Notes on Redox Reactions (Chapter - 8)

Oxidation and Reduction

Oxidation‐reduction (redox) reactions

At different times, oxidation and reduction (redox) have had different, but complimentary, definitions. Compare the following definitions:

Oxidation is:                                                 Reduction is:

=>Gaining oxygen                                           =>Losing oxygen

=>Losing hydrogen                                         =>Gaining hydrogen

=>Losing electrons (+ charge increases)                =>Gaining electrons (+ charge decreases)

Oxidation and reduction are opposite reactions. They are also paired reactions: in order for one to occur, the other must also occur simultaneously.

The oxidation number describes the oxidation state of the element in a compound, and these numbers are assigned following a relatively simple set of rules.

Rules for assigning oxidation numbers

1.      The oxidation number of an element in its elemental form is 0 (zero).

2.      The oxidation number of a simple ion is equal to the charge on the ion. Both the size and the polarity of the charge are part of the oxidation number: an ion can have a +2 oxidation number or a -­‐2 oxidation number. The “+” and “-­‐“ signs are just as important as the number!

3.      The oxidation numbers of group 1A and 2A (group 1 and group 2) elements are +1 and +2 respectively.

4.      In compounds, the oxidation number of hydrogen is almost always +1. The most common exception occurs when hydrogen combines with metals; in this case the oxidation number of hydrogen is typically –1.

5.      In compounds, the oxidation number of oxygen is almost always –2. The most common exception is in peroxides, when the oxidation number is –1. Peroxides are compounds having two oxygen atoms bonded together. 

      For example, hydrogen peroxide is H‐O‐O‐H. In hydrogen peroxide, each oxygen atom has a ‐1 oxidation number. When oxygen is bonded to fluorine, as in hypofluorous acid (HOF), the oxidation number of oxygen is 0. Oxygen- fluorine compounds are relatively rare and not too terribly important for our studies.

6.      In compounds, the oxidation number of fluorine is always –1. The oxidation number of other halogens (Cl, Br, I) is also –1, except when they are combined with oxygen. The oxidation number of halides (except fluorine) combined with oxygen is typically positive. For example, in ClO^-, chlorine’s oxidation number is +1.

7.  For a complex ion, the sum of the positive and negative oxidation numbers of all elements in the ion equals the charge on the ion.

8.      For an electrically neutral compound, the sum of the positive and negative oxidation numbers of all elements in the compound equals zero.

Identifying redox reactions

Now that you can assign proper oxidation numbers to the elements in substances, we can use changes in the oxidation numbers to identify oxidation and reduction reactions.

Consider the following chemical equation:

Zn(s) + 2HCl(aq) → Zn⁺²(aq) + 2C^­‐(aq) + H₂ (g)

The oxidation number for elemental zinc is 0. The oxidation number for zinc ion is +2. The oxidation number for hydrogen in hydrogen chloride is +1. The oxidation number for elemental hydrogen, H2, is 0. 

The oxidation number for chlorine in hydrogen chloride is –1. The oxidation number for chloride ion is –1.

Zinc has lost two electrons, and therefore developed a +2 charge. Zinc has been oxidized – its oxidation number has become more positive (from 0 to +2).

Hydrogen has gained an electron, and its positive charge has decreased. Hydrogen has been reduced – its oxidation number has become less positive (from +1 to 0).

Chloride has not changed its oxidation state. It is a spectator ion in this reaction.

Figure 1 may be useful in deciding if an element has been oxidized or reduced. If an elements oxidation number increases (moves towards the right), then the element is oxidized. If an elements oxidation number decreases (moves towards the left), then the element is reduced.

 NOTE: an element doesn’t have to become positive or negative for oxidation or reduction to occur. Instead, the element has to become more positive or more negative. A change from -­‐3 to -­‐1 is still oxidation, while a change from +3 to +1 is still reduction.

Substances that cause changes in the oxidation state are called oxidizing agents or reducing agents. An oxidizing agent causes oxidation number to occur.

How does the oxidizing agent cause oxidation?

To increase the positive oxidation number of an element, the oxidizing agent must take one or more electrons from the element. As the element being oxidized loses electron(s), its oxidation number becomes more positive.

 However, the electrons don’t disappear! The oxidizing agent has taken these electrons, and therefore the oxidizing agent becomes more negative – it is reduced!

In our reaction above, hydrogen in hydrogen chloride takes an electron from the zinc metal. The zinc metal becomes more positive; it is oxidized. 

By taking the electron from the zinc metal, the hydrogen in hydrogen chloride becomes less positive; it is reduced. Hydrogen chloride is the oxidizing agent, because it contains the element that causes oxidation to occur.

Similarly, the zinc metal donates electrons to the hydrogen in hydrogen chloride, causing the oxidation state of hydrogen to decrease from +1 to 0. By providing the electrons necessary to reduce the oxidation number of hydrogen, the oxidation number of zinc increases from 0 to +2; zinc is oxidized. While being oxidized, zinc reduced the oxidation number of hydrogen. Therefore, zinc is a reducing agent.

Oxidizing agents are substances containing the element(s) that accept electrons, allowing another element(s) to be oxidized. By accepting electrons, the element(s) in the oxidizing agent are reduced.

Reducing agents are substances containing the element(s) that donate electrons, allowing another element(s) to be reduced. By donating electrons, the element(s) in the reducing agent are oxidized.

As you can see, there is a great deal of symmetry between oxidizing and reducing agents, and between oxidation and reduction. 

Whenever one process happens, the other process MUST also happen, because we are transferring electrons from one material to another material. Electrons must come from someplace, and must go someplace; electrons cannot simply appear and disappear.

There are two general considerations to keep in mind when discussing oxidizing/reducing agents.

First, in most cases there will only be one element in each agent that is oxidized or reduced. In our reaction, only hydrogen in hydrogen chloride was reduced – the chloride wasn’t changed. 

Second, the oxidizing or reducing agent is the compound containing the element that is oxidized or reduced.

The reducing agent is hydrogen chloride, not just hydrogen ion.

 Similarly, the oxidizing agent is the compound containing the element that is reduced. Oxidizing and reducing agents are ALWAYS reactants, NEVER products!

Balancing redox reactions.

We can use our knowledge of redox reactions to help us balance chemical reactions. The simplest process to use is the half-­‐reaction method. Consider the

following chemical reaction (unbalanced):

MnO4^‐ + C2O4^‐2 → Mn⁺² + CO2

First, we assign oxidation numbers to all elements shown in the reaction:

+7  ‐2	+3  ‐2	+2	+4 -2
MnO4-   +	C2O4‐	→  Mn+2   +	CO2

Comparing oxidation numbers, we see that manganese goes from +7 to +2, and has been reduced, while carbon goes from +3 to +4 and has been oxidized. 

Oxygen doesn’t change its oxidation state. The oxidizing agent is permanganate, and the reducing agent is oxalate.

The half‐reaction method involves balancing the oxidation reaction as if it were an isolated reaction. Then the reduction half-reaction is balanced as if it were an isolated reaction. Finally, the two half‐reactions are combined.

The oxidation half‐reaction is:

C₂O₄^‐2 → CO₂

The first step is to balance elements, other than oxygen and hydrogen, using the normal methods of balancing chemical equations. We have two carbons in oxalate, so we need a 2 in front of carbon dioxide to balance the carbons:

C₂O₄^‐2 → 2CO₂

By comparing reactants and products, we see that we have the same number of carbon and oxygen atoms on each side of the reaction. There aren’t any other elements present, so we don’t need to change any more coefficients.

Charge balance requires that the net charge of reactants and products must be equal. We equalize the charges by adding 2 electrons to the products side:

C₂O₄^‐2 → 2CO₂ + 2e^‐

All elements are identical on both sides, the number of atoms of each element is the same on both sides, and the total charge is the same on both sides: ‐2 for oxalate and ‐2 for the two electrons. This is a balanced oxidation half‐reaction.

The reduction half‐reaction is:

MnO₄^‐ → Mn⁺²

The manganese atoms are balanced, but what are we to do about oxygen? To balance oxygen, we add water molecules. Four oxygen atoms in permanganate require four water molecules as products:

MnO₄^‐ → Mn⁺² + 4H₂O

Adding water molecules balanced oxygen, but now we have hydrogen atoms. We balance 8 hydrogen atoms from 4 water molecules by adding 8 hydrogen ions to the reaction:

8H⁺ + MnO₄^‐ → Mn⁺² + 4H₂O

Now we have the same number and kinds of atoms on both sides. However, charge balance requires that the net charge of reactants and products must be equal. We have a net electrical charge of +7 for the reactants, and a net electrical charge of +2 for the products. We add 5 electrons to the reactant side of the equation:

8H⁺ + MnO₄^‐ + 5e^- → Mn⁺² + 4H₂O

This is a balanced reduction half-reaction.

(NOTE: in an oxidation half‐ reaction, electrons are produced, while in a reduction half‐reaction, electrons are consumed.)

We are ready to combine the two half‐reactions. To combine these reactions, the same number of electrons must be produced as are consumed.

 When different numbers of electrons are produced and consumed, you need to find the least common multiple. 

Since 5 electrons are consumed in the reduction reaction, and 2 electrons are produced in the oxidation reaction, the least common multiple is 10. We need to multiply the reduction reaction by 2, and the oxidation reaction by 5:

16H⁺ + 2MnO₄^‐ + 10e^‐ → 2Mn⁺² + 8H₂O

5C₂O₄^‐2 → 10CO₂ + 10e^-

Now that we have the same number of electrons produced and consumed, and we can add the reactions:

16H⁺ + 2MnO₄^- + 10e^- + 5C₂O₄^‐2 → 2Mn⁺² + 8H₂O + 10CO₂ + 10e^-

We eliminate substances that are identical on both sides of the equation:

16H⁺ + 2MnO₄^- + 5C₂O₄^‐2 → 2Mn⁺² + 8H₂O + 10CO₂

Finally, we check the equation to insure that mass and charge balance have been achieved.

Reactants        Products

16 H        16 H

2 Mn        2 Mn

28 O        28 O

10 C       10 C

+4       +4

This is an example of balancing a redox reaction in acid solution, by the half-­‐ reaction method. We can also balance the reaction in basic solution. 

The initial work is the same as shown above. Once the reaction is balanced in acid solution, we neutralize hydrogen ion by adding an equal number of hydroxide ions to both sides of the reaction:

16OH^- + 16H⁺ + 2MnO₄^- + 5C₂O₄^‐2 → 2Mn⁺² + 8H₂O + 10CO₂ + 16OH^‐

Hydroxide ion combines with hydrogen ion, forming water molecules, and we eliminate equal numbers of water molecules from both sides of the equation:

16H₂O + 2MnO₄^- + 5C₂O₄^-2 → 2Mn⁺² + 8H₂O + 10CO₂ + 16 OH^-

8H₂O + 2MnO₄^- + 5C₂O₄^-2 → 2Mn⁺² + 10CO₂ + 16OH^-

Finally, we check to insure mass and charge balance.

Reactants       Products

16 H       16 H

2 Mn       2 Mn

36 O       36 O

10 C      10 C

‐12    ‐12

Solved Short Questions on S-Block Elements (Chapter 10)

After reading the chapter and understanding the concepts of topics, it becomes necessary to solve questions on topics of chapter. Therefore, We are giving some "Solved Short Questions on S-Block Elements" to practice and to know your Grasping capacity of concepts for this chapter.

Q.1.  Why the elements of second raw (first short period) show a number of differences in properties from other members of their respective families.

Ans.1. The difference in the properties of the first member of a group from those of the other members is due to

(i) Smaller size of the atom

(ii) Presence of one inner shell of only two electrons and

(iii) Absence of d-Orbitals.

Q.2. What is diagonal relationship due to?

Ans.2. Elements having similar size of atoms or ions electronegativity or similar polarizing power show diagonal relationship.

For example, Li and Mg;  Be and Al have many similar properties.

Q.3. What is the general name for elements of group 1?

Ans.3. Elements of group 1 are called “Alkali Metals” because their oxides dissolve in water to form hydroxides which are strong bases or alkalies.

Q.4. What are s- block elements?

Ans.4. S-block elements are those in which the last electron enters the outermost s-orbital.

As the s-orbital can accommodate only two electrons, two groups (1 & 2) belong to the s-block of the Periodic Table.

Q.5. Which elements of s- block are largely found in biological fluids & what is its importance?

Ans.5. Monovalent sodium and potassium ions and divalent magnesium and calcium ions are found in large proportions in biological fluids. These ions perform important biological functions such as maintenance of ion balance and nerve impulse conduction.

Q.6. Why alkali metals are highly electropositive & they are not found in free state ?

Ans.6. The loosely held s-electron in the outermost valence shell of these elements makes them the most electropositive metals. They readily lose electron to give monovalent M⁺ ions. Hence they are never found in free state in nature.

Q.7. Give reason for the colour imparted to the flame by alkali metals

Ans.7. The alkali metals and their salts impart characteristic colour to an oxidizing flame. This is because the heat from the flame excites the outermost orbital electron to a higher energy level. When the excited electron comes back to the ground state, there is emission of radiation in the visible region.

Q.8. What happens when alkali metals react with dihydrogen?

Ans.8. The alkali metals react with dihydrogen at about 673K (lithium at1073K) to form hydrides. All the alkali metal hydrides are ionic solids with high melting points.

2 M   +    H₂   ===>    2M⁺H^-

Q.9. Give reason for the low solubility of LiF & CsI in water.

Ans.9.  The low solubility of LiF in water is due to its high lattice enthalpy whereas the low solubility of CsI is due to smaller hydration enthalpy of its two ions. Other halides of lithium are soluble in ethanol, acetone and ethyl acetate; LiCl is soluble in pyridine also.

Q.10. Write any four Uses of washing soda.

Ans.10. i) It is used in water softening, laundering and cleaning.

(ii) It is used in the manufacture of glass, soap, borax and caustic soda.

(iii) It is used in paper, paints and textile industries.

(iv) It is an important laboratory reagent both in qualitative and quantitative analysis.

Q.11. How does the atomic and Ionic Radii of alkaline earth metals vary in comparison to alkali metals

Ans.11. The atomic and ionic radii of the alkaline earth metals are smaller than those of the corresponding alkali metals in the same periods. This is due to the increased nuclear charge in these elements. Within the group, the atomic and ionic radii increase with increase in atomic number.

Q.12. Alkali and alkaline earth metals cannot be obtained by chemical reduction method. Explain.

Ans.12. Since they are themselves stronger reducing agents than majority of the common reducing agents used.

Q.13. Why cannot sodium and potassium be prepared by the electrolysis of their aqueous solutions?

Ans.13.  The electrode potential, i.e., reduction potential of Na ( –2.71 V) or K (- 2.92) is much lower than that of H₂O (-0.83 V), therefore, upon electrolysis, water gets reduced in preference of Na⁺ or K⁺ ions. In other words, sodium and potassium cannot be obtained by electrolytic reduction of Na⁺ or K⁺ ions in aqueous solutions.

Q.14. Give reason . the sulphate of Be & Mg are soluble in water .

Ans.14. The sulphates of the alkaline earth metals are all white solids and stable to heat. BeSO₄, and MgSO₄ are readily soluble in water; the solubility decreases from CaSO₄ to BaSO₄. The greater hydration enthalpies of Be²⁺ and Mg²⁺ ions overcome the lattice enthalpy factor and therefore their sulphates are soluble in water.

Q.15. Name the gas liberated when alkali metals react with dil acid?

Ans.15. The alkaline earth metals readily react with acids liberating dihydrogen gas.

M    +     2HCl      →    MCl₂      +    H₂

This was the 1^st part of "Solved Short Questions on S-Block Elements (Chapter 10)". Check out more post from below "Related Post" Section and other chapter's help on right hand side. Comment below for any other Query or Help related to Chemistry.

Conceptual Questions on S-Block Elements - 1

Topic: Group 1 Elements – Alkali Metals
Q.1. Alkali metals are paramagnetic but their salts are diamagnetic. Explain.
Ans. Alkali metals contain one unpaired electron (ns¹) and hence are paramagnetic. However, during salt formation, this unpaired electron is transferred to the non-metallic atom forming its anion. As a result, salt has paired electrons both in the cation as well as in the anion and hence alkali metal salts are diamagnetic in nature.

Q.2. Which out of sodium or potassium has higher melting point?
Ans. On going from Na to K, the size of the atom increases and hence the metallic bonding weakens. In other words, sodium has higher melting point than potassium because of stronger metallic bonding.

Q.3. Among alkali metals in aqueous solution, Li⁺ ion has the lowest mobility. Why?
Ans. Because of smallest size among alkali metals, Li⁺ ions are most highly hydrated in aqueous solution. As a result, among alkali metals, mass of hydrated lithium ion is the highest and thus lowest ionic mobility.

Q.4. Why alkali metals impart color to the flame?
Ans. Alkali metals have low ionization enthalpies. Their valence electrons easily absorb energy from the flame and are excited to higher energy levels. When these excited electrons return to the ground state, the absorbed energy is emitted in form of light.

Q.5. Why can caesium be used in photoelectric cell while lithium cannot?
Ans. Caesium has the lowest while Lithium has the highest ionization enthalpy. Hence, Cs can lose electrons very easily whilie Li cannot.

Q.6. Lithium has highest ionization enthalpy in group 1 elements, yet it is the strongest reducing agent, Why?
Ans. In aqueous solutions, the tendency of an element to lose electrons does not entirely depend upon its ionization enthalpy. It also depend upon its enthalpy of Sublimation and the enthalpy of Hydration of the ion left after the loss of an electron. The combined effect of these factors is measured in terms of electrode potential. Since lithium has the most negative (-3.04 V) electrode potential of group 1 elements, therefore, lithium is the strongest reducing agent.

Q.7. Sodium fire in the laboratory should not be extinguished by pouring water. Why?
Ans.7. The reaction of sodium with H₂O is so exothermic that the H₂ produced catches fire.
2  Na     +     2 H₂O    ===>    2 NaOH    +    H₂
As a result, the fire spreads rather than being extinguished. Therfore, H₂O should not be used for extinguishing sodium fire. Instead pyrene (CCl₄) should be used. When CCl₄ is sprinked over fire, it being volatile (b.p. 350K) immediately forms vapors. These vapours being non-inflammable and heavy, surround the fire. As a result, supply of air is cut off and the fire gets extinguished.

Q.8.Why is it that on being heated in excess supply of air, K, Rb and Cs form superoxides in preference to oxides and peroxides?
Ans.8. K⁺, Rb⁺, and Cs+ are large cations and superoxide ion (O₂)^- is larger than oxide (O^2-) and peroxide (O₂)^2- ions. Since due to higher lattice energies, a large cation stabilizes a large anion, therefore, these metals form superoxides in preference to oxides and peroxides.

Q.9. Why superoxides of alkali metals are paramagnetic while normal oxides are diamagnetic?                          
Ans.9. Superoxieds contain the ion (O₂)^- which has the structure   containing a three electron bond, i.e., it has one unpaired electron and hence is paramagnetic. Normal oxides containing the ion   do not have any unpaired electron and hence are diamagnetic.

Q.10. Why alkali metals are normally kept in kerosene oil?
Ans. This is because in the air they are very easily oxidized to oxides which may dissolve in the moisture of the air to form hydroxides or they also combine directly with water vapors present in moisture to form hydroxides. The hydroxides, in turn, may combine with CO2 of the atmosphere to form carbonates.
4 M    +    O₂   ====>    2 M₂O
M₂O   +    H₂O    ====> 2 MOH
2 MOH    +    CO₂    ====>    M₂CO₃     +     H₂O
Therefore, to protect them from air and moisture, they are kept under kerosene oil.

Q.11. What makes lithium to show properties uncommon to the rest of the alkali metals?
Ans.11. The unusual properties of lithium as compared to other alkali metals is due to (i) the exceptionally small size of its atom and ion, (ii) high polarizing power (charge/size ratio), (iii) high ionization enthalpy and (iv) absence of d-orbitals.

Q.12. When is cation highly polarizing ? Which alkali metal cation has the highest polarizing power ?
Ans. A cation is highly polarizing if its charge/size ratio is very high. Because of its small size, Li+ has the highest polarizing power among the alkali metal ions.

Q.13. Why does table salt get wet in rainy season?
Ans.13. Pure NaCl is not hygroscopic but table salt is impure NaCl containing impurities of Mg2SO₄, CaSO₂, MgCl₂ and CaCl₂. All of these being hygroscopic absorb moisture from air in rainy season. As a result, table salt get wet.

Notes on S-Block Elements => Chapter 10

Topic => Alkali Metals (Group-1) 
1. The s-block elements of the periodic table are those in which the last electron enters the outermost s-orbital.

2. Elements of group1 are called alkali metals and elements of group 2 are called alkaline earth metal.

3. Group1 elements: Alkali metals
(i) Group 1 elements are called alkali metals because they form hydroxides on reaction with water which are strongly alkaline in nature
(ii) The general electronic for alkali metals is [noble gas] ns¹
Electronic configuration of alkali metals:

Element	Symbol	Atomic Number	Electronic Configuration
Lithium	Li	3	[He]2s1
Sodium	Na	11	[Ne]3s1
Potassium	K	19	[Ar]4s1
Rubidium	Rb	37	[Kr]5s1
Caesium	Cs	55	[Xe]6s1
Francium	Fr	87	[Rn]7s1

(iii)Trends in properties of alkali metals:
1. Atomic and ionic radii: Alkali metals have the largest atomic and ionic radii in their respective periods of the periodic table. On moving down the group, the atomic and ionic radii increase
Explanation: As we move in a period, the atomic radius and ionic radius tend to decrease due to increase in the effective nuclear charge. Therefore alkali metals have largest atomic and ionic radii in their respective group. On moving down the group there is increase in the number of shells .Thus there is an increase in distance between nucleus and outermost electrons which eventually increases atomic and ionic radii.

2. Ionisation enthalpies: Alkali metals have the lowest ionization enthalpy in each period. Within the group, the ionization enthalpies of alkali metals decrease down the group.

Explanation:Atoms of alkali metals are largest in their respective periods and therefore, the valence electrons are loosely held by the nucleus. By losing the valence electron, they acquire stable noble gas configuration. This accounts for their ease to lose electrons and hence they, have low ionization enthalpies.

3. The second ionization enthalpies of alkali metals are very high.
Explanation: When an electron is removed from the alkali metals, they form monovalent cations which have very stable electronic configurations (same as that of noble gases).Therefore it becomes difficult to remove the second electron from the stable noble gas configurations and hence their second ionization enthalpy values (IE₂) are very high.

4. Melting and boiling points: Alkali metals are soft and have low melting and boiling points.
Explanation: Alkali metals have only one valence electron per metal atom and therefore, the energy binding the atoms in the crystal lattice of the metal is low. Thus, the metallic bonds in these metals are not very strong and consequently, their melting and boiling points decrease on moving down from Li to Cs.

5. Density: Densities of alkali metals are quite low as compared to other metals. The densities increase on moving down the group. But K is lighter than Na
Explanation: The densities of metallic elements depend upon the type of packing of atoms in metallic state and also on their size. The alkali metals have close packing of metal atoms in their lattice and because of the large size of their atoms, they have low densities.
On moving down the group from Li to Cs, there is increase in atomic size as well as atomic mass. But the increase in atomic mass is more and compensates the increase in atomic size. As a result, the densities (mass/volume) of alkali metals gradually increase from Li to Cs. K is lighter than Na due to increase in atomic size of K.

6. Electropositive or metallic character: All the alkali metals are strongly electropositive or metallic in character
Explanation: The electropositive character of an element is expressed in terms of the tendency of its atom to release electrons:
M   =====>    M⁺ + e^-
As alkali metals have low ionization enthalpies, their atoms readily lose their valence electron. These elements are, therefore, said to have strong electropositive or metallic character.
Since, the ionization energies decrease down the group, the electron releasing tendency or electropositive character is expected to increase down the group.

7. Oxidation states: All the alkali metals predominantly exhibit an oxidation state of +1 in their compounds.

Explanation: Alkali metals have only one electron in their valence shell and therefore can lose the single valence electron readily to acquire the stable configuration of a noble gas. Since the second ionization energies are very high, they cannot form divalent ions. Thus, alkali metals are univalent and form ionic compounds

8. Characteristic flame colouration: All the alkali metals and their salts impart characteristic flame colouration.
Explanation: Alkali metals have very low ionization enthalpies. The energy from the flame of Bunsen burner is sufficient to excite the electrons of alkali metals to higher energy levels.
Excited state is quite unstable and therefore when these excited electrons come back to their original energy levels, they emit extra energy, which falls in the visible region of the electromagnetic spectrum and thus appear coloured.
Characteristic flame colouration by different alkali metals can be explained on the basis of difference in amount of energy absorbed for excitation of the valence electron.

9. Photoelectric effect: Phenomenon of ejection of electrons when electromagnetic radiation of suitable frequency strikes metal surface is called photoelectric effect. Alkali metals exhibit photoelectric effect.
Explanation: Alkali metals have low ionization enthalpies and therefore, the electrons are easily ejected when exposed to light. Among alkali metals, caesium has lowest ionization enthalpy and hence it shows photoelectric effect to the maximum extent.

10. Hydration of alkali metal ions: Alkali metal ions are highly hydrated.
Explanation: Smaller the size of the ion, the greater is the degree of hydration. Thus, Li⁺ ion gets much more hydrated than Na⁺ ion which is more hydrated than K ⁺ ion and so on. Therefore, the extent of hydration decreases from Li⁺ to Cs⁺. As a result of larger hydration of Li⁺ ion than Na⁺ ion, the effective size of hydrated Li⁺ ion is more than that of hydrated Na⁺ ion.
Hydrated ionic radii decrease in the order: Li⁺ > Na⁺ > K ⁺ > Rb⁺ > Cs⁺
Due to extensive hydration, Li⁺ ion has lowest mobility in water.

11. Reducing Nature: Alkali metals are strong reducing agents. This is due to their greater ease to lose electrons.
Explanation: Lithium is the strongest reducing agent. Tendency to act as reducing agent depends on energy requirement involved in three processes i.e. sublimation, ionization and hydration.

Lithium being small in size has high ionization enthalpy. On the other hand because of small size it is extensively hydrated and has very high hydration enthalpy.
This high hydration enthalpy compensates the high energy needed to remove electron (in second step). Thus Li has greater tendency to lose electrons in solution than other alkali metals. Thus Li is the strongest reducing agent.

12. Solutions in liquid ammonia: Alkali metals dissolve in liquid ammonia giving deep blue solutions which are conducting in nature
Explanation: In solution the alkali metal atom readily loses the valence electron. Both the cation and the electron combine with ammonia to form ammoniated cation and ammoniated electron.
The blue colour of the solution is due to the ammoniated electron which absorbs energy in the visible region of light and thus imparts blue colour to the solution.


13. Basic strength of hydroxides: Hydroxides of alkali metals are strongly basic and basic strength increases down the group
Explanation: The Metal—OH bond in the hydroxides of alkali metals is very weak and it can easily ionize to form M⁺ and OH^– ions. This accounts for their basic character. Since the ionization energy decreases down the group, the bond between metal and oxygen becomes weak.
Therefore, the basic strength of the hydroxides increases accordingly.
(iv) Important Compounds of alkali metals:

Name of Compound	Name of process and Brief about the process	Related Chemical Equation
Sodium Carbonate(Washing, Soda), Na2CO3.10H2O	Solvay Process: When CO2 gas is passed through a brine solution saturated with ammonia, sodium bicarbonate is formed. Sodium bicarbonate on heating forms sodium Carbonate.	2NH3 + H2O CO2 => (NH4)2CO3 (NH4)2CO3 + H2O + CO2 => 2NH4HCO3 NH4HCO3 + NaCl => NH4Cl + NaHCO3 2NaHCO3 => Na2CO3 + CO2 + H2O
Sodium Chloride, NaCl	Evaporation of sea water: Evaporation of sea water gives crude salt which contains impurities of CaSO4, Na2SO4 etc. To obtain pure Sodium Chloride,the crude salt is dissolved in minimum amount of water and filtered to remove insoluble impurities. The solution is then saturated with HCl gas. Crystals of pure sodium chloride separate out.
Sodium Hydroxide (Caustic Soda), NaOH	Electrolysis of NaCl in Castner-Kellner cell: A brine solution is elecrolysed using a mercury cathode. Na metal discharged at the cathode combines with mercury to form sodium amalgam. Chlorine gas is evolved at the anode. Sodium amalgam on treatment with water forms sodium hydroxide	At Cathode:   Na+  +  e-    ==Hg ==>    Na-amalgam At Anode:     Cl-  ===>  1/2 Cl2   +    e- 2Na-amalgam  +  2H2O   ==>   2NaOH  +   2Hg  +  H2
Sodium HydrogenCarbonate(Baking Soda), NaHCO3	NaHCO3 is made by saturating a solution of sodium carbonate with carbon dioxide	Na2CO3  +  H2O  + CO2  ===> 2 NaHCO3

Topic => Alkaline earth metals (Group-2) 

(i) Group 2 elements are called alkaline earth metals because their oxides and hydroxides are alkaline in nature and these metal oxides are found in the earth’s crust
(ii)The general electronic configuration for alkaline earth metals is [noble gas] ns²
Electronic configuration of alkaline earth metals:

Element	Symbol	Atomic Number	Electronic Configuration
Beryllium	Be	4	[He]2s2
Magnesium	Mg	12	[Ne]3s2
Calcium	Ca	20	[Ar]4s2
Strontium	Sr	38	[Kr]5s2
Barium	Ba	56	[Xe]6s2

Radium

Ra

88

[Rn]7s2

(iii) Trends in properties of alkaline metals and their comparison with alkali metals:
1. Atomic and ionic radii: Atomic and ionic radii of alkaline earth metals increases down the group and are smaller than the corresponding members of the alkali metals
Explanation: Alkaline earth metals have a higher nuclear charge and therefore, the electrons are attracted more towards the nucleus. As a result, their atomic and ionic radii are smaller than those of alkali metals. Atomic and ionic radii increase on moving down the group, due to screening effect and increases in the number of the shells.

2. Ionisation enthalpies: Alkaline earth metals have low ionization enthalpies due to their large size
Explanation: Although IE₁ values of alkaline earth metals are higher than those of alkali metals, the IE₂ values of alkaline earth metals are much smaller than those of alkali metals
In case of alkali metals (for e.g. Na) the second electron is to be removed from a cation which has already acquired a noble gas configuration. However in the alkaline earth metals (for e.g. Mg), the second electron is to be removed from a monovalent cation, i.e. Mg⁺: (1s²2s²p⁶3s¹) which still has one electron in the outermost shell. Thus, the second electron in Mg can be removed easily.

3. Melting and boiling points: Alkaline earth metals have higher melting and boiling points than the corresponding alkali metals
Explanation: Atoms of alkaline earth metals have smaller size as compared to alkali metals. Due to two valence electrons in alkaline earth metals, metallic bond is stronger than alkali metals. Hence alkaline earth metals have higher melting and boiling points than corresponding alkali metals

4. Electropositive or metallic character: The electropositive character increases down the group i.e., from Be to Ba but alkaline earth metals are not as strongly electropositive as the alkali metals
Explanation: Because of the low ionization enthalpies of alkaline earth metals, they are strongly electropositive in nature. However, these are not as strongly electropositive as the alkali metals of group 1 because of comparatively higher ionization energies

5. Oxidation states: All the alkaline earth metal exhibits an oxidation state of +2 in their compounds
Explanation: Alkaline earth metals exhibit oxidation state of +2 as they can lose two electrons and form bivalent ions. Following are the reason for the easy removal of two electrons
(i) Divalent cations have stable noble gas configuration
(ii)In solution, the +2 ions of alkaline earth metals are extensively hydrated and the high hydration energies of E²⁺ ions make then more stable than E⁺ ions.
It is observed that the amount of energy released when E²⁺ ion is dissolved in water is much more that that for E⁺ ions. This large amount of extra energy released in the hydration of +2 ions is more and compensates the second ionization enthalpy required for the formation of such E²⁺ ions
(iii)In the solid state, the divalent cations form stronger lattices than monovalent cations and therefore, a lot of energy called lattice enthalpy is released. It is the greater lattice enthalpy of E²⁺ ion which compensates for the high second ionization enthalpy and is responsible for its greater stability as compared to E⁺ ion

6. Hydration enthalpy: Hydration enthalpies of alkaline earth metal ions decreases down the group and are larger than those of the corresponding alkali metals ions
Explanation: This is due to smaller size of alkaline earth metal ions as compared to corresponding alkali metal ions of the same period. Due to smaller size, the alkaline earth metal ions are strongly attracted by polar water molecules and hence have larger hydration enthalpies. The hydration enthalpies of alkaline earth metal ions decrease with increase in ionic size down the group.
Be²⁺ > Mg²⁺ > Ca²⁺ > Sr²⁺ > Ba²⁺

7. Basic strength of hydroxides: Hydroxides of alkaline earth metals are less basic than alkali metals of the corresponding period.
Explanation: Lesser basic strength of hydroxides of alkaline earth metal hydroxides is due to the their high ionization enthalpies, small ionic size and dipositive charge on the ions
As a result, the Metal—O bond in these hydroxides is relatively stronger than that of corresponding alkali metals and therefore, does not break easily. Therefore, they are less basic than corresponding alkali metals.

(iii) Important Compounds of alkaline earth metals:

Name of Compound	Name of process & Brief about the process	Related chemical equations
Calcium Oxide or Quick Lime, CaO	It is prepared by heating limestone (CaCO3) in a rotary kiln at 1070-1270K. The carbon dioxide is removed as soon as it is produced to enable the reaction to proceed to completion.	CaCO3 + heat ⇌ CaO  + CO2
Calcium Hydroxide (Slaked lime), Ca(OH)2	It is prepared by adding water to quicklime, CaO	CaO + H2O ===>  Ca(OH)2
1. Calcium Carbonate, CaCO3	1. It is prepared by passing carbon dioxide through slaked lime. Excess of carbon dioxide should be avoided since this leads to the formation of water soluble calcium hydrogencarbonate	Ca(OH)2  +  CO2  ==>  CaCO3  +  H2O
2. Calcium Carbonate, CaCO3	2. It is prepared by the additionof sodium carbonate to calcium chloride.	CaCl2  +  Na2CO3  ===>  CaCO3  +  2 NaCl
Calcium Sulphate (Plaster of Paris), CaSO4.1/2H2O	It is obtained when gypsum, CaSO4.2H2O, is heated to 393K	2(CaSO4.2 H2O)  ====>  2(CaSO4).H2O  +  3 H2O

5. Diagonal relationship: It is observed that some elements of second period show similarities with the elements of third period present diagonally to each other though belonging to different groups. This is called diagonal relationship
Cause of diagonal relationship: Reason is the similarity in properties such as electronegativity, ionization enthalpy size, charge/radius ratio, etc. between the diagonal elements.
On moving from left to right across a period, for example, the electronegativity increases and while moving down group electronegativity decreases.
Therefore, on moving diagonally, the two opposing tendencies almost cancel out and the electronegativity values remain almost same as we move diagonally. Thus, the diagonal pairs have many similar properties.
Following pairs exhibit diagonal similarity:
(i) Lithium – Magnesium
(ii) Beryllium – Aluminium
(iii) Boron – Silicon

6. Some important reasons:
(1) Lithium carbonate decomposes at a lower temperature whereas sodium carbonate decomposes at a higher temperature
Explanation: Lithium carbonate decomposes at a lower temperature whereas sodium carbonate decomposes at a higher temperature because lithium carbonate is unstable to heat. Lithium ion is smaller in size than sodium ion and thus polarizes the carbonate ion to a greater extent than sodium ion leading to the formation of more stable lithium oxide and carbon dioxide.

(2) Potassium carbonate cannot be prepared by Solvay process
Explanation: Unlike NaHCO₃, the intermediate KHCO₃ formed during reaction in Solvay process, is highly soluble in water and thus cannot be taken out from solution to obtain K₂CO₃.Hence, K₂CO₃ cannot be prepared by Solvay process.

(3) Alkali and alkaline earth metals cannot be obtained by chemical reduction methods
Explanation: Alkali and alkaline earth metals are themselves very strong reducing agents and therefore cannot be reduced by chemical reduction methods

(4) An aqueous solution of carbonates and bicarbonates of alkali metals is alkaline
Explanation: The aqueous solution of carbonates and bicarbonates of alkali metals is alkaline due to the hydrolysis of these salts which produces hydroxide ions.
(HCO₃)^-    +    H₂O   ===>  H₂CO₃   +    OH^-
BiCarbonates
(CO₃)^-     +    H₂O   ===>   HCO₃      +    OH^-
Carbonates

(5) LiCl is more covalent than KCl
Explanation: According to Fajan’s rule smaller the size of cation and larger the size of anion greater is the covalent character of ionic bond. Li is small in size than K, thus Li⁺ has a high charge density. Thus polarizing power of Li⁺ is higher than K⁺, hence LiCl is more covalent than KCl.

(6) BaO is soluble but BaSO₄ is insoluble in water
Explanation: Size of O ^2- ion is smaller than SO₄^2-. Since a bigger anions stabilizes bigger cation more than a smaller cation stabilizes a bigger anion,

10
lattice enthalpy of BaO is smaller than BaSO₄. BaO is soluble as hydration energy is more than lattice energy but BaSO₄ (as hydration energy is less than lattice energy) is insoluble in water.

(7) Solubility of alkaline earth metal carbonates and sulphates in water decrease down the group
Explanation: The size of anions being much larger compared to cations, the lattice enthalpy will remain almost constant within a particular group. Since the hydration enthalpies decrease down the group, solubility will decrease as found for alkaline earth metal carbonates and sulphates.

(8) Alkali metals are not found in nature
Explanation: Alkali metals are highly reactive because of low ionization enthalpy value and therefore are not found in nature. They are present in combined state only in form of halides, oxides etc.

(9) Sodium is less reactive than potassium
Explanation: Ionization Energy of potassium is less than sodium because of large size or less effective nuclear charge. Thus, potassium is more reactive than sodium

(10) NaOH is a stronger base than LiOH
Explanation: The M-OH bond in hydroxides of alkali metal is very weak and can easily ionize to form M ⁺ ions and OH^- ions. This accounts for the basic character. Since ionization energy decreases down the group bond between metal and oxygen becomes weak. Therefore basic strength of hydroxides increases accordingly. Thus NaOH is a stronger base than LiOH

(11) Alkali metals are kept in paraffin or kerosene
Explanation: Alkali metals react explosively with water forming metal hydroxides along with hydrogen. Hydrogen gas released immediately catches fire .Thus alkali metals are highly sensitive towards air and water and hence are kept therefore in kerosene or paraffin oil

(12) Except for Be and Mg, the alkaline earth metal salts impart characteristic colours to the flame
Explanation: Beryllium and magnesium atoms are comparatively smaller and their ionization energies are very high. Hence, the energy of the flame is not sufficient to excite their electrons to higher energy levels. These elements, therefore, do not give any colour in Bunsen flame.