Group VA and VIA elements Solved Exercise

September 24, 2024 by samreen

Q4. Short questions

i) How does nitrogen differ from other elements of its group?

Answer: Nitrogen differs from other Group 15 elements due to its small size, high electronegativity, and ability to form multiple bonds (N≡N). It shows maximum covalency of four, while heavier elements like phosphorus and arsenic can show higher covalency. Nitrogen also forms a stable diatomic molecule (N₂), whereas other elements do not. Additionally, nitrogen does not have vacant d-orbitals, limiting its ability to expand its octet, unlike other Group 15 elements.

ii) Why does aqua regia dissolve gold and platinum?

Answer: Aqua regia, a mixture of concentrated hydrochloric acid (HCl) and concentrated nitric acid (HNO₃), dissolves gold and platinum due to the formation of chloroauric acid (HAuCl₄) or chloroplatinic acid (H₂PtCl₆). Nitric acid oxidizes the metal to form metal ions, while hydrochloric acid provides chloride ions to stabilize these metal ions in solution as complex ions. The overall reactions for gold and platinum are:
Au + 3HNO₃ + 4HCl → HAuCl₄ + 3NO₂ + 2H₂O
Pt + 4HNO₃ + 6HCl → H₂PtCl₆ + 4NO₂ + 2H₂O

iii) Why do the elements of Group VIA other than oxygen show more than two oxidation states?

Answer: Group VIA elements like sulfur, selenium, and tellurium have access to vacant d-orbitals, allowing them to show higher oxidation states (+4, +6) in addition to the common -2 oxidation state. Oxygen, due to its small size and absence of d-orbitals, is limited to -2 and rarely shows other oxidation states.

iv) Write down a comparison of the properties of oxygen and sulphur.

Answer:
Oxygen is a diatomic gas (O₂) at room temperature, while sulfur exists as a solid (S₈) with puckered rings.
Oxygen is more electronegative (3.44) than sulfur (2.58).
Oxygen forms strong hydrogen bonds in water, making water a liquid, while sulfur does not form hydrogen bonds.
Oxidation States: Oxygen generally shows -2 oxidation states, while sulfur shows a range of oxidation states (-2, +4, +6).
Chemical Reactivity: Oxygen is more reactive than sulfur, forming oxides with almost all elements, while sulfur is less reactive and primarily reacts at higher temperatures.

v) Write down the equation for the reaction between conc. H₂SO₄ and copper and explain what type of reaction it is.

Answer:
The reaction between concentrated sulfuric acid (H₂SO₄) and copper (Cu) is a redox reaction where sulfuric acid acts as an oxidizing agent. The equation is:
Cu + 2H₂SO₄ (conc) → CuSO₄ + SO₂ + 2H₂O
Copper is oxidized to Cu²⁺, and sulfur in H₂SO₄ is reduced from +6 in H₂SO₄ to +4 in SO₂.

Q5.

(a) Explain the Birkeland and Eyde’s process for the manufacture of nitric acid.

Answer: The Birkeland and Eyde process is an older method for producing nitric acid (HNO₃) by oxidizing nitrogen gas (N₂) from the atmosphere to nitric oxide (NO) using an electric arc. The reaction proceeds as follows:
N₂ + O₂ → 2NO
The nitric oxide is further oxidized to nitrogen dioxide (NO₂), which dissolves in water to produce nitric acid:
2NO + O₂ → 2NO₂
3NO₂ + H₂O → 2HNO₃ + NO
This process was replaced by the Ostwald process, which is more efficient.

(b) Which metals evolve hydrogen upon reaction with nitric acid? Illustrate along with chemical equations.

Answer: Most metals do not evolve hydrogen when reacting with nitric acid because nitric acid is a strong oxidizing agent, and it reduces to nitrogen oxides instead. However, metals like magnesium (Mg) and manganese (Mn) can release hydrogen when reacting with dilute nitric acid under specific conditions. For example:
Mg + 2HNO₃ (dil) → Mg(NO₃)₂ + H₂
Mn + 2HNO₃ (dil) → Mn(NO₃)₂ + H₂

(c) What is meant by fuming nitric acid?

Answer: Fuming nitric acid refers to concentrated nitric acid that contains dissolved nitrogen dioxide (NO₂), giving it a red-brown color and producing fumes. It is more corrosive and reactive than regular concentrated nitric acid and is typically used in nitration reactions.

Q6.

(a) Sulphuric acid is said to act as an acid, an oxidizing agent, and a dehydrating agent. Describe two reactions in each case to illustrate the truth of this statement.

Answer:
As an Acid:
1. Reaction with metals:
  Zn + H₂SO₄ → ZnSO₄ + H₂
  Sulphuric acid reacts with metals like zinc, liberating hydrogen gas.
2. Reaction with bases:
  H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
  Neutralization reaction with sodium hydroxide (NaOH).
As an Oxidizing Agent:
1. Reaction with copper:
  Cu + 2H₂SO₄ (conc) → CuSO₄ + SO₂ + 2H₂O
  Concentrated H₂SO₄ oxidizes copper, releasing sulfur dioxide (SO₂).
2. Reaction with carbon:
  C + 2H₂SO₄ (conc) → CO₂ + 2SO₂ + 2H₂O
  Sulfuric acid oxidizes carbon to carbon dioxide (CO₂) and sulfur dioxide (SO₂).
As a Dehydrating Agent:
1. Dehydration of sugar:
  C₁₂H₂₂O₁₁ (sugar) + H₂SO₄ → 12C + 11H₂O
  Sulfuric acid removes water from sugar, leaving behind carbon.
2. Dehydration of ethanol:
  C₂H₅OH → C₂H₄ + H₂O
  Concentrated sulfuric acid dehydrates ethanol to form ethene (C₂H₄).

(b) Give the advantages of the contact process for the manufacture of sulphuric acid.

Answer:

Higher Efficiency: The contact process is highly efficient and capable of producing large quantities of sulfuric acid.
Purity: The acid produced by the contact process is highly pure (around 98% concentration).
Economic Viability: The process is cost-effective due to the recycling of raw materials like sulfur dioxide.
Environmental Benefits: The contact process emits fewer pollutants compared to older methods.

Q.7 (a)

The industrial preparation of sulfuric acid involves the Contact Process. The steps are as follows:

1. Sulfur is burned in the air to produce sulfur dioxide (SO₂).

2. The sulfur dioxide is then oxidized to sulfur trioxide (SO₃) using a vanadium oxide (V₂O₅) catalyst in the presence of excess oxygen.

3. The sulfur trioxide is dissolved in concentrated sulfuric acid to produce oleum (H₂S₂O₇).

4. Finally, oleum is diluted with water to form sulfuric acid (H₂SO₄).

Q.7 (b)

SO₃ is dissolved in H₂SO₄ instead of water because it reacts violently with water, producing a mist of sulfuric acid that is difficult to handle. The reaction with sulfuric acid is more controlled, producing oleum which is later diluted with water.

Q.7 (c)

Sulfuric acid is a strong acid and acts as both an oxidizing agent and a dehydrating agent. It reacts with metals to form metal sulfate and hydrogen gas. For example:

Zn + H₂SO₄ → ZnSO₄ + H₂

Q.8

NO₂ can be prepared by heating concentrated nitric acid with copper:

Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O

NO₂ is a reddish-brown gas and is a major component of air pollution.

Q.9

PCl₃ and PCl₅ are used as chlorinating agents in organic chemistry. They are used to convert alcohols to alkyl chlorides and carboxylic acids to acyl chlorides.

For example:

CH₃CH₂OH + PCl₅ → CH₃CH₂Cl + POCl₃ + HCl

Q.10 (i)

The ‘Ring test’ for nitrates involves adding concentrated sulfuric acid and iron(II) sulfate to a nitrate solution. A brown ring forms at the interface, indicating the presence of nitrate ions.

Q.10 (ii)

NO₂ is a strong oxidizing agent. It can oxidize metals such as copper and non-metals such as carbon:

C + 2NO₂ → CO₂ + 2NO

Q.10 (iii)

When HNO₃ reacts with arsenic and antimony, it forms arsenic and antimony oxides and nitrogen dioxide gas.

For example:

As + 5HNO₃ → H₃AsO₄ + NO₂ + H₂O

Q.10 (iv)

Phosphorus trichloride (PCl₃) is prepared by reacting phosphorus with chlorine:

P₄ + 6Cl₂ → 4PCl₃

Q.10 (v)

Phosphorus pentoxide (P₂O₅) is a powerful dehydrating agent. It can be used to dehydrate acetic acid to ketene:

CH₃COOH → CH₂=C=O + H₂O

Q.11

Complete and balance the following chemical equations:

i) P + NO → P₂O₃

ii) NO + Cl₂ → NOCl

iii) H₂S + NO → S + H₂O + N₂

iv) Pb(NO₃)₂ → PbO + NO₂ + O₂

v) NO₂ + H₂O → HNO₃ + NO

vi) NO + H₂SO₄ → NO₂ + H₂O

vii) HNO₃ + HI → I₂ + H₂O + NO₂

viii) HNO₃ + (COOH)₂ → CO₂ + H₂O + NO₂

ix) KNO₃ + H₂SO₄ → HNO₃ + K₂SO₄

Q.12

Phosphorus pentoxide is prepared by burning white phosphorus in excess oxygen:

P₄ + 5O₂ → P₄O₁₀

Phosphorus pentoxide is a powerful dehydrating agent and is used in organic synthesis for removing water molecules.

Q.13

The Group VIA elements (chalcogens) show the following trends in physical properties:

1. Atomic size increases down the group as the number of electron shells increases.

2. Ionization energy decreases down the group due to increased atomic size and shielding effect.

3. Electronegativity decreases as atomic size increases.

4. Melting and boiling points increase down the group, except for oxygen.

Group IIIA and IVA elements Complete Exercise Solved

September 17, 2024 by samreen

Q6. Why is aluminium not found as a free element? Explain the chemistry of the borax bead test.

Answer: Aluminium is highly reactive and readily combines with oxygen to form aluminium oxide (Al₂O₃), which is a stable compound. This prevents aluminium from being found in a free, metallic state in nature.
The borax bead test is used for detecting metal ions. Borax (Na₂B₄O₇·10H₂O) melts into a clear glassy bead at high temperatures. When certain metal oxides are heated in the bead, they react with the borax to produce characteristic colors. Aluminium doesn’t form distinct colors in this test.

Q7. How does orthoboric acid react with:

(a) Sodium hydroxide
(b) Ethyl alcohol

Answer:
(a) With Sodium Hydroxide (NaOH): Orthoboric acid reacts with NaOH to form sodium borate (Na₂B₄O₇), as shown in the reaction:
B(OH)₃ + NaOH → NaBO₂ + 2H₂O
(b) With Ethyl Alcohol: Orthoboric acid reacts with ethyl alcohol in the presence of concentrated sulfuric acid to form triethyl borate:
B(OH)₃ + 3C₂H₅OH → B(OC₂H₅)₃ + 3H₂O

Q8. How will you convert boric acid into borax and vice versa?

Answer:
Boric acid to borax: Boric acid (H₃BO₃) can be converted into borax by heating it with sodium carbonate (Na₂CO₃), which forms sodium tetraborate (borax) and water:
4H₃BO₃ + Na₂CO₃ → Na₂B₄O₇ + 6H₂O + CO₂
Borax to boric acid: Borax reacts with a strong acid like hydrochloric acid (HCl) to produce boric acid:
Na₂B₄O₇ + 2HCl + 5H₂O → 4H₃BO₃ + 2NaCl

Q9. Why are liquid silicones preferred over ordinary organic lubricants?

Answer: Liquid silicones are preferred over organic lubricants due to their higher thermal stability, resistance to oxidation, chemical inertness, water repellency, and low-temperature fluidity. They also have low vapor pressure, making them suitable for high-temperature applications.

Q10. Explain:

(a) CO₂ is non-polar in nature:

CO₂ has a linear molecular geometry with equal bond dipoles in opposite directions. The symmetry of the molecule causes the dipoles to cancel out, making CO₂ non-polar.
(b) CO₂ is acidic in character:
When CO₂ dissolves in water, it reacts to form carbonic acid (H₂CO₃), which dissociates to release hydrogen ions (H⁺), demonstrating its acidic nature:
CO₂ + H₂O → H₂CO₃ → H⁺ + HCO₃⁻

Q11. Why is CO₂ a gas at room temperature while SiO₂ is a solid?

Answer: CO₂ is a small, simple molecule with weak van der Waals forces between molecules, allowing it to remain in the gaseous state at room temperature. In contrast, SiO₂ forms a giant covalent network structure with strong Si-O bonds, making it a solid with a high melting point.

Q12. Give the names and the formulas of different acids of boron.

Answer:
Boric acid (H₃BO₃)
Tetrahydroxyboric acid (H₄BO₄)
Metaboric acid (HBO₂)

Q13. What is the importance of oxides of lead in paints?

Answer: Lead oxides, such as lead(II) oxide (PbO) and lead(IV) oxide (Pb₃O₄), are used in paints because they enhance durability, increase resistance to corrosion, and provide brilliant color to the pigments. However, due to toxicity, their use is now limited.

Q14. Give the names, electronic configurations, and occurrence of Group-IIIA elements of the periodic table.

Answer: The Group-IIIA elements are:
Boron (B): 1s² 2s² 2p¹
Aluminium (Al): [Ne] 3s² 3p¹
Gallium (Ga): [Ar] 3d¹⁰ 4s² 4p¹
Indium (In): [Kr] 4d¹⁰ 5s² 5p¹
Thallium (Tl): [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p¹
These elements are found in various ores and minerals such as borates, bauxite (Al), and zinc ores (Ga and In).

Q15. Discuss the peculiar behavior of boron with respect to the other members of Group-IIIA elements.

Answer: Boron exhibits several unique properties compared to other Group-IIIA elements (Al, Ga, In, Tl):
Non-metallic Character: Boron is a metalloid, while the others are metals.
Covalency: Boron forms covalent bonds, whereas the other elements can form ionic compounds.
High Ionization Energy: Boron has a much higher ionization energy compared to the heavier Group-IIIA elements.
Lack of Metallic Conductivity: Unlike the metals in its group, boron is a poor conductor of electricity.
Allotropes: Boron exists in various allotropic forms, unlike aluminium and other members.

Q16. (a) What is borax?

(b) Describe its commercial preparation.
(c) Outline the principal uses of borax.
(d) How does borax serve as a water-softening agent?

Answer:
(a) Borax is a sodium salt of boric acid with the formula Na₂B₄O₇·10H₂O.
(b) Commercial Preparation: Borax is prepared by extracting borate minerals from lake beds and then treating them with soda ash (Na₂CO₃) to form borax.
Na₂CO₃ + CaB₄O₇ + H₂O → Na₂B₄O₇ + CaCO₃
(c) Uses of Borax:
Used in glass and ceramic production.
As a flux in metallurgy.
In detergents and soaps as a water softener.
As an antiseptic and preservative.
(d) Water Softening: Borax helps soften water by binding to calcium and magnesium ions, preventing them from interfering with soap action. This allows soap to lather better in hard water.

Q17. (a) What is boric acid?

(b) How is boric acid prepared in the laboratory?
(c) Give properties and uses of boric acid.

Answer:
(a) Boric Acid (H₃BO₃) is a weak monobasic acid derived from boron.
(b) Laboratory Preparation: Boric acid can be prepared by adding hydrochloric acid to a hot, concentrated solution of borax:
Na₂B₄O₇ + 2HCl + 5H₂O → 4H₃BO₃ + 2NaCl
(c) Properties:
It is a white crystalline solid, soluble in water.
Boric acid is a weak acid, antiseptic, and has a low vapor pressure.
Uses:
In antiseptic solutions for minor cuts.
In the manufacture of glass and ceramics.
In insecticides to kill ants and roaches.
As a pH buffer in swimming pools.

Q18. (a) Give the names along with the formulas of three important ores of aluminium.

(b) How and under what conditions does aluminium react with the following:
i) Oxygen ii) Hydrogen iii) Halogens iv) Acids v) Alkalies

Answer:
(a) Ores of Aluminium:
Bauxite: Al₂O₃·2H₂O
Cryolite: Na₃AlF₆
Corundum: Al₂O₃
(b) Reactions of Aluminium:
With Oxygen: Aluminium reacts readily with oxygen, forming a protective layer of aluminium oxide:
4Al + 3O₂ → 2Al₂O₃
With Hydrogen: Aluminium does not react with hydrogen directly.
With Halogens: Aluminium reacts with halogens to form aluminium halides:
2Al + 3Cl₂ → 2AlCl₃
With Acids: Aluminium dissolves in acids to release hydrogen gas:
2Al + 6HCl → 2AlCl₃ + 3H₂
With Alkalies: Aluminium reacts with alkalis to form aluminates and hydrogen gas:
2Al + 2NaOH + 6H₂O → 2Na[Al(OH)₄] + 3H₂

Q19. Give the names, electronic configurations, and occurrence of Group-IVA elements of the periodic table.

Answer: The Group-IVA elements are:
Carbon (C): 1s² 2s² 2p²
Silicon (Si): [Ne] 3s² 3p²
Germanium (Ge): [Ar] 3d¹⁰ 4s² 4p²
Tin (Sn): [Kr] 4d¹⁰ 5s² 5p²
Lead (Pb): [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p²
These elements are found in nature in various forms, such as carbon in coal, silicon in sand (SiO₂), and lead in galena (PbS).

Q20. Discuss the peculiar behavior of carbon with respect to the other members of Group-IVA of the periodic table.

Answer:
Catenation: Carbon uniquely forms long chains and rings of atoms due to its strong C-C bonds. Other Group-IVA elements do not show such extensive catenation.
Allotropes: Carbon exists in various allotropes like graphite, diamond, and fullerenes, whereas silicon and others have fewer allotropic forms.
Multiple Bonds: Carbon forms stable double and triple bonds (C=C and C≡C), unlike heavier elements, which rarely form multiple bonds.
Non-metallic Nature: Carbon is a non-metal, whereas the others in the group (like Sn and Pb) exhibit metallic properties.

Q21. (a) What are silicones?

(b) Give a brief summary of the principal properties of silicones.
(c) Outline the uses of silicones.
(d) What are silicates?
(e) Describe the structures of silicates.

Answer:
(a) Silicones are synthetic polymers made up of repeating units of siloxane (Si-O-Si) with organic side groups attached to the silicon atoms.
(b) Properties of Silicones:
High thermal stability
Water repellency
Electrical insulating properties
Flexibility and low toxicity
(c) Uses of Silicones:
Used as sealants and adhesives
In medical devices and implants
As lubricants and insulating materials
In cosmetics and personal care products
(d) Silicates: Silicates are minerals containing silicon and oxygen, commonly combined with other elements like magnesium, iron, and aluminium. Examples include quartz (SiO₂) and feldspar.
(e) Structures of Silicates: Silicates have a tetrahedral structure, where one silicon atom is surrounded by four oxygen atoms. These tetrahedra can link in various ways to form different types of silicates, such as isolated, chain, sheet, and framework silicates.

Periodic Classification of Elements

September 15, 2024 by samreen

Periodic classification of elements and periodicity (Solved exercise PTB)

Q4:
What are the improvements made in Mendeleev’s periodic table?

Mendeleev left gaps for undiscovered elements and predicted their properties accurately.
He arranged elements based on increasing atomic mass, but later corrections were made based on atomic number.
Grouping of elements with similar chemical properties was better explained by the periodic table developed by Moseley using atomic number.

Q5:
How does the classification of elements in different blocks help in understanding their chemistry?

Elements are classified into s-block, p-block, d-block, and f-block based on their valence electron configurations, which helps in understanding their chemical behavior.
s-block elements are typically metals, p-block elements include non-metals, metalloids, and some metals, d-block elements are transition metals, and f-block contains lanthanides and actinides with complex chemistry.

Q6:
How do you justify the position of hydrogen at the top of various groups?

Hydrogen is placed at the top of Group 1 due to its ability to form a +1 oxidation state like alkali metals, but it can also form a -1 oxidation state (like halogens in Group 17).
Its properties resemble both alkali metals (in forming H⁺ ions) and halogens (in forming H⁻ ions), which justifies its unique position.

Q7:
Why is the ionic radii of negative ions larger than the size of their parent atoms?

When an atom gains an electron to form a negative ion (anion), the repulsion between the electrons increases, causing the electron cloud to expand. This results in the anion having a larger radius than the parent atom.

Q8:
Why does ionization energy decrease down the group and increase along a period?

Down a group: Ionization energy decreases because the atomic size increases, and the outermost electrons are farther from the nucleus, reducing the attraction.
Across a period: Ionization energy increases because atomic size decreases and nuclear charge increases, making it harder to remove an electron.

Q9:
Why is the second value of electron affinity of an element usually shown with a positive sign?

The second electron affinity is positive because after the addition of one electron, the atom becomes negatively charged. Adding another electron to a negative ion requires energy to overcome the repulsion between the added electron and the negative ion.

Q10:
Why does metallic character increase from top to bottom in a group of metals?

As you move down a group, the atomic size increases, and the outermost electrons are farther from the nucleus, making them easier to lose. This enhances the metallic character, which is associated with the ease of losing electrons.

Q11:
Explain the variation in melting points along the short periods.

Melting points tend to increase across a period until reaching Group 14, where they reach a maximum and then decrease. This is due to the increasing strength of metallic bonding or covalent bonding in elements like silicon and then weaker van der Waals forces in non-metals.

Q12:
Why is the oxidation state of noble gases usually zero?

Noble gases have a completely filled valence shell, making them highly stable and inert. They do not easily lose or gain electrons, which is why their oxidation state is generally zero.

Q13:
Why is diamond a non-conductor and graphite a good conductor?

Diamond: In diamond, each carbon atom is tetrahedrally bonded to four other carbon atoms, and all valence electrons are used in bonding. There are no free electrons to conduct electricity.
Graphite: In graphite, each carbon atom is bonded to three other carbon atoms, with one delocalized electron per atom. These free electrons can move through the layers, allowing graphite to conduct electricity.

Q14:
Give a brief reason for the following:

(a) d- and f-block elements are called transition elements.

They are called transition elements because they exhibit properties that are transitional between s- and p-block elements. They have partially filled d- or f-orbitals, which gives them unique chemical and physical properties.

(b) Lanthanide contraction controls the atomic sizes of elements of the 6th and 7th periods.

Lanthanide contraction refers to the steady decrease in atomic and ionic sizes of the lanthanides as atomic number increases. This contraction influences the sizes of elements in the 6th and 7th periods, particularly in d- and f-block elements.

(c) The melting and boiling points of elements increase from left to the middle of the s- and p-block elements and decrease onward.

From left to the middle of the s- and p-block, elements exhibit stronger metallic or covalent bonding, resulting in higher melting and boiling points. After the middle, the elements become non-metals with weaker van der Waals forces, leading to lower melting and boiling points.

(d) The oxidation states vary in a period but remain almost constant in a group.

Across a period, elements can exhibit multiple oxidation states due to the availability of different numbers of valence electrons for bonding. Within a group, the number of valence electrons remains the same, so the oxidation states are more consistent.

(e) The hydration energies of the ions are in the following order: Al³⁺ > Mg²⁺ > Na⁺.

Hydration energy is higher for ions with a greater charge and smaller size. Al³⁺ has the smallest size and highest charge, followed by Mg²⁺, and then Na⁺, leading to the given order.

(f) Ionic character of halides decreases from left to right in a period.

As you move across a period, the electronegativity of the elements increases. The greater the electronegativity difference between the metal and the halide, the higher the ionic character. This difference decreases across a period, reducing the ionic character.

(g) Alkali metals give ionic hydrides.

Alkali metals react with hydrogen to form ionic hydrides (e.g., NaH) where the hydrogen is present as a hydride ion (H⁻) due to the large difference in electronegativity between alkali metals and hydrogen.

(h) Although sodium and phosphorus are present in the same period of the periodic table, their oxides are different in nature: Na₂O is basic while P₄O₁₀ is acidic.

Sodium, being a metal, forms a basic oxide (Na₂O), which reacts with water to produce a strong base (NaOH). Phosphorus, being a non-metal, forms an acidic oxide (P₄O₁₀), which reacts with water to form phosphoric acid (H₃PO₄).

s block elements- Solve Exercise (Punjab Boards)

September 15, 2024September 15, 2024 by samreen

Q4 (a):
Names, electronic configurations, and occurrence of s-block elements:

Elements: The s-block elements include Group 1 (alkali metals: Li, Na, K, Rb, Cs, Fr) and Group 2 (alkaline earth metals: Be, Mg, Ca, Sr, Ba, Ra).
Electronic configurations:
Alkali metals: General configuration is (ns^1).
- Li: (1s² 2s¹), Na: (1s^2 2s^2 2p^6 3s^1), etc.
Alkaline earth metals: General configuration is (ns^2).
- Be: (1s^2 2s^2), Mg: (1s^2 2s^2 2p^6 3s^2), etc.
Occurrence: These elements are found in the Earth’s crust in minerals, salts, and ores (e.g., NaCl, CaCO₃, MgCO₃).

(b):
Peculiar behavior of lithium with respect to other alkali metals:

Lithium has smaller atomic and ionic size compared to other alkali metals.
It forms a stable oxide (Li₂O) and reacts with nitrogen to form lithium nitride (Li₃N).
Lithium salts have more covalent character due to its higher polarization ability.
It is less reactive with water than other alkali metals.
Shows diagonal relationship with magnesium, resembling its properties more.

Q5:
Trends in the chemical properties of oxides, hydroxides, carbonates, nitrates, and sulfates of Group IA (alkali metals) and Group IIA (alkaline earth metals):

Oxides: Alkali metal oxides (e.g., Li₂O) are basic, while alkaline earth metal oxides (e.g., MgO, CaO) are also basic but less soluble.
Hydroxides: Alkali metal hydroxides (e.g., NaOH) are strong bases, while alkaline earth metal hydroxides (e.g., Ca(OH)₂) are less soluble and weaker bases.
Carbonates: Alkali metal carbonates (e.g., Na₂CO₃) are soluble in water, while alkaline earth metal carbonates (e.g., CaCO₃) are insoluble.
Nitrates: Both alkali and alkaline earth nitrates (e.g., NaNO₃, Ca(NO₃)₂) decompose on heating to give oxides, but alkaline earth nitrates decompose more readily.
Sulfates: Alkali metal sulfates (e.g., Na₂SO₄) are soluble in water, while alkaline earth metal sulfates (e.g., BaSO₄) are less soluble.

Q6:
Comparison of chemical behavior of lithium with magnesium:

Both lithium and magnesium form oxides (Li₂O and MgO) and hydroxides (LiOH and Mg(OH)₂).
They form nitrides when reacted with nitrogen (Li₃N, Mg₃N₂).
Lithium and magnesium carbonates decompose on heating to form oxides and CO₂.
Both do not form peroxides or superoxides, unlike other members of their groups.
Both show covalent bonding due to their smaller size and higher ionization energies.

Q7 (a):
Properties of beryllium that make it different from its family members:

Beryllium does not react with water, unlike other alkaline earth metals.
Its salts are generally covalent rather than ionic.
Beryllium forms complexes, e.g., [BeF₄]²⁻.
It has a high melting point and does not show characteristic flame coloration.
Beryllium oxide (BeO) is amphoteric, unlike other Group II metal oxides, which are basic.

Q8 (b):
Why is an aqueous solution of Na₂CO₃ alkaline?

Na₂CO₃ is a salt of a strong base (NaOH) and a weak acid (H₂CO₃). In water, it hydrolyzes to produce OH⁻ ions, making the solution alkaline:
[ Na₂CO₃ + H₂O → 2Na⁺ + CO₃^{2-} ]
[ CO₃^{2-} + H₂O → HCO₃^- + OH⁻ ]

Q9 (a):
Describe with a diagram the manufacture of sodium by Down’s cell:

Down’s cell is an electrolytic process used to manufacture sodium by the electrolysis of molten NaCl.
The cathode is made of iron and the anode is made of graphite.
Na⁺ ions are reduced at the cathode to form sodium metal, and Cl⁻ ions are oxidized at the anode to form chlorine gas:
[ 2NaCl → 2Na + Cl₂ ]

(b):
Three advantages of the Down’s process:

High-purity sodium is obtained.
It is an economical and continuous process.
Chlorine gas, a useful byproduct, is obtained.

Q10:
Compare the physical and chemical properties of alkali metals with those of alkaline earth metals:

Physical properties:
Alkali metals are softer and have lower melting and boiling points than alkaline earth metals.
Alkali metals are more reactive than alkaline earth metals.
Alkali metals have lower densities compared to alkaline earth metals.
Chemical properties:
Alkali metals form monovalent cations (M⁺), whereas alkaline earth metals form divalent cations (M²⁺).
Alkali metals react more vigorously with water than alkaline earth metals.
Alkali metal oxides are more basic than alkaline earth metal oxides.

Here are the answers to the given questions based on the image:

Q9 (b):
What happens when:

Lithium carbonate is heated?

Lithium carbonate (Li₂CO₃) decomposes upon heating to form lithium oxide (Li₂O) and carbon dioxide (CO₂):
[ Li₂CO₃ → heat Li₂O + CO₂ ]

Lithium hydroxide is heated to red heat?

Lithium hydroxide (LiOH) decomposes to form lithium oxide (Li₂O) and water (H₂O):
[ 2LiOH → Li₂O + H₂O ]

Beryllium is treated with sodium hydroxide?

Beryllium reacts with sodium hydroxide (NaOH) to form sodium beryllate (Na₂BeO₂) and hydrogen gas (H₂):
[ Be + 2NaOH + 2H₂O → Na₂BeO₂ + H₂ ]

Lithium hydride is treated with water?

Lithium hydride (LiH) reacts with water to form lithium hydroxide (LiOH) and hydrogen gas (H₂):
[ LiH + H₂O→ LiOH + H₂ ]

Q10:
Give the formulas of the following minerals:

Dolomite:
[ \text{CaMg(CO₃)₂} ]
Asbestos:
[ \text{Mg₃Si₂O₅(OH)₄} ] (Chrysotile, the most common form)
Halite:
[ \text{NaCl} ] (Sodium chloride)
Natron:
[ \text{Na₂CO₃·10H₂O} ] (Sodium carbonate decahydrate)
Beryl:
[ \text{Be₃Al₂Si₆O₁₈} ] (Beryllium aluminum cyclosilicate)
Sylvite:
[ \text{KCl} ] (Potassium chloride)
Phosphorite:
[ \text{Ca₅(PO₄)₃F} ] or [ \text{Ca₃(PO₄)₂} ] (Fluorapatite or phosphate rock)
Chile saltpeter:
[ \text{NaNO₃} ] (Sodium nitrate)

Q11:
Answer the following questions briefly:

(a) Why are alkali and alkaline earth metals among the most reactive elements of the periodic table?

Alkali metals (Group 1) and alkaline earth metals (Group 2) have low ionization energies and readily lose electrons to form cations. Their reactivity increases down the group due to the decreasing ionization energy.

(b) Why does lime water turn milky with CO₂ but becomes clear with excess CO₂?

When CO₂ is bubbled through lime water (Ca(OH)₂), it reacts to form calcium carbonate (CaCO₃), which is insoluble and causes the milky appearance:
[ Ca(OH)₂ + CO₂ &→ CaCO₃ + H₂O ]
On passing excess CO₂, calcium carbonate reacts with water and CO₂ to form soluble calcium bicarbonate (Ca(HCO₃)₂), causing the solution to clear:
[ CaCO₃ + CO₂ + H₂O → Ca(HCO₃)₂ ]

Gypsum (CaSO₄·2H₂O) is heated at around 150°C to 180°C, losing water and converting into plaster of Paris (CaSO₄·½H₂O):
[ CaSO₄·2H₂O →{\text{150°C-180°C}} CaSO₄·½H₂O + 1½H₂O ]

(d) Why is 2% gypsum added in cement?

Gypsum is added to cement to slow down the setting process by controlling the hydration reaction of the cement. It prevents the cement from setting too quickly during the mixing and application phase.

(e) Why is lime added to acidic soil?

Lime (CaO or Ca(OH)₂) neutralizes the excess acidity in soil by reacting with the acids, thus increasing the pH and improving soil fertility.

(f) How are lime and sand used to make glass?

Lime (CaO) and sand (SiO₂) are melted together at high temperatures to form calcium silicate (CaSiO₃), a component of glass. Lime acts as a stabilizer, preventing the glass from being too soluble in water.

(g) How is lime mortar prepared?

Lime mortar is prepared by mixing slaked lime (Ca(OH)₂) with sand and water. When exposed to air, the lime reacts with carbon dioxide (CO₂) to form calcium carbonate, hardening the mortar:
[ Ca(OH)₂ + CO₂ → CaCO₃ + H₂O ]

Periodic Classification of Elements

s block elements- Solve Exercise (Punjab Boards)

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