Hydrogen exists in diatomic form rather than monoatomic form under normal conditions -

1. Due to high ionization enthalpy

2. Due to low ionization enthalpy

3. Due to high electron gain enthalpy

4. Due to low electron gain enthalpy

The role of an electrolyte in the  preparation of H₂ by electrolytic method is-

1. It lowers the availability of ions available in the process for the conduction of electricity.

2. It makes the ions available in the process for conduction of electricity.

3. It cools down the temperature.

4. None of the above

The products obtained on the completion of the following reactions respectively are -

$$\begin{gathered} \mathrm{i}. {\mathrm{C}}_3{{\mathrm{H}}_8}_{\left(\mathrm{g}\right)}+3{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}\underset{\mathrm{catalyst}}{\overset{∆}{\to }} \\ \mathrm{ii}. {\mathrm{Zn}}_{\left(\mathrm{s}\right)} + {\mathrm{NaOH}}_{\left(\mathrm{aq}\right)}\stackrel{\mathrm{heat}}{\to } \end{gathered}$$

$$\begin{gathered} 1. \mathrm{i}=\mathrm{CO}, \mathrm{and} {\mathrm{H}}_2 ; \mathrm{ii} ={\mathrm{Na}}_2{\mathrm{ZnO}}_2 \\ 2. \mathrm{i} = {\mathrm{CO}}_2 ; \mathrm{ii} = {\mathrm{Na}}_4{\mathrm{ZnO}}_4 \\ 3. \mathrm{i} = {\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{OH} ; \mathrm{ii} = \mathrm{ZnO} \\ 4. \mathrm{None} \mathrm{of} \mathrm{the} \mathrm{above} \end{gathered}$$

The consequences of high enthalpy of H–H bond in dihydrogen is /are -

1. Hydrogen has a low tendency to form ${\mathrm{H}}^{+}$ions.

2. It forms diatomic molecules ($H_2$), and a large number of covalent hydrides.

3. Hydrogen does not possess metallic characteristics.

4. All of the above.

The correct statement about electron-deficient hydride is that it-
1. Does not have sufficient electrons to form a regular bond.
2. Cannot be represented by conventional Lewis structures.
3. Accept electrons.
4. All of the above.

The hydrides of carbon (C_nH_{2n + 2}) can act as -

1. Lewis acid.

2. Lewis base.

3. Both 1 and 2

4. None of the above

“Non-stoichiometric hydrides” are -

1. Hydrogen-deficient compounds formed by the reaction of a hydrogen atom with d-block and f-block elements.

2. Hydrogen-deficient compounds formed by the reaction of dihydrogen with d-block and f-block elements.

3. Electron-deficient compounds formed by the reaction of dihydrogen with d-block and f-block elements.

4. Hydrogen-deficient compounds formed by the reaction of dihydrogen with s-block and p-block elements.

Metallic hydrides are useful for hydrogen storage because : 

1. Metallic hydrides are oxygen-rich.

2. Metallic hydrides are oxygen-deficient.

3. Metallic hydrides are hydrogen-rich.

4. Metallic hydrides are hydrogen-deficient.

Atomic hydrogen or oxy-hydrogen torch is used for cutting and welding purposes because-

1. Atomic hydrogen converts into molecular hydrogen and generates a large amount of energy

2. Atomic hydrogen converts into molecular hydrogen and generates a low amount of energy

3. Molecular hydrogen converts into atomic hydrogen and generates a large amount of energy

4. Molecular hydrogen converts into atomic hydrogen and generates a low amount of energy

 The correct order of hydrogen bonding among ${\mathrm{NH}}_3$, H2O and HF, is 

1. HF > ${\mathrm{H}}_2\mathrm{O}$ > ${\mathrm{NH}}_3$

2. ${\mathrm{H}}_2\mathrm{O}$ > HF > ${\mathrm{NH}}_3$

3. HF > ${\mathrm{NH}}_3$ > ${\mathrm{H}}_2\mathrm{O}$

4. ${\mathrm{NH}}_3$ > HF > ${\mathrm{H}}_2\mathrm{O}$