Question:

What are the oxidation states of phosphorus in the following: (i) H₃​PO₃ (ii) PCl₃ (iii) Ca₃P₂ (iv) Na₃PO₄ (v) POF₃?

Answer:

(i) H₃PO₃: Oxidation state of phosphorus is +3

(ii) PCl₃: Oxidation state of phosphorus is +3

(iii) Ca₃P₂: Oxidation state of phosphorus is +5

(iv) Na₃PO₄: Oxidation state of phosphorus is +5

(v) POF₃: Oxidation state of phosphorus is +3

Question:

Explain why inspite of nearly the same electronegativity, nitrogen forms hydrogen bonding while chlorine does not.

Answer:

Step 1: Explain what electronegativity is. Electronegativity is the tendency of an atom to attract electrons towards itself when it is part of a chemical bond.

Step 2: Explain why nitrogen forms hydrogen bonding. Nitrogen forms hydrogen bonding because it has a lone pair of electrons that can be shared with a hydrogen atom. This creates an electrostatic attraction between the two atoms, which is known as a hydrogen bond.

Step 3: Explain why chlorine does not form hydrogen bonding. Chlorine does not form hydrogen bonding because it does not have a lone pair of electrons that can be shared with a hydrogen atom. The electronegativity of chlorine is slightly higher than that of nitrogen, so the chlorine atom will not be able to attract the electrons from the hydrogen atom.

Question:

Write the reactions of F₂ and Cl₂​ with water.

Answer:

1. F₂ + H₂O –> HF + O₂

2. Cl₂ + H₂O –> HCl + HOCl

Question:

Can PCl₅ act as an oxidising as well as a reducing agent? Justify.

Answer:

Answer: Yes, PCl₅ can act as an oxidising as well as a reducing agent. This is because PCl₅ is a Lewis acid, which means it can accept an electron pair from another molecule, thus acting as an oxidising agent. It can also donate an electron pair to another molecule, thus acting as a reducing agent.

Question:

What inspired N. Bartlett for carrying out reaction between Xe and PtF₆?

Answer:

1. Research the background of N. Bartlett and his research interests.

2. Investigate any reports or papers written by N. Bartlett about reaction between Xe and PtF₆.

3. Read any reviews or comments written by other scientists about N. Bartlett’s work on the reaction between Xe and PtF₆.

4. Interview N. Bartlett to find out what inspired him to carry out the reaction between Xe and PtF₆.

Question:

Why does NH₃​ form hydrogen bond but PH₃​ does not?

Answer:

Step 1: Understand the definition of hydrogen bonding. Hydrogen bonding is the electrostatic attraction between two molecules that contain a hydrogen atom bonded to a highly electronegative atom, such as nitrogen, oxygen, or fluorine.

Step 2: Compare the molecular structures of NH₃ and PH₃. NH₃ contains a nitrogen atom bonded to three hydrogen atoms, while PH₃ contains a phosphorus atom bonded to three hydrogen atoms.

Step 3: Understand that nitrogen is more electronegative than phosphorus. This means that the hydrogen atoms in NH₃ are more likely to be attracted to other electronegative atoms, resulting in hydrogen bonding. The hydrogen atoms in PH₃, however, are less likely to be attracted to other electronegative atoms, resulting in no hydrogen bonding.

Step 4: Conclude that NH₃ forms hydrogen bonds because the hydrogen atoms are more likely to be attracted to other electronegative atoms, while PH₃ does not form hydrogen bonds because the hydrogen atoms are less likely to be attracted to other electronegative atoms.

Question:

Why does nitrogen show catenation properties less than phosphorus?

Answer:

Step 1: Catenation is the ability of an element to form bonds with itself.

Step 2: Nitrogen has a small atomic radius and low electronegativity, which makes it difficult for it to form strong bonds with itself.

Step 3: On the other hand, phosphorus has a larger atomic radius and higher electronegativity, which allows it to form strong bonds with itself.

Step 4: Therefore, nitrogen shows catenation properties less than phosphorus.

Question:

List the uses of neon and argon gases.

Answer:

1. Neon: a. Neon is used in advertising signs, as its bright red-orange color is highly visible. b. It is also used in high-voltage indicators and in television tubes. c. Neon is used in cryogenic refrigeration systems. d. It is also used in helium-neon lasers.

2. Argon: a. Argon is used to fill incandescent and fluorescent light bulbs. b. It is also used in arc welding. c. Argon is used in the production of titanium and other reactive elements. d. It is used to create an inert atmosphere for growing silicon and germanium crystals. e. Argon is used in the production of semiconductors and other electronic components.

Question:

Give the resonating structures of NO₂​ and N₂O₅

Answer:

NO₂: 1.

N₂O₅: 1.

Question:

How are xenon fluorides XeF₂, XeF₄ and XeF₆ obtained?

Answer:

1. Xenon fluorides XeF₂, XeF₄, and XeF₆ are typically obtained through the reaction of xenon gas with fluorine gas.

2. The reaction is typically carried out at high temperatures, usually between 200 and 600 degrees Celsius.

3. The reaction yields a mixture of the three xenon fluorides, which must then be separated through fractional distillation.

4. The distillation process relies on the different boiling points of the various xenon fluorides, with XeF₂ boiling at the lowest temperature and XeF₆ boiling at the highest temperature.

Question:

Why do noble gases have comparatively large atomic sizes?

Answer:

1. Noble gases are elements in the group 18 of the periodic table.
2. These elements have full outer shells of electrons, making them relatively stable.
3. As a result, the atoms of noble gases have to increase in size to accommodate the additional electrons.
4. This causes the atomic size of noble gases to be larger than other elements.

Question:

Which aerosols deplete ozone?

Answer:

1. Research the effects of aerosols on the ozone layer.

2. Identify which types of aerosols are known to cause ozone depletion.

3. Learn more about the specific chemical reactions that cause ozone depletion when these aerosols are present in the atmosphere.

Question:

Knowing the electron gain enthalpy values for O→O^- and O→O^2- as −141 and 702 kJ mol^-1 respectively, how can you account for the formation of a large number of oxides having O₂^- species and not O^-?

Answer:

1. The electron gain enthalpy of O→O^- is -141 kJ mol^-1, which is relatively low compared to O→O^2-, which is 702 kJ mol^-1. This means that it is more energetically favorable for oxygen atoms to form O₂^- species rather than O^-.

2. This explains why a large number of oxides contain O₂^- species and not O^-, as it is more energetically favorable to form O₂^- species.

Question:

How is ammonia manufactured industrially?

Answer:

Step 1: Ammonia is manufactured industrially by the Haber-Bosch process.

Step 2: This process involves the reaction of nitrogen gas (N) and hydrogen gas (H₂) at high temperatures and pressures in the presence of a catalyst, usually iron.

Step 3: The reaction produces ammonia (NH₃) which is then purified and collected.

Question:

How can you prepare Cl₂​ from HCl and HCl from Cl₂​? Write reactions only.

Answer:

1. HCl + Cl₂ –> 2HCl

2. HCl –> Cl₂ + H₂

Question:

Explain why NH₃​ is basic while BiH₃​ is only feebly basic.

Answer:

1. NH₃​ is a protonic base, meaning it can accept protons from other molecules. This is because it has a lone pair of electrons that can accept a proton.

2. BiH₃​ is an electron-deficient compound, meaning it does not have a lone pair of electrons that can accept a proton. Therefore, it can only act as a weak base.

Question:

Give the formula and describe the structure of a noble gas species which is isostructural with: (i) ICl4​^- (ii) IBr₂^- (iii) BrO₃​^-

Answer:

(i) Noble gas species isostructural with ICl4^-: XeF₄^2- The structure of XeF₄^2- is square planar with a central Xe atom surrounded by four F atoms arranged at the corners of a square.

(ii) Noble gas species isostructural with IBr₂^-: XeBr₂^2- The structure of XeBr₂^2- is linear with a central Xe atom surrounded by two Br atoms arranged at either end of the linear structure.

(iii) Noble gas species isostructural with BrO₃^-: XeO₃^2- The structure of XeO₃^2- is trigonal planar with a central Xe atom surrounded by three O atoms arranged at the corners of a triangle.

Question:

Nitrogen exists as diatomic molecule and phosphorus as P4​. Why?

Answer:

1. Nitrogen is an element that exists as a diatomic molecule because it has five valence electrons in its outer shell, which allows it to form a covalent bond with another nitrogen atom to form a stable N molecule.

2. Phosphorus, on the other hand, has five valence electrons in its outer shell, but it also has a vacant 3d orbital which allows it to form a more stable P4 molecule, which consists of four phosphorus atoms covalently bonded together.

Question:

The HNH angle value is higher than HPH,HAsH and HSbH angles. Why?

Answer:

1. The HNH angle is an angle between two atoms of hydrogen (H) and one atom of nitrogen (N).

2. The HPH angle is an angle between two atoms of hydrogen (H) and one atom of phosphorus (P).

3. The HAsH angle is an angle between two atoms of hydrogen (H) and one atom of arsenic (As).

4. The HSbH angle is an angle between two atoms of hydrogen (H) and one atom of antimony (Sb).

5. The HNH angle is larger than the HPH, HAsH, and HSbH angles because nitrogen is a larger atom than phosphorus, arsenic, and antimony. This larger atom size causes the HNH angle to be larger than the other angles.

Question:

Explain why fluorine forms only one oxoacid, HOF.

Answer:

Step 1: Begin by explaining what an oxoacid is. An oxoacid is a compound composed of oxygen and hydrogen atoms, along with another atom, such as sulfur, nitrogen, or fluorine.

Step 2: Explain why fluorine forms only one oxoacid. Fluorine has a high electronegativity, meaning it has a strong affinity for electrons. This causes it to form strong bonds with other atoms, such as oxygen, and makes it difficult for fluorine to form multiple oxoacids. Therefore, fluorine forms only one oxoacid, HOF (hydrogen fluoride).

Question:

Write balanced equations for the following: (i) NaCl is heated with sulphuric acid in the presence of MnO₂​. (ii) Chlorine gas is passed into a solution of NaI in water

Answer:

(i) 2NaCl + H₂SO₄ + MnO₂ –> Na2SO₄ + 2HCl + MnSO₄

(ii) Cl₂ + 2NaI + 2H₂O –> 2NaCl + 2HI + H₂O

Question:

Illustrate how copper metal can give different products on reaction with HNO₃​.

Answer:

1. Copper metal reacts with HNO₃ to form copper nitrate, Cu(NO₃)2, and nitric oxide, NO.
2. When copper metal is heated with concentrated HNO₃, it forms copper oxide, CuO, and nitric acid, HNO₃.
3. When copper metal is heated with fuming HNO₃, it forms copper nitrate, Cu(NO₃)2, and nitrogen dioxide, NO₂.
4. When copper metal is heated with a mixture of HNO₃ and H₂SO₄, it forms copper sulfate, CuSO₄, and nitric acid, HNO₃.

Question:

Discuss the general characteristics of Group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionisation enthalpy and electronegativity

Answer:

1. Electronic Configuration: Group 15 elements have 5 valence electrons in their outermost shell. This means they can form three covalent bonds with other elements.

2. Oxidation State: Group 15 elements typically have an oxidation state of -3, which is the most common oxidation state for these elements.

3. Atomic Size: Group 15 elements generally have larger atomic radii than other elements in the same period. This is due to the extra electron in their outermost shell.

4. Ionisation Enthalpy: Group 15 elements generally have high ionisation enthalpy values, which means they are relatively difficult to remove electrons from.

5. Electronegativity: Group 15 elements generally have relatively low electronegativity values compared to other elements in the same period. This is due to the extra electron in their outermost shell, which makes them less likely to attract electrons.

Question:

How is SO₂​ an air pollutant?

Answer:

1. SO₂ is an air pollutant because it is a gas released into the atmosphere from burning fossil fuels, such as coal and petroleum, and from other industrial processes.

2. SO₂ reacts with other compounds in the atmosphere to form sulfuric acid, which is a major component of acid rain.

3. Acid rain can damage vegetation and aquatic ecosystems, corrode buildings and monuments, and cause respiratory problems in humans.

Question:

With what neutral molecule is ClO^- isoelectronic? Is that molecule a Lewis base?

Answer:

Answer: ClO^- is isoelectronic with the chlorine monoxide (ClO) molecule. No, ClO is not a Lewis base because it does not contain a lone pair of electrons.

Question:

Arrange the following in order of property indicated for each set: a. F₂,Cl₂​,Br₂,I₂ - increasing bond dissociation enthalpy. b. HF,HCl,HBr,HI - increasing acid strength. c. NH₃​,PH₃​,AsH₃​,SbH₃​,BiH₃​ - increasing base strength.

Answer:

a. I₂,Br₂,Cl₂​,F₂ b. HI,HBr,HCl,HF c. NH₃​,PH₃​,AsH₃​,SbH₃​,BiH₃​

Question:

Write two uses of ClO₂​.

Answer:

1. ClO₂ is used as a disinfectant in water treatment plants to kill bacteria, viruses, and other microorganisms.

2. ClO₂ is also used in industrial settings to disinfect surfaces and equipment, and to remove odors from the air.

Question:

Give reasons for the following: R₃​P=O exists but R₃​N=O does not, R is an alkyl group?

Answer:

Answer:

1. R₃P=O exists because phosphorus can form a stable covalent bond with oxygen, forming a phosphate group.

2. R₃N=O does not exist because nitrogen cannot form a stable covalent bond with oxygen, so the nitrogen-oxygen bond would be too weak to form a stable structure.

3. R is an alkyl group because it is an organic group composed of carbon and hydrogen atoms. Alkyl groups are commonly found in organic compounds and are characterized by their carbon-hydrogen bonds.

Question:

Give the disproportionation reaction of H₃​PO₃.

Answer:

H₃PO₃ → H₂PO₂^- + H₃PO₄

The disproportionation reaction of H₃PO₃ is:

2H₃PO₃ → H₂PO₂^- + H₃PO₄

Question:

How are XeO₃​ and XeOF₄ prepared?

Answer:

1. XeO₃ is prepared by reacting xenon with ozone in an oxygen-rich environment.

2. XeOF₄ is prepared by reacting xenon with fluorine gas in an oxygen-rich environment.

Question:

Discuss the trends in chemical reactivity of group 15 elements.

Answer:

1. Research the chemical reactivity of group 15 elements.

2. Analyze the data to identify any trends in chemical reactivity.

3. Identify any factors that may be influencing the trends in chemical reactivity.

4. Discuss the trends in chemical reactivity of group 15 elements, including any factors that may be influencing the trends.

Question:

How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions involved.

Answer:

1. Nitrogen can be prepared in the laboratory by the Haber Process, which involves the reaction of nitrogen gas (N) with hydrogen gas (H₂) to form ammonia (NH₃).

Chemical equation: N + 3H₂ → 2NH₃

2. The Haber Process is typically conducted at high temperatures and pressures, and in the presence of a catalyst such as iron (Fe).

Chemical equation: N + 3H₂ → 2NH₃ + Fe

Question:

Why are halogens coloured?

Answer:

1. Halogens are coloured because they absorb light in the visible spectrum.

2. This absorption of light is due to the electronic transitions that occur in the halogen atoms.

3. These electronic transitions cause the electrons to move from one energy level to another, releasing energy in the form of light.

4. The colour of the light emitted depends on the energy of the electrons involved in the transition.

Question:

Why is dioxygen a gas but sulphur a solid?

Answer:

Step 1: Oxygen is composed of two atoms of oxygen, which are held together by a strong covalent bond.

Step 2: Sulphur is composed of eight atoms of sulphur, which are held together by van der Waals forces.

Step 3: The strong covalent bond in oxygen molecules makes it difficult for the molecules to break apart and remain in the gas phase.

Step 4: The weaker van der Waals forces in sulphur molecules allow them to break apart more easily and remain in the solid phase.

Question:

Write main differences between the properties of white phosphorus and red phosphorus.

Answer:

Step 1: Understand the terms ‘white phosphorus’ and ‘red phosphorus’.

Step 2: Research the properties of both white phosphorus and red phosphorus.

Step 3: Compare the properties of both white phosphorus and red phosphorus.

Step 4: Identify the main differences between the properties of white phosphorus and red phosphorus.

Step 5: Write down the main differences between the properties of white phosphorus and red phosphorus.

Question:

Why are halogens strong oxidising agents?

Answer:

1. Halogens are strong oxidising agents because they have a high electron affinity, meaning they are highly electronegative and have a strong tendency to gain electrons.

2. This makes them highly reactive, and they readily form compounds with other elements by transferring electrons.

3. As a result, halogens are able to oxidise other elements by taking electrons away from them, and this is why they are considered strong oxidising agents.

Question:

Why does the reactivity of nitrogen differ from phosphorus?

Answer:

Step 1: Understand the terms “reactivity” and “nitrogen” and “phosphorus”.

Reactivity is a measure of how easily an element or compound interacts with other elements or compounds. Nitrogen is a non-metallic element found in the atmosphere, while phosphorus is a non-metallic element found in minerals.

Step 2: Research the differences between nitrogen and phosphorus.

Nitrogen is a relatively unreactive element due to its strong triple bond, while phosphorus is more reactive due to its ability to form multiple bonds with other elements. Additionally, nitrogen has five electrons in its outer shell, while phosphorus has five valence electrons, making it more likely to form bonds.

Step 3: Analyze the differences between nitrogen and phosphorus.

The differences in reactivity between nitrogen and phosphorus can be attributed to the differences in their electron configuration. Nitrogen’s triple bond and five outer shell electrons make it relatively unreactive, while phosphorus’s ability to form multiple bonds and five valence electrons make it more reactive.

Question:

Justify the placement of O,S,Se,Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation state and hydride formation.

Answer:

1. O,S,Se,Te and Po are placed in the same group of the periodic table because they all have the same outermost electron configuration (ns²np⁴).

2. This common electron configuration allows them to show similar oxidation states and form hydrides. For example, they all show +2 and +6 oxidation states and can form hydrides such as H₂O, H₂S, H₂Se, H₂Te, and H₂Po.

3. Furthermore, the elements in this group can form covalent bonds with other elements by sharing electrons, which is due to their common outermost electron configuration.

Question:

Which one of the following does not exist? (i) XeOF₄ (ii) NeF₂ (iii) XeF₂ (iv) XeF

Answer:

Answer: (i) XeOF₄

Question:

Describe the manufacture of H₂​SO₄​ by contact process?

Answer:

Step 1: The contact process involves the oxidation of SO₂ in the presence of a catalyst.

Step 2: The SO₂ is oxidized to SO₃ using oxygen in the presence of a catalyst.

Step 3: The SO₃ is then reacted with water to form H₂SO₄.

Step 4: The reaction is exothermic and is carried out at high temperatures (around 400-450°C) and pressures (around 6-10 atmospheres).

Step 5: The H₂SO₄ produced is then cooled and purified before being used in various applications.