MCQ
${M^{ + 3}}$ ion loses $3{e^ - }$. Its oxidation number will be
- A$0$
- B$+3$
- ✓$+6$
- D$-3$
$2 \times 3{e^ - } = + 6$.
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|
List $I$ (Conversion) |
List $II$ (Number of Faraday required) |
| $A$. $1 \mathrm{~mol}$ of $\mathrm{H}_2 \mathrm{O}$ to $\mathrm{O}_2$ | $l$. $3 \mathrm{~F}$ |
| $B$. $1 \mathrm{~mol}$ of $\mathrm{MnO}_4^{-}$to $\mathrm{Mn}^{2+}$ | $II$. $2 F$ |
| $C$. $1.5 \mathrm{~mol}$ of $\mathrm{Ca}$ from molten $\mathrm{CaCl}_2$ | $III$. $1F$ |
| $D$. $1 \mathrm{~mol}$ of $\mathrm{FeO}$ to $\mathrm{Fe}_2 \mathrm{O}_3$ | $IV$. $5 \mathrm{~F}$ |
Choose the correct answer from the options given below:
$BF _{3}+ NaH \stackrel{450 K }{\longrightarrow} A + NaF$
$A + NMe _{3} \rightarrow B$