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

Ice melts at a temperature lower than its usual melting point when the pressure is increased. This is because-

1. Ice has less density than water

2. Pressure generates heat

3. The chemical bonds break under pressure

4. Ice is not a true solid

An ionic compound is dissolved simultaneously in heavy water and ordinary water, its solubility is –

1. Larger in heavy water

2. Smaller in heavy water

3. Solubility is same in both waters

4. Smaller in simple water

The method used for the preparation of ${\mathrm{H}}_2{\mathrm{O}}_2$ is -

1. Cold H₂SO₄ + BaO₂.8H₂O                             

2. HCl + BaO₂

3. Conc. H₂SO₄ + Na₂O₂                          

4. H₂ + O₂

The oxide that gives hydrogen peroxide $\left({\mathrm{H}}_2{\mathrm{O}}_2\right)$ on the treatment with a dilute acid $\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)$ is –

(A) ${\mathrm{MnO}}_2$                                  

(B) ${\mathrm{PbO}}_2$

(C) ${\mathrm{Na}}_2{\mathrm{O}}_2$                                 

(D) ${\mathrm{TiO}}_2$ 

The oxidising property of ${\mathrm{H}}_2{\mathrm{O}}_2$ is best explained by assuming that two oxygen atoms in its molecule are.

(1) Bonded differently

(2) Bonded similarly

(3) Bonded covalently

(4) Bonded by hydrogen bonds

${\mathrm{H}}_2\mathrm{O}$ has higher boiling point than ${\mathrm{H}}_2\mathrm{S}$ because.

(1) ${\mathrm{H}}_2\mathrm{O}$ is a smaller molecule and hence more closely packed

(2) The bond angle in ${\mathrm{H}}_2\mathrm{O}$ is more than in ${\mathrm{H}}_2\mathrm{S}$ and hence the former molecules are more tightly packed

(3) Of intermolecular hydrogen bonding in liquid ${\mathrm{H}}_2\mathrm{O}$

(4) The latent heat of vaporisation is higher for ${\mathrm{H}}_2\mathrm{O}$ than for ${\mathrm{H}}_2\mathrm{S}$