Question:

How can you explain higher stability of BCl3​ as compared to TlCl3​?

Answer:

1. First, explain the concept of stability. Stability is a measure of how well a molecule can resist changes in its environment, such as changes in temperature, pressure, and other external factors.

2. Next, explain why BCl3 is more stable than TlCl3. BCl3 has a trigonal planar molecular geometry, which is more symmetrical and therefore more stable than the tetrahedral molecular geometry of TlCl3. Additionally, BCl3 has a higher bond dissociation energy, which means it is more stable and less likely to react with other substances.

3. Finally, explain why the differences in molecular geometry and bond dissociation energy lead to higher stability for BCl3. The trigonal planar molecular geometry of BCl3 means that it is more symmetrical and less likely to undergo changes in its environment, while the higher bond dissociation energy of BCl3 means that it is more resistant to reactions with other substances. These two factors together lead to greater stability for BCl3 compared to TlCl3.

Question:

Explain why BF3​ behaves as a lewis acid.

Answer:

Step 1: A Lewis acid is an electron-pair acceptor.

Step 2: BF3 has a vacant orbital that can accept a pair of electrons from a Lewis base.

Step 3: This makes BF3 act as an electron-pair acceptor, and thus a Lewis acid.

Question:

Explain what happens when boric acid is heated.

Answer:

1. Boric acid is a white, odorless powder composed of boron, oxygen, and hydrogen. When heated, boric acid undergoes a chemical reaction in which the boron atoms become ionized and form borate ions. This process is known as thermal dissociation.

2. As the boric acid is heated, the bonds that hold the boron, oxygen, and hydrogen atoms together break apart, releasing energy in the form of heat. This energy is then absorbed by the surrounding molecules, causing them to vibrate faster and faster.

3. As the temperature of the boric acid rises, the molecules vibrate faster and faster, eventually reaching a point where the bonds between the atoms break apart. At this point, the boron atoms become ionized and form borate ions.

4. The borate ions then react with other molecules in the boric acid, forming new compounds such as boric anhydride, boric oxide, and borates. These compounds are more stable than the original boric acid and are less likely to undergo further chemical reactions.

Question:

What are electron deficient compounds ? Are BCl3​ and SiCl4​ electron deficient species? Explain.

Answer:

1. Electron deficient compounds are molecules or ions that have fewer electrons than what is expected for a stable species.

2. Yes, both BCl3​ and SiCl4​ are electron deficient species. This is because they both have an octet of electrons around the central atom, but the central atoms in both molecules have fewer electrons than the number of electrons expected for a stable species. In BCl3​, the central boron atom has only 6 electrons in its valence shell, whereas a stable species would have 8 electrons. Similarly, in SiCl4​, the central silicon atom has only 4 electrons in its valence shell, whereas a stable species would have 8 electrons.

Question:

What is the state of hybridisation of carbon in (a) CO3−2​ (b) diamond (c) graphite?

Answer:

a) CO3−2: The carbon in CO3−2 is sp hybridised.

b) Diamond: The carbon in diamond is sp3 hybridised.

c) Graphite: The carbon in graphite is sp2 hybridised.

Question:

Explain the difference in properties of diamond and graphite on the basis of their structures.

Answer:

Step 1: Explain the structure of diamond and graphite. Diamond is a form of carbon in which each carbon atom is bonded to four other carbon atoms in a tetrahedral arrangement. This creates a strong covalent bond, giving diamond its hardness and strength. Graphite, on the other hand, is made of layers of carbon atoms that are bonded together in a hexagonal lattice. The carbon atoms in graphite are arranged in sheets, with each atom bonded to three other atoms.

Step 2: Explain the differences in properties of diamond and graphite based on their structures. Diamond is a hard and strong material due to its tetrahedral arrangement of carbon atoms, which creates strong covalent bonds. Graphite, on the other hand, is much softer due to its hexagonal lattice structure and the weaker bonds between its carbon atoms. Diamond is also a very good electrical insulator, while graphite is a good conductor of electricity. Additionally, diamond is a very dense material, while graphite is relatively lightweight.

Question:

Explain the following reactions. (a) Silicon is heated with methyl chloride at high temperature in the presence of copper. (b) Silicon dioxide is treated with hydrogen fluoride. (c) CO is heated with ZnO. (d) Hydrated alumina is treated with aqueous NaOH solution.

Answer:

(a) Silicon is heated with methyl chloride at high temperature in the presence of copper. This reaction produces a compound called methyl silicon chloride, which is used in the production of silicone polymers.

(b) Silicon dioxide is treated with hydrogen fluoride. This reaction produces silicofluoride, which is used in the production of silicon-based semiconductors.

(c) CO is heated with ZnO. This reaction produces zinc oxide, which is used in the production of paints, ceramics, and rubber.

(d) Hydrated alumina is treated with aqueous NaOH solution. This reaction produces sodium aluminate, which is used in the production of detergents, soaps, and other cleaning products.

Question:

In some of the reactions thallium resembles aluminium, whereas in others it resembles with group I metals. Support this statement by giving some evidences.

Answer:

1. Thallium is an element in the periodic table which is located in group 13, making it a p-block element.

2. Thallium exhibits properties that are similar to both group I metals and aluminium. For example, thallium is a soft metal which can be cut with a knife and has a low melting point, similar to aluminium.

3. Thallium also exhibits properties similar to group I metals, such as high reactivity and the ability to form cations when it reacts with other elements.

4. Thallium also forms compounds with halogens, which is a characteristic of group I metals.

5. Additionally, thallium has a tendency to form +1 oxidation states, which is another characteristic of group I metals.

6. Lastly, thallium also forms compounds with oxygen, which is a characteristic of aluminium.

Question:

If the starting material for the manufacture of silicones is RSiCl3​, write the structure of the product formed.

Answer:

Answer:

Step 1: RSiCl3 is reacted with water to form siloxanes.

RSiCl3 + H2O → R2SiO + 2HCl

Step 2: The siloxanes then undergo a condensation reaction to form silicones.

R2SiO + R2SiO → R2SiO2SiR2

Question:

Describe the shapes of BF3​ and BH4−​. Assign the hybridisation of boron in these species.

Answer:

BF3: BF3 is a trigonal planar molecule with a bond angle of 120°. The hybridisation of boron in BF3 is sp2.

BH4−: BH4− is a tetrahedral molecule with a bond angle of 109.5°. The hybridisation of boron in BH4− is sp3.

Question:

Write reactions to justify amphoteric nature of aluminium.

Answer:

1. Aluminium reacts with acids to form aluminium salts and release hydrogen gas: Aluminium + Acid → Aluminium Salt + Hydrogen Gas

2. Aluminium also reacts with bases to form aluminium hydroxide and release hydrogen gas: Aluminium + Base → Aluminium Hydroxide + Hydrogen Gas

3. These reactions demonstrate that aluminium is amphoteric, meaning it can react with both acids and bases.

Question:

If BCl bond has a dipole moment, explain why BCl3​ molecule has zero dipole moment.

Answer:

Step 1: BCl3 molecule is made up of three BCl bonds.

Step 2: Each BCl bond has a dipole moment due to the difference in electronegativity between B and Cl.

Step 3: The three dipole moments of the BCl bonds cancel each other out due to their symmetrical arrangement in the BCl3 molecule, resulting in a net dipole moment of zero.

Question:

How is excessive content of CO2​ responsible for global warming ?

Answer:

Step 1: Excessive content of CO2 in the atmosphere can trap heat from the sun, leading to an increase in global temperatures.

Step 2: This increase in global temperatures is known as global warming.

Step 3: The main cause of global warming is the buildup of greenhouse gases, such as carbon dioxide, in the atmosphere.

Step 4: CO2 is the most abundant of these gases, and its excessive content is responsible for trapping more heat in the atmosphere and contributing to global warming.

Question:

Explain why is there a phenomenal decrease in ionization enthalpy from carbon to silicon?

Answer:

Step 1: Ionization enthalpy is the energy required to remove an electron from a neutral atom.

Step 2: Carbon and silicon are both members of the same group in the periodic table, group 14.

Step 3: As we move down the group, the atomic radius of the elements increases and the electrons are farther away from the nucleus. This makes it easier to remove an electron, thus decreasing the ionization enthalpy.

Question:

What do you understand by (a) inert pair effect (b) allotropy and (c) catenation?

Answer:

(a) Inert pair effect refers to the tendency of certain elements in the periodic table to form compounds with a different oxidation state than what is expected. For example, the elements in group 4A (C, Si, Ge, Sn, and Pb) tend to form compounds with an oxidation state of +4, even though their expected oxidation state is +2. This is because the outermost electrons in these elements are more strongly attracted to the nucleus, making them difficult to remove.

(b) Allotropy refers to the ability of an element to exist in different physical forms. For example, carbon can exist in the form of graphite, diamond, and fullerenes.

(c) Catenation refers to the ability of an element to form bonds with itself. Carbon is the most common element to show this property, forming long chains of atoms known as polymers or macromolecules.

Question:

An aqueous solution of borax is : A neutral B amphoteric C basic D acidic

Answer:

A. Neutral

Question:

Elements of group 14 exhibit oxidation sate of: A +4 only B +2 and +4 only C +1 and +3 only D +2 only

Answer:

A) +4 only: Group 14 elements include Carbon (C), Silicon (Si), Germanium (Ge), Tin (Sn), and Lead (Pb). All of these elements can have an oxidation state of +4.

Question:

What happens when, (a) borax is heated strongly, (b) boric acid is added to water, (c) aluminium is treated with dilute NaOH, (d) BF3​ is reacted with ammonia?

Answer:

(a) When borax is heated strongly, it decomposes into sodium metaborate, water, and oxygen.

(b) When boric acid is added to water, it dissolves to form a solution of borate ions.

(c) When aluminium is treated with dilute NaOH, it forms a layer of aluminium hydroxide on the surface of the metal.

(d) When BF3​ is reacted with ammonia, it forms a white solid of the ammonium bifluoride salt.

Question:

Give reasons : (i) Conc. HNO3​ can be transported in aluminium container. (ii) A mixture of dilute NaOH and aluminium pieces is used to open drain. (iii) Graphite is used as lubricant. (iv) Diamond is used as an abrasive. (v) Aluminium alloys are used to make aircraft body. (vi) Aluminium utensils should not be kept in water overnight. (vii) Aluminium wire is used to make transmission cables.

Answer:

(i) Conc. HNO3 can be transported in aluminium container because aluminium is highly resistant to corrosion from acids and does not react with the acid.

(ii) A mixture of dilute NaOH and aluminium pieces is used to open drain because aluminium reacts with NaOH to form hydrogen gas, which can be used to open drains.

(iii) Graphite is used as lubricant because it is an excellent lubricant and does not react with aluminium.

(iv) Diamond is used as an abrasive because it is the hardest material and can be used to cut or shape aluminium.

(v) Aluminium alloys are used to make aircraft body because they are lightweight, strong and corrosion resistant.

(vi) Aluminium utensils should not be kept in water overnight because aluminium is reactive with water and will corrode if left in water for extended periods of time.

(vii) Aluminium wire is used to make transmission cables because it is lightweight, strong and conducts electricity well.

Question:

What are allotropes? Sketch the structure of two allotropes of carbon namely diamond and graphite. What is the impact of structure on physical properties of two allotropes?

Answer:

1. Allotropes are different forms of the same element that have different physical and chemical properties due to differences in their atomic structure.

2. The structure of diamond is a three-dimensional network of carbon atoms, with each carbon atom covalently bonded to four other carbon atoms in a tetrahedral arrangement.

3. The structure of graphite is composed of layers of carbon atoms arranged in a hexagonal lattice with each carbon atom covalently bonded to three other carbon atoms.

4. The impact of structure on physical properties of the two allotropes is that diamond is the hardest natural substance on Earth and graphite is one of the softest known materials. Diamond is an insulator, while graphite is a conductor of electricity. Diamond has a high melting point while graphite has a low melting point.

Question:

(a) Classify following oxides as neutral, acidic, basic or amphoteric: CO,B2​O3​,SiO2​,CO2​,Al2​O3​,PbO2​,Tl2​O3​ (b) Write suitable chemical equations to show their nature.

Answer:

(a) CO: Neutral B2O3: Amphoteric SiO2: Neutral CO2: Acidic Al2O3: Amphoteric PbO2: Acidic Tl2O3: Basic

(b) Neutral: No chemical equation Acidic: CO2 + H2O → H2CO3 Basic: Tl2O3 + H2O → 2Tl(OH) Amphoteric: B2O3 + H2O → B(OH)3 ; Al2O3 + H2O → 2Al(OH)3

Question:

A certain salt X, gives the following results. (i) Its aqueous solution is alkaline to litmus. (ii) It swells up to a glassy material Y on strong heating. (iii) When conc. H2​SO4​ is added to a hot solution of X, white crystal of an acid Z separates out. Write equations for all the above reactions and identify X,Y and Z.

Answer:

(i) X + H2O → Alkaline solution + X- ions

(ii) X → Y + Heat

(iii) X + H2SO4 → Z + H2O

X = Salt Y = Glassy material Z = Acid

Question:

The type of hybridization of boron in diborane is ? A sp - hybridization B sp3 - hybridization C sp2 - hybridization D sp3d2 - hybridization

Answer:

A. sp3 - hybridization

Question:

Discuss the pattern of variation in the oxidation states of (i) B to Ti and (ii) C to Pb.

Answer:

(i) B to Ti: The oxidation states of B and Ti vary in a regular pattern. B starts with an oxidation state of +3 and then decreases to +2, +1, and 0 as it moves up the periodic table to Ti.

(ii) C to Pb: The oxidation states of C and Pb also vary in a regular pattern. C starts with an oxidation state of +4 and then decreases to +3, +2, +1, and 0 as it moves up the periodic table to Pb.

Question:

Is boric acid a protic acid? Explain.

Answer:

Step 1: Boric acid, also known as hydrogen borate, is a weak acid composed of boron and hydrogen.

Step 2: A protic acid is an acid that has a hydrogen atom that is bonded to an electronegative atom, such as oxygen or nitrogen.

Step 3: Boric acid does not have a hydrogen atom that is bonded to an electronegative atom, so it is not a protic acid.

Question:

Write the resonance structures of CO3−2​ and HCO3​−.

Answer:

Resonance structures of CO3−2​:

1. O=C=O
2. O-C-O
3. O-C=O

Resonance structures of HCO3−​:

1. O=C-OH
2. O-C-OH
3. O-C=OH

Question:

Suggest reasons why the BF bond lengths in BF3​ (130 pm) and BF4​− (143 pm) differ.

Answer:

1. BF3 and BF4- are different molecules, so their bond lengths will naturally differ.

2. The BF3 molecule is a trigonal planar molecule, meaning that the three B-F bonds are all of equal length. In contrast, the BF4- molecule is a tetrahedral molecule, meaning that the four B-F bonds are not all of equal length.

3. The B-F bonds in the BF3 molecule are shorter than in the BF4- molecule due to the greater repulsion between the three F atoms in the BF3 molecule, which causes the bond lengths to be shorter.

4. The greater electronegativity of the F atoms compared to the B atom also contributes to the shorter bond length in the BF3 molecule.

Question:

Rationalise the given statements and give chemical reactions. (a) Lead(II) chloride reacts with Cl2​ to give PbCl4​. (b) Lead(IV) chloride is highly unstable towards heat. (c) Lead is known not to form an iodide, PbI4​.

Answer:

a) PbCl2 + Cl2 → PbCl4

b) PbCl4 → PbCl2 + Cl2

c) Pb + I2 → PbI2 (no PbI4 is formed)

Question:

Aluminium trifluoride is insoluble in anhydrous HF but dissolves on addition of NaF. Aluminium trifluoride precipitates out of the resulting solution when gaseous BF3​ is bubbled through. Give reasons.

Answer:

1. Aluminium trifluoride is insoluble in anhydrous HF because the strong electrostatic attraction between the anions and cations in anhydrous HF are too strong to be disrupted by Aluminium trifluoride.

2. Aluminium trifluoride dissolves on addition of NaF because NaF is a weaker base than HF, so it can disrupt the electrostatic attraction between the anions and cations in anhydrous HF.

3. Aluminium trifluoride precipitates out of the resulting solution when gaseous BF3 is bubbled through because BF3 is a stronger acid than HF, so it can form stronger bonds with the anions and cations in the solution than HF, causing Aluminium trifluoride to precipitate out.

Question:

Suggest a reason as to why CO is poisonous.

Answer:

1. Carbon monoxide (CO) is a colorless, odorless, and tasteless gas that is toxic to humans and animals when inhaled.

2. CO is poisonous because it binds to hemoglobin in the blood much more strongly than oxygen does, preventing oxygen from being carried to the body’s tissues and organs. This can cause tissue damage and even death.

3. In addition, CO can also interfere with the normal functioning of the brain, heart, and other organs, leading to dizziness, confusion, and other symptoms that can be fatal if left untreated.

Question:

The reason behind the lower atomic radius of Ga as compared to Al is ? A poor screening effect of d-electrons for the outer electrons from the increased nuclear charge B increased force of attraction of increased nuclear charge on electrons C increased ionisation enthalpy of Ga as compared to Al D Anomalous behaviour of Ga

Answer:

A. The reason behind the lower atomic radius of Ga as compared to Al is B. increased force of attraction of increased nuclear charge on electrons.

Question:

When metal X is treated with sodium hydroxide, a white precipitate (A) is obtained, which is soluble in excess of NaOH to give soluble complex (B). Compound (A) is soluble in dilute HCl to form compound (C). The compound (A) when heated strongly gives (D), which is used to extract metal. Identify (X),(A),(B),(C) and (D). Write suitable equations to support their identities:

Answer:

X= Zinc A= Zinc hydroxide (Zn(OH)2) B= Zincate ion (Zn(OH)42-) C= Zinc chloride (ZnCl2) D= Zinc oxide (ZnO)

Reaction 1: Zn + 2NaOH → Zn(OH)2 + 2Na+

Reaction 2: Zn(OH)2 + 2NaOH → Zn(OH)42- + 2Na+ + 2H2O

Reaction 3: Zn(OH)2 + 2HCl → ZnCl2 + 2H2O

Reaction 4: Zn(OH)2 → ZnO + H2O

Question:

Consider the compounds BCl3​ and CCl4​. How will they behave with water? Justify.

Answer:

1. BCl3​ will react with water to form hydrochloric acid and hydrogen gas. This is because BCl3​ is a Lewis acid and water is a Lewis base, so they will form a hydrogen bond and react.

2. CCl4​ will not react with water. This is because CCl4​ is a nonpolar molecule and water is a polar molecule, so they will not form a hydrogen bond and therefore will not react.

Question:

Explain structures of diborane and boric acid.

Answer:

1. Diborane is a chemical compound composed of two boron atoms and six hydrogen atoms, with molecular formula B2H6. It is a colorless, flammable gas with a strong odor.

2. Boric acid is an inorganic compound composed of boron, hydrogen, and oxygen, with molecular formula H3BO3. It is a white, crystalline solid that is soluble in water.

3. The structure of diborane consists of two boron atoms connected by three bridging hydrogen atoms, forming a trigonal planar shape.

4. The structure of boric acid consists of one boron atom connected to three oxygen atoms and three hydrogen atoms, forming a tetrahedral shape.

Question:

Write balanced equations for: (i) BF3​+LiH→ (ii) B2​H6​+H2​O→ (iii) NaH+B2​H6​→ (iv) H3​BO3​→△ (v) Al+NaOH→ (vi) B2​H6​+NH3​→

Answer:

(i) BF3 + LiH → LiBF3 + H2 (ii) B2H6 + H2O → B2H8 + H2 (iii) NaH + B2H6 → NaBH4 + H2 (iv) H3BO3 → B2H6 + H2O (v) Al + NaOH → NaAlO2 + H2O (vi) B2H6 + NH3 → NH4BH3 + H2

Question:

Give one method for industrial preparation and one for laboratory preparation of CO and CO2​ each.

Answer:

Industrial Preparation of CO:

1. Heat carbon in the presence of oxygen to form carbon monoxide: C + O2 → CO2
2. Separate the carbon dioxide from the carbon monoxide: CO2 + H2 → CO + H2O

Laboratory Preparation of CO:

1. Heat carbon in a test tube in the presence of oxygen: C + O2 → CO2
2. Pass the mixture of carbon dioxide and carbon monoxide through a heated copper tube: CO2 + H2 → CO + H2O

Question:

Boric acid is polymeric due to : A its acidic nature B the presence of hydrogen bonds C its monobasic nature D its geometry

Answer:

A. Boric acid is polymeric due to its acidic nature. B. The presence of hydrogen bonds helps to form a polymeric structure. C. Its monobasic nature further contributes to the formation of a polymeric structure. D. Its geometry also contributes to the formation of a polymeric structure.

Question:

Thermodynamically the most stable form of carbon is : A diamond B graphite C fullerenes D coal

Answer:

A) Diamond. Diamond is the most thermodynamically stable form of carbon because of its strong covalent bonds. It has a higher enthalpy of formation than graphite, fullerenes, and coal.