In the reaction, 2 Cu${}^{+}$ $\to$ Cu + Cu${}^{2+}$, the equivalent weight of Cu${}^{+}$ is (M is the mol. wt. of Cu${}^{+}$)

1. $\mathrm{M}$

2.$\frac{\mathrm{M}}{2}$

3. $\frac{\mathrm{M}}{4}$

4. $2\mathrm{M}$

The oxidation number of S in Na${}_2$S${}_4$O${}_6$ is
1. 2.5

2. +2 and +3

3. Zero and +5

4. +1 and +4

For the reaction,\(\mathrm{MnO}_4^{2-} \rightarrow \mathrm{MnO}_4^{-}+\mathrm{MnO}_2\)  The number of moles of \(\mathrm{MnO}_4^{-} \text {and } \mathrm{MnO}_2\) respectively yielded upon the disproportionation of one mole of \(\mathrm{MnO}_4^{2-}\) are:

1. $\frac{2}{3}$and $\frac{1}{3}$

2. $\frac{1}{3}$ and $\frac{2}{3}$

3. $\frac{1}{2}$ and $\frac{1}{2}$

4. $2$ and $1$

How many moles of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ are required to oxidise one mole of ${\mathrm{FeSO}}_4$ in acidic medium?

1. 6

2. 1/2

3. $\frac{1}{3}$

4. $\frac{1}{6}$

${\mathrm{KMnO}}_4$ is a strong oxidising agent in acidic medium. To provide acidic medium, ${\mathrm{H}}_2{\mathrm{SO}}_4$ is used instead of HCl. This is because

1. ${\mathrm{H}}_2{\mathrm{SO}}_4$ is stronger acid than HCl
2. HCl is oxidised by ${\mathrm{KMnO}}_4$ to ${\mathrm{Cl}}_2$
3. ${\mathrm{H}}_2{\mathrm{SO}}_4$ is dibasic acid
4. Rate is faster in the presence of ${\mathrm{H}}_2{\mathrm{SO}}_4$

What is the stoichiometric coefficient of sulfur (\(S\)) in the balanced equation for the following reaction in acidic medium?
\(\mathrm{H}_2 \mathrm{~S}(a q)+\mathrm{HNO}_3(a q) \rightarrow \mathrm{NO}(g)+\mathrm{S}(s)+\mathrm{H}_2 \mathrm{O}(l)\)

1. 1

2. 2

3. 3

4. 4

Equal volumes of 1M each of KMnO${}_4$ and ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ are used to oxidise Fe(II) solution in acidic medium.
The amount of Fe oxidised will be

1. More with KMnO${}_4$

2. More with ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$

3. Equal with both oxidising agents

4. Cannot be determined

The oxidation number of Pt in ${\left[\mathrm{Pt}\left({\mathrm{C}}_2{\mathrm{H}}_4\right){\mathrm{Cl}}_3\right]}^{-}$ is:

1. +1

2. +2

3. +3

4. +4

Oxidation state of nitrogen is incorrectly given for:      

A, B and C are three elements forming a part of
compound in oxidation states of +2, +5 and –2
respectively. What could be the compounds?

1. ${\mathrm{A}}_2{\left(\mathrm{BC}\right)}_2$

2. ${\mathrm{A}}_2{\left({\mathrm{BC}}_4\right)}_3$

3. ${\mathrm{A}}_3{\left({\mathrm{BC}}_4\right)}_2$

4. ABC