When neutral or faintly alkaline ${\mathrm{KMnO}}_4$ is treated with potassium iodide, the iodide ion is converted into 'X'. The formula of 'X' is:

1. ${\mathrm{I}}_2$

2. ${\mathrm{IO}}_4^{-}$

3. ${\mathrm{IO}}_3^{-}$

4. ${\mathrm{IO}}^{-}$

MnO₂ when fused with KOH and oxidized in air gives a dark green compound X. In acidic solution, X undergoes disproportionation to give an intense purple compound Y and MnO₂. The compounds X and Y, respectively are

1. K₂MnO₄ and KMnO₄

2. Mn₂O₇ and KMnO₄

3. K₂MnO₄ and Mn₂O₇

4. KMnO₄ and K₂MnO₄

Among the following metals, the strongest reducing agent is 

1. Ni

2. Cu

3. Zn

4. Fe

The number of moles of Br₂ produced when two moles of potassium permanganate are treated with excess potassium bromide in aqueous acidic medium is 

1. 1

2. 3

3. 2

4. 5

The number of moles of KMnO₄ required to oxidize one equivalent of KI in the presence of sulfuric acid is 

1. 5

2. 2

3. 1/2

4. 1/5

One of the constituents of German silver is 

1. Ag

2. Mg

3. Cu

4. Al

Lanthanide contraction is caused due to 

1. The appreciable shielding on outer electrons by 4f electrons from the nuclear charge

2. The appreciable shielding on outer electrons by 5d electrons from the nuclear charge

3. The same effective nuclear charge from Ce to Lu

4. The imperfect shielding on outer electrons by 4f electrons from the nuclear charge

Europium (Eu) and Terbium (Tb) attain stable 4f configuration by exhibiting oxidation states of 

1. +2 and +4

2. +3 and +4

3. +2 and +3

4. +1 and +3

Four statements for Cr and Mn are given below:

The correct statements are 

1. (I), (III),(IV)

2. (I) , (II)

3. (I), (II) and (IV)

4. (I), (IV)

Silver nitrate solution when added to a colourless aqueous solution E forms a white precipitate which dissolves in excess of E. If the white precipitate is heated with water, it turns black and the supernatant solution gives a white precipitate with acidified barium nitrate solution. Therefore, E is 

1. Na₂S

2. Na₂S₂O₃

3. Na₂SO₃

4. Na₂SO₄