MCQ
In $N{H_4}N{O_2},$ the oxidation number of nitrogen will be
- A$3$
- B$5$
- ✓$-3$ and $+ 3$
- D$+ 3 $ and $+ 5$
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$E ^0\left( Fe ^{3+} . Fe ^{2+}\right)=+0.77 V $
$E ^0\left( Fe ^{2+} . Fe \right)=-0.44 V $
$E ^{\circ}\left( Cu ^{2+} . Cu \right)=+0.34 V $
$E ^0\left( Cu ^{+} . Cu \right)=+0.52 V $
$E ^{\circ}\left( O _2( g )+4 H ^{+}+4 e ^{-} \rightarrow 2 H _2 O \right)=+1.23 V $
$E ^{\circ}\left( O _2( g )+2 H _2 O +4 e ^{-} \rightarrow 4 OH \right)=+0.40 V $
$E ^0\left( Cr ^{3+} . Cr \right)=-0.74 V $
$E ^{\circ}\left( Cr ^{2+} . Cr \right)=-0.91 V$
Match $E ^{\circ}$ of the rebox pair in List $I$ with the values given in List $II$ and select the correct answer using the code given below the lists:
| List $I$ | List $II$ |
| $P.$ $\quad E ^{\circ}\left( Fe ^{3+}, Fe \right)$ | $1.$ $\quad-0.18 V$ |
| $Q.$ $\quad E ^{\circ}\left(4 H _2 O \rightleftharpoons 4 H ^{+}+4 OH ^{-}\right)$ | $2.$ $\quad-0.4 V$ |
| $R.$ $\quad E ^{\circ}\left( Cu ^{2+}+ Cu \rightarrow 2 Cu ^{+}\right)$ | $3.$ $\quad-0.04 V$ |
| $S.$ $\quad E ^{\circ}\left( Cr ^{3+}, Cr ^{+2}\right)$ | $4.$ $\quad-0.83 V$ |
Codes: $ \quad P \quad Q \quad R \quad S $