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

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

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

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

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

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.

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

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.

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

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.

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

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.

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

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.
RBSE Solutions for Class 11 Chemistry Chapter 10 s-Block Elements 1

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.

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

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
RBSE Solutions for Class 11 Chemistry Chapter 10 s-Block Elements 2
(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.
RBSE Solutions for Class 11 Chemistry Chapter 10 s-Block Elements 3
The carbon dioxide required for this reaction is produced by heating (calcination) of the lime stone at 950-1100o 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.
RBSE Solutions for Class 11 Chemistry Chapter 10 s-Block Elements 4

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.
RBSE Solutions for Class 11 Chemistry Chapter 10 s-Block Elements 5
(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 :
RBSE Solutions for Class 11 Chemistry Chapter 10 s-Block Elements 6

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

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.
RBSE Solutions for Class 11 Chemistry Chapter 10 s-Block Elements 7

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)
RBSE Solutions for Class 11 Chemistry Chapter 10 s-Block Elements 8
∵ Bond order is zero so it does not exist as molecule.

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

Question 27.
What are oxides, peroxides and superoxides ? Explain with example.
Answer:
Oxides : Oxides have O2- 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.

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

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.
RBSE Solutions for Class 11 Chemistry Chapter 10 s-Block Elements 9
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.

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

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.

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

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

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

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

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

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.
RBSE Solutions for Class 11 Chemistry Chapter 10 s-Block Elements 12
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.
RBSE Solutions for Class 11 Chemistry Chapter 10 s-Block Elements 13
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

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

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 LiNOforms nitrites with the evolution of oxygen. On heating form respective oxides NOand O2 Except      Li2COthey are stable towards heat. They decompose on heating to form oxides with the evolution of CO2 Thermally stable Thermally Stable

RBSE Solutions for Class 11 Chemistry