MCQ
Highest $(+7) $ oxidation state is shown by
- A$Co$
- B$Cr$
- C$V$
- ✓$Mn$
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$Cr_2O_7^{2-}$ ( $aq$ $1\,M$ ) $+ 14H^+(aq) + 6e^-$ $\rightarrow 2Cr^{+3} (aq,\,1\,M) + 7H_2O (l)$

The structure of $C$ would be
(Given atomic number: $\mathrm{Sc}: 21, \mathrm{Ti}: 22, \mathrm{~V}: 23$, $\mathrm{Cr}: 24, \mathrm{Mn}: 25, \mathrm{Fe}: 26, \mathrm{Co}: 27, \mathrm{Ni}: 28, \mathrm{Cu}: 29$, $\mathrm{Zn}: 30)$