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. $I{O}^{-}$

 Given lanthanoid ions ,the tendency to form the complexes is highest for:

1. Ce⁺³

2. Pm⁺²

3. Lu⁺³

4. Eu⁺²

A colored ion in aqueous solutions among the following is:

1. La³⁺ (Z=57)

2. Ti³⁺ (Z=22)

3. Lu³⁺ (Z=71)

4. Sc³⁺ (Z=21)

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)

${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\mathrm{X}<mspace\; width="1em"></mspace>\stackrel{{\mathrm{H}}^{+}}{\to }<mspace\; width="1em"></mspace>{\mathrm{Cr}}^{3+}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{\mathrm{H}}_2\mathrm{O}<mspace\; width="1em"></mspace>+$ oxidised product of X
X in the above reaction can not be:
 

1. ${\mathrm{C}}_2{\mathrm{O}}_4^{2-}$

2. ${\mathrm{Fe}}^{2+}$

3. ${\mathrm{SO}}_4^{2-}$

4. ${\mathrm{S}}^{2-}$

The manganate and permanganate ions are tetrahedral, due to:

The reason for the lanthanoid contraction is:

1. negligible screening effect of 'f' orbitals.

2. increasing nuclear charge.

3. decreasing nuclear charge.

4. decreasing screening effect.

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