The manganate and permanganate ions are tetrahedral, due to:

When neutral or faintly alkaline $KMn{O}_4$ is treated with potassium iodide, iodide ion is converted into 'X'. 'X' is:

1. $I_2$

2. $I{O}_4^{-}$

3. $I{O}_3^{-}$

4. $I{O}^{-}$

A gas that can readily decolorize an acidified KMnO₄ solution is:

1. ${\mathrm{SO}}_2$

2. ${\mathrm{NO}}_2$

3. ${\mathrm{P}}_2{\mathrm{O}}_5$

4. ${\mathrm{CO}}_2$

SO₂ is passed through an acidified K₂Cr₂O₇ solution.
The correct statement among the following regarding the above experiment is:

1. The solution gets decolorized.

2. SO₂ gets reduced.

3. Green coloured Cr₂(SO₄)₃ gets formed.

4. The solution turns blue.

In acidic medium, H₂O₂ changes Cr₂O₇^–2 to CrO₅ which has two (–O – O–) bonds.
The oxidation state of Cr in CrO₅ is:

1. +5

2. +3

3. +6

4. -10