It is found that v forms a double salt isomorphous with Mohr's salt. The oxidation number of V in this compound is:

1. +3

2. +2

3. +4

4. -4

The correct order of reducing power of halide ions is:

1. Cl^- > Br^- > I^- > F^-

2. Cl^- >  I^- > Br^- > F^-

3.  Br^- > Cl^- > I^- > F^-

4.  I^- > Br^- > Cl^- > F^-

When H₂SO₃ is converted into H₂SO₄ the change in the oxidation state of sulfur is from:

1. 0 to +2

2. +2 to +4

3. +4 to +2

4. +4 to +6

The reaction during which nitrogen gets oxidised is-

1. ${\mathrm{NH}}_4^{+}$ $\to$N₂

2. ${\mathrm{NO}}_3^{-} \to$NO

3. NO₂ $\to$${\mathrm{NO}}_2^{-}$

4. ${\mathrm{NO}}_3^{-}$$\to$N${\mathrm{H}}_4^{+}$ 

Reaction that shows hydrogen as an oxidising agent is:

1. With iodine to give hydrogen iodide

2. With lithium to give lithium hydride

3. With nitrogen to give ammonia

4. With sulphur to give hydrogen sulphide

Fluorine is a strong oxidising agent because:

1. it has several isotopes

2. it is very small and has 7 electrons in valency shell

3. its valency is one

4. it is the first member of the halogen series

The oxidation state of Ni in Ni(CO)₄ is:

1. zero

2. +4

3. +8 

4. +2

Milliequivalents of a solute in a solution can be given by:

1. Mzeq_. = M x V_{in mL}

2. Meq. = N x V_{in mL}

3. Mzeq. = $\frac{mass}{Eq. mass}\times$1000

4. both 2 and 3

HBr and HI reduce H₂SO₄, HCl can reduce KMnO₄ and HF can reduce:

1. H₂SO₄

2. K₂Cr₂O₇

3. KMnO₄ 

4. none of these

Sulphurous acid can be used as:

1. Oxidising agent

2. Reducing agent

3. Bleaching agent

4. All of the above