RBSE Solutions for Class 11 Chemistry Chapter 10 s-Block Elements

December 17, 2021

Rajasthan Board RBSE Class 11 Chemistry Chapter 10 s-Block Elements

RBSE Class 11 Chemistry Chapter 10 Text Book Questions

RBSE Class 11 Chemistry Chapter 10 Multiple Choice Questions

Question 1.
Which of the following alkaline earth metal is most stable to heat ?
(a) MgCOg
(b) CaCOg
(c) SrCOg
(d) BaCOg
Answer:
(d) BaCOg

Question 2.
Which of the following compound is obtained as a bye-product in Solvay Ammonia Process ?
(a) Carbon dioxide
(b) Ammonia
(c) Calcium chloride
(d) Calcium carbonate
Answer:
(c) Calcium chloride

Question 3.
Which of the alkali metals halides have lowest lattice energy ?
(a) LiF
(b) NaCl
(c) KBr
(d) Csl
Answer:
(d) Csl

Question 4.
Which of the following do not give flame test ?
(a) Be
(b) K
(c) Sr
(d) Na
Answer:
(a) Be

Question 5.
Which of the following metal has lowest melting point ?
(a) Na
(b) K
(c) Rb
(d) Cs
Answer:
(d) Cs

RBSE Class 11 Chemistry Chapter 10 Very Short Answer Type Questions

Question 6.
Why Group I elements are known as alkali metals ?
Answer:
Group I elements are known as alkal metals because they form hydroxides on reaction with water which are strongly alkaline in nature.

Question 7.
In a period, melting points of alkali metals are lower than alkaline eath metals, Why ?
Answer:
In a period, melting points of alkali metals are lower than earth metals bacause alkali metals have weak metallic bond between them due to presence of only one valence electron.

Question 8.
Alkali metals are strong electropositive in nature, Why ?
Answer:
Alkali metals are strong electropositive in nature because they have low ionization energies and their atoms readily lose the valence electron.

Question 9.
Which metal is responsible for clotting of blood in our body ?
Answer:
Calcium ion is responsible for clotting of blood in our body.

Question 10.
Write the electronic configuration of alkali metals.
Answer:
The general electronic configuration of alkali metals is ns1.

Question 11.
Calculate the oxidation state of Na in Na₂O₂.
Answer:
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.

Question 12.
Sodium is less reactive than potassium. Why ?
Answer:
In potassium, the outer most electron is better shielded form the attractive force of the nucleus. It follows that this outermost electron is more easily lost than it is in sodium and potassium can be converted to ionic form more readily then sodium. Hence, sodium is less reactive than potassium.

Question 13.
Why alkali metals and alkaline earth metals are not obtained by chemical reduction method ?
Answer:
Alkali metals and alkaline earth metals are not obtained by chemical reduction method because they are very strong reducing agent and highly reactive metals. Due to this reason, they cannot be reduced by any other substance very easily.

Question 14.
Potassim carbonate cannot be prepared by Solvay process. Why ?
Answer:
Potassium carbonate cannot be prepared by solvay process because unlike sodium bicarbonate, potassium bicarbonate is fairly soluble in water and does not precipitate out.

Question 15.
What reaction takes place when quick lime is heated with silica ?
Answer:
When quick lime is heated with silica, it form calcium silicate.
CaO + siO₂ ➝ CasiO₃

Question 16.
Why second ionization energy of alkali metals are higher than first ionization energy ?
Answer:
The second ionization energy of alkali metals are higher than first ionization energy is due to following reason.

The electrons configuration of M⁺ ion is ns² np⁶ i.e. stable noble gas configuration.
The size of M⁺ ions is smaller due to increase in nuclear size.

Question 17.
Why lithium compounds are covalent in nature ?
Answer:
Lithium compounds are covalent in nature because lithium is the smallest atom in group 1 so the attraction between the outer electrons and the nucleus is greater. So, it is very hard for it to lose electrons to another element to form a compound by an ionic bond.

Question 18.
How is lithium aluminum hydride formed ?
Answer:
Lithium aluminium hydride is formed by the reaction between lithium hydride and aluminium chloride.
4LiH + AlCl₃ ➝ LiAlH₄ + 3LiCl

Question 19.
What is Hydrolith? How it reacts with water ?
Answer:
Calcium hydride (CaH2) is hydrolith. It reacts with water to form calcium hydroxide.

Question 20.
Which of the following NaOH or Mg (OH)2 is strong base ?
Answer:
NaOH is strong base, as it is completely hydrolysed in solution.

RBSE Class 11 Chemistry Chapter 10 Short Answer Type Questions

Question 21.
Compare the alkali metals and alkaline earth metals with respect to

ionisation enthalpy
atomic and ionic radius

Answer:

Ionisation enthalpy : The first ionisation enthalpies of alkaline earth metals are higher than those of alkali metals due to increases nuclear charge. However second ionisation enthalpies of alkaline earth metals are lower then those of corresponding alkali metals.
Atomic and ionic radius :
Atomic and ionic radii of alkaline earth metals are lesser than those of alkali metals due to increases nuclear charge.

Question 22.
Why potassium and calcium are used in photoelectric all in place of Lithium ?
Answer:
Potassium and calcium are used in photoelectric cell in place of lithium due to their ion-ionization energies. They are used as electrodes in photoelectric call.

Question 23.
Starting with sodium chloride how would proceed to prepare—
(i) sodium hydroxide
(ii) sodium carbonate ?
Answer:
(i) Sodium hydroxide is prepared by the electrolysis of sodium chloride solution in castner-kellne cell. A brine solution (1 % NaCl solution) is electrolysed Using a mercury cathode and a carbon anode.
NaCl ➝ Na⁺ + Cl^–
H₂O ➝ H⁺ + OH^–
At cathode Na⁺ + e^– ➝ Na
Na + Hg ➝ Na – Hg (amalgam)
At anode Cl^– ➝ Cl + e^–
Cl + Cl ➝ Cl₂
NaOH ➝ Na⁺ + OH^–
H₂O ➝ H⁺ + OH^–
Sodium and hydrogen ion migrate towards cathode while OH^– and Hg migrate towards anode.
At Cathode, 2H⁺ + 2e^– ➝ H₂
At anode, OH^– ➝ OH + e^–
Na-Hg + OH ➝ NaOH + Hg
Over all reaction

(ii) Sodium Carbonate is prepared from sodium chloride by solvay process. Common salt and lime stone are used as raw materials in this process. It is formed in following steps—

Step I — Carbon dioxide passes through a concentrated aqueous solution of sodium chloride and ammonia.

The carbon dioxide required for this reaction is produced by heating (calcination) of the lime stone at 950-1100^o C.
CaCO₃ ➝ CaO + CO₂

Step II— The sodium bi carbonate (NaHCO₃) that precipitates out in above step is filtered out from the hot ammonium chloride (NH₄ Cl) solution and the solution is then reacted with the quick lime (CaO) formed in above reaction.
2NH₄Cl + CaO ➝ 2NH₃ + CaCl₂ + H₂O
CaO makes a strong basic solution. The ammonia formed in this reaction is recycled back to the initial brine solution of first reaction.

Step III— Sodium bicarbonate that precipitate out in step I. It is converted to the product (sodium carbonate) by calcination at 160°-230°C.

Question 24.
Explain the structure of the following —
(a) BeCl₂ (vapour state)
(b) BeCl₂ (solid state)
Answer:
(a) In the vapour state, BeCl₂ tends to form a chlorobridged dimer which dissociates into the linear monomer at high temperature of the order of 1200K.

(b) In the solid state, BeCl₂ has a chain structure. In this structure, each Be atom is surrounded by chlorine atoms i. e., two Cl-atom by covalent bonds and other two Cl-atoms by co-ordinate covalent bond. The chain structure is shown as below :

Question 25.
Why alkali metals given blue colour with liquid ammonia ?
Answer:
Alkal metlals give blue colour with liquid ammonia due to formation of ammoniated electron which absorbs energy in the visible region of light and therefore imparts blue colour to the solution.

Question 26.
H₂ molecule exists but He₂ molecule does not. Why?
Answer:
H₂ molecule exists because of it bond order, which is equal to one where as He₂ molecule does not exist due to zero bond order. Bond order of these molecules can be calculated as follow :
H₂ Electronic configuration ₁H = 1s¹
₁H = 1s¹
Molecular orbital configuration σ(1s²)

∵ Bond order is zero so it does not exist as molecule.

Question 27.
What are oxides, peroxides and superoxides ? Explain with example.
Answer:
Oxides : Oxides have O^2- ions. Oxides are binary compounds of oxygen with another elements e.g. Na₂O,Li₂O, etc.
Peroxides : The peroxides are the compounds containing oxygen-oxygen (O-O) single bond or peroxide ion $\mathrm{O}_{2}^{2-}$ . The oxidation state of oxygen in peroxides is -1.
These are strong oxidising agents e.g. Na₂O₂, H₂O₂ etc.
Superoxides : These are the compounds having superoixde ion (O₂). The state a oxygen in superoxides is -1/2. These are paramagenetic in nature due to presence of paired electrons,
e.g. KO₂, RbO₂ etc.

Question 28.
Dehydrated calcium chloride is used as a dehydration agent. Why ?
Answer:
The dehydrated calcium chloride is used a dehydration agent because of high solvation energy of calcium ion. Due to this, it has a tendency to form hygrate when react with water.
CaCl₂ +2H₂O ➝ Ca(OH)₂ + 2HCl

Question 29.
LiF is almost insoluble in water where as Licl is soluble only in water but also in acetone. Explain the reason.
Answer:
LiF is almost insoluble in water. This is due to its high lattice energy where as LiCl is soluble in water as well as acetone due to low lattice energy.

Question 30.
The solution of sodium carbonate is alkaline. Why?
Answer:
Aqueous solution of sodium carbonate is alkaline in nature because it forms sodium hydroxide which is strong base and carbonic acid, which is weak acid.

Question 31.
Write the uses of following-
(a) Lime stone
(b) Sodium carbonate
Answer:
(a) Uses of Lime Stone —

It is used as building material in the form of marble.
It is used in the manufacture of quick lime.
It is used as flux along with magnesium carbonate in the extraction of metals such as iron.
It is used as an antacid, abrassive in tooth paste etc.

(b) Uses of Sodium Carbonate

It is used in paper, paints and textile industries.
It is an important laboratory reagent both in qualitative and analysis.
It is used in the manufacture of glass, soap, borex and caustic soda.
It is used in water softening.

Question 32.
Explain the following :
BeO is almost insoluble but BeSO4 is soluble in water.
Answer:
BeO is almost insoluble in water due to high lattice energy. BeS04 is soluble in water due to high hydration energy of Be²⁺ion. The high hydration energy of Be²⁺ in BeSO₄ overcomes the lattice energy factor and therefore BeSO₄ is soluble in water.

Question 33.
Write balanced equation for relations between
(a) Be₂ C and water
(b) KO₂ and water
(c) Lithium and nitrogen
Answer:
(a) Reaction between Be2C and water
Be₂C + 4H₂O ➝ 2Be(OH)₂ + CH₄
(b) Reaction between KO₂ and water :
4KO₂ + 2H₂O ➝ 4KOH + 3O₂
(c) Reaction between lithium and nitrogen
6Li + N₂ ➝ 2Li₂N

Question 34.
What is diagonal relationship ? How berglium shows similarity with Aluminium ?
Answer:
Diagonal relationship— A diagonal relation ship is said to exist between certain pairs of diagonally adjacent elements in the second and third periods of the periodic table.
e.g. Li and Mg, Be and Al, B and Si etc.
Beryllium shows similarity with aluminium because these elements exhibit diagonal relationship. The similarities are –

Beryllium hydroxide and aluminium hydroxide react with excess alkali to form respective ions.
Both these elements have the capacity to withstand the acid attack due to presence of an oxide film on the surface of the metal.

Question 35.
Lithium does not show similarity with other elements of the group. Explain the reasons.
Answer:
Lithium does not show similarity with other elements of the group due to following reasons –

exceptionally small size of its atom and ion.
high polarizing power (i.e. charge /radius ratio).
high ionization energy and least electropositive character.
absence of d-orbitals.

RBSE Class 11 Chemistry Chapter 10 Long Answer Type Questions

Question 36.
Write the properties of hydrides of Group I and Group II elements and draw explain the structure of Beryllium hydride.
Answer:
Properties of hydrides of Group I:

All the alkali metal hydrides are ionic solids with high melting points. Ionic character increases from Li H to CsH.
These hydrides are strong reducing agents and their reducing character increases on moving down the group.
On moving down the group, atomic size increases and M-H bond strength in hydrides decreases so, thermal stability of hydrides decreases.

Properties of hydrides of Group II:

BeH₂ and MgH₂ are electron deficient covalent compounds. They have polymeric structure due to electron deficiency.
CaH₂ and SrH₂ are ionic in nature.
Hydrides of alkaline earth metals react with water to librate hydrogen gas. So, they actions as reducing agents.

Structure of Beryllium Hydride :
Beryllium hydride (BeH₂) is electron deficient covalent compound. It has polymeric structure. It has three-centre two-electron bond with chemical formula (BeH₂)_n.

Question 37.
Explain the importance of Na, K, Mg and Ca in biological fluids.
Answer:
The biological importance of Na, K, Mg and Ca in biological fluids is given as below –

An adult body contains 90 g of Na, 170 g of K, 25 g of Mg and 1200 of Ca.
Na+ and K+ ions participate in transmission of nerve signals in regulating the flow of water across the cell membrane and in transport of sugar and amino acids into the cell.
K+ Ions activate many enzymes to participate in the oxidation of glucose to produce ATP and with sodium are responsible for the tranmission of nerve signals and also help in protein synthesis.
Sodium ions are found primarily on the outside of cells, being coded in blood plasma and in the interstitial fluid which surrounds the cell whereas potassium ions are most abundant cations with in the cell fluids.
M²⁺ions catalyse many enzymatic reaction. All enzymes that utilize ATP in phosphate transfer require magnesium as the cofactor.
The main pigment for obserbtion of light in plant is chlorophyll which contains magnesium.
About 99% to body calcium is present in bones and teeth.
Calcium plays an important role in neuromuscular function, inter neuronal transmission, cell membrane integrity and blood coagulation.

Question 38.
Explain the commercial method of preparation of sodium carbonate.
Answer:
Preparation of sodium carbonate by solvay process—Sodium carbonate is generally prepared by Solvay process. In this process, Brine (NaCl solution) is converted to sodium carbonate. Common salt and lime stone are used as raw material. The main steps of the process are :
1. In this process advantage is taken of the low solubility of sodium hydrogen Carbonate whereby it gets precipitated in the reaction of sodium chloride with ammonium hydrogen carbonate. The latter is prepared by passing CO₂ to a concentrated solution of sodium chloride saturated with ammonia, where ammonium carbonate followed by ammonium hydrogen carbonate are formed. The equations for the complete process may be written as
2NH₃ + H₂O + CO₂ ➝ (NH₄ )₂ CO₃
(NH₄ )₂ CO₃ + H₂O + CO₂ ➝ 2NH₄HCO₃
NH₄HCO₃ +NaCl ➝ NH₄Cl +NaHCO₃
Sodium hydrogen carbonate is insoluble due to common ion effect. It can be obtained by filtration.
2. Sodium hydrogen carbonate crystal separates. These are heated at 1500°C to given sodium carbonate.
2NaHCO₃ ➝ Na₂CO₃ + CO₂ + H₂O
3. In this process NH3 solution containing NH₄Cl is treated with Ca (OH)₂. Calcium chloride is obtained as a by-product.

CaCO₃ ➝ CaO + CO₂
CaO + H₂O ➝ Ca(OH)₂
2NH₄Cl + Ca(OH)₂ ➝ 2NH₃ + CaCl₃ + 2H₂O
It may be mentioned here that solvay process cannot be extended to the manufacture of potassium carbonate because potassium hydrogen carbonate is to soluble to be precipitated by the addition of ammonium hydrogen carbonate to a saturated solution of potassium chloride.

Question 39.
Explain the chemical properties of alkali metals and explain inspite of highest ionisation enthalpy energy in the group, Lithium is a strong reducing agent why ?
Answer:
Chemical Properties of Alkali Metals :
1. Reaction with Oxygen — The alkali metals ternish in dry air due to the formation of their oxides. They form different of oxides like oxides peroxides and superoxides.

2. Reaction with Water— The alkali metals react with water to form hydroxides and release dihydrogen.
3. Reaction with Dihydrogen—The alkali metals react with dihydrogen at about 673 k (Li at 1073 K) to from hydrides.

4. Reaction with Halogens—The alkali metals readily react vigorusly with halogens to form ionic halides,
e.g. 2M + X₂ ➝ 2MX
(Where M = Alkal metal like Na, K etc.)
5. Reaction with Liquid Ammonia—The Alkali metals dissolve in liquid ammonia to give deep blue solutions.
M + (x + y)NH₃ ➝ [M (NH₃)_x] ⁺ + [e[(NH₃)_y ]^–
Reducing nature of Lithium
Inspite of highest ionization enthalpy energy in the Group, lithium is a strong reducing agent because of the following reasons-

Maximum hydration energy of lithium
Small size of lithium

Question 40.
Compare the solubility and thermal stability of the following compounds of the alkali metals with
those of the alkaline earth metals
(a) nitrates
(b) carbonates
(c) sulphates.
Answer:
Comparison of stability and thermal stability sulphate of alkali metals with those of alkaline earth metals

Property	Nitrates Alkali metals	Alkaline earth metals	Alkali metals	Alkaline earth metals	Alkali metals	Alkaline earth metals
Solubility in water	Soluble	Soluble	Soluble except Li₂CO₃	Insoluble	Soluble	Less soluble
Thermal stability	On heating nitrates alkali metals except LiNO₃forms nitrites with the evolution of oxygen.	On heating form respective oxides NO₂and O₂	Except Li₂CO₃they are stable towards heat.	They decompose on heating to form oxides with the evolution of CO₂

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