MCQ
Oxidation number of $Fe$ in ${K_3}[Fe{(CN)_6}]$ is
- A$+2$
- ✓$+3$
- C$+1$
- D$+4$
$1 \times 3 + x + ( - 1 \times 6) = 0$
$3 + x - 6 = 0$; $x = + 3$.
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$CH_3-CH_2-OH$ $\mathop {\xrightarrow{{(i)\,KMn{O_4}/\mathop O\limits^\Theta H/\Delta }}}\limits_{(ii)\,{H^ \oplus }} (A)\mathop {\xrightarrow{{(i)\,SOC{l_2}}}}\limits_{(ii)\,N{H_3}/\Delta } (B)$ $\xrightarrow{{B{r_2}/KOH}}(C)$
$(C)$ will be :
(Given atomic number: $\mathrm{Sc}=21, \mathrm{Ti}=22, \mathrm{~V}=23$, $\mathrm{Cr}=24, \mathrm{Mn}=25$ and $\mathrm{Zn}=30)$
Reason : In mercury cell, the electrolyte is a paste of $KOH$ and $ZnO$.