $3$
- ✓$6$ and $3$
✓
Answer
Correct option: D.
$6$ and $3$
d
$\mathrm{en},$ $\begin{array}{*{20}{c}} {C{H_2} - N{H_2}} \\ {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ {C{H_2} - N{H_2}} \end{array}$ is bidentate ligand
$\mathrm{C}_{2} \mathrm{O}_{4}^{2-}$ is also bidentate ligand. Hence, coordination number of the element $E$ in the given complete comes out to be $6 .$ The complex can be ionised as
$\mathrm{en},$ $\begin{array}{*{20}{c}} {C{H_2} - N{H_2}} \\ {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ {C{H_2} - N{H_2}} \end{array}$ is bidentate ligand
$\mathrm{C}_{2} \mathrm{O}_{4}^{2-}$ is also bidentate ligand. Hence, coordination number of the element $E$ in the given complete comes out to be $6 .$ The complex can be ionised as
$\left[E(e n)_{2}\left(C_{2} O_{4}\right)\right] N O_{2} \longrightarrow$ $\left[E(e n)_{2}\left(C_{2} O_{4}\right)\right]^{+} N O_{2}^{-}$
Oxidation number $x+0+(-2)=+1$
$\Rightarrow$ Oxidation number, $x=3$
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