NCERT Solutions Class 11 Chemistry - The p-Block Elements

Q.7.3: Why is N₂ less reactive at room temperature?
Ans: N₂ has a very strong triple bond between the two nitrogen atoms (one σ and two π bonds). The bond dissociation energy of the N≡N bond is very high, so a large input of energy is required to break it. As a result, molecular nitrogen is relatively inert at room temperature.

Q.7.4: List the uses of Neon and argon gases.
Ans: Uses of neon gas:
(i) It is mixed with helium to protect electrical equipments from high voltage.
(ii) It is filled in discharge tubes with characteristic colours.
(iii) It is used in beacon lights.

Uses of Argon gas: 
(i) Argon along with nitrogen is used in gas-filled electric lamps. This is because Ar is more inert than N.
(ii) It is usually used to provide an inert temperature in a high metallurgical process.
(iii) It is also used in laboratories to handle air-sensitive substances.

Q.7.5: How does ammonia react with a solution of Cu²⁺?
Ans: NH₃ acts as a Lewis base; it donates its lone pair on nitrogen to metal ions. With Cu²⁺ it forms coordination complexes (for example, the deep blue [Cu(NH₃)₄]²⁺), via sequential ligand addition:

Cu²⁺ + 4 NH₃ → [Cu(NH₃)₄]²⁺

Q.7.6: What is the covalence of nitrogen in N₂O₅?
Ans:

From the structure of N₂O₅ (two NO₂ groups linked by an oxygen), each nitrogen forms four covalent bonds (one N-O single to the bridging O and two bonds to O in the NO₂ unit, counting double-bond character via resonance). Thus, the covalence of nitrogen in N₂O₅ is 4.

Q.7.7: Bond angle in is higher than that in PH₃. Why?
Ans: In PH₃, P is sp³ hybridized. Three orbitals are involved in bonding with three hydrogen atoms and the fourth one contains a lone pair. As lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion, the tetrahedral shape associated with sp³ bonding is changed to pyramidal. PH₃ combines with a proton to form in which the lone pair is absent. Due to the absence of lone pair in, there is no lone pair-bond pair repulsion. Hence, the bond angle in is higher than the bond angle in PH₃.

Q.7.8: What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO₂?
Ans: White phosphorus dissolves in boiling concentrated NaOH (under CO₂ or an inert atmosphere) to produce phosphine (PH₃) along with hypophosphite salts. The simplified reaction is:

Q.7.9: What happens when PCl₅ is heated?
Ans: PCl₅ has trigonal bipyramidal geometry with three stronger equatorial P-Cl bonds and two weaker axial P-Cl bonds. On strong heating, PCl₅ decomposes by cleavage of the weaker bonds to give PCl₃ and Cl₂ (or undergoes dissociation to PCl₃ and Cl₂ depending on conditions):

Q.7.10: Write a balanced equation for the hydrolytic reaction of PCl₅ in heavy water.
Ans:

Q.7.11: What is the basicity of H₃PO₄?
Ans: H₃PO₄

Orthophosphoric acid, H₃PO₄, has three hydroxyl groups (-OH) that can donate protons. Therefore its basicity (number of replaceable H⁺) is three; it is a tribasic acid.

Q.7.12: What happens when H₃PO₃ is heated?
Ans: H₃PO₃, on heating, undergoes disproportionation reaction to form PH₃ and H₃PO₄. The oxidation numbers of P in H₃PO₃, PH₃, and H₃PO₄ are 3, −3, and 5 respectively. As the oxidation number of the same element is decreasing and increasing during a particular reaction, the reaction is a disproportionation reaction.

Q.7.13: List the important sources of sulphur.
Ans: Sulphur mainly exists in combined form in the earth’s crust primarily as sulphates [gypsum (CaSO₄.2H₂O), Epsom salt (MgSO₄.7H₂O), baryte (BaSO₄)] and sulphides [galena (PbS), zinc blends (ZnS), copper pyrites (CuFeS₂)].

Q.7.14: Write the order of thermal stability of the hydrides of Group 16 elements.
Ans: The thermal stability of hydrides decreases on moving down the group. This is due to a decrease in the bond dissociation enthalpy (H−E) of hydrides on moving down the group.
Therefore,

Q.7.15: Why is H₂O a liquid and H₂S a gas?
Ans: H₂O has oxygen as the central atom. Oxygen has smaller size and higher electronegativity as compared to sulphur. Therefore, there is extensive hydrogen bonding in H₂O, which is absent in H₂S. Molecules of H₂S are held together only by weak Van der Waal’s forces of attraction.
Hence, H₂O exists as a liquid while H₂S as a solid.

Q.7.16: Which of the following does not react with oxygen directly? 
Zn, Ti, Pt, Fe
Ans: Pt is a noble metal and does not react very easily. All other elements, Zn, Ti, Fe, are quite reactive. Hence, oxygen does not react with platinum (Pt) directly.

Q.7.17: Complete the following reactions:
(i) C₂H₄  + O₂ →        
(ii) 4Al + 3O₂ →
Ans:

Q.7.18: Why does O₃ act as a powerful oxidising agent?
Ans: Ozone is not a very stable compound under normal conditions and decomposes readily on heating to give a molecule of oxygen and nascent oxygen. Nascent oxygen, being a free radical, is very reactive.

Therefore, ozone acts as a powerful oxidising agent.

Q.7.19: Why do noble gases have comparatively large atomic sizes?
Ans: Noble gases do not form molecules. In case of noble gases, the atomic radii corresponds to Van der Waal’s radii. On the other hand, the atomic radii of other elements correspond to their covalent radii. By definition, Van der Waal’s radii are larger than covalent radii. It is for this reason that noble gases are very large in size as compared to other atoms belonging to the same period.

Q.7.20: What happens when sulphur dioxide is passed through an aqueous solution of Fe(III) salt?
Ans: SO₂ acts as a reducing agent when passed through an aqueous solution containing Fe(III) salt. It reduces Fe(III) to Fe(II) i.e., ferric ions to ferrous ions.

Q.7.21: Comment on the nature of two S−O bonds formed in SO₂ molecule. Are the two S−O bonds in this molecule equal?
Ans: The electronic configuration of S is 1s² 2s² 2p⁶ 3s² 3p⁴.
During the formation of SO₂, one electron from 3p orbital goes to the 3d orbital and S undergoes sp² hybridization. Two of these orbitals form sigma bonds with two oxygen atoms and the third contains a lone pair. p-orbital and d-orbital contain an unpaired electron each. One of these electrons forms pπ- pπ bond with one oxygen atom and the other forms pπ- dπ bond with the other oxygen. This is the reason SO₂ has a bent structure. Also, it is a resonance hybrid of structures I and II.

Both S−O bonds are equal in length (143 pm) and have a multiple bond character.

Q.7.22: Give the formula and describe the structure of a noble gas species which is isostructural with:
(i) 
(ii)   
(iii)
Ans: (i)
XeF₄ is isoelectronic with and has square planar geometry.

(ii)
XeF₂ is isoelectronic to and has a linear structure.

(iii) 
XeO₃ is isostructural to and has a pyramidal molecular structure.

Q.7.23: Which one of the following does not exist?
(i) XeOF₄
(ii) NeF₂  
(iii) XeF₂ 
(iv) XeF₆
Ans: NeF₂ does not exist. Neon is extremely inert and its compounds are not observed under normal conditions.

Q.7.24: Write two uses of ClO₂.
Ans: Uses of ClO₂:
(i) It is used for purifying water.
(ii) It is used as a bleaching agent.

Q.7.25: How are XeO₃ and XeOF₄ prepared?
Ans: (i) XeO₃ can be prepared in two ways as shown.

(ii) XeOF₄ can be prepared using XeF₆.

Q.7.26: With what neutral molecule is ClO⁻ isoelectronic? Is that molecule a Lewis base?
Ans: ClO⁻ is isoelectronic to ClF. Also, both species contain 26 electrons in all as shown.
Total electrons ClO⁻ = 17 + 8 + 1 = 26
In ClF = 17+9 = 26
ClF acts like a Lewis base as it accepts electrons from F to form ClF₃.

Q.7.27: Give two examples to show the anomalous behaviour of fluorine.
Ans: Anomalous behaviour of fluorine
(i) It forms only one oxoacid as compared to other halogens that form a number of oxoacids.
(ii) Ionisation enthalpy, electronegativity, and electrode potential of fluorine are much higher than expected.

Q.7.28: Sea is the greatest source of some halogens. Comment.
Ans: Sea water contains chlorides, bromides, and iodides of Na, K, Mg, and Ca. However, it primarily contains NaCl. The deposits of dried up sea beds contain sodium chloride and carnallite, KCl.MgCl₂.6H₂O. Marine life also contains iodine in their systems. For example, sea weeds contain upto 0.5% iodine as sodium iodide. Thus, sea is the greatest source of halogens.

Q.7.29: Give the reason for bleaching action of Cl₂.
Ans: When chlorine reacts with water, it produces nascent oxygen. This nascent oxygen then combines with the coloured substances present in the organic matter to oxide them into colourless substances.

Coloured substances + [O] → Oxidized colourless substance

Q.7.30: Name two poisonous gases which can be prepared from chlorine gas.
Ans: Two poisonous gases that can be prepared from chlorine gas are
(i) Phosgene (COCl₂)
(ii) Mustard gas (ClCH₂CH₂SCH₂CH₂Cl)

Q.7.31: Why is ICl more reactive than I₂?
Ans: ICl is more reactive than I₂ because I−Cl bond in ICl is weaker than I−I bond in I₂.

Q.7.32: Why is helium used in diving apparatus?
Ans: Air contains a large amount of nitrogen and the solubility of gases in liquids increases with increase in pressure. When sea divers dive deep into the sea, large amount of nitrogen dissolves in their blood. When they come back to the surface, solubility of nitrogen decreases and it separates from the blood and forms small air bubbles. This leads to a dangerous medical condition called bends. Therefore, air in oxygen cylinders used for diving is diluted with helium gas. This is done as He is sparingly less soluble in blood.

Q.7.33: Balance the following equation: XeF₆ + H₂O → XeO₂F₂ + HF
Ans: Balanced equation
XeF₆ + 2 H₂O → XeO₂F₂  + 4 HF

Q.7.34: Why has it been difficult to study the chemistry of radon?
Ans: It is difficult to study the chemistry of radon because it is a radioactive substance having a half-life of only 3.82 days. Also, compounds of radon such as RnF₂ have not been isolated. They have only been identified.

Q.7.35: Why has it been difficult to study the chemistry of radon?
Ans: It is difficult to study the chemistry of radon because it is a radioactive substance having a half-life of only 3.82 days. Also, compounds of radon such as RnF₂ have not been isolated. They have only been identified.

Q.7.36: Why does the reactivity of nitrogen differ from phosphorus?
Ans: Nitrogen is chemically less reactive. This is because of the high stability of its molecule, N₂. In N₂, the two nitrogen atoms form a triple bond. This triple bond has very high bond strength, which is very difficult to break. It is because of nitrogen’s small size that it is able to form pπ−pπ bonds with itself. This property is not exhibited by atoms such as phosphorus. Thus, phosphorus is more reactive than nitrogen.

Q.7.37: Discuss the trends in chemical reactivity of group 15 elements.
Ans: General trends in chemical properties of group − 15
(i) Reactivity towards hydrogen:
The elements of group 15 react with hydrogen to form hydrides of type EH₃, where E = N, P, As, Sb, or Bi. The stability of hydrides decreases on moving down from NH₃ to BiH₃.
(ii) Reactivity towards oxygen:
The elements of group 15 form two types of oxides: E₂O₃ and E₂O₅, where E = N, P, As, Sb, or Bi. The oxide with the element in the higher oxidation state is more acidic than the other. However, the acidic character decreases on moving down a group.
(iii) Reactivity towards halogens:
The group 15 elements react with halogens to form two series of salts: EX₃ and EX₅. However, nitrogen does not form NX₅ as it lacks the d-orbital. All trihalides (except NX₃) are stable.
(iv) Reactivity towards metals:
The group 15 elements react with metals to form binary compounds in which metals exhibit −3 oxidation states.

Q.7.38: Why does NH₃ form hydrogen bond but PH₃ does not?
Ans: Nitrogen is significantly more electronegative and much smaller than phosphorus. In NH₃ the N-H bond is strongly polar and the lone pair on nitrogen is concentrated in a small volume, allowing strong hydrogen bonding between molecules. Phosphorus is larger with lower electronegativity; PH₃ has very weak polarity and the lone pair is more diffuse, so appreciable hydrogen bonding does not occur.

Q.7.39: How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions involved.
Ans: An aqueous solution of ammonium chloride is treated with sodium nitrite.

NO and HNO₃ are produced in small amounts. These are impurities that can be removed on passing nitrogen gas through aqueous sulphuric acid, containing potassium dichromate.

Q.7.40: How is ammonia manufactured industrially?
Ans: Ammonia is prepared on a large-scale by the Haber’s process.

The optimum conditions for manufacturing ammonia are:
(i) Pressure (around 200 × 10⁵ Pa)
(ii) Temperature (4700 K)
(iii) Catalyst such as iron oxide with small amounts of Al₂O₃ and K₂O

Fig: Haber's processQ.7.41: Illustrate how copper metal can give different products on reaction with HNO₃.
Ans: Concentrated nitric acid is a strong oxidizing agent. It is used for oxidizing most metals. The products of oxidation depend on the concentration of the acid, temperature, and also on the material undergoing oxidation.

Q.7.42: Why are halogens coloured?
Ans: Almost all halogens are coloured. This is because halogens absorb radiations in the visible region. This results in the excitation of valence electrons to a higher energy region. Since the amount of energy required for excitation differs for each halogen, each halogen displays a different colour.

Q.7.43: The HNH angle value is higher than HPH, HAsH and HSbH angles. Why? [Hint: Can be explained on the basis of sp³ hybridisation in NH₃ and only s−p bonding between hydrogen and other elements of the group].
Ans: Hydride NH₃ PH₃ AsH₃ SbH₃
H−M−H angle 107° 92° 91° 90°
The above trend in the H−M−H bond angle can be explained on the basis of the electronegativity of the central atom. Since nitrogen is highly electronegative, there is high electron density around nitrogen. This causes greater repulsion between the electron pairs around nitrogen, resulting in maximum bond angle. We know that electronegativity decreases on moving down a group. Consequently, the repulsive interactions between the electron pairs decrease, thereby decreasing the H−M−H bond angle.

Q.7.44: Why does R₃P=O exist but R₃N=O does not (R = alkyl group)?
Ans: N (unlike P) lacks the d-orbital. This restricts nitrogen to expand its coordination number beyond four. Hence, R₃N=O does not exist.

Q.7.45: 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.
Ans: (i)
(ii) 

Q.7.46: What are the oxidation states of phosphorus in the following:
(i) H₃PO₃ 
(ii) PCl₃  
(iii) Ca₃P₂ 
(iv) Na₃PO₄ 
(v) POF₃?
Ans: Let the oxidation state of p be x.
(i) H₃PO₃


(ii) PCl₃


(iii) Ca₃P₂


(iv) Na₃PO₄


(v) POF₃


Q.7.47: Write main differences between the properties of white phosphorus and red phosphorus.
Ans:

Q.7.48: Why does nitrogen show catenation properties less than phosphorus?
Ans: Catenation (ability to form chains of the same element) is less common for nitrogen because the N-N single bond is relatively weak compared with the P-P single bond. Nitrogen's small size leads to greater lone-pair-lone-pair repulsion when N atoms are bonded, weakening the N-N bond and making long nitrogen chains unstable. Phosphorus, being larger, forms stronger P-P single bonds and shows greater catenation.

Q.7.49: Give the disproportionation reaction of H₃PO₃.
Ans: On heating, orthophosphorus acid (H₃PO₃) disproportionates to give orthophosphoric acid (H₃PO₄) and phosphine (PH₃). The oxidation states of P in various species involved in the reaction are mentioned below.

Q.7.50: Can PCl₅ act as an oxidising as well as a reducing agent? Justify.
Ans: Phosphorus in PCl₅ is in its highest oxidation state (+5). Since it can be reduced (its oxidation state can decrease), PCl₅ acts as an oxidising agent (it oxidises other species while itself being reduced). It cannot act as a reducing agent (it cannot be further oxidised under normal conditions because +5 is the maximum for P).

Fig: Structure of PCl₅
Q.7.50: 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.
Ans: The elements of group 16 are collectively called chalcogens.
(i) Elements of group 16 have six valence electrons each. The general electronic configuration of these elements is ns² np⁴, where n varies from 2 to 6.
(ii) Oxidation state: As these elements have six valence electrons (ns² np⁴), they should display an oxidation state of −2. However, only oxygen predominantly shows the oxidation state of −2 owing to its high electronegativity. It also exhibits the oxidation state of −1 (H₂O₂), zero (O₂), and +2 (OF₂). However, the stability of the −2 oxidation state decreases on moving down a group due to a decrease in the electronegativity of the elements. The heavier elements of the group show an oxidation state of +2, +4, and +6 due to the availability of d-orbitals.

(iii) Formation of hydrides: These elements form hydrides of formula H₂E, where E = O, S, Se, Te, PO. Oxygen and sulphur also form hydrides of type H₂E₂. These hydrides are quite volatile in nature.

Q.7.51: What inspired N. Bartlett for carrying out reaction between Xe and PtF₆?
Ans: Neil Bartlett initially carried out a reaction between oxygen and PtF₆. This resulted in the formation of a red compound, .

Later, he realized that the first ionization energy of oxygen (1175 kJ/mol) and Xe (1170 kJ/mol) is almost the same. Thus, he tried to prepare a compound with Xe and PtF₆. He was successful and a red-coloured compound, was formed.

Q.7.52: How can you prepare Cl₂ from HCl and HCl from Cl₂? Write reactions only.
Ans: (i) Cl₂ can be prepared from HCl by Deacon’s process.

(ii) HCl can be prepared from Cl₂ on treating it with water.

Q.7.53: Which aerosols deplete ozone?
Ans: Freons or chlorofluorocarbons (CFCs) are aerosols that accelerate the depletion of ozone. In the presence of ultraviolet radiations, molecules of CFCs break down to form chlorine-free radicals that combine with ozone to form oxygen.

Q.7.54: Describe the manufacture of H₂SO₄ by contact process?
Ans: Sulphuric acid is manufactured by the contact process. It involves the following steps:

Fig: Contact ProcessStep (i):
Sulphur or sulphide ores are burnt in air to form SO₂.
Step (ii):
By a reaction with oxygen, SO₂ is converted into SO₃ in the presence of V₂O₅ as a catalyst.

Step (iii):
SO₃ produced is absorbed on H₂SO₄ to give H₂S₂O₇ (oleum).

This oleum is then diluted to obtain H₂SO₄ of the desired concentration.

In practice, the plant is operated at 2 bar (pressure) and 720 K (temperature). The sulphuric acid thus obtained is 96-98% pure.

Q.7.55: How is SO₂ an air pollutant?
Ans: Sulphur dioxide causes harm to the environment in many ways:
1. It combines with water vapour present in the atmosphere to form sulphuric acid. This causes acid rain. Acid rain damages soil, plants, and buildings, especially those made of marble.

2. Even in very low concentrations, SO₂ causes irritation in the respiratory tract. It causes throat and eye irritation and can also affect the larynx to cause breathlessness.

3. It is extremely harmful to plants. Plants exposed to sulphur dioxide for a long time lose colour from their leaves. This condition is known as chlorosis. This happens because the formation of chlorophyll is affected by the presence of sulphur dioxide.

Q.7.56: Write the reactions of F₂ and Cl₂ with water.
Ans:

(i)

Fluorine reacts vigorously with water to produce oxygen-containing and acidic products (and HF). The reaction is highly exothermic and can produce oxygen and HF among other products.

(ii)

Chlorine reacts with water as follows:

Cl₂ + H₂O ⇌ HCl + HOCl

Hypochlorous acid (HOCl) is a mild oxidising and disinfecting agent.

Q.7.57: Explain why fluorine forms only one oxoacid, HOF.
Ans: Fluorine's very high electronegativity and small size destabilise other possible oxoacids. Only hypofluorous acid (HOF) is isolable under special conditions; other higher oxoacids that heavier halogens form are not possible for fluorine because of its inability to stabilise positive oxidation states on the halogen or form extended O-X frameworks.

Q.7.58: Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not.
Ans: Although oxygen and chlorine have similar electronegativity values, oxygen is much smaller in size, resulting in higher electron density in its valence region. This concentrated lone pair density can interact strongly with hydrogen bonded to electronegative atoms, enabling hydrogen bonding. Chlorine's lone pairs are more diffuse (larger atom), reducing the directional lone-pair interaction necessary for hydrogen bonding, so chlorine does not participate significantly in hydrogen bonding.