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 $3$
- D$6$ and $3$
✓
Answer
In the given complex we have two bidentate ligands
(i.e $en$ and $C_2O_4$ ), so coordination number of $E$ is $6$
$(2 \times 2 + 1 \times 2 = 6)$
Let the oxidation state of $E$ in complex be $x$, then
$[x + (-2) = 1]$ or $x -2 = 1$
or $x = + 3$, so its oxidation state is $+ 3$ Thus option $(d)$ is correct.
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