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 > ${\mathrm{H}}_2\mathrm{O}$

CO2  is used as a fire extinguisher because -

1. ${\mathrm{CO}}_2$ has high lattice energy.

2. ${\mathrm{CO}}_2$ will be effective in isolating the burning surface from dihydrogen and dioxygen.

3. ${\mathrm{CO}}_2$ is lighter than dioxygen.

4. ${\mathrm{CO}}_2$ is lighter than dihydrogen.

The term ’auto-protolysis’ of water means -

1. Physical reaction in which two water molecules react to produce a hydroxide ion (OH^–) and a hydronium ion (${\mathrm{H}}_3{\mathrm{O}}^{+}$).

2. Chemical reaction in which five water molecules react to produce a hydroxide ion (OH^–) and a hydronium ion (${\mathrm{H}}_3{\mathrm{O}}^{+}$).

3. Chemical reaction in which two water molecules react to produce a hydroxide ion (OH^–) and a hydronium ion (${\mathrm{H}}_3{\mathrm{O}}^{+}$).

4. Chemical reaction in which two water molecules react to produce three hydroxide ions (OH^–)