Explore topic-wise InterviewSolutions in .

Both hydrogen and halogen are in short of one electron for the completion of the outermost orbital containing electrons.

Both LiCl and MgCl₂ are covalent compounds and dissolve in alcohol. This is due to high polarizing power of Li⁺ and Mg²⁺ ions.

Alkaline earth metals are higher due to the completely filled orbitals than alkali metals. 

The answer is (d) BaCOg

(c) Calcium chloride

Sodium carbonate

Solvay process is not applicable for the preparation of potassium carbonate because potassium bicarbonate is soluble in water and it doesn‘t precipitates out like sodium bicarbonate.

A suspension of slaked lime in water is known as milk of lime.

Calcium nitrate, when heated, undergoes decomposition to form calcium oxide.

2Ca(NO₃)_2(s) → 2CaO_(s) + 4NO_2(g) + O_2(g)

Milk of lime reacts with chlorine to form bleaching powder. 2 Ca(OH)₂+Cl₂ → CaCl₂ + Ca(OCl)₂ +2H₂O

Bleaching powder is formed when chlorine is made to react with slaked lime.

Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O

1. Sodium bicarbonate is used as antacid 

2. Slaked lime is used for white washing, 

3. NaOH is used in manufacture of soap.

(i) CaCO₃ \(\overset{heat}{\longrightarrow}\) CaO + CO₂ (decomposition/calcinations)

(ii) 2CaSO₄.2H₂O \(\overset{heat}{\underset{strongly}{\longrightarrow}}\) (CaSO₄) (Dead buront plaster)(Plaster of paris) + 2H₂O + 3H₂O

Sodium, cesium, lithium, francium, potassium, rubidium all together comprises of alkali metals. they consist of only one electron on its valence shell, which gets loosed easily due to their low ionization enthalpy.

They have strong metallic bonds due to smaller size and have more number of valence electrons.

Let, the oxidation state of Na = x
2x + 2(-1) = 0
2x – 2 = 0
2x = 2
x = \(\frac{2}{2}\) = 1
∴ The oxidation state of Na in Na₂O₂ is +1.

It is because of low ionisation energy. They can lose electrons easily, that is why they are strong reducing agents.

K has lower ionisation energy than sodium due to bigger atomic size, therefore, it is more reactive.

The sizes of Be²⁺ and O^2- are small and are highly compatible with each other. Due to this, a high amount of lattice energy is released during its formation. The hydration energy, when it is made to dissolve in water, is not enough to overcome the lattice energy. Thus, BeO is barely soluble in water.

Whereas the size of an SO₄^2- is large compared to Be²⁺ and there is lesser compatibility and lattice energy which can be easily overcome by the hydration energy. Thus, BeSO₄ is easily soluble in water.

Sodium carbonate is a salt of strong base and weak acid so it on hydrolysis gives NaOH which is a strong base so Sodium carbonate solution is alkaline.

Calcium is used for drying alcohol and non-acidic gases because Ca does not react with alcohol.

(i) 3Ca + N₂ → Ca₃N₂ 

(ii) 2Ca + SO₂ → 2CaO + S

(iii) Ca(OH)₂ + 2NH₄Cl \(\overset{heat}{\longrightarrow}\) CaCl₂ + 2H₂O + 2NH₃

(iv) 2Ca + CO₂ → CaO + C

1. Ca + H₂O → CaO + H₂O 

2. Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O 

3. BeO + NaOH → Be(OH)₂ + Na₂O

The hydration energy of Ca²⁺ over comes the second ionization energy of Ca, that is why Ca forms CaCl₂ and not CaCl. Ca⁺ is not stable.

Ca O is quick lime. When we add water to it slaked limes Ca (OH)₂ is formed.

Atomic size increases with increase in atomic number, therefore, nuclear force of attraction between valence electrons and nucleus decreases, hence ionisation energy decreases down the group.

It is due to largest atomic size, they can lose electrons easily.

Alkali metals show oxidation state of +1. With the loss of valence electron it attains the stable configuration of nearest inert gas. Its second ionization potential is high. Hence an alkali metal does not show variable oxidation states.

It is because their hydroxides are soluble bases called alkalies. Secondly their ashes are alkaline in nature.

Li⁺ > Ma⁺ > K⁺ > > Rb⁺ > Cs⁺

Li⁺ forms largest hydrated cations because it has highest hydration energy. It has smallest size therefore, it is most hydrated.

Ca²⁺ is smaller in size than Na⁺ and also it has higher charge, therefore, its hydration energy is more than that of Na⁺.

Because of its complex forming behaviour.

Because sodium is more electropositive.

The answer is (a) Be

The answer is (d) Cs

Li has highest and Fr has lowest ionization enthalpy.

The answer is (d) Csl

1. BeSO₄ is soluble in water while BaSO₄ is not. 

2. Be(OH)₂ dissolves in NaOH while Ba(OH)₂ is insoluble.

Second ionization energy of group 2 elements is less than group 1 elements because more energy is required to release second electron from alkali metals. i.e. group 1 metal ions.

It is due to their high second ionization enthalpy and stability of their ions [Na⁺ K⁺]

Beryllium, Magnesium, Calcium, Strontium, Barium and Radium.

It is due their high second ionization enthalpy and stability of their ions [Na⁺,K⁺].

Because of their largest size in their respective periods, solitary electron present in the valence shell can be removed by supplying small amount of energy.

The atomic and ionic radii of the alkaline earth metals are smaller than those of the corresponding alkali metals in the same period because of the increased nuclear charge in there elements.

The alkaline earth metals have low iodization enthalpies due to fairly large size of atoms.

The higher value of second ionization enthalpy is more than compensated by the higher enthalpy of hydration of Ca²⁺. Therefore formation of CaCl₂ becomes more favorable than CaCl energetically.

The alkali metals dissolve in liquid ammonia giving deep blue solutions

M +(x +y)NH₃= [M(NH₃ )x ]⁺ + [e (NH₃ )y ]^-

The blue colour of the solution is due to the ammoniated electron which absorbs energy in the visible region of light and thus imparts blue colour to the solution.

The solvated electron, [e(NH₃)]^– or ammoniated electron is responsible for blue colour of alkali metal solution in NH₃. It absorbs light from visible region and radiates complementary colour, (in the equation am = ammoniated)

Na⁺(am) + e^–(am) + NH₂(l) → NaNH₂(am) + \(\frac{1}{2}\)H₂(g)

When CO₂ gas is passed in lime water, it turns milky due to the formation of calcium carbonate.

Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

On passing excess of CO₂, milkiness disappears because the precipitate dissolves to form calcium hydrogen carbonate.

CaCO₃ + CO₂ + H₂O → Ca(HCO₃)₂