Question 1

Q1. Why are the elements of Group 15 called p-block elements?

Accepted Answer

The elements of Group 15 (Nitrogen family) are called p-block elements because in their atoms, the last electron enters one of the p-orbitals of their outermost shell. Their general outer electronic configuration is ns²np³.

Question 2

Q2. Discuss the general characteristics of Group 15 elements with reference to their electronic configuration, oxidation states, atomic radii, ionisation enthalpy and electronegativity

Accepted Answer

The elements of Group 15 are Nitrogen (N), Phosphorus (P), Arsenic (As), Antimony (Sb), and Bismuth (Bi). They exhibit a range of properties from non-metallic (N, P) to metalloid (As, Sb) to metallic (Bi). Their general characteristics are discussed below:

### 1. Electronic Configuration

**General Outer Electronic Configuration:** ns²np³

*   This means they have five electrons in their outermost shell (two in s-subshell and three in p-subshell). *   The half-filled p-orbitals (p³) and completely filled s-orbitals (s²) in the valence shell give them extra stability. *   **Specific Configurations:**
    *   Nitrogen (N): [He] 2s²2p³
    *   Phosphorus (P): [Ne] 3s²3p³
    *   Arsenic (As): [Ar] 3d¹⁰4s²4p³
    *   Antimony (Sb): [Kr] 4d¹⁰5s²5p³
    *   Bismuth (Bi): [Xe] 4f¹⁴5d¹⁰6s²6p³

### 2. Oxidation States

*   **Common Oxidation States:** The most common oxidation states are -3, +3, and +5.
*   **-3 Oxidation State:** Due to the presence of three half-filled p-orbitals, these elements can readily accept three electrons to achieve a stable noble gas configuration. The tendency to exhibit -3 oxidation state decreases down the group due to increasing atomic size and metallic character.
    *   Nitrogen forms stable nitrides (e.g., Mg₃N₂).
    *   Bismuth rarely forms compounds in the -3 oxidation state.
*   **+3 Oxidation State:** This is achieved by losing the three p-electrons. The stability of the +3 oxidation state increases down the group due to the 'inert pair effect'. The inert pair effect refers to the reluctance of the outermost s-electrons to participate in bonding.
    *   For example, BiCl₃ is more stable than BiCl₅.
*   **+5 Oxidation State:** This is achieved by losing all five valence electrons (two s and three p electrons). The stability of the +5 oxidation state decreases down the group due to the inert pair effect.
    *   Nitrogen and Phosphorus can form compounds with +5 oxidation state (e.g., N₂O₅, PCl₅).
    *   Bismuth in +5 oxidation state is highly oxidizing and less stable (e.g., BiF₅).
*   **Other Oxidation States:** Nitrogen, being highly electronegative, can exhibit a wide range of oxidation states from -3 to +5 in its compounds (e.g., N₂O (+1), NO (+2), N₂O₃ (+3), NO₂ (+4)).

### 3. Atomic and Ionic Radii

*   **Trend:** Atomic and ionic radii generally increase down the group.
*   **Reason:** This is due to the addition of new electron shells at each successive element.
*   **Anomaly:** There is a small increase in covalent radius from Arsenic to Antimony, and from Antimony to Bismuth. This is attributed to the presence of completely filled d and f orbitals in heavier elements, which causes poor shielding of the nuclear charge and a slightly greater effective nuclear charge than expected, leading to a smaller increase in size.

### 4. Ionisation Enthalpy

*   **Trend:** Ionisation enthalpy generally decreases down the group.
*   **Reason:** This is due to the increasing atomic size and increasing shielding effect of inner electrons, which reduces the attraction between the nucleus and the outermost electrons, making them easier to remove.
*   **Anomaly:** The ionisation enthalpies of Group 15 elements are higher than those of Group 14 elements in the corresponding periods. This is because of the extra stability of the half-filled p-orbitals and smaller size of Group 15 elements.

### 5. Electronegativity

*   **Trend:** Electronegativity generally decreases down the group.
*   **Reason:** This is due to the increasing atomic size and increasing distance of the valence electrons from the nucleus, which reduces the attraction for shared electron pairs.
*   **Values:** Nitrogen is the most electronegative element in the group (3.0 on Pauling scale), while Bismuth is the least electronegative. The decrease in electronegativity is not very regular from Arsenic to Bismuth due to the presence of d and f orbitals.

Question 3

Q3. Why does the reactivity of nitrogen differ from phosphorus?

Accepted Answer

The reactivity of nitrogen differs significantly from phosphorus due to several key factors:

1.  **Bond Enthalpy and Multiple Bonding:** Nitrogen exists as a diatomic molecule (N₂) with a very strong triple bond (N≡N). The bond dissociation enthalpy of N≡N (941.4 kJ/mol) is exceptionally high, making N₂ very unreactive at room temperature. Phosphorus, on the other hand, forms single P-P bonds in its elemental forms (e.g., P₄ tetrahedral structure in white phosphorus). The P-P single bond is much weaker than the N≡N triple bond, making phosphorus more reactive than nitrogen.

2.  **Atomic Size and Electronegativity:** Nitrogen is much smaller and more electronegative than phosphorus. Its small size prevents it from forming stable pπ-pπ multiple bonds with larger atoms like phosphorus does. Phosphorus, being larger, cannot form effective pπ-pπ bonds with itself or other elements, but it can form dπ-pπ or dπ-dπ bonds due to the availability of vacant d-orbitals.

3.  **Availability of d-orbitals:** Nitrogen (2nd period element) does not have vacant d-orbitals in its valence shell. This limits its maximum covalency to 4 (e.g., NH₄⁺). Phosphorus (3rd period element) and other heavier elements of Group 15 have vacant d-orbitals. This allows phosphorus to expand its octet and exhibit a covalency of 5 or 6 (e.g., PCl₅, [PCl₆]⁻), leading to different reaction pathways and product stabilities compared to nitrogen.

4.  **Allotropy:** Phosphorus exists in several allotropic forms (white, red, black phosphorus), each with different reactivities. White phosphorus is highly reactive due to angle strain in the P₄ tetrahedron. Nitrogen exists primarily as the unreactive N₂ molecule.

Question 4

Q4. Discuss the anomalous behaviour of nitrogen.

Accepted Answer

Nitrogen, the first element of Group 15, exhibits anomalous behaviour compared to the other elements of its group (phosphorus, arsenic, antimony, and bismuth). This distinct behaviour is primarily due to its:

1.  **Small Size:** Nitrogen has the smallest atomic and ionic radii in the group.
2.  **High Electronegativity:** It is the most electronegative element in Group 15.
3.  **High Ionisation Enthalpy:** It has the highest ionisation enthalpy in the group.
4.  **Absence of d-orbitals:** Unlike other elements in the group, nitrogen does not have vacant d-orbitals in its valence shell.

These factors lead to the following anomalous properties:

### 1. Ability to form pπ-pπ Multiple Bonds
*   Nitrogen has a unique ability to form stable pπ-pπ multiple bonds with itself (N≡N) and with other small, highly electronegative elements like carbon (C≡N) and oxygen (N=O). This is due to its small size and high electronegativity, which allows for effective sideways overlap of p-orbitals.
*   The N₂ molecule is diatomic and contains a very strong triple bond (N≡N), which makes it highly stable and unreactive at room temperature.
*   Heavier elements of the group (P, As, Sb, Bi) do not form stable pπ-pπ multiple bonds because their atomic orbitals are too large and diffuse to allow for effective sideways overlap. Instead, they prefer to form single bonds (e.g., P₄ exists as discrete tetrahedral molecules with P-P single bonds).

### 2. Maximum Covalency
*   Due to the absence of vacant d-orbitals, nitrogen cannot expand its octet. Its maximum covalency is restricted to four (e.g., in NH₄⁺, where it forms three covalent bonds and one coordinate bond).
*   In contrast, phosphorus and other heavier elements have vacant d-orbitals in their valence shell. This allows them to expand their octet and exhibit a covalency of five (e.g., PCl₅) or even six (e.g., [PCl₆]⁻).

### 3. Formation of Hydrides
*   Nitrogen forms ammonia (NH₃), which is a gas and exhibits hydrogen bonding due to the high electronegativity of nitrogen and the presence of lone pair electrons. This leads to a higher boiling point for NH₃ compared to PH₃.
*   Ammonia is basic due to the availability of a lone pair of electrons on nitrogen, which can accept a proton.
*   Hydrides of other elements (PH₃, AsH₃, SbH₃, BiH₃) are less stable, less basic, and do not exhibit significant hydrogen bonding.

### 4. Reactivity
*   The high bond dissociation enthalpy of the N≡N triple bond makes N₂ relatively inert at room temperature, requiring high temperatures or catalysts to react.
*   Phosphorus, existing as P₄ with weaker P-P single bonds, is much more reactive than nitrogen.

### 5. Allotropy
*   Nitrogen does not exhibit allotropy under normal conditions (exists as N₂ gas).
*   Phosphorus, arsenic, and antimony exist in various allotropic forms (e.g., white, red, black phosphorus), each with distinct properties and reactivities.

Question 5

Q5. What is the covalency of nitrogen in N₂O₅?

Accepted Answer

The covalency of an atom is the number of bonds it forms with other atoms. To determine the covalency of nitrogen in N₂O₅, we need to consider its structure.

N₂O₅ has the following structure:

O
‖
O — N — O — N — O
‖
O

In this structure, each nitrogen atom is bonded to four oxygen atoms. One nitrogen atom forms a double bond with one oxygen, a single bond with another oxygen (bridging oxygen), and a coordinate bond with a third oxygen. The other nitrogen atom forms a double bond with one oxygen, a single bond with the bridging oxygen, and a coordinate bond with a third oxygen.

Counting the bonds formed by each nitrogen atom:
*   One double bond (counts as 2 covalent bonds)
*   One single bond (counts as 1 covalent bond)
*   One coordinate bond (counts as 1 covalent bond)

Total bonds = 2 + 1 + 1 = 4.

Therefore, the covalency of nitrogen in N₂O₅ is **4**.

Question 6

Q6. Why are pentahalides more covalent than trihalides?

Accepted Answer

Pentahalides are more covalent than trihalides due to the higher oxidation state of the central atom in pentahalides. In pentahalides, the central atom (e.g., P in PCl5) is in a +5 oxidation state, whereas in trihalides (e.g., P in PCl3), it is in a +3 oxidation state.

According to Fajan's rules, a higher positive charge on the central metal ion leads to greater polarizing power. This increased polarizing power distorts the electron cloud of the halide ion more effectively, resulting in a greater degree of electron sharing and thus enhanced covalent character. Therefore, the M-X bond in pentahalides has a higher covalent character compared to trihalides.

Question 7

Q7. Why is Bi(V) a stronger oxidising agent than Sb(V)?

Accepted Answer

Bi(V) is a stronger oxidising agent than Sb(V) primarily due to the 'inert pair effect'.

1.  **Inert Pair Effect:** As we move down Group 15, the inert pair effect becomes more prominent. This effect refers to the reluctance of the outermost s-electrons to participate in bonding. For heavier elements like Bismuth (Bi), the +3 oxidation state becomes significantly more stable than the +5 oxidation state due to the poor shielding effect of the intervening d and f electrons, which causes the ns² electrons to be held more tightly by the nucleus.
2.  **Stability of Oxidation States:** For Antimony (Sb), both +3 and +5 oxidation states are relatively stable, with +5 being more common in compounds like SbF5. However, for Bismuth (Bi), the +3 oxidation state is much more stable than the +5 oxidation state. For example, BiCl3 is stable, while BiCl5 is unstable and decomposes readily.
3.  **Oxidising Nature:** An oxidising agent is a substance that gains electrons and gets reduced. Since Bi(V) is highly unstable and readily tends to convert to the more stable Bi(III) state by gaining two electrons, it acts as a strong oxidising agent. Sb(V), being more stable, has a lesser tendency to gain electrons and convert to Sb(III), making it a weaker oxidising agent compared to Bi(V).

Question 8

Q8. Explain why nitrogen forms N2, but phosphorus forms P4.

Accepted Answer

The difference in the allotropic forms of nitrogen and phosphorus can be explained by their atomic size and ability to form multiple bonds:

**1. Nitrogen (N2):**
*   **Small Size and High Electronegativity:** Nitrogen is the first element of Group 15 and has a very small atomic size and high electronegativity.
*   **Ability to form pπ-pπ multiple bonds:** Due to its small size, nitrogen atoms can come close enough to form effective sideways overlap of their p-orbitals, leading to the formation of strong pπ-pπ multiple bonds. A nitrogen atom forms three covalent bonds. In N2, one N-N sigma bond and two N-N pi bonds are formed, resulting in a very stable triple bond (N≡N).
*   **Diatomic Molecule:** This strong triple bond makes the N2 molecule highly stable and unreactive at room temperature. Thus, nitrogen exists as a diatomic gas, N2.

**2. Phosphorus (P4):**
*   **Larger Size and Lower Electronegativity:** Phosphorus is larger than nitrogen and has lower electronegativity.
*   **Weak pπ-pπ overlap:** Due to its larger atomic size, the p-orbitals of phosphorus atoms are more diffuse and cannot overlap effectively sideways to form strong pπ-pπ multiple bonds. The interatomic distance would be too large for effective pi bonding.
*   **Preference for single bonds:** Instead, phosphorus prefers to form strong P-P single bonds. To satisfy its valency of three, each phosphorus atom forms three single covalent bonds with three other phosphorus atoms.
*   **Tetrahedral Structure:** This leads to the formation of a discrete tetrahedral P4 molecule (white phosphorus), where each phosphorus atom is bonded to three other phosphorus atoms. The P-P-P bond angle in P4 is 60°, which introduces some strain in the molecule, but it is still more stable than forming weak pi bonds.

In summary, nitrogen's ability to form strong pπ-pπ multiple bonds leads to the stable N≡N diatomic molecule, while phosphorus's inability to form effective pπ-pπ bonds leads to the formation of P-P single bonds, resulting in the tetrahedral P4 molecule.

Question 9

Q9. Give the disproportionation reaction of H3PO3.

Accepted Answer

Phosphorous acid (H3PO3) undergoes disproportionation when heated. In a disproportionation reaction, an element in a particular oxidation state is simultaneously oxidised and reduced.

In H3PO3, the oxidation state of phosphorus is +3. When heated, it disproportionates to phosphine (PH3), where phosphorus is in the -3 oxidation state (reduced product), and orthophosphoric acid (H3PO4), where phosphorus is in the +5 oxidation state (oxidised product).

**Reaction:**
4H3PO3(l) → 3H3PO4(l) + PH3(g)

**Explanation of Oxidation States:**
*   In H3PO3: Oxidation state of P = +3
*   In H3PO4: Oxidation state of P = +5 (Oxidation)
*   In PH3: Oxidation state of P = -3 (Reduction)

Question 10

Q10. Can PCl5 act as an oxidising as well as a reducing agent? Justify.

Accepted Answer

To determine if PCl5 can act as an oxidising agent, a reducing agent, or both, we need to look at the oxidation state of phosphorus in PCl5 and its possible range of oxidation states.

1.  **Oxidation State of Phosphorus in PCl5:**
    *   Chlorine typically has an oxidation state of -1.
    *   Let the oxidation state of phosphorus be 'x'.
    *   x + 5(-1) = 0
    *   x - 5 = 0
    *   x = +5
    So, phosphorus in PCl5 is in the +5 oxidation state.

2.  **Possible Oxidation States for Phosphorus:**
    *   Phosphorus belongs to Group 15 and has a valence electron configuration of 3s²3p³.
    *   Its common oxidation states range from -3 to +5.
    *   The maximum oxidation state for phosphorus is +5.
    *   The minimum oxidation state for phosphorus is -3.

3.  **Ability to Act as Oxidising or Reducing Agent:**
    *   **As an Oxidising Agent:** Since phosphorus in PCl5 is already in its maximum possible oxidation state (+5), it cannot be further oxidised (cannot lose more electrons). However, it can gain electrons and get reduced to a lower oxidation state (e.g., +3 in PCl3, 0 in P4, or -3 in PH3). Therefore, PCl5 can act as an oxidising agent.
        *   *Example:* PCl5 + H2O → POCl3 + 2HCl (P remains +5, but in other reactions it can reduce)
        *   *Example of reduction:* PCl5 + Sn → PCl3 + SnCl2 (P goes from +5 to +3)
    *   **As a Reducing Agent:** Since phosphorus in PCl5 is already in its maximum possible oxidation state (+5), it cannot be further oxidised (cannot lose more electrons). Therefore, PCl5 cannot act as a reducing agent.

**Conclusion:** PCl5 can only act as an **oxidising agent** because phosphorus is in its maximum oxidation state (+5) and can only undergo reduction to a lower oxidation state. It cannot be oxidised further.

Question 11

Q11. Why are pentahalides more covalent than trihalides?

Accepted Answer

Pentahalides are more covalent than trihalides due to the higher oxidation state of the central atom in pentahalides. According to Fajan's rules, as the positive charge (oxidation state) on the central metal ion increases, its polarizing power increases. This increased polarizing power leads to a greater distortion of the electron cloud of the halide ion, thereby enhancing the covalent character of the bond. For example, in Group 15 elements, the central atom in a pentahalide (e.g., PCl5) is in a +5 oxidation state, while in a trihalide (e.g., PCl3), it is in a +3 oxidation state. The +5 oxidation state has a greater polarizing power than the +3 oxidation state, making pentahalides more covalent.

Question 12

Q12. Why is BiH3 the strongest reducing agent amongst all the hydrides of Group 15 elements?

Accepted Answer

BiH3 is the strongest reducing agent among the hydrides of Group 15 elements (NH3, PH3, AsH3, SbH3, BiH3) because of the decreasing bond dissociation enthalpy of the M—H bond down the group. As we move down Group 15 from nitrogen to bismuth:

1.  **Increase in Atomic Size**: The atomic size of the central atom (M) increases significantly.
2.  **Increase in M—H Bond Length**: Consequently, the M—H bond length increases.
3.  **Decrease in M—H Bond Strength**: Longer bonds are generally weaker bonds. This leads to a decrease in the bond dissociation enthalpy of the M—H bond.

A weaker M—H bond means that it requires less energy to break the bond and release hydrogen. A substance that readily donates hydrogen (or gets oxidized by losing hydrogen) acts as a strong reducing agent. Therefore, BiH3, having the weakest Bi—H bond, can most easily release hydrogen and thus acts as the strongest reducing agent.

Question 13

Q13. Explain the following observations:
(i) Phosphorus has a greater tendency for catenation than nitrogen.
(ii) Dinitrogen is unreactive at room temperature.
(iii) The N—N bond ene

Accepted Answer

Let's explain each observation:

### (i) Phosphorus has a greater tendency for catenation than nitrogen.

**Explanation:** Catenation is the ability of an atom to form bonds with other atoms of the same element, forming chains or rings. While both nitrogen and phosphorus belong to Group 15, phosphorus exhibits a much greater tendency for catenation than nitrogen due to two main reasons:

1.  **Weaker N—N bond vs. Stronger P—P bond**: The N—N single bond is significantly weaker (bond energy ~160 kJ/mol) than the P—P single bond (bond energy ~209 kJ/mol). This weakness in the N—N bond arises from the small size of nitrogen atoms. The lone pair of electrons on each nitrogen atom in an N—N single bond experiences strong interelectronic repulsion due to their close proximity. This repulsion destabilizes the N—N bond.
2.  **Larger size of Phosphorus**: Phosphorus atoms are larger than nitrogen atoms. This larger size reduces the interelectronic repulsion between the lone pairs of electrons on adjacent phosphorus atoms, allowing for the formation of stronger and more stable P—P single bonds. Consequently, phosphorus can form extensive chains and rings (e.g., P4, P8, polymeric phosphorus structures), while nitrogen's catenation is limited to only a few atoms (e.g., N2H4, N3-).

### (ii) Dinitrogen is unreactive at room temperature.

**Explanation:** Dinitrogen (N2) is remarkably unreactive at room temperature due to the exceptionally high bond dissociation enthalpy of the nitrogen-nitrogen triple bond (N≡N). The N≡N bond has a bond energy of approximately 941 kJ/mol, which is one of the highest for any diatomic molecule.

1.  **Strong Triple Bond**: The presence of three covalent bonds (one sigma and two pi bonds) between the two nitrogen atoms makes the molecule very stable.
2.  **High Activation Energy**: Breaking this strong triple bond requires a large amount of energy (high activation energy). At room temperature, the available thermal energy is insufficient to overcome this energy barrier and break the N≡N bond. Therefore, dinitrogen does not readily participate in chemical reactions under normal conditions.

### (iii) The N—N bond energy is less than the P—P bond energy.

**Explanation:** This observation is directly related to the explanation for catenation tendency (part i). The N—N single bond energy (~160 kJ/mol) is indeed less than the P—P single bond energy (~209 kJ/mol) primarily due to:

1.  **Interelectronic Repulsion in Nitrogen**: Nitrogen is a small atom. When two nitrogen atoms form a single bond, the lone pair of electrons on each nitrogen atom are in close proximity. The strong interelectronic repulsion between these lone pairs significantly weakens the N—N single bond.
2.  **Reduced Repulsion in Phosphorus**: Phosphorus atoms are larger. The increased atomic size means that the lone pairs on adjacent phosphorus atoms are further apart, leading to much weaker interelectronic repulsion. This allows for the formation of a stronger and more stable P—P single bond.

### (iv) Sulphur has a greater tendency for catenation than oxygen.

**Explanation:** Similar to the nitrogen-phosphorus comparison, sulphur exhibits a much greater tendency for catenation than oxygen, both belonging to Group 16. This is due to:

1.  **Weaker O—O bond vs. Stronger S—S bond**: The O—O single bond is relatively weak (bond energy ~146 kJ/mol) compared to the S—S single bond (bond energy ~226 kJ/mol). The weakness of the O—O bond is attributed to the small size of oxygen atoms. The lone pairs of electrons on each oxygen atom experience significant interelectronic repulsion due to their close proximity, which destabilizes the O—O bond.
2.  **Larger size of Sulphur**: Sulphur atoms are larger than oxygen atoms. This larger size effectively reduces the interelectronic repulsion between the lone pairs on adjacent sulphur atoms, allowing for the formation of stronger and more stable S—S single bonds. Consequently, sulphur can form extensive chains and rings (e.g., S8 rings, polymeric sulphur), while oxygen's catenation is limited to a few atoms (e.g., O3, H2O2).

Question 14

Q14. How does ammonia react with a solution of Cu2+?

Accepted Answer

When ammonia (NH3) is added to a solution containing Cu2+ ions (e.g., copper(II) sulfate solution), it reacts in two stages:

1.  **Formation of Copper(II) Hydroxide**: Initially, if the ammonia solution is dilute, it acts as a weak base and precipitates copper(II) hydroxide, which is a pale blue precipitate.
    CuSO4(aq) + 2NH4OH(aq) → Cu(OH)2(s) + (NH4)2SO4(aq)
    or, more simply, considering ammonia in water:
    Cu2+(aq) + 2NH3(aq) + 2H2O(l) → Cu(OH)2(s) + 2NH4+(aq)

2.  **Formation of Tetraamminecopper(II) Complex**: Upon adding excess ammonia solution, the precipitated copper(II) hydroxide dissolves to form a deep blue coloured soluble complex ion, tetraamminecopper(II) ion, [Cu(NH3)4]2+.
    Cu(OH)2(s) + 4NH3(aq) → [Cu(NH3)4]2+(aq) + 2OH-(aq)

The overall reaction can be written as:
    Cu2+(aq) + 4NH3(aq) → [Cu(NH3)4]2+(aq)

In this reaction, ammonia acts as a ligand, donating its lone pair of electrons to the central Cu2+ ion to form a coordination complex. The formation of the deep blue colour is a characteristic test for Cu2+ ions.

Question 15

Q15. What is the covalency of nitrogen in N2O5?

Accepted Answer

The covalency of an atom is defined as the number of covalent bonds (including coordinate bonds) formed by that atom. To determine the covalency of nitrogen in N2O5, we need to consider its structure.

The structure of N2O5 is:
O=N—O—N=O
  ↓     ↓
  O     O

Each nitrogen atom in N2O5 is bonded to four oxygen atoms:
1.  One N—O single bond (to the bridging oxygen).
2.  One N=O double bond.
3.  One N→O coordinate bond (dative bond).

Counting these bonds, each nitrogen atom forms a total of **four** bonds. Therefore, the covalency of nitrogen in N2O5 is **4**.

Question 16

Q16. Write the structures of the following compounds:
(i) H3PO3
(ii) H2S2O7
(iii) XeF4
(iv) HClO4

Accepted Answer

The structures of the given compounds are as follows:

**(i) H3PO3 (Phosphorous acid)**
Central atom: Phosphorus (P)
Hybridization: sp³
Geometry around P: Tetrahedral
Bonds: One P=O double bond, two P-OH single bonds, and one P-H single bond.
Basicity: 2 (diprotic acid, as only H atoms attached to oxygen are ionizable).

**(ii) H2S2O7 (Pyrosulphuric acid or Oleum)**
Central atoms: Two Sulphur (S) atoms linked by an oxygen atom (S-O-S linkage).
Hybridization: sp³ for each S atom.
Geometry around each S: Tetrahedral.
Bonds: Each S atom is bonded to two oxygen atoms via double bonds (S=O), one oxygen atom via a single bond (S-O-S bridge), and one hydroxyl group (S-OH).

**(iii) XeF4 (Xenon Tetrafluoride)**
Central atom: Xenon (Xe)
Hybridization: sp³d²
Geometry: Square planar (due to two lone pairs occupying axial positions, minimizing repulsion with bond pairs).
Bonds: Four Xe-F single bonds.

**(iv) HClO4 (Perchloric acid)**
Central atom: Chlorine (Cl)
Hybridization: sp³
Geometry around Cl: Tetrahedral.
Bonds: One Cl-OH single bond and three Cl=O double bonds.
Acidity: Strongest oxyacid of chlorine.

Question 17

Q17. How will you prepare Cl2 from HCl and HCl from Cl2? Write reactions only.

Accepted Answer

Here are the reactions for the preparation of Cl2 from HCl and HCl from Cl2:

**1. Preparation of Cl2 from HCl:**
Chlorine (Cl2) can be prepared from hydrochloric acid (HCl) by oxidation, as HCl contains chloride ions (Cl⁻) which can be oxidized to Cl2.

*   **By heating with a strong oxidizing agent (e.g., MnO2):**
    MnO2(s) + 4HCl(aq) → MnCl2(aq) + Cl2(g) + 2H2O(l)

*   **Deacon's Process (catalytic oxidation):**
    4HCl(g) + O2(g)  --(CuCl2 catalyst, 723 K)-->  2Cl2(g) + 2H2O(g)

*   **By heating with Potassium Permanganate (KMnO4):**
    2KMnO4(s) + 16HCl(aq) → 2KCl(aq) + 2MnCl2(aq) + 5Cl2(g) + 8H2O(l)

*   **Electrolytic Process (from aqueous HCl):**
    2HCl(aq)  --(electrolysis)-->  H2(g) + Cl2(g)

**2. Preparation of HCl from Cl2:**
Hydrochloric acid (HCl) can be prepared from chlorine (Cl2) by reacting it with hydrogen.

*   **Direct combination of Hydrogen and Chlorine:**
    H2(g) + Cl2(g)  --(diffused sunlight / UV light)-->  2HCl(g)

Question 18

Q18. What is the basicity of H3PO4?

Accepted Answer

The basicity of an acid is defined as the number of ionizable hydrogen atoms (protons) that it can furnish per molecule in an aqueous solution. For oxyacids, these are typically the hydrogen atoms directly bonded to oxygen atoms (O-H bonds).

In the case of H3PO4 (Orthophosphoric acid), its structure is:

O
|| 
H—O—P—O—H
| 
O—H

From the structure, it is clear that all three hydrogen atoms in H3PO4 are bonded to oxygen atoms (P-O-H bonds). Therefore, all three hydrogen atoms are ionizable.

Thus, the basicity of H3PO4 is **3**.

Question 19

Q19. What happens when H3PO3 is heated?

Accepted Answer

When H3PO3 (phosphorous acid) is heated, it undergoes a disproportionation reaction. In this reaction, phosphorus in H3PO3 (oxidation state +3) is simultaneously oxidized to a higher oxidation state (+5 in H3PO4) and reduced to a lower oxidation state (-3 in PH3).

The balanced chemical equation for the reaction is:

**4H3PO3(l)  --(heat)-->  3H3PO4(l) + PH3(g)**

**Explanation:**
*   **H3PO3:** Oxidation state of P is +3.
*   **H3PO4 (Orthophosphoric acid):** Oxidation state of P is +5 (oxidation product).
*   **PH3 (Phosphine):** Oxidation state of P is -3 (reduction product).

This reaction is a classic example of a disproportionation reaction, where a single element in an intermediate oxidation state is both oxidized and reduced.

Question 20

Q20. Account for the following:
(i) SF6 is known but SCl6 is not.
(ii) H3PO3 is a diprotic acid.
(iii) Noble gases have very low boiling points.

Accepted Answer

Let's account for each of the given statements:

### (i) SF6 is known but SCl6 is not.

This can be explained by considering two main factors: steric hindrance and electronic factors.

1.  **Steric Hindrance:**
    *   Fluorine (F) is a much smaller atom compared to chlorine (Cl). Six fluorine atoms can easily be accommodated around a central sulfur atom (which is also relatively small) without significant steric repulsion, forming a stable SF6 molecule with octahedral geometry.
    *   Chlorine atoms are significantly larger than fluorine atoms. Attempting to accommodate six large chlorine atoms around a relatively small sulfur atom would lead to severe steric repulsion between the bulky chlorine atoms. This steric crowding makes the formation of SCl6 highly unfavorable and unstable.

2.  **Electronic Factors (Electronegativity):**
    *   Fluorine is the most electronegative element. Its high electronegativity allows it to effectively withdraw electron density from the sulfur atom, stabilizing the high +6 oxidation state of sulfur in SF6.
    *   Chlorine is less electronegative than fluorine. While it can stabilize higher oxidation states, its ability is not as pronounced as fluorine's. However, the primary reason for the non-existence of SCl6 is steric hindrance, as sulfur can form SCl4, indicating it can achieve a +4 oxidation state with chlorine.

Therefore, the combination of the large size of chlorine atoms leading to steric hindrance around the sulfur atom, along with the relatively lower electronegativity of chlorine compared to fluorine, prevents the formation of SCl6, while SF6 is a stable and well-known compound.

### (ii) H3PO3 is a diprotic acid.

The basicity (or proticity) of an acid is determined by the number of ionizable hydrogen atoms present in its molecule. For oxyacids, only the hydrogen atoms directly bonded to oxygen atoms (O-H bonds) are acidic and ionizable.

Let's look at the structure of H3PO3 (Phosphorous acid):

O
||
H—O—P—O—H
|
H

In the structure of H3PO3:
*   There are two P-O-H bonds. The hydrogen atoms in these bonds are acidic and can be released as protons (H⁺) in an aqueous solution.
*   There is one P-H bond. The hydrogen atom directly bonded to phosphorus is not acidic and does not ionize in water because the P-H bond is covalent and non-polar, making it difficult to break and release H⁺.

Since only two hydrogen atoms are attached to oxygen and are ionizable, H3PO3 can furnish two protons per molecule. Hence, H3PO3 is a **diprotic acid**.

### (iii) Noble gases have very low boiling points.

Noble gases (Group 18 elements like He, Ne, Ar, Kr, Xe, Rn) are monoatomic and exist as individual atoms. Their atoms have completely filled valence electron shells, making them extremely stable and unreactive. This stability means they do not form chemical bonds with other atoms (including other noble gas atoms) under normal conditions.

The only forces of attraction that exist between noble gas atoms are very weak intermolecular forces, specifically **London dispersion forces (or instantaneous dipole-induced dipole forces)**.

*   **Nature of London Dispersion Forces:** These forces arise from temporary fluctuations in the electron distribution around an atom, creating instantaneous dipoles. These instantaneous dipoles then induce dipoles in neighboring atoms, leading to weak, short-lived attractive forces.
*   **Weakness of Forces:** London dispersion forces are the weakest type of intermolecular forces. They are significant only at very short distances and are much weaker than covalent or ionic bonds.
*   **Monoatomic Nature:** Since noble gases are monoatomic, there are no other stronger intermolecular forces like dipole-dipole interactions or hydrogen bonding present.

Due to these extremely weak London dispersion forces, very little energy is required to overcome these attractions and separate the atoms from each other to transition from the liquid to the gaseous state. Consequently, noble gases have exceptionally low boiling points, often close to absolute zero (e.g., Helium has the lowest boiling point of any element at 4.2 K or -268.9 °C).

Question 21

Q21. How is the manufacturing of sulphuric acid by contact process advantageous over the lead chamber process?

Accepted Answer

The Contact Process is significantly more advantageous than the Lead Chamber Process for the manufacturing of sulphuric acid due to several factors related to the purity, concentration, yield, and environmental impact of the product. The Lead Chamber Process, an older method, produced a less concentrated and less pure acid, often contaminated with oxides of nitrogen. In contrast, the Contact Process, which utilizes a vanadium pentoxide catalyst, yields highly concentrated and pure sulphuric acid, making it suitable for a wider range of industrial applications.

**Advantages of Contact Process over Lead Chamber Process:**

1.  **Purity of Acid:** The Contact Process produces sulphuric acid of very high purity (typically 98-99%). The Lead Chamber Process, on the other hand, yields acid that is often contaminated with impurities like lead sulfate and oxides of nitrogen, requiring further purification for many applications.

2.  **Concentration of Acid:** The Contact Process can produce highly concentrated sulphuric acid (up to 98% or even oleum, H₂S₂O₇). The Lead Chamber Process typically yields acid with a maximum concentration of about 78-80%, which limits its utility for applications requiring concentrated acid.

3.  **Yield:** The Contact Process offers a much higher yield of sulphuric acid (typically over 99%) due to the efficient catalytic conversion of SO₂ to SO₃. The Lead Chamber Process generally has a lower yield, making it less efficient.

4.  **Environmental Impact:** The Contact Process is generally more environmentally friendly. The Lead Chamber Process releases significant amounts of nitrogen oxides (NOx) into the atmosphere, contributing to air pollution and acid rain. The Contact Process has better control over emissions.

5.  **Cost-Effectiveness:** Although the initial setup cost for the Contact Process might be higher, its higher yield, purity, and concentration of the product, along with lower operational costs (less waste, no need for extensive purification), make it more cost-effective in the long run.

6.  **Versatility:** The highly concentrated and pure acid produced by the Contact Process is suitable for a vast array of industrial applications, including the manufacture of fertilizers, detergents, dyes, explosives, and in petroleum refining. The lower concentration and purity of acid from the Lead Chamber Process restrict its applications.

Question 22

Q22. What are the important characteristics of group 15 elements?

Accepted Answer

Group 15 elements, also known as the Pnictogens, include Nitrogen (N), Phosphorus (P), Arsenic (As), Antimony (Sb), and Bismuth (Bi). They exhibit a wide range of properties, transitioning from non-metallic to metallic character down the group. Here are their important characteristics:

**1. Electronic Configuration:**
*   The general outer electronic configuration for Group 15 elements is ns²np³. This indicates that they have five valence electrons. The presence of half-filled p-orbitals (p³) makes their electronic configuration exceptionally stable, contributing to their high ionization enthalpies and reluctance to form ions easily.

**2. Atomic and Ionic Radii:**
*   **Atomic Radii:** Atomic radii increase down the group from N to Bi. This is due to the addition of new electron shells with each successive element. However, there is a small increase in atomic radius from As to Sb and from Sb to Bi, which is attributed to the poor shielding effect of d- and f-electrons in heavier elements.
*   **Ionic Radii:** Ionic radii (e.g., for M³⁻ or M³⁺ ions) also increase down the group due to the same reason of increasing number of electron shells.

**3. Ionisation Enthalpy:**
*   Ionisation enthalpy generally decreases down the group from N to Bi. This is because the atomic size increases, and the outermost electrons are further away from the nucleus, experiencing less nuclear attraction. However, Group 15 elements have higher ionization enthalpies compared to Group 14 elements in the same period due to their stable half-filled p-orbitals and smaller size.

**4. Electronegativity:**
*   Electronegativity decreases down the group from N to Bi. Nitrogen is the most electronegative element in the group. This trend is due to the increasing atomic size and increasing distance of valence electrons from the nucleus, which reduces the attraction for shared electron pairs.

**5. Metallic Character:**
*   There is a gradual increase in metallic character down the group. Nitrogen and Phosphorus are non-metals. Arsenic and Antimony are metalloids (exhibiting properties of both metals and non-metals). Bismuth is a typical metal. This trend is consistent with the decrease in ionization enthalpy and electronegativity down the group.

**6. Oxidation States:**
*   The common oxidation states exhibited by Group 15 elements are -3, +3, and +5.
    *   **-3 Oxidation State:** This is observed for elements like Nitrogen (e.g., NH₃) and Phosphorus (e.g., PH₃) when they react with more electropositive elements. The tendency to exhibit -3 oxidation state decreases down the group due to increasing metallic character and increasing atomic size, making it difficult to accommodate three extra electrons.
    *   **+3 and +5 Oxidation States:** These are the most common positive oxidation states. The stability of the +5 oxidation state decreases down the group, while the stability of the +3 oxidation state increases down the group. This is due to the 'inert pair effect', where the ns² electrons become increasingly reluctant to participate in bonding for heavier elements (As, Sb, Bi). For example, Bi(V) compounds are strong oxidising agents, while Bi(III) compounds are stable.

**7. Allotropy:**
*   All elements of Group 15, except Bismuth, exhibit allotropy.
    *   **Nitrogen:** Exists as N₂ (diatomic gas).
    *   **Phosphorus:** Exists in several allotropic forms, most notably white, red, and black phosphorus.
    *   **Arsenic:** Exists as yellow, grey, and black arsenic.
    *   **Antimony:** Exists as yellow, explosive, and metallic antimony.

**8. Chemical Reactivity:**
*   **Towards Hydrogen (Formation of Hydrides, EH₃):** All elements form hydrides of the type EH₃ (e.g., NH₃, PH₃, AsH₃, SbH₃, BiH₃). The stability of these hydrides decreases down the group (NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃) due to decreasing bond dissociation enthalpy. The reducing character increases down the group.
*   **Towards Oxygen (Formation of Oxides, E₂O₃ and E₂O₅):** They form two types of oxides: E₂O₃ and E₂O₅. The acidic character of the oxides decreases down the group (N₂O₃, P₂O₃ are acidic; As₂O₃, Sb₂O₃ are amphoteric; Bi₂O₃ is basic). Similarly, E₂O₅ oxides show a decreasing acidic character down the group.
*   **Towards Halogens (Formation of Halides, EX₃ and EX₅):** They form trihalides (EX₃) and pentahalides (EX₅). Nitrogen does not form pentahalides due to the absence of d-orbitals in its valence shell. The stability of pentahalides decreases down the group due to the inert pair effect.
*   **Towards Metals:** These elements react with metals to form binary compounds in which they exhibit a -3 oxidation state (e.g., Ca₃N₂, Ca₃P₂).

Question 23

Q23. Discuss the general characteristics of Group 16 elements with reference to their electronic configuration, oxidation states, atomic radii, ionisation enthalpy and electronegativity

Accepted Answer

Group 16 elements, also known as Chalcogens, include Oxygen (O), Sulphur (S), Selenium (Se), Tellurium (Te), and Polonium (Po). They show a gradual transition from non-metallic to metallic character down the group. Polonium is radioactive. Here are their general characteristics:

**1. Electronic Configuration:**
*   The general outer electronic configuration for Group 16 elements is ns²np⁴. This means they have six valence electrons. They need two more electrons to achieve a stable noble gas configuration, which explains their common tendency to form M²⁻ ions or share two electrons.

**2. Atomic and Ionic Radii:**
*   **Atomic Radii:** Atomic radii increase down the group from O to Po. This is due to the addition of new electron shells with each successive element, leading to an increase in the distance of the outermost electrons from the nucleus. The increase is less significant from Se to Te and Te to Po due to the poor shielding effect of d- and f-electrons in heavier elements.
*   **Ionic Radii:** The ionic radii (e.g., for M²⁻ ions) also increase down the group, following the same trend as atomic radii.

**3. Ionisation Enthalpy:**
*   Ionisation enthalpy generally decreases down the group from O to Po. This is because the atomic size increases, and the outermost electrons are further away from the nucleus, experiencing less nuclear attraction. However, Group 16 elements have lower ionization enthalpies compared to Group 15 elements in the same period. This is because Group 16 elements have a p⁴ configuration, meaning one p-orbital contains two electrons, leading to inter-electronic repulsion, making it easier to remove an electron compared to the stable half-filled p³ configuration of Group 15 elements.

**4. Electronegativity:**
*   Electronegativity decreases down the group from O to Po. Oxygen is the second most electronegative element in the entire periodic table (after Fluorine). This trend is due to the increasing atomic size and increasing distance of valence electrons from the nucleus, which reduces the attraction for shared electron pairs.

**5. Oxidation States:**
*   The common oxidation states exhibited by Group 16 elements are -2, +2, +4, and +6.
    *   **-2 Oxidation State:** Oxygen predominantly shows a -2 oxidation state (except in peroxides, superoxides, and OF₂). Other elements like S, Se, Te also show -2 oxidation state when reacting with more electropositive elements. The tendency to exhibit -2 oxidation state decreases down the group due to increasing metallic character and increasing atomic size.
    *   **+2, +4, +6 Oxidation States:** These positive oxidation states are observed when Group 16 elements react with more electronegative elements (like F, O). Oxygen, due to the absence of d-orbitals, cannot expand its octet and thus does not show +4 or +6 oxidation states (except in OF₂ where it is +2). Sulphur, Selenium, and Tellurium commonly show +2, +4, and +6 oxidation states (e.g., SF₆, H₂SO₄, SO₂). The stability of the +6 oxidation state decreases down the group (due to the inert pair effect), while the stability of the +4 oxidation state increases down the group. For example, Te(IV) compounds are more stable than Te(VI) compounds.

Question 24

Q24. Discuss the general characteristics of Group 17 elements with reference to their electronic configuration, oxidation states, atomic radii, ionisation enthalpy and electronegativity

Accepted Answer

Group 17 elements, known as Halogens, include Fluorine (F), Chlorine (Cl), Bromine (Br), Iodine (I), and Astatine (At). They are highly reactive non-metals, with reactivity decreasing down the group. Astatine is radioactive. Here are their general characteristics:

**1. Electronic Configuration:**
*   The general outer electronic configuration for Group 17 elements is ns²np⁵. This indicates that they have seven valence electrons. They are one electron short of a stable noble gas configuration, which makes them highly reactive and strong oxidizing agents, with a strong tendency to gain one electron to form X⁻ ions.

**2. Atomic and Ionic Radii:**
*   **Atomic Radii:** Atomic radii increase down the group from F to At. This is due to the addition of new electron shells with each successive element, leading to an increase in the distance of the outermost electrons from the nucleus. The increase is less significant for heavier elements due to the poor shielding effect of d- and f-electrons.
*   **Ionic Radii:** The ionic radii (for X⁻ ions) also increase down the group, following the same trend as atomic radii. The X⁻ ions are significantly larger than their corresponding neutral atoms due to increased electron-electron repulsion and decreased effective nuclear charge.

**3. Ionisation Enthalpy:**
*   Ionisation enthalpy generally decreases down the group from F to At. This is because the atomic size increases, and the outermost electrons are further away from the nucleus, experiencing less nuclear attraction. Halogens have very high ionization enthalpies due to their small atomic size and high effective nuclear charge, making it difficult to remove an electron.

**4. Electronegativity:**
*   Electronegativity decreases down the group from F to At. Fluorine is the most electronegative element in the entire periodic table. This trend is due to the increasing atomic size and increasing distance of valence electrons from the nucleus, which reduces the attraction for shared electron pairs. Their high electronegativity is responsible for their strong oxidizing nature.

**5. Oxidation States:**
*   The common oxidation states exhibited by Group 17 elements are -1, +1, +3, +5, and +7.
    *   **-1 Oxidation State:** All halogens exhibit a -1 oxidation state. This is their most common and stable oxidation state, especially for Fluorine, which always shows a -1 oxidation state in its compounds (due to its highest electronegativity and absence of d-orbitals). The tendency to exhibit -1 oxidation state decreases down the group as metallic character increases.
    *   **Positive Oxidation States (+1, +3, +5, +7):** Chlorine, Bromine, and Iodine can exhibit positive oxidation states (+1, +3, +5, +7) when they react with more electronegative elements (like Oxygen or Fluorine) or in interhalogen compounds. This is possible due to the presence of vacant d-orbitals in their valence shell, allowing for the expansion of their octet. Fluorine, lacking d-orbitals, cannot expand its octet and thus does not show positive oxidation states.
    *   The stability of higher oxidation states (+5, +7) decreases down the group (due to the inert pair effect), while the stability of lower positive oxidation states (+1, +3) tends to increase.

Question 25

Q25. Write the main differences between the properties of white phosphorus and red phosphorus.

Accepted Answer

White phosphorus and red phosphorus are two important allotropes of phosphorus, exhibiting significant differences in their physical and chemical properties due to their distinct structural arrangements. Here are the main differences:

**Main Differences between White Phosphorus and Red Phosphorus**

**Conclusion:** The significant differences in properties stem primarily from their structural variations. White phosphorus, with its discrete P₄ tetrahedra and high strain, is highly reactive and toxic, while red phosphorus, with its polymeric network, is much more stable and less reactive.

Question 26

Q26. Why does nitrogen show catenation very rarely?

Accepted Answer

Catenation is the ability of an element to form bonds with atoms of itself to form long chains or rings. Nitrogen shows catenation very rarely compared to other elements in its group (like phosphorus) or carbon due to the following reasons:

1.  **Weak N-N Bond Strength:** The single N-N bond energy (163 kJ/mol) is significantly weaker than the C-C bond energy (348 kJ/mol) and even weaker than the P-P bond energy (209 kJ/mol). This weak bond makes nitrogen chains unstable and prone to breaking.
2.  **Small Size and High Electron-Electron Repulsion:** Nitrogen atoms are very small. When two nitrogen atoms form a single bond, the lone pair of electrons on each nitrogen atom are in close proximity. This leads to strong inter-electronic repulsion between the non-bonding electron pairs (lone pairs) on adjacent nitrogen atoms. This repulsion destabilizes the N-N single bond, making catenation unfavorable.
3.  **High Electronegativity:** Nitrogen is highly electronegative. While this doesn't directly prevent catenation, it contributes to the overall electronic environment that makes N-N bonds less stable compared to elements with lower electronegativity where electron density can be more evenly distributed in chains.

Question 27

Q27. Write the reactions of HNO₃ with the following:
(i) Cu (dilute)
(ii) Cu (conc.)
(iii) Zn (dilute)
(iv) Zn (conc.)

Accepted Answer

Nitric acid (HNO₃) is a strong oxidizing agent, and its reactions with metals vary depending on the concentration of the acid and the nature of the metal. Here are the reactions:

**(i) Reaction of Copper (Cu) with Dilute Nitric Acid (HNO₃)**

When copper reacts with dilute nitric acid, it gets oxidized to copper(II) nitrate, and nitric oxide (NO) gas is produced. Nitric oxide is a colorless gas that turns reddish-brown in air due due to oxidation to NO₂.

**Reaction:**
3Cu(s) + 8HNO₃(dilute) → 3Cu(NO₃)₂(aq) + 2NO(g) + 4H₂O(l)

**(ii) Reaction of Copper (Cu) with Concentrated Nitric Acid (HNO₃)**

When copper reacts with concentrated nitric acid, it is oxidized to copper(II) nitrate, and nitrogen dioxide (NO₂) gas is produced. Nitrogen dioxide is a reddish-brown gas.

**Reaction:**
Cu(s) + 4HNO₃(conc.) → Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(l)

**(iii) Reaction of Zinc (Zn) with Dilute Nitric Acid (HNO₃)**

Zinc is a more reactive metal than copper. With dilute nitric acid, the reduction product of nitric acid depends on the dilution. With moderately dilute HNO₃, nitrous oxide (N₂O) is formed. With very dilute HNO₃, ammonium nitrate (NH₄NO₃) is formed.

*   **With moderately dilute HNO₃ (e.g., 1:2 or 1:3 dilution):**
    4Zn(s) + 10HNO₃(dilute) → 4Zn(NO₃)₂(aq) + N₂O(g) + 5H₂O(l)

*   **With very dilute HNO₃ (e.g., 1:10 dilution):**
    4Zn(s) + 10HNO₃(very dilute) → 4Zn(NO₃)₂(aq) + NH₄NO₃(aq) + 3H₂O(l)
    (Note: The ammonium nitrate formed can further react with excess acid, but this is the primary reduction product of nitrogen.)

**(iv) Reaction of Zinc (Zn) with Concentrated Nitric Acid (HNO₃)**

When zinc reacts with concentrated nitric acid, it is oxidized to zinc(II) nitrate, and nitrogen dioxide (NO₂) gas is produced, similar to the reaction with concentrated copper.

**Reaction:**
Zn(s) + 4HNO₃(conc.) → Zn(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(l)

Question 28

Q28. Give reasons:
(i) H₃PO₃ is a diprotic acid.
(ii) Although nitrogen is less electronegative than chlorine, it exhibits positive oxidation state only in oxoacids.
(iii) Bismuth is

Accepted Answer

**(i) H₃PO₃ is a diprotic acid.**

**Reason:** The protic (ionizable) nature of an acid depends on the number of hydrogen atoms directly bonded to oxygen atoms (O-H bonds). These are the hydrogens that can be released as H⁺ ions in an aqueous solution. In the structure of H₃PO₃ (phosphorous acid), there are two P-OH bonds and one P-H bond.

*   The two hydrogen atoms attached to oxygen atoms (P-O-H) are acidic and can be ionized.
*   The hydrogen atom directly attached to phosphorus (P-H) is not acidic and does not ionize in water.

Therefore, H₃PO₃ can donate only two protons (H⁺ ions) per molecule, making it a diprotic acid. Its structure is O=P(H)(OH)₂.

**Reactions:**
H₃PO₃(aq) + H₂O(l) ⇌ H₃O⁺(aq) + H₂PO₃⁻(aq)
H₂PO₃⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + HPO₃²⁻(aq)

**(ii) Although nitrogen is less electronegative than chlorine, it exhibits positive oxidation state only in oxoacids.**

**Clarification and Reason:** The statement 'it exhibits positive oxidation state only in oxoacids' is factually incorrect. Nitrogen exhibits positive oxidation states in various compounds, including its oxides (e.g., N₂O (+1), NO (+2), N₂O₃ (+3), NO₂ (+4), N₂O₅ (+5)) and halides (e.g., NCl₃ (+3)). In NCl₃, nitrogen has a positive oxidation state (+3) because chlorine is more electronegative than nitrogen (Pauling electronegativity: Cl = 3.16, N = 3.04). Therefore, the premise of the question regarding 'only in oxoacids' is flawed.

However, if the question intends to ask why nitrogen's *highest* positive oxidation state (+5) is typically observed in oxoacids (like HNO₃) and not in halides (like NCl₅, which doesn't exist), the reason is as follows:

*   **Absence of d-orbitals:** Nitrogen, being a second-period element, lacks vacant d-orbitals in its valence shell. This limits its maximum covalency to four (one s and three p orbitals). Therefore, it cannot expand its octet to accommodate more than four bonds.
*   **Formation of NCl₅ is not possible:** Unlike phosphorus, which can form PCl₅ by utilizing its vacant 3d-orbitals to expand its octet and achieve a +5 oxidation state and five covalent bonds, nitrogen cannot form NCl₅. In NCl₅, nitrogen would need to form five covalent bonds, which requires an expanded octet, not possible for nitrogen.
*   **Stability in Oxoacids:** In oxoacids like HNO₃, nitrogen achieves its +5 oxidation state by forming multiple bonds with oxygen (e.g., N=O and N→O coordinate bonds), where the total number of bonds does not exceed four (e.g., one double bond and two single bonds, or one triple bond and one single bond, or four single bonds with one being a coordinate bond). The highly electronegative oxygen atoms effectively draw electron density, allowing nitrogen to achieve a high positive oxidation state without exceeding its maximum covalency of four.

So, while nitrogen does exhibit positive oxidation states in compounds other than oxoacids, its highest positive oxidation state (+5) is most prominently observed and stable in oxoacids like HNO₃ due to its inability to expand its octet and form five single bonds, a limitation not faced by elements with vacant d-orbitals.

**(iii) Bismuth is a strong oxidising agent in pentavalent state.**

**Reason:** Bismuth (Bi) is the last element in Group 15. As we move down a group, the stability of the lower oxidation state increases, and the stability of the higher oxidation state decreases. This phenomenon is known as the **inert pair effect**.

*   **Inert Pair Effect:** For heavier p-block elements, the outermost s-electrons (the 'inert pair') tend to remain paired and do not participate in bonding as readily as the p-electrons. This is due to the poor shielding effect of intervening d and f electrons, which causes the ns² electrons to be held more tightly by the nucleus.
*   **Stability of Oxidation States:** For Group 15 elements, the common oxidation states are +3 and +5. Due to the inert pair effect, the +3 oxidation state becomes more stable than the +5 oxidation state for heavier elements like bismuth. Conversely, the +5 oxidation state becomes less stable.
*   **Oxidizing Nature:** Since the +5 oxidation state of bismuth (Bi⁵⁺) is less stable than its +3 oxidation state (Bi³⁺), Bi⁵⁺ has a strong tendency to gain two electrons and convert to the more stable Bi³⁺ state. An agent that gains electrons readily is an oxidizing agent. Therefore, bismuth in its pentavalent state (e.g., in BiF₅ or NaBiO₃) acts as a strong oxidizing agent, readily getting reduced to the +3 oxidation state.

Question 29

Q29. Why are halogens strong oxidising agents?

Accepted Answer

Halogens are strong oxidising agents due to their high electronegativity and high electron gain enthalpy. They have a strong tendency to accept an electron to achieve a stable noble gas configuration (ns²np⁶). By accepting an electron, they themselves get reduced, and in the process, oxidise other substances. The oxidising power generally decreases down the group from fluorine to iodine, with fluorine being the strongest oxidising agent among all halogens.

Question 30

Q30. Explain the bleaching action of Cl₂.

Accepted Answer

The bleaching action of chlorine (Cl₂) is due to its oxidising property, which occurs in the presence of moisture. When chlorine gas is passed through water, it reacts to form hydrochloric acid (HCl) and hypochlorous acid (HOCl).

Cl₂(g) + H₂O(l) → HCl(aq) + HOCl(aq)

Hypochlorous acid (HOCl) is unstable and readily decomposes to produce nascent oxygen (atomic oxygen).

HOCl(aq) → HCl(aq) + [O]

This nascent oxygen is a powerful oxidising agent. It reacts with coloured organic substances, oxidising them to colourless products. This makes the bleaching action of chlorine permanent, as the colour is destroyed by oxidation rather than simply being masked.

Colouring matter + [O] → Colourless matter

Question 31

Q31. Name two poisonous gases which can be prepared from chlorine gas.

Accepted Answer

Two poisonous gases that can be prepared from chlorine gas are:

1.  **Phosgene (Carbonyl chloride, COCl₂)**:
    It is formed by the reaction of carbon monoxide (CO) with chlorine (Cl₂) in the presence of sunlight or activated charcoal as a catalyst.
    CO(g) + Cl₂(g) → COCl₂(g)
    Phosgene is a highly toxic gas used in chemical warfare and as a precursor in organic synthesis.

2.  **Mustard gas (Bis(2-chloroethyl) sulfide, ClCH₂CH₂SCH₂CH₂Cl)**:
    It is prepared by the reaction of ethene (ethylene) with sulfur monochloride (S₂Cl₂).
    2CH₂=CH₂(g) + S₂Cl₂(l) → ClCH₂CH₂SCH₂CH₂Cl(l) + S(s)
    Mustard gas is a potent blistering agent and a chemical warfare agent.

Question 32

Q32. Write the reactions of F₂ and Cl₂ with water.

Accepted Answer

The reactions of fluorine (F₂) and chlorine (Cl₂) with water are significantly different due to their varying oxidising powers and electronegativities.

1.  **Reaction of Fluorine (F₂) with water:**
    Fluorine is an extremely strong oxidising agent. It oxidises water to oxygen (O₂) or ozone (O₃). In this reaction, fluorine itself gets reduced to hydrofluoric acid (HF).
    2F₂(g) + 2H₂O(l) → 4HF(aq) + O₂(g)
    (or) 3F₂(g) + 3H₂O(l) → 6HF(aq) + O₃(g)
    This reaction is vigorous and often explosive.

2.  **Reaction of Chlorine (Cl₂) with water:**
    Chlorine reacts with water to form hydrochloric acid (HCl) and hypochlorous acid (HOCl). This reaction is a disproportionation reaction, where chlorine is simultaneously oxidised (to +1 in HOCl) and reduced (to -1 in HCl).
    Cl₂(g) + H₂O(l) ⇌ HCl(aq) + HOCl(aq)
    The equilibrium lies to the left, but in the presence of sunlight, HOCl decomposes to HCl and nascent oxygen, which drives the reaction forward and is responsible for the bleaching action of chlorine water.
    HOCl(aq) → HCl(aq) + [O]

Question 33

Q33. Write the balanced chemical equations for the following reactions:
(i) NaCl is heated with concentrated H₂SO₄ in the presence of MnO₂.
(ii) Chlorine gas is passed into a solution

Accepted Answer

(i) When sodium chloride (NaCl) is heated with concentrated sulphuric acid (H₂SO₄) in the presence of manganese dioxide (MnO₂), chlorine gas (Cl₂) is produced. This is a common laboratory method for the preparation of chlorine.
2NaCl + MnO₂ + 2H₂SO₄ (conc) → Na₂SO₄ + MnSO₄ + 2H₂O + Cl₂

(ii) When chlorine gas (Cl₂) is passed into a solution of sodium iodide (NaI) in water, chlorine, being a stronger oxidizing agent than iodine, oxidizes iodide ions (I⁻) to iodine (I₂). Sodium chloride (NaCl) is formed as a byproduct.
Cl₂ (g) + 2NaI (aq) → 2NaCl (aq) + I₂ (aq)

(iii) When hydrogen iodide (HI) is heated with concentrated sulphuric acid (H₂SO₄), HI acts as a strong reducing agent and reduces concentrated H₂SO₄ to sulphur dioxide (SO₂). HI itself gets oxidized to iodine (I₂).
2HI + H₂SO₄ (conc) → I₂ + SO₂ + 2H₂O

Question 34

Q34. Why are halogens strong oxidising agents?

Accepted Answer

Halogens (Group 17 elements) are strong oxidising agents due to their strong tendency to accept an electron and achieve a stable noble gas configuration (ns²np⁶). This tendency arises from a combination of the following factors:

1.  **High Electronegativity:** Halogens have very high electronegativity values. This means they have a strong attraction for electrons, enabling them to readily pull an electron from another species.

2.  **High Electron Gain Enthalpy (highly negative):** Halogens possess a very high negative electron gain enthalpy. This indicates that a significant amount of energy is released when a halogen atom gains an electron to form a halide ion (X⁻). This exothermic process makes the acceptance of an electron energetically favourable.

3.  **Low Bond Dissociation Enthalpy (relative to electron gain enthalpy and hydration enthalpy):** Although halogens exist as diatomic molecules (X₂), the energy required to break the X-X bond (bond dissociation enthalpy) is relatively low. This allows the molecule to readily dissociate and accept electrons.

4.  **Formation of Stable Halide Ions:** By gaining one electron, halogens achieve the stable octet configuration of the nearest noble gas, forming stable halide ions (F⁻, Cl⁻, Br⁻, I⁻).

Among the halogens, fluorine is the strongest oxidising agent. This is attributed to its exceptionally high electronegativity, low F-F bond dissociation enthalpy, and the very high hydration enthalpy of the small F⁻ ion. Fluorine can oxidise other halide ions and even water.

In essence, by readily accepting electrons, halogens cause the oxidation of other substances, thereby acting as strong oxidising agents.

Question 35

Q35. Bleaching of wood pulp is done by using Cl₂. Explain the principle.

Accepted Answer

The bleaching of wood pulp by using chlorine (Cl₂) is based on the strong oxidizing property of chlorine. The principle involves the following steps:

1.  **Formation of Hypochlorous Acid:** When chlorine gas is dissolved in water, it reacts to form hypochlorous acid (HOCl) and hydrochloric acid (HCl).
    Cl₂ (g) + H₂O (l) → HOCl (aq) + HCl (aq)

2.  **Release of Nascent Oxygen:** Hypochlorous acid is an unstable compound and readily decomposes to produce hydrochloric acid and highly reactive nascent oxygen ([O]).
    HOCl (aq) → HCl (aq) + [O]

3.  **Oxidation of Coloring Matter:** The nascent oxygen released is a powerful oxidizing agent. It reacts with the colored organic substances (chromophores) present in the wood pulp, oxidizing them into colorless compounds. This process removes the color from the wood pulp, making it white.
    Colored matter + [O] → Colorless matter

The bleaching action of chlorine is permanent because the coloring matter is chemically destroyed by oxidation.

Question 36

Q36. Name two poisonous gases which can be prepared from chlorine gas.

Accepted Answer

Two poisonous gases that can be prepared from chlorine gas are:

1.  **Phosgene (Carbonyl chloride), COCl₂**
    *   **Preparation:** Phosgene is prepared by the reaction of carbon monoxide (CO) with chlorine gas (Cl₂) in the presence of sunlight or activated charcoal as a catalyst.
        CO (g) + Cl₂ (g) \xrightarrow{	ext{Sunlight/Charcoal}} COCl₂ (g)
    *   **Nature:** It is a highly toxic gas, known for its use as a chemical warfare agent.

2.  **Chloropicrin (Trichloronitromethane), CCl₃NO₂**
    *   **Preparation:** Chloropicrin can be prepared by the reaction of chloroform (CHCl₃) with nitric acid (HNO₃) in the presence of chlorine, or more directly by the chlorination of nitromethane (CH₃NO₂) or picric acid. A common industrial method involves the reaction of nitromethane with chlorine in the presence of a base.
        CH₃NO₂ + 3Cl₂ \xrightarrow{	ext{Base}} CCl₃NO₂ + 3HCl
    *   **Nature:** Although technically a liquid at room temperature, its vapor is a potent tear gas and a highly poisonous substance, also used as a chemical warfare agent and a fumigant.

Chapter 7: The p-Block Elements