The hydrides ion ${\mathrm{H}}^{-}$ is a stronger base than its hydroxide ion ${\mathrm{OH}}^{-}$. Which of the following reactions will occur if sodium hydride $\left(\mathrm{NaH}\right)$ is dissolved in water –

(A) ${\mathrm{H}}^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}_3{\mathrm{O}}^{-}\left(\mathrm{aq}\right)$

(B) ${\mathrm{H}}^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathcal{l}\right)\to {\mathrm{OH}}^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right)$

(C) ${\mathrm{H}}^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathcal{l}\right)\to \mathrm{No}<mspace\; width="1em"></mspace>\mathrm{reaction}$

(D) $\mathrm{None}<mspace\; width="1em"></mspace>\mathrm{of}<mspace\; width="1em"></mspace>\mathrm{these}$

Blackened oil planting can be restored into original form by the action of:

1. Chlorine

2. ${\mathrm{BaO}}_2$

3. ${\mathrm{H}}_2{\mathrm{O}}_2$

4. ${\mathrm{MnO}}_2$

Which of the following statement regarding ${\mathrm{H}}_2{\mathrm{O}}_2$ is wrong.

(1) It is stable in acid medium

(2) It acts as oxidising as well as reducing agent

(3) It has zero dipole moment

(4) Pure ${\mathrm{H}}_2{\mathrm{O}}_2$ is slightly acidic

Polyphosphates are used as water softening agents because they –

(1) form soluble complexes with anionic species

(2) precipitate anionic species

(3) form soluble complexes with cationic species

(4) precipitate cationic species

Permanent hardness of water can be removed, by adding –

(1) Soda lime

(2) Sodium bicarbonate

(3) Washing soda

(4) Sodium chloride

The correct increasing order of the acidity of ${\mathrm{CO}}_2,<mspace\; width="1em"></mspace>{\mathrm{H}}_2\mathrm{O}<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{H}}_2{\mathrm{O}}_2$ is $-$

(1) ${\mathrm{CO}}_2<{\mathrm{H}}_2{\mathrm{O}}_2<{\mathrm{H}}_2\mathrm{O}$                         

(2) ${\mathrm{H}}_2\mathrm{O}<{\mathrm{H}}_2{\mathrm{O}}_2<{\mathrm{CO}}_2$

(3) ${\mathrm{H}}_2\mathrm{O}<{\mathrm{H}}_2{\mathrm{O}}_2>{\mathrm{CO}}_2$                         

(4) ${\mathrm{H}}_2{\mathrm{O}}_2>{\mathrm{CO}}_2>{\mathrm{H}}_2\mathrm{O}$

At absolute zero.

(1) Only para hydrogen exists

(2) Only ortho hydrogen exists

(3) Both para and ortho hydrogen exist

(4) None of these

Metal which does not react with cold water but evolves H₂ with steam is –

(1) Na                           

(2) K

(3) Pt                             

(4) Fe

The oxidation states of the most electronegative element in the products of the reaction between ${\mathrm{BaO}}_2$ with dil. ${\mathrm{H}}_2{\mathrm{SO}}_4$ are–

1. 0, and –1                                   

2. –1, and –2

3. –2, and 0                                     

4. –2, and + 1

Hardness of water is due to presence of salts of –

(1) ${\mathrm{Na}}^{+}<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{K}}^{+}$                                  

(2) ${\mathrm{Ca}}^{2+}<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{Mg}}^{2+}$

(3) ${\mathrm{Ca}}^{2+}<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{K}}^{+}<mspace\; width="1em"></mspace>$                                

(4) ${\mathrm{Ca}}^{2+}<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{Na}}^{+}$