JEE MAIN - Chemistry (2008 - No. 18)
The coordination number and the oxidation state of the element ‘E’ in the complex [E(en)2(C2O4)]NO2 (where (en) is ethylene diamine) are, respectively,
6 and 2
4 and 2
4 and 3
6 and 3
Explanation
In the given complex we have two bidentate ligands (i.e. $$en$$ and $${C_2}{O_4}$$ ), so coordination number of $$E$$ is $$6$$ $$\left( {2 \times 2 + 1 \times 2 = 6} \right)$$
Let the oxidation state of $$E$$ in complex be $$x,$$
then $$\left[ {x + \left( { - 2} \right) = 1} \right]$$ or $$x - 2 = 1$$
or $$\,\,\,\,$$ $$x=+3,$$ so its oxidation state is $$+3$$
Thus option $$(d)$$ is correct.
Let the oxidation state of $$E$$ in complex be $$x,$$
then $$\left[ {x + \left( { - 2} \right) = 1} \right]$$ or $$x - 2 = 1$$
or $$\,\,\,\,$$ $$x=+3,$$ so its oxidation state is $$+3$$
Thus option $$(d)$$ is correct.
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