The reducing power of a metal depends on various factors. Suggest the factor which makes Li, the strongest reducing agent in aqueous solution.

1. Sublimation enthalpy

2. Ionisation enthalpy

3. Hydration enthalpy

4.  Electron-gain enthalpy

Metal carbonates decompose on heating to give metal oxide and carbon dioxide. Which of the metal carbonates is most stable thermally?

1. ${\mathrm{MgCO}}_3$

2. ${\mathrm{CaCO}}_3$

3. ${\mathrm{SrCO}}_3$

4. ${\mathrm{BaCO}}_3$

Which of the carbonates given below is unstable in air and is kept in CO₂ atmosphere to avoid decomposition?

1. ${\mathrm{BeCO}}_3$

2. ${\mathrm{MgCO}}_3$

3. ${\mathrm{CaCO}}_3$

4. ${\mathrm{BaCO}}_3$

The least basic metal hydroxide among the following is-

1.  $\mathrm{Mg}{\left(\mathrm{OH}\right)}_2$

2.  $\mathrm{Ca}{\left(\mathrm{OH}\right)}_2$

3.  $\mathrm{Sr}{\left(\mathrm{OH}\right)}_2$

4.  $\mathrm{Ba}{\left(\mathrm{OH}\right)}_2$

Some of the Group 2 metal halides are covalent and soluble in organic solvents. Among the following metal halides, the one which is soluble in ethanol is

1. BeCl₂ 

2.  MgCl₂ 

3.  CaCl₂ 

4.  SrCl₂ 

The order of decreasing ionisation enthalpy in alkali metals is

1. Na>Li>K>Rb

2. Rb<Na<K<Li

3. Li>Na>K>Rb

4. K<Li<Na<Rb

The solubility of metal halides depends on their nature, lattice enthalpy and hydration enthalpy of the individual ions. Amongst fluorides of alkali metals, the lowest solubility of LiF in water is due to -

1. Ionic nature of lithium fluoride

2. High lattice enthalpy

3. High hydration enthalpy for lithium ion

4. Low ionisation enthalpy of lithium ion

Amphoteric hydroxides react with both alkalies with acids. Which of the following Group 2 metal hydroxides is soluble in sodium hydroxide?

1. $\mathrm{Be}{\left(\mathrm{OH}\right)}_2$

2. $\mathrm{Mg}{\left(\mathrm{OH}\right)}_2$

3. $\mathrm{Ca}{\left(\mathrm{OH}\right)}_2$

4. $\mathrm{Ba}{\left(\mathrm{OH}\right)}_2$

In the synthesis of sodium carbonate, the recovery of ammonia is done by treating ${\mathrm{NH}}_4\mathrm{Cl}$ with $\mathrm{Ca}{\left(\mathrm{OH}\right)}_2$. The by-product obtained in this process is

1. ${\mathrm{CaCl}}_2$

2. $\mathrm{NaCl}$

3. $\mathrm{NaOH}$

4. ${\mathrm{NaHCO}}_3$

When sodium is dissolved in liquid ammonia, a solution of deep blue color is obtained. The color of the solution is due to

1. Ammoniated electron

2. Sodium ion

3. Sodium amide

4. Ammoniated sodium ion