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 $3$
(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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Reason : $V_2O_5$ or $Pt$ is used in the preparation of $H_2SO_4$ by contact process.
$(I)\, [Co(NO_2)_3(NH_3)_3]$
$(II)$ cis $-[ RhCl_2 (NH_3)_4]^+$
$(III)\, [Cr(OX)_3]^{-3}$
$(IV)$ cis $ -[PtCl_2 (en)]$
$(V)$ trans $-[Cr(en)_2 Br_2]^+$
$(VI)$ cis $-[Cr(en)_2 Br_2]^+$
If all products formed in the above reaction undergo reaction with $Br_2(CCl_4)$ individually then number of products formed will be :