1st PUC Chemistry Question Bank Chapter 11 p-Block Elements

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Karnataka 1st PUC Chemistry Question Bank Chapter 11 p-Block Elements

Question 1.
What are p-block elements?
Answer:
The elements in which the last electron enters into outer most p-orbital are called p-block elements.

Question 2.
Write the general valence shell electronic configuration of p-block elements.
Answer:
ns² np^1-6 (except for He, 1s²)

Question 3.
Discuss the general properties of p-block elements.
Answer:
1. Atomic and ionic radii: Atomic and ionic radii of ‘p’ block elements decrease on moving from left to right and increase from top to bottom in any group.

2. Ionization enthalpy: Ionization enthalpy of ‘p’ block elements increases along a period and decreases down the group.

3. Electron affinity: Along a period, electron affinity of ‘p’ block elements increases from left to right due to decrease in atomic size. Down a group, electron affinity decreases due to increase in atomic size.

4. Electronegativity: Electronegativity of ‘p’ block elements increases along a period and decreases down the group. Decrease is due to the increase in atomic size.

5. Metallic character: Metallic character decreases along a period and increases down a group. The • increase in metallic character down a group is due to increase in atomic size and decrease in ionization
enthalpy. It is interesting to note that the nonmetals and metalloids exist only in the p-block of the periodic table. The non-metallic character decreases down the group.

6. Oxidation states: The ‘p’ block elements exhibit a variety of positive and negative oxidation states. The maximum oxidation state formed by ‘p’ block elements is (X-10) where X is the group number. It is called group oxidation state. The group oxidation state some time attributed to the inert pair effect. The occurrences of oxidation states two unit less than the group oxidation states are sometime attributed to the ‘inert pair effect’.

Group 13 Elements: The Boron Family

Question 1.
Give the occurrence of boron and aluminium.
Answer:
Boron occurs as orthoboric acid (H₃BO₃), borax (Na₂B₄O₇.10H₂O), and kernite (Na₂B₄O₇.4H₂O)
Aluminium occurs as bauxite (Al₂O₃.2H₂O) and cryolite (Na₃AlF₆)
Note: Aluminium is the most abundant metal in the earth’s crust (7.5-8.4%). Aluminum is the third most abundant element in the earth’s crust after oxygen (45.5%) and silicon (27.7%).

Question 2.
Write the valence shell electronic configuration of boron series of elements.
Answer:
ns²np₁

Question 3.
How does atomic radii of group -13 elements vary down the group? Give reason.
Answer:
The atomic and ionic radii of group 13 elements increases on moving down the group. This is due to the addition of new shells.

Question 4.
Atomic radius of Al is slightly more than gallium. Why?
Answer:
In gallium, shielding effect is less due to the presence of additional d¹⁰ electrons in its inner core electronic configuration. Therefore, gallium size is slightly less than Al.

Question 5.
Explain the variation of ionisation enthalpy in group – 13 elements.
Answer:
Ionisation enthalpy generally decreases down the group due to increase in size.
Ionisation enthalpy of gallium is higher than Al. This is because of poor shielding effect due to electrons of inner core electronic configuration.

Question 6.
Electronegativity of group – 13 elements decrease from B to Al then increases. Why?
Answer:
This is because of the discrepancies in atomic size of the elements.

Question 7.
Boron has very high melting point. Why?
Answer:
Due to strong crystalline lattice.

Question 8.
Lithium does not form + 3 ions. Why?
Answer:
This is because of its small size and sum of its first three ionisation enthalpies is very high.

Question 9.
Aluminium does not form + 1 ions. Why?
Answer:
Because sum of the first three ionisation enthalpies of Al is less and it forms only Al³⁺ ions.

Question 10.
The stability of + 1 oxidation state in group – 13 elements increases down the group. Why?
Answer:
Due to inner pair effect.

Question 11.
Boron compounds acts as lewis acids. Explain.
Answer:
In trivalent state, the number of electrons around the central atom in a molecules of the boron compounds will be only six. Hence they have tendency to accept a pair of electrons to achieve stable electronic configuration. Hence they act as Lewis acids.
Example:

Question 12.
Al forms trivalent ion (Al³⁺) but not boron. Explain.
Answer:
Due to small size of boron, the sum of its first three ionization enthalpies is very high. This prevents it to form +3 ions and forces it to form only covalent compounds. But as we move from B to Al, the sum of the first three ionization enthalpies of Al considerably decreases, and is therefore able to form Al³⁺ ions.

Question 13.
Write the dimeric structure of aluminium chloride.
Answer:

Question 14.
What is the hybridisation state of Al in [Al(H₂O)₆]³⁺ ion?
Answer:
sp³d²

Question 15.
Explain the reactivity of group – 13 elements towards air.
Answer:

• Boron does not reacts with air in crystalline form.
• Amorphous boron on heating in air form B₂O₃.
  2B + 3O₂ → B₂O₃
• Aluminium reacts with air to form thin oxide layer on the surface which protects the metal from further attack. This is called passivity.
• Aluminium form Al₂O₃ on heating with air.
  2Al + 3O₂ → Al₂ O₃
• Group – 13 elements react with dinitrogen at high temperature to form nitrides.
  2E + N₂ → 2EN
  where E = B, Al, Ga, In and Tl.

Question 16.
Explain the reactivity of group – 13 elements towards acids and alkalies.
Answer:
Boron does not react with acids and alkalies even at moderate temperature. But aluminium dissolves in mineral acids and aqueous alkalies and shows amphoteric character.
Aluminium dissolves in dilute HCl and liberates dihydrogen.
2Al(s) + 6HCl (aq) → 2Al³⁺ (aq) + 6Cl^– (aq) + 3H₂ (g)
But concentrated nitric acid renders aluminium passive by forming a protective oxide layer on the surface. Aluminium reacts with aqueous alkali to form sodium tetra hydroxo aluminate(III) and liberates dihydrogen.

Question 17.
White fumes appear around the bottle of anhydrous aluminium chloride. Give reason.
Answer:
Anhydrous aluminium chloride is partially hydrolysed with atmospheric moisture to liberate HCl gas. Moist HCl appears as white fumes.

Important Trends and Anomalous Properties of Boron

Question 1.
Why boron shows anomalous behaviour with other elements of the same group.
Answer:
Boron – the first member of group 13, shows anomalous behavior due to

• extremely small size
• high nuclear charge/size ratio (polarizing power)
• high electronegativity and
• non-availability of d orbitals.

Question 2.
Give some anomalous behaviour of boron with respect to other elements of the same group.
Answer:

• Boron is a nonmetal whereas other members are metals.
• Boron is a bad conductor of electricity, whereas other metals are good conductors.
• Boron shows allotropy and exists in two forms – crystalline and amorphous. Aluminum is a soft metal
  and does not exist in different forms.
• Like other non-metals, the melting point and boiling point of boron are much higher than those of other elements of group 13.
• Boron forms only covalent compounds whereas aluminum and other elements of group 13 form even some ionic compounds.
• The hydroxides and oxides of boron are acidic in nature whereas those of others are amphoteric and basic.
• The trihalides of boron (BX₃) exist as monomers. On the other hand, aluminum halides exist as dimers (Al₂X₆).
• Dilute acids have no action on boron but others liberate H₂ from them.
• Boron exhibits maximum covalency of four due to non availability of d-orbitals, while other members exhibit a maximum covalency of six due to availability of d-orbitals, example, [Al(OH)₆]^3-ion.

Question 3.
Boron is unable to form BF₆^3- ion. Explain.
Answer:
Due to non-availability of d orbitals, boron is unable to expand its octet. Therefore, the maximum covalence of boron cannot exceed 4.

Some Important Compounds of Boron

Question 1.
What is the chemical composition of borax?
Answer:
Na₂B₄O₇ 10H₂O

Question 2.
Write the correct formula of borax.
Answer:
Na₂[B₄O₅(OH)₄] . 8H₂O or Na₂B₄O₇ . 10H₂O

Question 3
Complete the following reaction. Na₂B₄O₇ + 7H₂O → 4 NaOH + _______
Answer:
4H₃BO₃

Question 4.
What is the action of heat on borax?
Answer:
On heating, borax loses its water of crystallization and swells up to form a fluffy mass. On further heating, it melts to give a clear liquid which solidifies to a transparent glassy bead consisting of sodium metaborate (NaBO₂) and boric anhydride (B₂O₃).

Question 5.
Explain borax bead test.
Answer:
Borax bead is a mixture of NaBO₂ and B₂O₃. B₂O₃ on heating combines readily with a number of coloured transition metal oxides such as Co, Ni, Cr, Cu, Mn, etc. to form the corresponding metaborates which possess characteristic colours. For example, when borax is heated in a Bunsen burner flame with CoO on a loop of platinum wire, a blue coloured Co(BO₂)₂ bead is formed.

Question 6.
Give the preparation of orthoboric acid (H₃BO₃).
Answer:
Boric acid can be prepared by reacting borax (sodium tetraborate decahydrate) with a mineral acid, such as hydrochloric acid.
Na₂B₄O₇.10H₂O + 2 HCl → 4 B(OH)₃ [or H₃BO₃] + 2 NaCl + 5 H₂O

Question 7.
Give the hydrolysis reaction of orthoboric acid.
Answer:
Boric acid is a Lewis acid which interacts with water molecules to form the tetrahydroxyborate ion.
B(OH)₃ + H₂O ⇌ [B(OH)₄]^– + H⁺

Question 8.
Explain the action of heat on orthoboric acid.
Answer:
When heated above 170°C, it dehydrates, forming metaboric acid (HBO₂).
H₃BO₃ → HBO₂ + H₂O
Metaboric acid melts at 236 °C, and when heated above 300 °C further dehydration takes place leading to the formation of tetraboric acid or pyroboric acid (H₂B₄O₇).
4 HBO₂ → H₂B₄O₇ + H₂O
Further heating leads to boron trioxide,
H₂B₄O₇ → 2 B₂O₃ + H₂O
Student’s illuminator

Question 9.
Write the structure of boric acid.
Answer:

Question 10.
Why is boric acid considered as a weak acid?
Answer:
Because it is not able to release H⁺ ions on its own. It receives OH^– ions from water molecule to complete its octet and in turn releases H⁺ ions.

Question 11.
Give three methods of preparation of diborane (B₂H₆)
Answer:
1. Diborane is prepared by treating boron trifluoride with lithium aluminium hydride in dimethyl ether.
4BF₃ + 3LiAlH₄ → 2B₂H₆ + 3LiAlF₄

2. Industrial method: The industrial synthesis involves the reduction of BF₃ by sodium hydride.
2BF₃ + 6NaH → B₂H₆ + 6NaF

3. Laboratory method: A convenient laboratory method for the preparation of diborane involves the oxidation of sodium borohydride with iodine.
2NaBH₄ + I₂ → B₂H₆ + 2NaI + H₂

Question 12.
What is the action of air on diborane?
Answer:
Diborane catches fire spontaneously upon exposure to air. It bums in oxygen releasing an enormous amount of energy.
B₂H₆ + 3O₂ → B₂O₃ + 3H₂O; Δ_CH° =-1976 kJ mol^-1

Question 13.
How does diborane undergoes hydrolysis? Give equation.
Answer:
Diborane readily undergoes hydrolysis with water to give boric acid.
B₂H₆(g) + 6H₂O(l) → 2B(OH)₃ (aq) + 6H₂ (g)

Question 14.
Give the reaction of diborane with ammonia and other ligands (Lewis bases).
Answer:
The direct reaction of ammonia with diborane gives B₂H₆.2NH₃ which is formulated as [(NH₃)₂BH₂]⁺ [BH₄]^– further heating gives borazine (B₃N₃H₆) (called inorganic benzene due to its ring structure).
2NH₃ + B₂H₆ → [(NH₃)₂BH₂]⁺ [BH₄]^– → 2B₃N₃H₆ + 12H₂
Diborane undergoes cleavage reactions with Lewis bases (L) to give borane adducts. BH₃.L
B₂H₆ + 2NMe₃ → 2BH₃ . NMe₃
B₂H₆ + 2CO → 2BH₃ CO

Question 15.
Write the structure of diborane.
Answer:

Question 16.
How do you prepare lithium and sodium tetra hydridoborates (borohydrides) from diborane?
Answer:
Lithium and sodium tetrahydridoborates, also known as borohydrides, are prepared by the reaction of metal hydrides with B₂H₆ in diethyl ether.
2MH + B₂H₆ → 2M⁺[BH₄]^– (M = Li or Na)

Question 17.
Give two uses of lithium and sodium tetrahydridobarates.
Answer:
Both LiBH₄ and NaBH₄ are used as reducing agents in organic synthesis. They are useful starting materials for preparing other metal borohydrides.

Question 18.
Explain bonding in diborane.
Answer:
In diborane, each B atom uses sp³ hybrids for bonding. Out of the four sp³ hybrids on each B atom, one is without an electron shown in broken lines. The terminal B-H bonds are normal 2-centre-2-electron bonds but the two bridge bonds are 3-centre-2-electron bonds. The 3-centre-2-electron bridge bonds are also referred to as banana bonds.

Uses of Boron and Aluminium and their Compounds

Question 1.
Give uses of boron and its compounds.
Answer:

• Boron is used to dope silicon and germanium semiconductors, modifying their electrical properties.
• Boron oxide (B₂O₃) is used in glassmaking and ceramics.
• Borax (Na₂B₄O₇.10H₂O) is used in making fiberglass, as a cleansing fluid, a water softener, insecticide, herbicide and disinfectant.
• Boric acid (H₃BO₃) is used as a mild antiseptic and as a flame retardant.
• Boron nitride has better thermal and chemical stability; hence boron nitride ceramics are used in high-temperature equipment.
• Boron carbide (B₄C) is used in tank armor and bullet proof vests.

Question 2.
Give uses of aluminum and its compounds.
Answer:
Aluminum is used in

• Transportation (automobiles, aircraft, trucks, railway cars, marine vessels, bicycles etc.)
• Packaging (cans, foil, etc.)
• Alum is used in water treatment .
• Construction (windows, doors, siding, building wire, etc.) .
• Cooking utensils
• Electrical transmission lines for power distribution
• Alnico magnets
• Super purity aluminum in used in electronics and CDs.
• Powdered aluminum is used in paint, and in pyrotechnics such as solid rocket fuels and thermite.
• Alloys like magnalium, duralumin, etc.

Group 14 Elements: The Carbon Family

Question 1.
Give the names of carbon family (group – 14) elements.
Answer:
The carbon family consists of one nonmetal (carbon), two metalloids (silicon and germanium) and two metals (tin and lead).

Question 2.
Give the occurrence of group -14 elements.
Answer:
1. Carbon is the seventeenth most abundant element by mass in the earth’s crust. It is widely distributed in nature in free as well as in the combined state. In elemental state it is available as coal, graphite and diamond and, in combined state it is present as metal carbonates, hydrocarbons and carbon dioxide gas (0.03%) in air. Naturally occurring carbon contains two stable isotopes: ¹²C and ¹³C. In addition to these, third isotope, ¹⁴C is also present. It is a radioactive isotope with half life 5770 years.

2. Silicon is the second (27.7 % by mass) most abundant element on the earth’s crust and is present in nature in the form of silica and silicates.

3. Germanium exists only in traces.

4. Tin occurs mainly as cassiterite, SnO₂ and lead as galena, PbS.

Question 3.
Write the general valence shell electronic configuration of group -14 elements.
Answer:
ns²np²

Question 4.
Give the variations of covalent radii, ionisation enthalpy and electronegativity in group -14 elements.
Answer:
Covalent radii: Covalent radius increases from C to Si but from Si to Pb, a small increase in radius is observed. This is due to the presence of completely filled d and/orbitals in heavier members.

Ionization Enthalpy: The first ionization enthalpy of group 14 members is higher than the corresponding members of group 13 due to the influence of inner core electrons. In general, the ionisation enthalpy decreases down the group. Small decrease in ionization enthalpy from Si to Ge to Sn and slight increase in ionization enthalpy from Sn to Pb is due to poor shielding effect of intervening d and/orbitals and increase in size of the atom.

Electronegativity: The electronegativity of carbon family elements are slightly more than group 13 elements due their small size. The electronegativity values for elements from Si to Pb are almost the same.

Question 5.
Explain oxidation states and trends in chemical reactivity of group -14 elements.
Answer:
1. The common oxidation states exhibited by group 14 elements are +4 and +2. Carbon also exhibits negative oxidation states.

2. Since the sum of the first four ionization enthalpies is very high, compounds in +4 oxidation state are generally covalent in nature.

3. In heavier members the tendency to show +2 oxidation state increases in the sequence Ge < Sn < Pb due to the inability of ns² electrons of valence shell to participate in bonding (Inert pair effect).

4. The relative stabilities of +4 and +2 oxidation states vary down the group. Carbon and silicon mostly show + 4 oxidation states. Germanium forms stable compounds in + 4 state and only few compounds in + 2 states. Tin forms compounds in both oxidation states (Sn in + 2 state is a reducing agent). Lead compounds in +2 state are stable and in +4 state are strong oxidising agents.

Question 6.
How does group – 14 elements reacts with oxygen? Give the properties of their oxides.
Answer:
All group 14 elements when heated in oxygen form two types of oxides; (i) monoxide (MO) and (ii) dioxide (MO₂). SiO exist only at high temperature. Oxides in higher oxidation states are generally more acidic than those in lower oxidation states. The dioxides such as CO₂, SiO₂ and GeO₂ are acidic, whereas SnO₂ and PbO₂ are amphoteric in nature. Among monoxides, CO is neutral, GeO is distinctly acidic whereas SnO and PbO are amphoteric.

Question 7.
Explain the reactivity of group -14 elements towards water.
Answer:
Carbon, silicon and germanium are not affected by water. Tin decomposes steam to form dioxide and dihydrogen gas.
Sn + 2H₂O → SnO + 2H₂
Lead is unaffected by water due to the formation of protective oxide film on its surface.

Question 8.
Explain the reactivity of group – 14 elements towards halogens.
Answer:
Group – 14 elements (except carbon) reacts directly with halogens (X₂) to form halides of the type MX₂ and MX₄ (where X = F, Cl, Br, I)

• Most of the MX₄ are covalent in nature. The central metal atom in these halides undergoes sp³ hybridisation and molecule in tetrahedral shape. SnF₄ and PbF₄ are ionic in nature.
• PbI₄ does not exist because Pb-I bond initially formed during the reaction does not release enough energy to unpair 6s² electrons and excite one of them to higher orbital to have four unpaired electrons around lead atom.
• Ge to Pb are able to form dihalides of the formula MX₂. Stability of dihalides increases down the group.
• Thermal and chemical stability of GeX₄ is more stable than GeX₂ but PbX₂ is more stable than PbX₄.
• Tetrahalides (except CCl₄) are easily hydrolysed by water because the central atom can accommodate the lone pair of electrons from oxygen atom of water molecule in d-orbital.

For example: SiCl₄ undergoes hydrolysis by accepting lone pair of electrons from water molecule in d-orbital of Si, finally leading to the formation of Si(OH)₄.

Question 9.
[SiF₆]^2- is known whereas [SiCl₆]^2- not. Give possible reasons.
Answer:
This is due to

• Six large Cl^– ions cannot be accommodated around Si⁴⁺ due to limitation of its size.
• Interaction between lone pair of Cl^– and Si⁴⁺ is not very strong.

Important Trends and Anomalous behavior of Carbon

Question 1.
Carbon shows anamolous behaviour with other elements of the same group. Give reasons.
Answer:
Carbon – the first member of group 14, shows anomalous behavior from rest of the members due to

• small size
• high electronegativity
• higher ionization enthalpy
• non-availability of d orbitals.

Question 2.
Give the main points of differences of carbon with respect to other elements of the same group.
Answer:
The main points of differences with respect to other elements of the same group are:
1. In carbon, only s and p orbitals are available for bonding and (therefore, it can accommodate only four pairs of electrons around it). This would limit the maximum covalence to four whereas other members can expand their covalence due to the presence of d orbitals.

2. Carbon forms pπ-pπ multiple bonds with itself and (with other atoms of small size and high electronegativity) where as heavier elements do not form pπ – pπ bonds (because their atomic orbitals are too large and diffuse to have effective overlapping).

3. Carbon atoms have the tendency to link with one another through covalent bonds to form chains and rings. This property is called catenation. (This is because C – C bonds are very strong). Catenation property decreases down the group due to their increased size. (The order of catenation is C > > Si > Ge > – Sn. Lead does not show catenation. Due to property of catenation and pπ – pπ bond formation, carbon is able to show allotropic forms).

Allotropes of Carbon

Question 1.
Name three allotropic forms of carbon.
Answer:
Graphite, Diamond and fullerenes.

Question 2.
Explain the structure of diamond.
Answer:
Each carbon in a diamond is bonded to four other carbon atoms making a giant macromolecular array (lattice). As each carbon has four single bonds, it is sp³ hybridised and has tetrahedral bond angles of 109° 28′.

Question 3.
Explain the properties of diamond based on its structure.
Answer:

• Each carbon atom forms four covalent bonds in a very rigid giant covalent structure, which makes diamond the hardest natural substance. This makes diamonds ideal as cutting tools.
• Due to strong covalent bonds in diamond, it is very rigid and has very high melting point.
• it doesn’t conduct electricity because it has no free electrons.

Question 4.
Explain the structure of graphite.
Answer:

In graphite, each carbon atom is covalently bonded to three other surrounding carbon atoms and graphite has a layered structure. Each carbon atom undergoes sp² hybridisation. The bond angle between the carbon atoms is 120°. This structure is made up of flat hexagons. These hexagons are joined together to form two-dimensional flat layers.

Therefore, each layer is a giant molecule, so the bond inside the hexagons within each layer is strong, but the bonds between the layers are weak. This weak bond between layers is called van der Waals’ forces.

Question 5.
Graphite is soft and slippery. Why?
Answer:
The weak bond between layers make the layers can slide over each other. Therefore, graphite is soft and slippery. So it is used as a lubricant.

Question 6.
Graphite is good conductor of electricity. Explain.
Answer:
In graphite, each carbon atom has one outer shell electron that is not used to form covalent bonds. These electrons are delocalised. These electrons can move along the layers from one carbon atom to the next. Hence, Graphite conducts electricity.

Question 7.
Write a note on fullerenes.
Answer:
These are small molecules of carbon in which the giant structure is closed over into spheres of atoms (bucky balls) or tubes (sometimes called nano-tubes). Fullerenes were first isolated from the soot of chimneys and extracted from solvents as red crystals.

The bonding has delocalised pi-molecular orbitals extending throughout the structure and the carbon atoms are a mixture of sp² and sp² hybridised systems.

Fullerenes are insoluble in water but soluble in methyl benzene. They are nonconductors as the individual molecules are only held to each other by weak van der Waal’s forces.

Question 8.
Explain the structure of C₆₀ fullerenes.
Answer:

The smallest fullerene has 60 carbon atoms arranged in pentagons and hexagons like a foot ball. This is called “Buckminster fullerene”. This name comes from the inventor “Richard buck minister Fuller” after the discovery of C₆₀ fullerene.
C₆₀ fullerene consists of 20 six membered rings and 12 five membered rings. All the carbon atoms are equal and they undergo sp² hybridisation. Each carbon atom forms three sigma bonds with other three carbon atoms. The remaining electron is delocalised in molecular orbital. This gives the aromatic character to molecule. C₆₀ also contains double bonds.
C – C single bond length = 143.5pm
C – C double bond length = 138.3pm

Question 9.
Give any three uses of carbon.
Answer:

• Graphite is used as electrodes in batteries and industrial electrodes.
• Carbon black is used in black ink and as filler in automobile tyres,
• Diamond is used jewellary.
• Coke is used as fuel.

Some Important Compounds of Carbon and Silicon

Question 1.
Give the methods of preparation of carbon monoxides.
Answer:
(1) Oxidation of carbon with limited supply of oxygen or air gives carbon monoxide.

(2) Pure carbon monoxide is prepared by the dehydration of formic acid using concentrated H₂SO₄ at 373K.

(3) Carbon monoxide can also be obtained in the form of producer gas and water gas.
(a) Producer gas: It is a mixture of carbon monoxide and nitrogen. It is obtained by passing air through heated coke at 1273K.

(b) Water gas (synthesis gas): It is a mixture of carbon monoxide and hydrogen. The gas is made by passing steam over a red-hot coke.

Question 2.
Carbon monoxide is used as reducing agent in metallurgy. Explain with examples.
Answer:
Carbon monoxide is a strong reducing agent and reduces metal oxides to metals.
For example.
Fe₂O₃ (s) + 3 CO (g) → 2 Fe (S) + 3 CO₂(g)

Question 3.
Explain the poisonous nature of carbon monoxide.
Answer:
Carbon monoxide inhibits the blood’s ability to carry oxygen to body tissues including vital organs such as the heart and brain. When CO is inhaled, it combines with the oxygen carrying hemoglobin of the blood to form carboxy haemoglobin (COHb). This complex is nearly 300 times more stable than haemoglobin. This prevents haemoglobin in the red blood corpuscles from carrying oxygen round the body and ultimately resulting in death.

Question 4.
Carbon monoxide forms metal carbonyls. Explain.
Answer:
Due to the presence of a lone pair on carbon (:C ≡ O:), carbon monoxide molecule acts as a electron pair donor and reacts with certain metals when heated to form metal carbonyls.

Question 5.
Give the methods of preparation of carbon dioxide.
Answer:
(1) CO₂ can be obtained by complete combustion of carbon and carbon containing fuels in excess of air.
C + O₂ → CO₂
CH₄ + 2O₂ → CO₂ + 2H₂O

(2) Laboratory method: In the laboratory, carbon dioxide is prepared by the action of dilute HCl on calcium carbonate.
CaCO₃(s) + 2HCl (aq) → CaCl₂ (aq) + CO₂ (g) + H₂O(1)

Question 6.
Explain the role of carbon dioxide in photosynthesis.
Answer:
Carbon dioxide is one of the most abundant gases (normally 0.03% by volume) in the atmosphere. Carbon dioxide plays an important role in plant and animal processes such as photosynthesis and respiration. Green plants convert carbon dioxide and water into food compounds, such as glucose, and oxygen. This process is called photosynthesis.
The reaction of photosynthesis is as follows:

Plants and animals, convert the food compounds by combining it with oxygen to release energy for growth and other life activities. This is the respiration process, the reverse of photosynthesis.
The respiration reaction is as follows:
C₆H₁₂O₆ + 6O₂ → 6 CO₂ + 6H₂O
Photosynthesis and respiration play an important role in the carbon cycle and are at equilibrium with one another. Photosynthesis dominates during the warmer part of the day and respiration dominates during the colder part of the day.

Question 7.
What is the effect of excess CO₂ in the atmosphere?
Answer:
It causes green house effect.

Question 8.
What is dry ice?
Answer:
Solid carbon dioxide is called dry ice.

Question 9.
Give the uses of carbon dioxide.
Answer:

• Photosynthesis – manufacture of food by plants using light energy.
• Fire extinguisher – carbon dioxide does not support combustion.
• Refrigeration: dry ice is solid carbon dioxide (-78°C) and is used to keep items cold without wetting them. Dry ice turns directly into a gas on heating, and it does not melt forming a liquid.

Question 10.
Write the three dimensional structure of SiO.
Answer:

Question 11.
What are the main sources of silica?
Answer:
Silica occurs in nature as sandstone, silica sand or quartzite.

Question 12.
Give the reaction of silicon dioxide in NaOH solution.
Answer:
Silicon dioxide dissolves in NaOH solution to form sodium silicate.
SiO₂ + 2NaOH → Na₂SiO₃ + H₂O

Question 13.
How does SiO₂ reacts with HF?
Answer:
SiO₂ is attacked by HF to form SiF₄
SiO₂ + 4HF → SiF₄ + 2H₂O
Therefore, HF cannot be stored in glass bottles which contain silica.

Question 14.
Give the uses of silica.
Answer:

• It is used in large amount to form mortar which is a building material.
• It is also used in the manufacture of glass and lenses.
• Silica gel is used as drying agent.
• Amorphous silica is used in chromatography.

Question 15.
What are silicones? Write their general formula.
Answer:
Polymeric organo-silicon compounds containing Si – O – Si bonds are called silicones. These have the general formula (R₂SiO)_n. Where R is CH_3^– group (alkyl substituted) or C₆H_5^– group (aryl substituted).

Question 16.
Write the partial structure of silicones.
Answer:

Question 17.
Give the preparation of silicones.
Answer:
The preparation of silicones is generally carried out by the hydrolysis of dialkyldichlorosilanes (R₂SiCl₂) or diaryldichlorosilanes (Ar₂SiCl₂), which are prepared by passing vapours of RCl or ArCl over silicon at 570 K with copper as a catalyst.

Question 18.
Give any two uses of silicones.
Answer:

• Silicones have been used for making water-proof papers, wools, textiles, wood, etc., after coating these articles with silicones.
• The viscosity of silicones do not change with changes in temperature. Therefore, these are used as all weather lubricants.
• As antifoaming agent in industrial processes.
• As a mould releasing agent in rubber industry and foundry.
• For making body implants in cosmetic surgery due to its inert nature.
• Due to their water repellent nature and high dielectric constant, silicones are used in electrical condensers.

Question 19.
What are silicates?
Answer:
A salt in which the anion contains both silicon and oxygen, especially [SiO₄]^2- ion is called silicate.

Question 20.
Give examples for silicates.
Answer:
Feld spar, zeolites, mica and asbestos.

Question 21.
What is the basic structural unit of silicates?
Answer:
SiO^4-₄

Question 22.
Write the tetrahedral structure of SiO^4-₄.
Answer:

Question 23.
Give examples for man made silicates.
Answer:
Glass and cement

Question 24.
What are zeolites?
Answer:
Zeolites are hydrated alumino-silicate minerals with an “honey comb” structure which can accommodate a wide variety of positive ions, like Na⁺, K⁺, Ca²⁺, Mg²⁺ and others.

Question 25.
Give any two uses of zeolites.
Answer:

1. Zeolites are widely used as a catalyst in petrochemical industries for cracking of hydrocarbons and isomerisation, e.g., ZSM-5 (A type of zeolite) used to convert alcohols directly into gasoline.
2. Hydrated zeolites are used as ion exchangers in softening of “hard” water.