The ease of adsorption of the hydrated alkali metal ions on an ion-exchange resins

follows the order:

1. Li⁺<K⁺<Na⁺<Rb⁺

2. Rb⁺<K⁺<Na⁺<Li⁺

3. K⁺<Na⁺<Rb⁺<Li⁺

4. Na⁺<Li⁺<K⁺<Rb⁺

In acidic medium, H₂O₂ changes ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$ to CrO₅ which has two (-O-O-) bonds. Oxidation

state of Cr in CrO₅ is

1. +5

2. +3

3. +6

4. -10

(I) ${\mathrm{H}}_2{\mathrm{O}}_2+{\mathrm{O}}_3 \stackrel{ }{\to }{\mathrm{H}}_2\mathrm{O} + 2{\mathrm{O}}_2$

(II) ${\mathrm{H}}_2{\mathrm{O}}_2 + {\mathrm{Ag}}_2\mathrm{O} \stackrel{ }{\to }2\mathrm{Ag} + {\mathrm{H}}_2\mathrm{O} + {\mathrm{O}}_2$

Role of hydrogen peroxide in the above reactions is respectively

1. oxidising in (I) and reducing in (II)

2. reducing in (I) and oxidising in (II)

3. reducing in (I) and  (II)

4. oxidising in(I) and (II)

The correct statement for the hydrogenation of oil is:

H₂O₂ restores the colour of old lead paintings by:

1. Converting PbO₂ to Pb

2. Oxidising PbS to PbSO₄

3. Converting PbCO₃ to Pb

4. Oxidizing PbSO₃ to PbSO₄

Which conversion is not done by H₂O₂ ?

1. ${\mathrm{Fe}}^{2+} \stackrel{ }{\to }{\mathrm{Fe}}^{3+}$(acidic medium)

2. $\mathrm{PbS}\stackrel{ }{\to }{\mathrm{PbSO}}_4$

3. ${\mathrm{Fe}}^{+} \stackrel{ }{\to }{\mathrm{Fe}}^{3+}$ (alkaline medium)

4. $\mathrm{KI} \stackrel{ }{\to }{\mathrm{I}}_2$

Poorest reducing agent among following is

1. atomic hydrogen

2. nascent hydrogen

3. dihydrogen

4. all have same reducing strength

Maximum concentration of ortho-$H_2$ in ordinary hydrogen is:

(1) 75% ortho-$H_2$ + 25% para-$H_2$

(2) 25% ortho-$H_2$ + 75% para-$H_2$

(3) 50% ortho-$H_2$ + 50% para-$H_2$

(4) 99% para-$H_2$ + 1% ortho-$H_2$

Hydrogen will not reduce$-$

(1) heated cupric oxide                                 

(2) heated ferric oxide

(3) heated stannic oxide                               

(4) heated aluminium oxide

The first ionization energy value $\left(\mathrm{kJ}\right)$ ${\mathrm{mol}}^{-1}$ for $\mathrm{H},$ $\mathrm{Li},$ $\mathrm{F},$ $\mathrm{and}$ $\mathrm{Na}$ 

is given below but not in a correct order

$1681,$ $520,$ $1312,$ $495.$

The ionization energy value of H would be -

1. 1681                                 

2. 1312

3. 520                                   

4. 485