Courses for Kids
Free study material
Offline Centres
Store Icon

Tthe S Block Elements Class 11 Notes CBSE Chemistry Chapter 10 (Free PDF Download)

Last updated date: 17th Jul 2024
Total views: 744k
Views today: 15.44k

Revision Notes for CBSE Class 11 Chemistry Chapter 10 (The s-Block Elements) - Free PDF Download

The group 1 and group 2 elements of the periodic table are referred to as the s-block elements. Further, group 1 elements are called alkali metals, and the group 2 elements are called alkaline earth metals. They are named so due to the alkaline nature of the oxides and hydroxides. Two s-electrons characterize the Alkaline earth metals, and the alkali metals are characterized by one s-electron in the valence shell of their atoms. Both these metals form a mono positive ion and dipositive ions and are extremely reactive.

Let us look at more about the topic, s-block elements, its subtopics, and more related things in brief.

Download CBSE Class 11 Chemistry Revision Notes 2024-25 PDF

Also, check CBSE Class 11 Chemistry revision notes for other chapters:

The s-Block Elements Chapter Related Important Study Materials
It is a curated compilation of relevant online resources that complement and expand upon the content covered in a specific chapter. Explore these links to access additional readings, explanatory videos, practice exercises, and other valuable materials that enhance your understanding of the chapter's subject matter.

Competitive Exams after 12th Science

The s-Block Elements Class 11 Notes Chemistry - Basic Subjective Questions

Section – A (1 Mark Question

1. Melting and boiling points of alkali metals is low. Explain?

Ans. The melting and boiling points of the alkali metals are low indicating weak metallic bonding due to the presence of only a single valence electron in them.

2. Explain the diagonal relationship in periodic table. 

Ans. The diagonal relationship is due to the similarity in ionic sizes and/or charge/radius ratio of the elements.

3. Why are lithium compounds soluble in organic solvents? 

Ans. Due to high polarizing power, there is increased covalent character of lithium compounds which is responsible for their solubility in organic solvents.

4. By which process we can prepapre sodium carbonate?

Ans. Sodium carbonate is generally prepared by Solvay’s process.

5. What is the formula of soda ash?

Ans. Na2CO3 is the formaula of soda ash.

6. Why do alkaline earth metals have low ionization enthalpy?

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

7. What is the nature of oxide formed by Be?

Ans. BeO is covalent and amphoteric while oxides of other elements are ionic and basic in nature.

8. Beryllium show similarities with Al. Why?

Ans. Because of their similarity in charge/radius ratios. 

(Be2+,2/31 = 0.064 and Al3+,3/50 = 0.66)

9. What happens when gypsum is heated to 390K?

Ans. When we heat gypsum at 390K ,Plaster of parts will form.


10. Whyanhydrous calcium sulphate cannot be used as plaster of Paris?

Ans. Because it does not have the ability to set like plaster of Paris.

Section – B (2 Marks Questions)

11. Why metals like potassium and sodium cannot be extracted by reduction of their oxides by carbon?

Ans. Potassium and sodium are strong electropositive metals and have great affinity for oxygen than that of carbon. Hence they cannot be extracted from their oxides by reduction with carbon.

12. Complete the reactions and balance them.

(a) Na2O2 and water

(b) KO2 and water

Ans. (a)2Na2O2 + 2H2O → 4NaOH + O2

(b) 2KO2 + 2H2O → 2KOH + H2O + O2

13. Potassium carbonate can not be prepared by the SOLVAY process. Give reason.

Ans. Potassium carbonate cannot be prepared by the SOLVAY process because potassium bicarbonate (KHCO3) is highly soluble in water, unlike NaHCO3 which was separated as crystals. Due to its high solubility KHCO3 cannot be precipitated by the addition of ammonium bicarbonate to a saturated solution of KCl.

14. Explain what happens when 

(i) Sodium hydrogen carbonate is heated.

(ii) Sodium with mercury reacts with water.

Ans. (i) $\underset{\text{Sodium hydrogen carbonate}}{2\text{NaHCO}_3}\xrightarrow[]{\text{Heat}}\underset{\text{Sodium carbonate}}{\text{Na}_2\text{CO}_3}+\text{H}_2\text{O}+\text{CO}_2\uparrow$ 

 (ii) 2Na-Hg+2H2O→2NaOH+H2↑+2Hg

15. S-block elements never occur in free state. Explain. What are their usual modes of occurrence?

Ans. The metals (Alkali & alkaline earth metals) are highly reactive because of their low ionization energy. They are highly electropositive forming positive ions. So, they are never found in a free state. They are widely distributed in nature in a combined state. They occur in the earth’s crust in the form of oxides, chlorides, silicates & carbonates.

16. Why second ionization enthalpy of Calcium is more than the first? How is that calcium forms CaCl2 and not CaCl . Explain

Ans. 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 favourable than CaCl energetically.

17. Solubility of alkaline metal hydroxides increases down the group . Explain.

Ans. If the anion and the cation are of comparable size, the cationic radius will influence the lattice energy. Since lattice energy decreases much more than the hydration energy with increasing ionic ionic size, solubility will increases as we go down the group. This is the case of alkaline earth metal hydroxides.

18. Solution of Na2CO3 is alkaline. Give reason.

Ans. The solution of Na2CO3 is alkaline in nature because when Na2CO3 is treated with water, it gets hydrolysed to form an alkaline solution:

CO32 + H2O → HCO3 + OH

19. Discuss the various reactions that occur in the solvay process

Ans. 2NH3 + H2O + CO2→ (NH4)2CO3

(NH4)CO3 + CO2 + H2O → 2NH4HCO

NH4HCO3 + NaCl → NH4Cl + NaHCO3

2NaHCO3→ Na2CO3 + CO2 + H2O

20. Give reason:

Lithium halides are covalent in nature.

Ans. Lithium halides are covalent because of the high polarization capability of lithium ion. The Li+ ion is very small in size and has high tendency to distort electron cloud around the negative halide ion.

PDF Summary - Class 11 Chemistry The s-Block Elements Notes (Chapter 12)

Alkali Metals (Group 1)

They have $n{s^1}$ electronic configuration and are highly reactive metals


Atomic Number

Electronic Configuration



















Physical Properties

Atomic Size

In their respective times, the atoms are at their largest. As you progress through the group, the atomic size grows larger.

Oxidation State

Group 1 elements has +1 oxidation state.


Alkali metals have low density due to their large size.

$Density = \dfrac{{Atomic\;mass}}{{Atomic\;volume}}$ 

The group's atomic weight increases from $Li$  to $Cs$ , as does its volume, but the atomic weight increase outweighs the volume increase. As a result, density rises from $Li$  to $Cs$

Exception: Density of sodium is more than that of potassium

Order: $Li < K < Na < Rb < Cs$ 

Nature of Bonds

When the values of electronegativity are low, they combine with other elements to form Ionic bonds.

Ionization Energy

The atoms in this group have lower initial ionisation energy than any other group in the periodic table. Because atoms are so big, the outer electron is only weakly bound by the nucleus, resulting in a low ionisation energy. As you move down the group, the ionisation energy drops.

Flame Test

When alkali metals are burned on a flame, the electrons in the valence shell migrate from a lower energy level to a higher energy level due to heat absorption from the flame. When they return to their original condition, they release the additional energy in the form of visible light, which gives the flame colour.






Golden yellow




Red Violet



Standard Oxidation Potential

The electrode potential of a metal in water is a measurement of its tendency to donate electrons. The standard electrode potential is defined as the concentration of metal ions being equal to one. Lithium has the largest ionisation potential, but due to its high hydration energy, it also has the highest electrode potential.

Hydration of Ions

The ions have a lot of water in them. The degree of hydration is proportional to the size of the ion. As a result, from Li+ to Cs+, the degree of hydration falls. As a result, electrical conductivity diminishes as hydration increases.

Lattice Energy

Ionic solids are alkali metal salts. The lattice energy of alkali metal salts with a common anion drops as one moves down the group.

Solubility in Liquid Ammonia

$M + nN{H_3} \to {[M{(N{H_3})_x}]^ + } + {e^ - }{(N{H_3})_y}$ 

$(n = x + y)$ 

The major species found in solvated metal ions and solvated electrons in dilute alkali metal solutions in liquid ammonia are solvated metal ions and solvated electrons. The colour diminishes until it disappears if the blue solution is left to stand due to the creation of metal amide. Because of the presence of solvated electrons, metal solutions in liquid conduct electricity. Because they include free electrons, the dilute solutions are paramagnetic.

Electronegativity Values

The electronegativity values are small which decrease from lithium to cesium.


The reactivity of alkali metals goes on increasing in the following order: $Li < Na < K < Rb < Cs$ 

Colourless and Diamagnetic Ions

The number of unpaired electrons present in an ion determines whether the ion is colourless or coloured. If an anion has unpaired electrons, these electrons can be stimulated by light energy and subsequently return to the ground state to show colour. Unpaired electron ions have magnetic properties, while paired electron ions cancel out each other's magnetic fields. Diamagnetic ions are such ions. The presence of unpaired electrons causes super oxides to be para magnetic and coloured.

Melting and Boiling Point

The cohesive energy is the force that holds the atoms or ions in a solid together. The cohesive energy is proportional to the number of electrons capable of bonding. Alkali metals contain only one valence electron that participates in bonding, and the outer bonding electron is big and diffuse, therefore the cohesive force reduces as the group gets smaller. As the atoms get bigger as you go down the group, the bonds get weaker, the cohesive energy drops, and the metal gets softer. As a result, the melting point drops as the group progresses. The boiling point also reduces the size of the group.

Chemical Properties

Some Common Reactions of Group 1 Metals



$M + {H_2}O \to MOH + {H_2}$ 

Hydroxides are strongest base known.

$Li + {O_2} \to L{i_2}O$ 

Monoxide formed by lithium and to a small extent by sodium.

$Na + {O_2} \to N{a_2}{O_2}$ 

Peroxide formed by sodium and to a small extent by lithium.

$K + {O_2} \to K{O_2}$ 

Superoxide formed by potassium, rubidium and cesium.

$M + {H_2} \to MH$ 

Ionic salt like hydrides.

$Li + {N_2} \to L{i_3}N$ 

Nitride formed only by lithium

$M + S \to {M_2}S$ 

All metals form sulphides.

$M + {X_2} \to MX$ 

All metals form halides.

$M + N{H_3} \to MN{H_2}$ 

All the metals form amides.

Reaction with Air

Group 1 elements are very reactive and tarnish quickly when exposed to air.

These metals form alkaline carbonates in moist air.

$2Na + {O_2} \to 2N{a_2}O$ 

$N{a_2}O + {H_2}O \to 2NaOH$ 

$2NaOH + C{O_2} \to N{a_2}C{O_3} + {H_2}O$ 

Reaction with ${O_2}$ 

Lithium forms $L{i_2}O$ , sodium forms two types of oxide $({M_2}O,\;{M_2}{O_2})$ and potassium, rubidium and cesium form superoxide $(M{O_2})$ .

Basic Nature, Ionic Nature of the Oxides

Because the size of the cation increases, the basic nature of oxides changes from lithium to cesium.

  1. From lithium to cesium, the cation size increases. The ionic nature of these oxides rises from lithium to cesium, according to Fajan's Rule.

  2. As the ionic nature of these metal oxides changes, solubility in water increases from lithium to cesium oxides.

Reaction with water

Group 1 metals react with water and liberates hydrogen and thus hydroxides are formed.

$2 \mathrm{Li}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{LiOH}+\mathrm{H}_{2}$

$2 \mathrm{Na}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{NaOH}+\mathrm{H}_{2}$

$2 \mathrm{~K}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{KOH}+\mathrm{H}_{2}$

Reaction with Hydrogen

Group 1 metals react with hydrogen and forms ionic hydrides. Thermal stability of $\mathrm{LiH}$ is high.

Stability of hydrides is in the order:


Reaction with Dilute acids

These metals react quickly with dilute acids due to their alkaline nature, and the rate of reaction increases from lithium to cesium as the basic character increases.

Compounds of Alkali Metal


Caustic soda is another name for sodium hydroxide. Because of its corrosive qualities, potassium hydroxide is known as caustic potash. The strongest base in aqueous solution is caustic alkali. The solubility of hydroxides rises as they progress through the group.

The bases react with acids to form salt and water

$\mathrm{KOH}+\mathrm{HCl} \rightarrow \mathrm{KCl}+\mathrm{H}_{2} \mathrm{O}$

$\mathrm{NaOH}+\mathrm{HCl} \rightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}$

$2 \mathrm{NaOH}+\mathrm{CO}_{2} \rightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}$

Ammonia is liberated by the bases from ammonium salts

$\mathrm{NaOH}+\mathrm{NH}_{4} \mathrm{Cl} \rightarrow \mathrm{NH}_{3}+\mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}$

$\mathrm{KOH}+\mathrm{NH}_{4} \mathrm{Cl} \rightarrow \mathrm{NH}_{3}+\mathrm{KCl}+\mathrm{H}_{2} \mathrm{O}$

In all of its reactions, potassium hydroxide is similar to sodium hydroxide. However, because potassium hydroxide is so much more expensive, it is rarely utilised. However, because potassium hydroxide is more soluble in alcohol, the equilibrium produces $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-}$ ions.

$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\mathrm{OH}^{-} \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-}+\mathrm{H}_{2} \mathrm{O}$

Oxides, Peroxides and Superoxides

Normal oxides – monoxide: Ionic monoxides are present. They are very basic oxides that create strong bases when they react with water.

$\mathrm{Na}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{NaOH}$

$\mathrm{K}_{2} \mathrm{O}+\mathrm{H}_{3} \mathrm{O} \rightarrow \mathrm{KOH}$



$2 \mathrm{Na}+\mathrm{O}_{2}(\mathrm{excess}) \underset{\longrightarrow}{300^{\circ} \mathrm{C}}{\longrightarrow} \mathrm{Na}_{2} \mathrm{O}_{2}$

$2 \mathrm{Na}_{2} \mathrm{O} \quad{ }^{7400^{\circ} \mathrm{C}}{\longrightarrow} \mathrm{Na}_{2} \mathrm{O}_{2}+\mathrm{Na}$ (vapour)


$\mathrm{Na}_{2} \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{dil}) \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}_{2}$

$\mathrm{Na}_{2} \mathrm{O}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{NaOH}+\mathrm{H}_{2} \mathrm{O}_{2}$

$\mathrm{Na}_{2} \mathrm{O}_{2}$ is a powerful oxidant as it reacts with carbon dioxides present in the air.

$\mathrm{Na}_{2} \mathrm{O}_{2}+\mathrm{CO} \rightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}$

$\mathrm{Na}_{2} \mathrm{O}_{2}+2 \mathrm{CO}_{2} \rightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{O}_{2}$

$\mathrm{Na}_{2} \mathrm{O}_{2}+\mathrm{Cr}^{3+} \rightarrow \mathrm{CrO}_{4}^{2-}$


Structure of peroxide ion

Oxygen atom is $\mathrm{sp}^{3}$ hybridized. Peroxide ion has 18 electrons which occupies the molecular orbital as shown:

$\sigma 1 \mathrm{~s}^{2}, \sigma^{*} 1 \mathrm{~s}^{2}, \sigma 2 \mathrm{~s}^{2}, \sigma^{*} 2 \mathrm{~s}^{2}, \sigma 2 \mathrm{p}_{z}^{2}, \pi 2 \mathrm{p}_{\mathrm{y}}^{2}=\pi 2 \mathrm{p}_{\mathrm{z}}^{2}, \pi^{*} 2 \mathrm{p}_{\mathrm{y}}^{2}=\pi^{*} 2 \mathrm{P}_{z}^{1}$

The bond order being 1 so, it is diamagnetic.


Superoxides are ionic oxides $\mathrm{M}^{+} \mathrm{O}_{2}^{-}$


$\mathrm{M}+\mathrm{O}_{2}($ excess $) \rightarrow \mathrm{MO}_{2}$ $(\mathrm{M}=\mathrm{K}, \mathrm{Rb}, \mathrm{Cs})$

Superoxides are stronger oxidizing agents than peroxides. The stability of these superoxides is in the order:



$\mathrm{KO}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{KOH}+\mathrm{H}_{2} \mathrm{O}_{2}+1 / 2 \mathrm{O}_{2}$

Because it creates $\mathrm{O}_{2}$ and eliminates $\mathrm{CO}_{2}$, $\mathrm{KO}_{2}$ is utilised in space capsules, submarines, and breathing masks.

$4 \mathrm{KO}_{2}+2 \mathrm{CO}_{2} \rightarrow 2 \mathrm{~K}_{2} \mathrm{CO}_{3}+3 \mathrm{O}_{2}$

$4 \mathrm{KO}_{2}+4 \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow 4 \mathrm{KHCO}_{3}+\mathrm{O}_{2}$

Sodium superoxide can be made by reacting sodium peroxide with oxygen at high temperatures and pressures, rather than by burning metal in oxygen.

$\mathrm{Na}_{2} \mathrm{O}+\mathrm{O}_{2} \rightarrow 2 \mathrm{NaO}_{2}$


Structure of superoxide ion

The paramagnetic property is explained by the existence of one unpaired electron in a three-electron bond. The superoxide molecule has 17 electrons and a bond order of1.5, occupying the molecular orbitals as illustrated.

$\sigma 1 \mathrm{~s}^{2}, \sigma^{*} 1 \mathrm{~s}^{2}, \sigma 2 \mathrm{~s}^{2}, \sigma^{*} 2 \mathrm{~s}^{2}, \sigma 2 \mathrm{p}_{\mathrm{x}}^{2}, \pi 2 \mathrm{p}_{\mathrm{y}}^{2}=\pi 2 \mathrm{p}_{\mathrm{z}}^{2}, \pi^{*} 2 \mathrm{p}_{\mathrm{y}}^{2}=\pi^{*} 2 \mathrm{P}_{\mathrm{z}}^{1}$

The stability of oxides is given as:

Normal oxide > peroxide > superoxide

Carbonates and Bicarbonates

Solid bicarbonates are formed by Group 1 metals $\left(\mathrm{MHCO}_{3}\right)$. $\mathrm{M}_{2} \mathrm{CO}_{3}$ carbonates are formed by all alkali metals. The carbonates and bicarbonates of alkali metals are extremely heat stable due to their electro positive character \[\mathbf{Li}_{2}\mathbf{CO}_{3}\] decomposes easily by heat).

The unusual behaviour of $\mathrm{Li}_{2} \mathrm{CO}_{3}$ can be explained by the fact that:

  1. Lithium's small size and high polarisation disrupts the electron cloud of the nearby oxygen atom of the massive $\mathrm{CO}_{3}^{2}$, weakening the carbon-hydrogen bond.

$\mathrm{Li}_{2} \mathrm{CO}_{3} \longrightarrow \mathrm{Li}_{2} \mathrm{O}+\mathrm{CO}_{2}$

  1. When a larger carbonate ion is replaced by a smaller carbonate ion, the lattice energy increases, favouring breakdown.

$\mathrm{M}_{2} \mathrm{CO}_{3} \longrightarrow \mathrm{M}_{2} \mathrm{O}+\mathrm{CO}_{2} \uparrow$

As a washing soda, $\mathrm{Na}_{2} \mathbf{C O}_{3}$is utilised. Baking soda is made from $\mathrm{NaHCO}_{3}$ . Both $\mathrm{NaHCO}_{3}$ and $\mathrm{KHCO}_{3}$ have hydrogen bonds in their crystal structures. The $\mathrm{HCO}_{3}^{-}$ in $\mathrm{NaHCO}_{3}$ forms an endless chain, whereas $\mathrm{KHCO}_{3}$ forms a dimeric anion.


$2 \mathrm{HNO}_{3}+\mathrm{K}_{2} \mathrm{CO}_{3} \rightarrow 2 \mathrm{KNO}_{3}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}$

$2 \mathrm{NaHCO}_{3} \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}$

$\mathrm{M}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons 2 \mathrm{M}^{+}+\mathrm{HCO}_{3}^{-}+\mathrm{OH}^{-}$


All of the metals in this group create MX halides. Because lithium ion is the smallest ion in the group, it is more likely than other metals to produce hydrated salts.


Alkali metal halides are excellent ionic compounds, as evidenced by the following features.

  1. With the exception of lithium fluoride, all alkali halides are easily soluble in water (Lithium fluoride is soluble in non-polar solvents).

  2. Their melting and boiling points are extremely high.

  1. The melting and boiling points of the same alkali metal drop in a predictable order.

Fluoride > chloride > bromide > iodide

This is described in terms of the metal halides' lattice energy*. The lattice energy of the same metal reduces when the halogen's electronegativity lowers.

  1. The melting point of lithium halides is lower than that of sodium halides for the identical halide ion. However, as we travel down the group from sodium to cesium, the melting points of halides decline. Lithium halides exhibit aberrant behaviour due to their covalent character, whereas sodium and other halides are ionic in nature. As we advance along the group of ionic halides, the melting point lowers as the lattice energy decreases.

$NaCl > KCl > RbCl > CsCl$ 

  1. Solubility of halides of alkali metals: Alkali metal halides have a range of solubilities. The solubility of alkali metal fluorides in water, for example, gradually increases from lithium to caesium. Lithium chloride has a far better solubility in water than sodium chloride when it comes to chlorides. This is owing to the lithium ion's tiny size and high hydration energy. However, when the lattice energy of the crystals decreases, solubility in water increases steadily from sodium chloride to cesium chloride.

  2.  In the fused condition, they are good conductors of electricity.

  3. They are made up of ionic crystals. Lithium halides, on the other hand, have a partially covalent character due to the polarising power of lithium ions.

  4. The lattice energy and polarising power are responsible for the structure and stability (solubility) of alkali metal halides.

  1. Lattice Energy: The energy produced during the production of a crystal lattice from gaseous cations and anions, or the energy necessary to split one mole of a solid ionic compound into its gaseous ions, is known as lattice energy. As a result, lattice energy (the force of attraction between ions) is a direct measure of ionic crystal stability; the higher the lattice energy of a molecule, the lower its solubility in water.

When an ionic compound crystal comes into contact with a polar solvent like water, the water molecule's hydrogen end (positive pole) is attracted to a negative ion, whereas the oxygen end (negative pole) is drawn to a positive ion. Solvation (or hydration, if the solvent is water) of the ions is the process of polar solvent molecules attaching to the ions. When ions are stabilised by solvation, a considerable amount of solvation energy (or hydration energy) is released, which, if it exceeds the crystal's lattice energy, causes the ionic compound to dissolve in the solvent. In the case of lithium fluoride, however, if the solvation energy is insufficient to oppose the lattice energy, the material stays insoluble. The combination of small lithium ions and small fluoride ions gives lithium fluoride its high lattice energy. The lattice energy of a particular ion increases as the size of the oppositely charged ion decreases.

  1. Polarising power and polarisability (Fajan’s Rule): Although an ionic bond in a chemical like $\mathrm{M}^{+} \mathrm{X}^{-}$ is thought to be 100 percent ionic, it is discovered to have significant covalent character in some circumstances (e.g., lithium halides). When two oppositely charged ions approach each other, the nature of the link between them, according to Fajan, is determined by the action of one ion on the other. 

When two oppositely charged ions come into contact, the positive ion attracts electrons from the anion's outermost shell while repelling the anion's positively charged nucleus. The anion is distorted, deformed, or polarised as a result of this. The ability of a cation to distort an anion is called polarisation power, and the anion's susceptibility to be polarised by the cation is called polarisability. If the degree of polarisation is modest, an ionic bond is established; however, if the degree of polarisation is considerable, electrons are attracted from the anion to the cation by electrostatic attraction, resulting in a higher electron density between the two ions and a covalent connection. In general, the stronger an ion's polarisation power or polarisability, the greater its proclivity to form covalent bonds. Because polarisation power grows as the size of the cation decreases and polarisability increases as the size of the anion increases, the polarisation in a compound containing big negative ions and small positive ions may be so strong that the connection becomes covalent. In nature, lithium iodide, which is made up of lithium ions (the smallest alkali metal ion) and iodide ions (the largest halide ion), is found to be highly covalent.

Other examples of such ionic-covalent compounds are $\mathrm{AlCl}_{3}, \mathrm{FeCl}_{3}, \mathrm{SnCl}_{4}$, etc.


Ionic polyhalides are formed when alkali metal halides react with halogen and interhalogen compounds.

$\mathrm{KI}+\mathrm{I}_{2} \rightarrow \mathrm{K}\left[\mathrm{I}_{3}\right]$

$\mathrm{KBr}+\mathrm{ICl} \rightarrow \mathrm{K}[\mathrm{Br} \mathrm{ICl}]$

$\mathrm{KF}+\mathrm{BrF}_{3} \rightarrow \mathrm{K}\left[\mathrm{BrF}_{4}\right]$


They form sulphates of type $\mathrm{M}_{2} \mathrm{SO}_{4}$

Anomalous Behaviour of Lithium

Although lithium has many of the properties of the group I elements, it also differs from them in a number of ways. The incredibly small size of the lithium atom and its ion causes this unusual behaviour. The high charge density of the lithium ion is due to its small size. As a result, out of all the alkali metal ions, lithium ion has the most polarising power. As a result, it has a significant distorting impact on a negative ion. As a result, the lithium ion has a strong proclivity for solvation and the creation of covalent bonds. It's also worth noting that the polarising power of the lithium ion is similar to that of the magnesium ion, making the two elements resemble very much in their properties.

  1. Lithium is substantially more difficult to work with than the other elements in Group I. (similarity with magnesium which is also a hard metal).

  2. Lithium has a high melting point and boiling point.

  3. Lithium, unlike the other elements in this group, is the least reactive, as shown by the following points.

  1. It is not influenced by air, unlike others.

  2. It, unlike others, decomposes water slowly (similar to magnesium).

  3. It rarely reacts with bromine, unlike many others.

  4. On burning in oxygen, it forms only the monoxide $\mathrm{Li}_{2} \mathrm{O}$ , while the others form peroxides too. 

  1. Unlike other elements, it forms nitride when it comes into contact with nitrogen, $\mathrm{Li}_{3} \mathrm{~N}$ (similarity with $\mathrm{Mg}$ )

  2. Lithium is much less electropositive and, therefore, several of its compounds $\left(\mathrm{Li}_{2} \mathrm{CO}_{3}\right.$ and $\left.\mathrm{LiOH}\right)$ are less stable (similarity with Mg). For example,

$2 \mathrm{LiOH} \rightarrow \mathrm{Li}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}$

$\mathrm{Mg}(\mathrm{OH})_{2} \rightarrow \mathrm{MgO}+\mathrm{H}_{2} \mathrm{O}$

  1. When heated, lithium nitrate produces nitrogen dioxide and oxygen, leaving lithium oxide behind (similar to magnesium nitrate), but sodium and potassium nitrates produce only oxygen, leaving nitrites.

  2. The majority of lithium salts (for example, hydroxide, carbonate, oxalate, phosphate, and fluoride) are water insoluble (similarity with magnesium.) The sodium and potassium salts that correspond to them are both water soluble.

  3. Lithium halides and lithium alkyls are soluble in organic solvents, but sodium and potassium halides and alkyls are not; $ \mathrm{MgCl}_{2}$  is soluble in alcohol as well.

  4. Lithium chloride, like magnesium chloride, undergoes some hydrolysis in hot water, but only to a little level; sodium chloride and potassium chloride do not.

  5. Lithium sulphate, unlike sulphates of other alkali metals, does not create alums.

  6. Partially covalent lithium compounds, particularly lithium halides, exist in nature. This is owing to lithium ions' proclivity for attracting electrons (polarising power). This explains why lithium compounds have a smaller dipole moment than expected.

  7. The ions and compounds of alkali metals are more hydrated than those of other alkali metals (similarity with magnesium).

Extraction of Sodium

Sodium is obtained on large scale by two process:

Castner’s process

The electrolysis of fused sodium hydroxide takes place at $330^{\circ} \mathrm{C}$ , with iron as the cathode and nickel as the anode.

$2 \mathrm{NaOH} \rightleftharpoons 2 \mathrm{Na}^{+}+2 \mathrm{OH}^{-}$

At cathode: \[2 \mathrm{Na}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Na}\]

At anode: $4 \mathrm{OH}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}+4 \mathrm{e}^{-}$

Oxygen and water are created during electrolysis. At the cathode, water generated at the anode is partially evaporated and partially broken down, resulting in hydrogen discharge.

$\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-}$

At cathode: \[2\mathbf{H}^{+}+2 \mathbf{e} \rightarrow 2 \mathbf{H} \rightarrow \mathbf{H}_{2}\uparrow\]

Down’s Process

Nowadays, the metal is produced using Down's technique. It includes employing iron as a cathode and graphite as an anode to electrolyze fused sodium chloride containing calcium chloride and potassium fluoride at roughly 600 degrees Celsius. The cell is made up of a steel tank with heat-resistant bricks lining it. In the centre of the cell, a circular graphite anode is installed, which is encircled by a cylindrical iron cathode. A steel gauze cylinder separates the anode and cathode, allowing fused charge to pass through. A dome-shaped steel hood protects the anode and provides an outlet for chlorine gas to escape. The molten metal that has been freed at the cathode rises and flows into the kerosene receiver.

Down’s Process

Reactions: $\mathrm{NaCl} \rightleftharpoons \mathrm{Na}^{+}+\mathrm{Cl}^{-}$

At Cathode: $2 \mathrm{Na}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Na}$

At Anode: $2 \mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_{2}+2 \mathrm{e}^{-}$

The sodium chloride when obtained from this method is $99.5 \%$ pure.

The electrolysis of pure sodium chloride has a number of drawbacks:

  1. Sodium chloride has a high fusion temperature of 803 degree celsius, which is difficult to sustain.

  2. Because sodium is volatile at this temperature, some of it vapourizes, forming a metallic fog.

  3. The electrolysis products, salt and chlorine, are corrosive at this temperature and may harm the cell's substance.

Pure sodium chloride is blended with calcium chloride and potassium fluoride to overcome the problems mentioned above. At the voltage used, calcium chloride and potassium fluoride do not breakdown, but they do lower the fusion temperature. A combination containing 40% sodium chloride, 60% calcium chloride, and a trace amount of potassium fluoride reaches a fusion temperature of around 600 degrees Celsius. In the electrolytic cell, this combination is electrolyzed at 600 degrees Celsius.

Example 1: Alkali metals are paramagnetic but their salts are diamagnetic. Explain.

Solution: The outermost energy shell in metals is solely occupied, whereas in cations, all orbitals are double occupied (inert gas configuration).

e.g., $\quad \mathrm{Na}, 1 \mathrm{~s}^{2}, 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6}, 3 \mathrm{~s}^{2} 3 \mathrm{p}^{6}$, 4s $^{1}$ paramagnetic

$\mathrm{Na}^{+} 1 \mathrm{~s}^{2}, 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6},\left(3 \mathrm{~s}^{2} 3 \mathrm{p}^{6}\right)$ Diamagnetic

Example 2: Alkali metals are good reducing agents. Explain.

Solution: Alkali metals are strong reducing agents because their low ionisation enthalpy values and high oxidation potential allow them to easily lose valence electrons.

Example 3: Which alkali metal ion has the maximum polarising power and why?

Solution: Among the alkali metal ions, the lithium ion has the most polarising power. This is owing to the lithium ion's small size.

Example 4: Lithium ion is far smaller than other alkali metal ions but it moves through a solution less rapidly that the others. Explain.


The conductance of lithium salts is less in comparison to the salts of other alkali metals. Explain.

Solution: Lithium ion has the highest degree of hydration because of its strong charge, which pulls numerous water molecules surrounding it. As a result, the size of the hydrated lithium ion is larger than that of the other alkali metal ions, affecting its mobility in solution and lowering conductance.

Size: $\quad[\mathrm{Li}(\mathrm{aq})]^{+}>[\mathrm{Na}(\mathrm{aq})]^{+}>[\mathrm{K}(\mathrm{aq})]^{+}$

Example 5: Sodium salts in aqueous solutions are either neutral or alkaline in nature. Explain.

Solution: Strong acids or weak acids produce the anions in sodium salts. There is no hydrolysis when anions come from strong acids, and aqueous solutions are neutral. When anions come from weak acids, however, they undergo hydrolysis, resulting in alkaline solutions. Solns. of sodium carbonate or bicarbonate, for example, are alkaline.

\[\mathrm{CO}_{3}^{2-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{HCO}_{3}^{-}+\mathrm{OH}^{-}\]

$\mathrm{HCO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}+\mathrm{OH}^{-}$

Example 6: Why do potassium, rubidium and cesium form superoxides in preference to oxides and peroxides on being heated in excess supply of air?

Solution: $\mathrm{K}^{+}, \mathrm{Rb}^{+}$and $\mathrm{Cs}^{+}$are large cations in size and superoxide ion $\left(\mathrm{O}_{2}^{-}\right)$is larger in size in comparison to oxide $\left(\mathrm{O}^{2-}\right)$ and peroxide $\left(\mathrm{O}_{2}^{2-}\right)$ ion. These metals form superoxides rather than oxides or peroxides because a larger cation can stabilise a large anion.

Example 7: Why is ${\text{K}}{{\text{O}}_{\text{2}}}$ paramagnetic?

Solution: The superoxide $\mathrm{O}_{2}^{-}$is paramagnetic because of one unpaired electron in $\pi * 2 \mathrm{p}$ molecular orbital.

$\mathrm{KK} \sigma(2 \mathrm{~s})^{2} \sigma^{*}(2 \mathrm{~s})^{2} \sigma\left(2 \mathrm{p}_{\mathrm{x}}\right)^{2} \pi\left(2 \mathrm{p}_{\mathrm{x}}\right)^{2}\left(\pi 2 \mathrm{p}_{\mathrm{y}}\right)^{2} \pi^{*}\left(2 \mathrm{p}_{\mathrm{x}}\right)^{2}$ $\pi^{*}\left(2 \mathrm{p}_{\mathrm{v}}\right)^{1}$

Example 8: Among the alkali metals which element has 

  1. Highest melting point

  2. Highest size of hydrated ion in solution 

  3. Strongest reducing agent in solution 

  4. Least electronegative

Solution: The elements which have:

  1. Highest melting point: Lithium

  2. Highest size of hydrated ion in solution: $[\mathrm{Li}(\mathrm{aq})]^{+}$

  3. Strongest reducing agent in solution: Lithium

  4. Least electronegative: Cesium

Example 9: What happens when following compounds are heated?

  1. $L{i_2}C{O_3} \to L{i_2}O + C{O_2}$ 

  2. $N{a_2}C{O_3}.10{H_2}O \to N{a_2}C{O_3} + 10{H_2}O$ 

  3. $4LiN{O_3} \to 2L{i_2}O + 4N{O_2} + {O_2}$ 

  4. $2NaN{O_3} \to 2NaN{O_2} + {O_2}$ 

Example 10: 

  1. Arrange $\mathrm{LiF}, \mathrm{NaF}, \mathrm{KF}, \mathrm{RbF}$ and CsF in order of increasing lattice energy.

  2. Arrange the following in order of the increasing covalent character. $\mathrm{MCl}, \mathrm{MBr}, \mathrm{MF}, \mathrm{MI}$ (where $\mathrm{M}=$ alkali metal)


  1. $\mathrm{CsF}<\mathrm{RbF}<\mathrm{KF}<\mathrm{NaF}<\mathrm{LiF}$

  2. $\mathrm{MF}<\mathrm{MCl}<\mathrm{MBr}<\mathrm{MI}$

With increasing size of the anion, covalent character increases.

Example 11: Why a standard solution of sodium hydroxide cannot be prepared by weighing ?

Solution: The material sodium hydroxide is deliquescent. It absorbs moisture and reacts with atmospheric carbon dioxide, both of which increase its mass. As a result, precise weighing is difficult.

Example 12: What happens when:

  1. Fused sodium reacts with dry ammonia.

  2. Sodium hydrogen carbonate is heated.

  3. Sodium hydroxide is heated with sulphur?


  1. $2 \mathrm{Na}+2 \mathrm{NH}_{3}+\rightarrow 2 \mathrm{NaNH}_{2}+\mathrm{H}_{2}$

  2. Sodium carbonate is formed. $2 \mathrm{NaHCO}_{3} \rightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}$

  3. Sodium thiosulphate is formed.

$4 \mathrm{~s}+6 \mathrm{NaOH} \rightarrow \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}+2 \mathrm{Na}_{2} \mathrm{~S}+3 \mathrm{H}_{2} \mathrm{O}$

Example 13: Give reasons for the following:

  1. $\mathrm{LiCl}$ is more covalent than $\mathrm{NaCl}$ 

  2.  Lithium Iodide has lower melting point then $\mathrm{LiCl}$ 

  3. $\mathrm{MgCl}_{2}$ is more covalent than $\mathrm{NaCl}$ 

  4. $\mathrm{CuCl}$ is more covalent than $\mathrm{NaCl}$


  1. Lithium ion is more polarising than sodium ion due to its smaller size, and so lithium chloride is more covalent than sodium chloride.

  2. Due to bigger size, $\mathrm{I}^{-}$ is more polarisable than $\mathrm{Cl}^{-}$ and hence lithium iodide is more covalent than lithium chloride. Therefore, lithium iodide has lower melting point than $\mathrm{LiCl}$.

  3. Magnesium ion is more polarising than sodium ion due to its greater charge, and so magnesium chloride is more covalent than sodium chloride.

  4. Copper ion is more polarising than sodium ion because to the pseudo inert gas configuration, and so copper chloride is more covalent than sodium chloride.

Example 14: Identify (A), (B), (C) and (D) and give their chemical formulae.

  1. (A) $+\mathrm{NaOH} \quad$ Heat ${\longrightarrow} \mathrm{NaCl}+\mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O}$ 

  2. $\mathrm{NH}_{3}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow(\mathrm{B})$

  3. $(\mathrm{B})+\mathrm{NaCl} \longrightarrow(\mathrm{C})+\mathrm{NH}_{4} \mathrm{Cl}$

  4. (C) Heat $\mathrm{Na}_{2} \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}+(\mathrm{D})$


  1. $\mathrm{NH}_{4} \mathrm{Cl}+\mathrm{NaOH}_{\longrightarrow}{\text { Heat }}{\longrightarrow} \mathrm{NH}_{3}+\mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}$

Compound A is ammonium chloride $\left(\mathrm{NH}_{4} \mathrm{Cl}\right)$.

  1. $\mathrm{NH}_{3}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{NH}_{4} \mathrm{HCO}_{3}$

Compound B is ammonium bicarbonate $\left(\mathrm{NH}_{4} \mathrm{HCO}_{3}\right)$.

  1. $\mathrm{NH}_{4} \mathrm{HCO}_{3}+\mathrm{NaCl} \longrightarrow \mathrm{NaHCO}_{3}+\mathrm{NH}_{4} \mathrm{Cl}$

Compound (C) is sodium bicarbonate $\left(\mathrm{NaHCO}_{3}\right)$.

  1. $2 \mathrm{NaHCO}_{3} \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}$

Compound (D) is carbon dioxide $\left(\mathrm{CO}_{2}\right)$.

Example 15: Arrange the following as specified:

  1. $\mathrm{MgO}, \mathrm{SrO}, \mathrm{K}_{2} \mathrm{O}$ and $\mathrm{Cs}_{2} \mathrm{O}$ (increasing order of basic character)

  2. $\mathrm{LiCl}$, ${\text{LiBr}}$ , ${\text{LiI}}$  (decreasing order of covalent character)

  3. $\mathrm{NaHCO}_{3}, \mathrm{KHCO}_{3}, \mathrm{Mg}\left(\mathrm{HCO}_{3}\right)_{2}, \mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}$ (decreasing solubility in water)

  4. $\mathrm{LiF}, \mathrm{NaF}, \mathrm{RbF}, \mathrm{KF}$ and CsF (in order of increasing lattice energy)

  5. $\mathrm{Li}, \mathrm{Na} \mathrm{K}$ (in order to decreasing reducing nature in solution)


  1. $\mathrm{MgO}<\mathrm{SrO}<\mathrm{K}_{2} \mathrm{O}<\mathrm{Cs}_{2} \mathrm{O}$}

  2. $\mathrm{LiI}>\mathrm{LiBr}>\mathrm{LiCl}$

  3. $\mathrm{NaHCO}_{3}<\mathrm{KHCO}_{3}<\mathrm{Mg}\left(\mathrm{HCO}_{3}\right)_{2}<\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}$

  4. $\mathrm{CsF}<\mathrm{RbF}<\mathrm{KF}<\mathrm{NaF}<\mathrm{LiF}$

  5. $\mathrm{Li}>\mathrm{K}>\mathrm{Na}$

Example 16: 

  1. What happens when $\mathrm{KO}_{2}$ reacts with water ? Give the balanced chemical equation.

  2. Predict giving reason the outcome of the reaction :

$\mathrm{LiI}+\mathrm{KF} \longrightarrow$


  1. When $\mathrm{KO}_{2}$ reacts with water, oxygen is evolved and an alkaline solution containing potassium hydroxide and $\mathrm{H}_{2} \mathrm{O}_{2}$ is formed:

$2\mathrm{KO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{KOH}+\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{O}_{2}$

  1. Lithium iodide reacts with potassium fluoride and anions are exchanged in this process:

$\mathrm{LiI}+\mathrm{KF} \longrightarrow \mathrm{LiF}+\mathrm{KI}$

As stable compounds form, the exchange takes place, with the bigger cation stabilising the larger anion and the smaller cation stabilising the smaller anion.

Alkaline Earth Metals


Group-II of the periodic table contains the elements beryllium, magnesium, calcium, strontium, barium, and radium.

All of these substances are metals. Calcium, strontium, and barium oxides were discovered far before the metals themselves, and they were dubbed alkaline earths because they were alkaline and found in the earth. Alkaline earth metals were given to the elements once they were found. Radium shares chemical properties with alkaline earth metals, but because it is a radioactive element, it is researched independently from the other radioactive elements.

Physical Properties

Atomic Size

On going down the group, atomic size of elements increases.

Oxidation State

+2 oxidation state is exhibited by group II elements.


Group II elements are smaller in size than group I elements, hence they have a higher density than group I elements. From beryllium to radon, density rises.

Exception: Calcium has a lower density than magnesium, whereas magnesium has a lower density than beryllium.

Nature of Bonds

Beryllium forms mainly covalent compound. The rest of the elements in group II form ionic bonds.

Hydration Energy

Because of their smaller size and higher charge, the hydration energies of group 2 ions are four to five times higher than those of group 1 ions. As the size of the ions grows larger, the hydration enthalpy falls.

Lattice Energy

The lattice energy of alkali metal salts with a common anion drops as one moves down the group.

Ionization Energy

Because the atoms in group 2 are smaller, the electrons are more firmly bound, requiring more energy to remove the initial electron (first ionisation energy) than in group 1. The amount of energy necessary to remove the second electron is approximately double that required to remove the first. As a result, the energy necessary to make divalent ions from group 2 elements is four times that required to make $\mathrm{M}^{+}$ from group 1 metals.

Flame Test

These elements' electrons are stimulated to higher energy levels when energy is delivered to them in a flame, as is the case with alkali metals under identical conditions. The excess energy generated by the electrons when they return to their original energy level produces visible light with distinct colours, as shown below:




Brick red


Crimson red


Grassy green



The atoms of beryllium and magnesium are smaller. As a result, the electrons in these atoms are more tightly bound. As a result, the energy of the flame does not stimulate them to higher energy levels. As a result, these ingredients do not produce any colour in the bunsen flame.

Standard Oxidation Potential

Standard Oxidation Potential of Alkaline Earth Metals


Oxidation Reaction

Standard Oxidation Potential (volt)


$\mathrm{Be} \rightarrow \mathrm{Be}^{2+}+2 \mathrm{e}^{-}$



$\mathrm{Mg} \rightarrow \mathrm{Mg}^{2+}+2 \mathrm{e}^{-}$



$\mathrm{Ca} \rightarrow \mathrm{Ca}^{2+}+2 \mathrm{e}^{-}$



$\mathrm{Sr} \rightarrow \mathrm{Sr}^{2+}+2 \mathrm{e}^{-}$



$\mathrm{Ba} \rightarrow \mathrm{Ba}^{2+}+2 \mathrm{e}^{-}$


Solubility in Liquid Ammonia

The metals, like group 1 metals, dissolve in liquid ammonia. Due to the creation of solvated electrons, dilute solutions are blue in colour. As the solution decomposes, amides form and hydrogen gas is released.

$2 \mathrm{NH}_{3}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{NH}_{2}^{-}+\mathrm{H}_{2}$

Electronegative Values

Group II element’s electronegativity values are low, although they are higher than group I element’s. The value of electronegativity diminishes as you progress through the group.

Colourless and Diamagnetism

The elements of the alkaline earth metal group generate $\mathrm{M}^{2+}$ ions, which are diamagnetic and colourless due to the lack of an unpaired electron.

Melting and Boiling Point

The melting point of elements in group II lowers as the cohesive force diminishes as the group progresses. 

Exception: Magnesium has the lowest melting point

Metallic Properties

Group-II elements have typical metallic characteristics. They have a nice metallic sheen and excellent electrical and thermal conductivity.

GROUP – I and II


Sodium Oxide


  1. It's made by heating sodium to 180 degrees Celsius in a small amount of air or oxygen and then distilling the surplus sodium away.

$2 \mathrm{Na}+1 / 2 \mathrm{O}_{2} \stackrel{180^{\circ}}{\longrightarrow} \mathrm{Na}_{2} \mathrm{O}$

  1. By heating sodium peroxide, nitrate or nitrate with sodium.

$N{a_2}{O_2} + 2Na \to 2N{a_2}O$ 

$2NaN{O_3} + 10Na \to 6N{a_2}O + {N_2}$ 

$2NaN{O_2} + 6Na \to 4N{a_2}O + {N_2}$ 


  1. It is a white amorphous mass.

  2. It gets decomposed at 400 degree Celsius into sodium peroxide and sodium.

$2 \mathrm{Na}_{2} \mathrm{O} \stackrel{400^{\circ} \mathrm{C}}{\longrightarrow} \mathrm{Na}_{2} \mathrm{O}_{2}+2 \mathrm{Na}$

  1. It gets dissolved violently in water and yields caustic soda.

$\mathrm{Na}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{NaOH}$

Sodium Peroxides


It is formed by heating the metal in excess of air or oxygen at 300 degrees Celsius in a dry, carbon dioxide-free environment.

$2 \mathrm{Na}+\mathrm{O}_{2} \longrightarrow \mathrm{Na}_{2} \mathrm{O}_{2}$


  1. It is a pale yellow solid, becoming white in air from the formation of a film of $\mathrm{NaOH}$ and $\mathrm{Na}_{2} \mathrm{CO}_{3}$ .

  2. In cold water, sodium peroxide produces $\mathrm{H}_{2} \mathrm{O}_{2}$ but at room temperature it produces oxygen. Sodium peroxide produces hydrogen peroxide in ice-cold mineral acids.

$\mathrm{Na}_{2} \mathrm{O}_{2}+2 \mathrm{H}_{2} \mathrm{O} \stackrel{\sim 0^{\circ} \mathrm{C}}{\longrightarrow} 2 \mathrm{NaOH}+\mathrm{H}_{2} \mathrm{O}_{2}$

$2 \mathrm{Na}_{2} \mathrm{O}_{2}+2 \mathrm{H}_{2} \mathrm{O} \stackrel{25^{\circ} \mathrm{C}}{\longrightarrow} 4 \mathrm{NaOH}+\mathrm{O}_{2}$

$\mathrm{Na}_{2} \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{SO}_{4} \stackrel{\sim 0^{\circ} \mathrm{C}}{\longrightarrow} \mathrm{Na}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}_{2}$

  1. It combines with carbon dioxide to produce sodium carbonate and oxygen, which is why it's used to purify air in small spaces. Example: submarine, ill-ventilated room.

$2 \mathrm{Na}_{2} \mathrm{O}_{2}+2 \mathrm{CO}_{2} \longrightarrow 2 \mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{O}_{2}$

  1. It is an oxidising agent and oxidises charcoal, $\mathrm{CO}, \mathrm{NH}_{3}$, $\mathrm{SO}_{2}$

$3 \mathrm{Na}_{2} \mathrm{O}_{2}+2 \mathrm{C} \longrightarrow 2 \mathrm{Na}_{2} \mathrm{CO}_{3}+2 \mathrm{Na}$

$\mathrm{Na}_{2} \mathrm{O}_{2}+\mathrm{CO} \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}$

$\mathrm{Na}_{2} \mathrm{O}_{2}+\mathrm{SO}_{2} \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}$

$3 \mathrm{Na}_{2} \mathrm{O}_{2}+2 \mathrm{NH}_{3}+\longrightarrow 6 \mathrm{NaOH}+\mathrm{N}_{2}$

  1. It contains peroxide ion 


  1. For preparing $\mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{O}_{2}$

  2. Oxygenating the air in submarines

  3. Oxidising agent in the laboratory.

Oxides of Potassium:


$\mathrm{K}_{2} \mathrm{O}$


$\mathrm{K}_{2} \mathrm{O}_{2}$


$\mathrm{K}_{2} \mathrm{O}_{3}$



Bright Yellow


Reddish Brown Needles


  1. $2KN{O_3} + 10K\xrightarrow{{heating}}6{K_2}O + {N_2}$ 

$\mathrm{K}_{2} \mathrm{O} \stackrel{\text { heating }}{\longrightarrow} \mathrm{K}_{2} \mathrm{O}$

${K_2}O + {H_2}O \to 2KOH$ 

  1. $2 \mathrm{~K}+\mathrm{O}_{2} \stackrel{\text { Controlled }}{\text { air at } 300^{\circ} \mathrm{C}} \mathrm{K}_{2} \mathrm{O}_{2}$

  2. Passage of $\mathrm{O}_{2}$ through a blue solution of $\mathrm{K}$ in liquid $\mathrm{NH}_{3}$ yields oxides $\mathrm{K}_{2} \mathrm{O}_{2}$ (white), $\mathrm{K}_{2} \mathrm{O}_{3}($ red $)$ and $\mathrm{KO}_{2}$ (deep yellow)

$\mathrm{K} \text { in liq. } \mathrm{NH}_{3} \longrightarrow \mathrm{K}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{K}_{2} \mathrm{O}_{3} \longrightarrow \mathrm{KO}_{2}$

$2 \mathrm{KO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{KOH}+\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{O}_{2}$

Magnesium Oxide:

It's also known as magnesia, and it's made from natural magnesite that's been heated.

$\mathrm{MgCO}_{3} \longrightarrow \mathrm{MgO}+\mathrm{CO}_{2}$


  1. It is present as a white powder

  2. The melting point of magnesium oxide is 2850 degree Celsius. So, it is used in the manufacturing of refractory bricks for furnances.

  3. It imparts alkaline reaction and it is very slightly soluble in water.

Calcium Oxide

It is manufactured by dissolving lime stone at a high temperature of roughly 1000 degrees Celsius and is known as fast lime or lime.

$CaC{O_3} \to CaO + C{O_2} + 4200cal$ 


  1. It is a white amorphous powder having a melting point of 2570 degree Celsius.

  2. When heated in an oxygen-hydrogen flame, it produces strong light (lime light).

  3. It is a basic oxide that reacts with acidic oxides, such as sulphur dioxide.

$\mathrm{CaO}+\mathrm{SiO}_{2} \longrightarrow \mathrm{CaSiO}_{3}$

$\mathrm{CaO}+\mathrm{CO}_{2} \longrightarrow \mathrm{CaCO}_{3}$

  1. On combination with water, it produces slaked lime

$\mathrm{CaO}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}$

Magnesium Peroxide and Calcium Peroxide:

These are obtained by passing $\mathrm{H}_{2} \mathrm{O}_{2}$ in a suspension of $\mathrm{Mg}(\mathrm{OH})_{2}$ and $\mathrm{Ca}(\mathrm{OH})_{2}$


Magnesium peroxide is a whitening agent and an antimicrobial in tooth paste.


Sodium Hydroxides


  1. Electrolysis of Brine:

$\mathrm{NaCl} \rightleftharpoons \mathrm{Na}^{+}+\mathrm{Cl}^{-}$

At Anode: $2 \mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2}+2 \mathrm{e}$

At Cathode: $ \mathrm{Na}+\mathrm{e}^{-} \longrightarrow \mathrm{Na}$ 

$2 \mathrm{Na}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{NaOH}+\mathrm{H}_{2}$

  1. Caustication of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ (Gossage's method):

$\mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{Ca}(\mathrm{OH})_{2} \rightleftharpoons 2 \mathrm{NaOH}+\mathrm{CaCO}_{3}$

Since the $\mathrm{K}_{\mathrm{sp}}\left(\mathrm{CaCO}_{3}\right)<\mathrm{K}_{\mathrm{sp}}\left(\mathrm{Ca}(\mathrm{OH})_{2}\right)$, the reaction shifts towards right.


  1. It's a crystalline white solid that's highly corrosive and deliquescent.

  2. It is resistant to heat.

  3. Its aqueous solution has an alkaline pH and feels soapy to the touch.

  4. $\mathrm{FeCl}{3}+3 \mathrm{NaOH} \longrightarrow \mathrm{Fe}(\mathrm{OH}){3} \downarrow+3 \mathrm{NaCl}$

$\mathrm{NH}{4} \mathrm{Cl}+\mathrm{NaOH}{ } \longrightarrow \mathrm{NaCl}+\mathrm{NH}{3} \uparrow+\mathrm{H}{2} \mathrm{O}$

$\mathrm{ZnCl}_{2}+2 \mathrm{NaOH} \longrightarrow Z n(\mathrm{OH}), \downarrow+2 \mathrm{NaCl}$

$\mathrm{Zn}(\mathrm{OH}), \downarrow+2 \mathrm{NaOH}{\text {Excess }}{\longrightarrow} \mathrm{Na}{1} \mathrm{ZnO}_{2}+2 \mathrm{H}, \mathrm{O}$

  1. Acidic and amphoteric oxidise gets dissolved easily, e.g:

$\mathrm{CO}_{2}+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}$

$\mathrm{Al}_{2} \mathrm{O}_{3}+2 \mathrm{NaOH} \longrightarrow 2 \mathrm{NaAlO}_{2}+\mathrm{H}_{2} \mathrm{O}$

  1. Aluminium and Zn metal gives $\mathrm{H}_{2}$ from $\mathrm{NaOH}$

$2 \mathrm{Al}+2 \mathrm{NaOH}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 3 \mathrm{H}_{2}+2 \mathrm{NaAlO}_{2}$

  1. Several non metals such as phosphorous, sulphur, calcium etc. yield a hydride instead of hydrogen e.g.

$4 \mathrm{P}+3 \mathrm{NaOH}+3 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{PH}_{3}+3 \mathrm{NaH}_{2} \mathrm{PO}_{2}$

Potassium Hydroxide

Preparation: It is prepared by the electrolysis of aqueous solution of potassium chloride.


  1. It is stronger base compared to sodium hydroxide

  2. Potassium hydroxide is more soluble in water as compared to sodium hydroxide.

  3. In alcohol, $\mathrm{NaOH}$ is sparingly soluble but $\mathrm{KOH}$ is highly soluble.

  4. As a reagent $\mathrm{KOH}$ is less frequently used but in absorption of $\mathrm{CO}_{2}, \mathrm{KOH}$ is preferably used compared to $\mathrm{NaOH}$. Because $\mathrm{KHCO}_{3}$ formed is soluble whereas $\mathrm{NaHCO}_{3}$ is insoluble.

Magnesium Hydroxide

It is found as the mineral brucite in nature.


It's made by combining a caustic soda solution with a magnesium sulphate or chloride solution.

$\mathrm{MgSO}_{4}+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}+\mathrm{Mg}(\mathrm{OH})_{2}$


  1. It can only be dried at temperatures up to 100 degrees Celsius; otherwise, at greater temperatures, it will break down into its oxide.

$\mathrm{Mg}(\mathrm{OH})_{2} \longrightarrow \mathrm{MgO}+\mathrm{H}_{2} \mathrm{O}$

  1. It is alkalinizing because it is mildly soluble in water.

  2. It dissolves in $\mathrm{NH}_{4} \mathrm{Cl}$ solution

$\mathrm{Mg}(\mathrm{OH})_{2}+2 \mathrm{NH}_{4} \mathrm{Cl} \longrightarrow \mathrm{MgCl}_{2}+2 \mathrm{NH}_{4} \mathrm{OH}$

Thus, $\mathrm{Mg}(\mathrm{OH})_{2}$ is not therefore precipitated from a solution of $\mathrm{Mg}^{+2}$ ions by $\mathrm{NH}_{4} \mathrm{OH}$ in presence of excess of $\mathrm{NH}_{4} \mathrm{Cl}$

Calcium Hydroxide

Preparation: It can be prepared easily by spraying water on quicklime.

$\mathrm{CaO}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}$


  1. The solubility in water of calcium hydroxide is very less.

  2. It has a lower solubility in hot water than in cold water. When a result, as the temperature rises, so does the solubility.

  3. Carbon dioxide is readily absorbed by calcium hydroxide and is used as a test for the gas.



  1. Leblanc Process:

$\mathrm{NaCl}+\mathrm{H}_{2} \mathrm{SO}_{4}$ (conc.) $\stackrel{\text { mild heating }}{\longrightarrow} \mathrm{NaHSO}_{4}+\mathrm{HCl}$

$\mathrm{NaCl}+\mathrm{NaHSO}_{4} \quad \stackrel{\text { Strongly }}{\text { heated }}{\longrightarrow} \mathrm{Na}_{2} \mathrm{SO}_{4}+\mathrm{HCl}$

$\mathrm{Na}_{2} \mathrm{SO}_{4}+4 \mathrm{C} \longrightarrow \mathrm{Na}_{2} \mathrm{~S}+4 \mathrm{CO} \uparrow$

$\mathrm{Na}_{2} \mathrm{~S}+\mathrm{CaCO}_{3} \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{CaS}$

  1. Solvay Process:

$\mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2} \longrightarrow \mathrm{NH}_{4} \mathrm{HCO}_{3}$

$\mathrm{NaCl}+\mathrm{NH}_{4} \mathrm{HCO}_{3} \longrightarrow \mathrm{NaHCO}_{3}+\mathrm{NH}_{4} \mathrm{Cl}$

$2 \mathrm{NaHCO}_{3} \stackrel{150^{\circ} \mathrm{C}}{\longrightarrow} \mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}$


  1. Soda ash is anhydrous sodium carbonate that does not disintegrate when heated but melts at 852 degrees Celsius.

  2. Sodium carbonate forms a number of hydrates.

  3. Hydrated $\mathrm{Na}_{2} \mathrm{CO}_{3}$ is called washing soda $\left(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot 10\right.$ $\left.\mathrm{H}_{2} \mathrm{O}\right)$ and is prepared by Le Blanc process solvay process and electrolytic process.

  4. Sodium carbonate absorbs carbon dioxide and produces sodium bicarbonate, which can be calcined to obtain pure sodium carbonate at 250 degrees Celsius.

$\mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2} \underset{\sum_{250^{\circ} \mathrm{C}}} 2 \mathrm{NaHCO}_{3}$

  1. It was causticized by lime after being dissolved in acid with effervescence of carbon dioxide.

$\mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{HCl} \longrightarrow 2 \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}$

$\mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{Ca}(\mathrm{OH})_{2} \longrightarrow 2 \mathrm{NaOH}+\mathrm{CaCO}_{3}$

Uses: Sodium carbonate is widely used as a smelter in glass making

Potassium Carbonate

It can be made by the leblanc method, but not by the solvay process since potassium carbonate is water soluble.

Properties: It resembles with $\mathrm{Na}_{2} \mathrm{CO}_{3}$, its melting point is $900^{\circ} \mathrm{C}$ but a mixture of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ and $\mathrm{K}_{2} \mathrm{CO}_{3}$ melts at $712^{\circ} \mathrm{C}$.

Uses: Potassium carbonate is used in glass manufacturing.

Calcium Carbonate

Marble, limestone, chalk, and calcite are examples of natural calcite. It's made by dissolving marble or limestone in $\mathrm{HCl}$, eliminating any iron or aluminium, precipitating with $\mathrm{NH}_{3}$, and adding $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}$to the solution.

$\mathrm{CaCl}_{2}+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3} \longrightarrow \mathrm{CaCO}_{3}+2 \mathrm{NH}_{4} \mathrm{Cl}$


  1. It is dissociates above 1000 degree Celsius as follows:

$\mathrm{CaCO}_{3} \longrightarrow \mathrm{CaO}+\mathrm{CO}_{2}$

  1. It dissolves in carbon dioxide-containing water to produce calcium bicarbonate, but boiling separates it from the solution.

\[\mathrm{CaCO}_{3} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}\;\underset{boiling} {Ca}(HCO_3)_2\]

Magnesium Carbonate

It is found in nature as magnesite, which is isomorphic to calcite. It is precipitated as a white by adding sodium bicarbonate to a solution of a magnesium salt, although only basic carbonate, known as magnesia alba, with the approximate composition $\mathrm{MgCO}_{3} . \mathrm{Mg}(\mathrm{OH})_{2} .3 \mathrm{H}_{2} \mathrm{O}$, is precipitated.

Properties: The properties of magnesium carbonate is same as calcium carbonate.


Sodium Bicarbonates

Preparation: By absorption of $\mathrm{CO}_{2}$ in $\mathrm{Na}_{2} \mathrm{CO}_{3}$ solution.

$\mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2} \rightleftharpoons 2 \mathrm{NaHCO}_{3}$

Uses: It is used in pharmaceutical industries and also as baking powder.

Potassium bicarbonates

Preparation: Same as $\mathrm{NaHCO}_{3}$

Properties: Same as $\mathrm{NaHCO}_{3}$

But it is more alkaline and more soluble in water compared to $\mathrm{NaHCO}_{3}$.

Magnesium Bicarbonate

$\mathrm{MgCO}_{3}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} \stackrel{\text { boiling }}{\rightleftharpoons} \mathrm{Mg}\left(\mathrm{HCO}_{3}\right)_{2}$

Calcium Bicarbonate

$\mathrm{CaCO}_{3}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}$


Sodium Chloride

It is prepared by the method of brine which contains 25% sodium chloride.


  1. It is nonhygroscopic, whereas ordinary salt contains magnesium chloride, which makes it hygroscopic.

  2. It's used to make freezing mixture in the lab (freezing mixture is ice-common salt mixture with a temperature of-25 degrees Celsius).

  3. To melt ice and snow from the road.

Potassium Chloride

It also occurs in nature as sylvite $(\mathrm{KCl})$ or carnallite $\mathrm{KCl} . \mathrm{MgCl}_{2} .6 \mathrm{H}_{2} \mathrm{O}$.

Uses: Potassium chloride is used as a fertiliser.

Magnesium Chloride

Preparation: It is prepared by dissolving $\mathrm{MgCO}_{3}$ in dilute hydrochloric acid.

$\mathrm{MgCO}_{3}+2 \mathrm{HCl} \longrightarrow \mathrm{MgCl}_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}$


  1. It crystallises as hexahydrate. $\mathrm{MgCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}$.

  2. It is deliquescent solid.

  3. This hydrate undergoes hydrolysis as follows :

$\mathrm{MgCl}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Mg}(\mathrm{OH}) \mathrm{Cl}+\mathrm{HCl}$ 

$\mathrm{Mg}(\mathrm{OH}) \mathrm{Cl} \longrightarrow \mathrm{MgO}+\mathrm{HCl}$

  1. Anhydrous $\mathrm{MgCl}_{2}$ can be prepared by heating a double salt like. $\mathrm{MgClO}_{2} . \mathrm{NH}_{4} \mathrm{Cl} .6 \mathrm{H}_{2} \mathrm{O}$ as follows:

$MgC{l_2}.N{H_4}C{l_2}.6{H_2}O\xrightarrow[\Delta ]{{ - {H_2}O}}MgC{l_2}N{H_4}Cl\xrightarrow[\Delta ]{{strong}}MgC{l_2} + N{H_3} + HCl$ 

Sorel Cement: It's a paste-like mixture of magnesium oxide and magnesium chloride that hardens when left to stand. This is utilised in dental fillings, flooring, and other applications.

Calcium Chloride

  1. In the solvay process, it is a by-product.

  2. The carbonate can alternatively be made by dissolving it with hydrochloric acid.

$\mathrm{CaCO}_{3}+2\mathrm{HCl}\longrightarrow\mathrm{CaCl}_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}$


  1. It's made out of deliquescent crystals.

  2. It gets hydrolysed like $\mathrm{MgCl}_{2}$ hence anhydrous $\mathrm{CaCl}_{2}$ cannot be prepared.

$\mathrm{CaCl}_{2}+\mathrm{H}_{2}\mathrm{O}\rightleftharpoons\mathrm{CaO}+2 \mathrm{HCl}$

  1. Anhydrous $\mathrm{CaCl}_{2}$ is used in drying gases and organic compounds but not $\mathrm{NH}$ or alcohol due to the formation of $\mathrm{CaCl}_{2} .8 \mathrm{NH}_{3}$ and $\mathrm{CaCl}_{2} \cdot 4 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$


Sodium Sulphate


It is created by heating common salt with sulphuric acid in the first step of the leblanc process.

$2 \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}+2 \mathrm{HCl}$

Thus the salt cake formed is crystallised out from its aqueous solution as $\mathrm{Na}_{2} \mathrm{SO}_{4} .10 \mathrm{H}_{2} \mathrm{O}$. This is called as Glauber's salt.


Sodium sulphate is reduced to $\mathrm{Na}_{2} \mathrm{S}$ when it is fused with carbon.

$\mathrm{Na}_{2} \mathrm{SO}_{4}+4 \mathrm{C} \longrightarrow \mathrm{Na}_{2} \mathrm{S}+4 \mathrm{CO}$

Uses: It is used in pharmaceutical industries.

Potassium Sulphate

It's found in stassfurt potash deposits as schonite and kainite, and it's made by dissolving it in water and crystallising it. It separates as crystals from the solution, whereas $\mathrm{Na}_{2} \mathrm{SO}_{4}$ is a decahydrate.

Magnesium Sulphate


  1. It's made by dissolving kieserite $\mathrm{MgSO}_{4} . \mathrm{H}_{2} \mathrm{O}$ in boiling water and then crystallising the resulting hepta hydrate solution. Epsom salt is what it's called.

  2. It is also obtained by dissolving magnesite in hot dilute $\mathrm{H}_{2} \mathrm{SO}_{4}$

$\mathrm{MgCO}_{3}+\mathrm{H}_{2}\mathrm{SO}_{4}\longrightarrow\mathrm{MgSO}_{4}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}$

  1. It is isomorphous with $\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}, \mathrm{ZnSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}$

Calcium Sulphate

It occurs as anhydrite $\mathrm{CaSO}_{4}$ and as the dihydrate $\mathrm{CaSO}_{4}.2\mathrm{H}_{2}\mathrm{O}$, gypsum, alabaster or satinspar.


  1. Calcium sulphate solubility increases till a certain point and then declines as temperature rises.

  2. Because of its porous nature, plaster of Paris is utilised in the construction of wood.

Example 17: 

  1. $\mathrm{Mg}_{3} \mathrm{N}_{2}$ when reacted with water, givesoff $\mathrm{NH_3}$. but HCl is not obtained from $\mathrm{MgCl}$ on reaction with water at room temperature. Why?

  2. The crystalline salts of alkaline earth metals contain more water of crystallization than corresponding alkali metal salts. Why?


  1. Because $\mathrm{Mg}_{3} \mathrm{N}_{2}$ is a salt of a strong base and a weak acid $\left(\mathrm{NH}_{3}\right)$ , it can be hydrolyzed.

$\mathrm{Mg}_{3} \mathrm{~N}_{2}+6 \mathrm{H}_{2} \mathrm{O}\longrightarrow 3 \mathrm{Mg}(\mathrm{OH})_{2}+2 \mathrm{NH}_{3}$

  1. In comparison to alkali metal ions, alkaline earth metal ions have a stronger tendency to hydrate due to their small size and high nuclear charge. As a result, alkaline earth metal salts contain more crystallisation water than alkali metal salts.

Example 18: What happens when:

  1. Beryllium carbide reacts with water. 

  2. Magnesium nitrate is heated.

  3. Quick lime is heated in electric furnace with powdered coke. 

  4. Sodium chloride solution is added to zinc chloride solution.


  1. A gas named methane is evolved

$\mathrm{Be}_{2} \mathrm{C}+4 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{Be}(\mathrm{OH})_{2}+\mathrm{CH}_{4}$

  1. A gas which is brown in colour, $\mathrm{NO}_{2}$, is evolved

$2 \mathrm{Mg} \mathrm{NO}_{3} \longrightarrow 2 \mathrm{MgO}+4 \mathrm{NO}_{2}+\mathrm{O}_{2}$

  1. Calcium carbide is formed

$CaO + 3C\xrightarrow[{Arc}]{{Electric}}Ca{C_2} + CO$ 

  1. A white precipitate of zinc hydroxide is evolved which forms sodium zincate on dissolving with excess of sodium hydroxide.

$\mathrm{ZnCl}_{2}+2\mathrm{NaOH}\longrightarrow\mathrm{Zn}(\mathrm{OH})_{2}+2 \mathrm{NaCl}$

$\mathrm{Zn}(\mathrm{OH})_{2}+2\mathrm{NaOH}\longrightarrow\mathrm{Na}_{2} \mathrm{ZnO}_{2}+2 \mathrm{H}_{2} \mathrm{O}$

Example 19: Draw a structure of:

  1. $BeC{l_2}$ in vapour state

  2. $BeC{l_2}$ in solid state


It features a chlorobridged dimer structural bond in the vapour state. At 1000 degrees Celsius, it dissociates into a linear monomer.

BeCl2 dimers and monomer

In the solid state, it has a polymeric structure with chlorobridges, in which a halogen atom connected to one beryllium atom forms a coordinate bond with a lone pair of electrons and a covalent binding with another atom.

Polymeric BeCl2 in solid state

Class 11 Chemistry Revision Notes for Chapter 10 - The s-Block Elements

About s-Block Elements Revision Notes

The s-block elements fall among the most popular and speak about the elements in chemistry. Thus in class 11, chapter 10 basically throws light on s-block elements, including their properties, characteristics, types, and the importance of different compounds. Aiming to help the students learn about these topics and study productively, a free CBSE revision notes for class 11, chapter 10 Chemistry - The s-block elements are provided at Vedantu. Students can access the revision notes here at this page and get easily accustomed to all the important topics given in the chapter. Also, students can use the revision notes at their convenience irrespective of the time or the place.

Periodic Trends in the Properties of s-block Elements

  • Alkali Metal - These are the metals of silvery-white, soft, low melting, and highly reactive. The ionic size and the atomic size as well increase down the group. On the other side, the ionization enthalpies decrease down the group. Mostly, these compounds are ionic. Their oxides and hydroxides dissolve in water to form strong alkalies. A few compounds of sodium are sodium hydrogen carbonate, sodium chloride, and more related. The Castner-Kellner process does the manufacturing of Sodium hydroxide, and the Solvay process achieves the manufacturing of sodium carbonate.

  • Alkaline Earth Metals - These are the metals having increased cationic charges. Their atomic size and the ionic size is reduced. Oxides and hydroxides of alkaline earth metals are less basic. A few of the compounds of calcium are calcium hydroxide, calcium carbonate, and more related.

  • Diagonal Relationship - The first element of group 1 and the second element of group 2 exhibits similar properties: the Lithium from group 1 and Beryllium from group 2. Thus, such similarities are known as diagonal relationships.

What are The s-block Elements?

The s-block of the periodic table contains Group 1 and Group 2 elements. Moreover, their hydroxides and oxides are alkaline in nature. Furthermore, they are characterized by the s- electron or electrons in their atom's valence shell. Besides, they are highly reactive metals that form mono and dipositive ions. Additionally, with the increasing atomic number, there is a regular trend in the alkali metal's physical and chemical properties.

Most notably, the first element of both Group 1 and Group 2 exhibits the same properties. In the periodic table, these similarities are referred to as the diagonal relationship. Moreover, the chemistry of alkaline earth metals is much similar to the alkali metals. Furthermore, there are some differences because of the ionic and atomic sizes.

Sub-Topics of The s- Block Elements

Let us look at the subtopics that fall under the topic, The s-block elements:

  • Anomalous Behaviour of Lithium - This topic explains about lithium, its nature, and similarities between the lithium and magnesium, including the difference between lithium with other alkali metals.

  • Beryllium, Magnesium, and Calcium - This topic gives an overview of magnesium, beryllium, and calcium.

  • Characteristics of the Compounds of Alkali Earth Metals - This topic speaks about the alkali earth metals, including their physical and chemical properties.

  • Characteristics of the Compounds of Alkali Metals - The topic describes the characteristics of compounds of alkali metals and also discusses a few other alkali metals.

  • Group 1 Elements, Alkali Metals - This topic highlights the group1 elements like sodium, lithium, potassium, and more.

  • Group 2 Elements, Alkali Earth Metals - This topic teaches us about the group 2 elements such as calcium, magnesium, and more related.

  • Some Important Compounds of Potassium and Sodium - This topic defines the important compound uses and sodium and potassium properties.

Why Choose Vedantu?

Vedantu is one of the biggest online learning platforms where students can get many benefits, where some of them are mentioned below:

  • Students can be updated on the latest information regarding their respective state boards

  • They can get various info on question papers, revision notes, last 5 years question papers, information on specific topics, and a lot more

  • By going through all these academic papers, students can gain a better understanding of every topic or chapter

  • They can also clear the doubts on a specific topic or chapter

  • Even the students can download and save these papers and keep with them, which can be used further at the time of their examinations

FAQs on Tthe S Block Elements Class 11 Notes CBSE Chemistry Chapter 10 (Free PDF Download)

1. Where can I find well-prepared Revision Notes for Class 11 Chemistry Chapter 10- The s-Block Elements?

Vedantu is a top-notch online learning platform that caters to chapter-wise  Revision Notes for Class 11 Chemistry.  This includes well-curated Revision Notes for Class 11 Chemistry Chapter 10 The s-Block Elements. These notes are prepared by subject matter experts to help students in revision. At Vedantu, you can find the best online notes for Class 11 Chemistry Chapter 10 which include the main topics of the chapter explained in a brief manner. This is designed with an emphasis on the latest syllabus. Vedantu’s notes are precise yet effective to understand and revise the chapter thoroughly.

2. Is Vedantu’s Revision Notes for Class 11 Chemistry Chapter 10 helpful in exam preparation?

Yes. Vedantu’s revision notes are highly beneficial in studies and exam preparation.  Online Revision Notes for Class 11 Chemistry Chapter 10  The s-Block Elements provided by experts is exceptional study material. As these materials are prepared by experts, the quality of the material is unparalleled to any other website. It facilitates effective revision. It covers all the aspects to be included regarding the key concepts related to the chapter. Hence, students can trust the material. Revision Notes for Class 11 Chemistry Chapter 10 are beneficial to revise the chapter and learn the fundamentals in less time.

3. What does Class 11 Chemistry Chapter 10 The s-Block Elements deal with?

Chapter 10 of Class 11 Chemistry deals with the concepts related to s-block elements of the periodic table. It helps students to understand the electronic configuration, physical and chemical properties, general characteristics of transition elements, etc. The chapter has a high weightage in competitive exams.

4. What are some examples of s-Block elements?

S-block consists of 14 elements. These are Hydrogen (H), Lithium (Li), Helium (He), Sodium (Na), Beryllium (Be), Potassium (K), Magnesium (Mg), Rubidium (Rb), Calcium (Ca), Caesium (Cs), Strontium (Sr), Francium (Fr), Barium (Ba) and Radium (Ra).

5. What are the basics of Class 11 Chemistry, Chapter 10- ‘The s-Block Elements?’

Class 11 Chemistry, Chapter 10- ‘The s-Block Elements,' as the name suggests, discusses the s-block Elements in the periodic table. The chapter, first, discusses the Alkali Metals which comprise the group one elements. The sub-topics under this are- Atomic and Ionic Radii,  Ionization Enthalpy and the Hydration Enthalpy of alkali metals. Then, we study the physical and chemical properties of Alkali metals and their uses of alkali metals. Similarly, we study and analyse the Group two elements i.e Alkaline Earth Metals. 

6. What are the best revision notes for NCERT Class 11 Chemistry, Chapter 10- ‘The s-Block Elements?’

Vedantu's Revision Notes is the best study material for Class 11 Chemistry, Chapter 10- ‘The s-Block Elements.' These notes are the best study material because they are crisp, well-structured, easy to understand notes prepared exclusively to crack the Chemistry exam. All important portions from the chapter are systematically covered in these notes. These notes are highly credible too as they are compiled by an expert faculty of the most experienced Chemistry teachers. Studying from these notes will surely help you ace the Chemistry exam. These notes are available on the Vedantu app and on the Vedantu website at free of cost.

7. Do I need to practice all the questions provided in Class 11 Chemistry, Chapter 10- ‘The s-Block Elements?’

All the NCERT questions given in Class 11 Chemistry Chapter 10- ‘The s-Block Elements’ are extremely important from an examination point of view, and must not be ignored to score well in the exam. The questions given in between the chapter and the back exercise questions at the end of the chapter must be attempted and practised at least a few times before appearing for the exam. Read NCERT thoroughly as questions can be framed from any portion of the chapter. A thorough understanding of the basics will also help you solve the application based questions from this chapter. 

8. What are the physical properties of the alkali metals? 

As discussed in the chapter, the metals that are soft, light and white in colour are alkali metals. They usually have a low density and as we move down the group of alkali metals, their density increases. The colour in the oxidising flame is due to the presence of alkali metals. Some other properties of alkali metals are that they can be easily cut with a knife and they generally have a low melting point. 

9. How can I score full marks in Class 11 Chemistry, Chapter 10- ‘The s-Block Elements?’

The first step to ace Class 11 Chemistry, Chapter 10- ‘The s-Block Elements' is to study it thoroughly from the standard NCERT textbook. Make extensive use of the periodic table while studying this chapter to grasp the topics easily. Never cram for a test instead, aim for conceptual understanding of this chapter. Clear all doubts with the help of online study materials such as Vedantu's Revision Notes for this chapter. Lastly, practice as many questions from the NCERT exercises and the previous year papers as much as you can to ace the Chemistry exam.