MCQ
Oxidation number of carbon in ${H_2}{C_2}{O_4}$ is
- A$+4$
- ✓$+3$
- C$+2$
- D$-2$
$2 + 2x - 2 \times 4 = 0$; $2x = 8 - 2 = 6$
$x = \frac{6}{2} = + 3$.
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$C{H_3} - CH = C{H_2}\xrightarrow[{CC{l_4}}]{{B{r_2}}}A\xrightarrow[{(3\,\,moles)}]{{\mathop N\limits^ \oplus a\mathop N\limits^\Theta {H_2}}}B\xrightarrow{{C{H_3} - Br}}C\xrightarrow[{{H_2}}]{{Pd + BaS{O_4}}}D$