For the four successive transition elements (Cr, Mn, Fe and Co), the correct order of stability of +2 oxidation state among the following is: 
(At. No Cr= 24, Mn= 25, Fe=26, Co= 27)

1.  Fe>Mn>Co>Cr

2.  Co>Mn>Fe>Cr

3.  Cr>Mn>Co>Fe

4.  Mn>Fe>Cr>Co

The acidified \(\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7\) solution turns green when \(\mathrm{Na}_2 \mathrm{SO}_3\) is added to it. This is due to the formation of:

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)

The species that are colourless:

(Z of Ti = 22, Co=27, Cu = 29, Ni= 28)

1. ${\mathrm{TiF}}_6^{2-}$ $\mathrm{and}$ ${\mathrm{CoF}}_6^{3-}$

2. ${\mathrm{Cu}}_2{\mathrm{Cl}}_2$ $\mathrm{and}$ ${\mathrm{NiCl}}_4^{2-}$

3. ${\mathrm{TiF}}_6^{2-}$ $\mathrm{and}$ ${\mathrm{Cu}}_2{\mathrm{Cl}}_2$

4. ${\mathrm{CoF}}_6^{3-}$ $\mathrm{and}$ ${\mathrm{NiCl}}_4^{2-}$

The incorrect statement among the following is:  

Which one of the following ions is the most stable in aqueous solution? 
(At. No. Ti = 22, V = 23, Cr = 24, Mn = 25)

1. $C{r}^{3+}$

2. $V^{3+}$

3. $T{i}^{3+}$

4. $M{n}^{3+}$

In which of the following pairs are both the ions coloured in aqueous solution? 
(At. no. : Sc = 21, Ti = 22, Ni = 28, Cu = 29, Co= 27)

1. $N{i}^{2+}, T{i}^{3+}$

2. $S{c}^{3+}, T{i}^{3+}$

3. $S{c}^{3+}, C{o}^{2+}$

4. $N{i}^{2+}, C{u}^{+}$

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