NCERT Solutions for Class 12 Chemistry Chapter 7 The p-Block Elements

Here we provide NCERT Solutions for Class 12 Chemistry Chapter 7 The p-Block Elements for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Solutions for Class 12 Chemistry Chapter 7 The p-Block Elements pdf, free NCERT solutions for Class 12 Chemistry Chapter 7 The p-Block Elements book pdf download. Now you will get step by step solution to each question.

NCERT Solutions for Class 12 Chemistry Chapter 7 The p-Block Elements

Class 12 Chemistry Chapter 7 explains the p block elements and their properties, The elements in group 15, 16 and 17 are discussed along with their properties. Various concepts such as electronegativity, chemical and physical properties, ionization therapy, etc. are discussed in detail.

This chapter is important from examination perspective and the students are advised to go through the NCERT Solutions for better practice. The solutions are provided along with the diagrams for better understanding.

NCERT IN-TEXT QUESTIONS

Question 1.
Why are pentahalides more covalent than trihalides in the members of the nitrogen family?
Answer:
The electronic configuration of the elements of nitrogen family (group 15) is ns²p³. Because of the inert pair effect, the valence s-electrons cannot be released easily for the bond formation. This means that the elements can form trivalent cation (E³⁺) by releasing valence p-electrons while it is difficult to form pentavalent cation (E⁵⁺). Under the circumstances, if all the five valence electrons are to be involved in the bond formation, the compounds showing pentavalency or +5 oxidation state must be of covalent nature. This is particularly the case in the higher members (Sb, Bi) of the family where the inert pair effect is quite prominent.

Question 2.
Why is BiH₃ the strongest reducing agent amongst all the hydrides of Group 15 elements?
Answer:
This is because as we move down the group, the size increases, as a result, length of E-H bond increases and its strength decreases, so that the bond can be broken easily to release H₂ gas. Hence, BiH₃ is the strongest reducing agent.

Question 3.
Why is N₂ less reactive at room temperature?
Answer:
In the nitrogen molecule (N₂), two atoms of nitrogen are linked by triple bond (N = N). Due to small atomic size of the element (atomic radius = 70
pm), the bond dissociation enthalpy is very high (946 kJ mol¹). This means that it is quite difficult to cleave or break the triple bond at room temperature. As a result, N₂ is less reactive at room temperature.

Question 4.
Mention the conditions required for the maximum yield of ammonia.
Answer:
In Haber’s process, ammonia is formed by the following reaction.

According to Le-chatelier’s principle, the favourable conditions for the maximum yield of ammonia are :
(i) Low temperature : But optimum temperature of 700 K is necessary to keep the forward reaction in progress.
(ii) High pressure : Pressure to the extent of about 200 atm is required.
(iii) Catalyst & promoter : In order to achieve the early attainment of equilibrium, iron oxide acts as catalyst. Along with that; K₂O, Al₂O₃ or Mo metal may act as the promoter to increase the efficiency of the catalyst.

Question 5.
How does ammonia react with blue solution having Cu²⁺ ions ?
Answer:
When ammonia gas is passed through blue solution containing Cu²⁺ ions to give a soluble complex with deep blue colour.

Question 6.
What is the covalency of nitrogen in N₂O₅?
Answer:
The covalent structure of nitrogen pentoxide (N₂O₅) is given. Since the nitrogen atom has shared four electron pairs, its covalency is four in the molecule of N₂O₅.

Question 7.
Bond angle in PH₄⁺ is higher than that in PH₃. Why?
Answer:
P in PH₃ is sp³-hybridized with 3 bond pairs and one lone pair around P. Due to stronger lp-bp repulsions than bp-bp repulsions, tetrahedral angle decreases from 109°28′ to 93.6°. As a result, PH₃ is pyramidal. In PH₄⁺, there are 4 bp’s and no lone pair. As a result, there are only identical bp-bp repulsions so that PH₄⁺ assumes tetrahedral geometry and the bond angle is 109°28′.Hence, bond angle of PH₄⁺ > bond angle of PH₃

Question 8.
What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO₂?
Answer:
A mixture of sodium hypophosphite and phosphine gas is formed upon heating the reaction mixture in an inert atmosphere.

Question 9.
What happens when PCl₅ is heated?
Answer:
Upon heating, PCl₅ dissociates to give molecules of PCl₃ and Cl₂. Actually, the two P—Cl (a) bonds with more bond length break away from the molecule leaving three P— Cl(e) bonds attached to the central P atom since these are more firmly linked
PCl₅ (underrightarrow { heat } ) PCl₃ + Cl₂

Question 10.
Write the balanced equation for the hydrolytic reaction of PCl₅ in heavy water.
Answer:
With heavy water (D₂O) ; phosphorus pentachloride (PCl₅ reacts as follows :

Question 11.
What is the basicity of H₃PO₄ ?
Answer:
The acid is tribasic since it has three P—OH bonds which can release H⁺ ions.

Question 12.
What happens when phosphorus acid (H₃PO₃) is heated ? (C.B.S.E. 2008)
Answer:
In phosphorus acid (H₃PO₃), central atom P is in +3 oxidation state. Upon heating, it gives a mixture PH ₃ (P in -3 oxidation state) and H₃PO₄ (P in +5 oxidation state). This means that phosphorus acid undergoes disproportionation reaction.

Question 13.
List the important sources of sulphur.
Answer:
Combined sulphur exists as sulphates, such as gypsum, epsom, baryte and sulphides such as galena, zinc blende, copper pyrites, etc. Traces of sulphur occur as hydrogen sulphide in volcanoes. Few organic materials like eggs, proteins, garlic, onion, mustard, hair and wool contain sulphur. 0.03 – 0.1% sulphur is present in the earth’s crust.

Question 14.
Write the order of the thermal stability of the hydrides of group 16 elements.
Answer:
The order of thermal stability of hydride is :
H₂O > H₂S > H₂Se > H₂Te
This is related to the bond dissociation enthalpies of the E—H bonds where E stands for the element.
E—H bond :                                           O—H S—H Se—H Te—H
Bond dissociation enthalpy : (kJ mol^-1) 463    347      276  238
Based on bond dissociation enthalpy, H₂O is maximum stable thermally while H₂Te is the least stable.

Question 15.
Why is H₂O a liquid and H₂S a gas?
Answer:

In H₂O, the electronegativity difference between 0(3·5) and H(2·1) is more than difference between S(2·5) and H(21) in H₂S. As a result, O—H bond is more polar than S—H bond. This leads to inter molecular hydrogen bonding in H₂0 molecules while it is almost absent in the molecules of H₂S. The H₂O molecules get associated and consequently exist as liquid (water). The association in H₂S molecules is negligible and it exists as a gas at room temperature.

Question 16.
Which of the following does not react with oxygen directly? Zn, Ti, Pt, Fe
Answer:
Pt being a noble metal does not react with oxygen directly. In contrast, Zn, Ti and Fe are active metals and hence they react with oxygen directly to form their oxides.

Question 17.
Complete the following reactions :
(i) C₂H₄ + O₂ →
(ii) Al + O₂ →
Answer:
(i) C₂H₄ + 3O₂ (underrightarrow { heat } ) 2CO₂ + 2H₂O
(ii) 4Al + 3O₂ (underrightarrow { heat } ) 2Al₂O₃

Question 18.
Why does O₃ act as a powerful oxidising agent ? (C.B.S.E. 2013)
Answer:
Upon heating, ozone (O₃) readily decomposes to give molecular oxygen (O₂) which is more stable along with nascent oxygen (O). The released nascent oxygen readily takes part in oxidation reactions. Therefore, ozone acts as a powerful oxidising agent.
O₃ (underrightarrow { heat } ) O₂ + O (Nascent)

Question 19.
How is ozone estimated quantitatively ?
Answer:
When ozone reacts with an excess of KI solution buffered with a borate buffer (pH = 9.2), iodine is liberated which can be titrated against standard solution of sodium thiosulphate. This is used as a method of estimation of ozone quantitatively.

Question 20.
What happens when sulphur dioxide gas is passed through an aqueous solution of Fe(III) salt ?
Answer:
The gas is a reducing agent and reduces a Fe(III) salt to Fe(II) salt.

Question 21.
Comment on the nature of two S—O bonds formed in SO₂ molecule. Are the two bonds in the molecule equal ?
Answer:
The two S—O bonds in the molecule are equal with bond length equal to 143 pm. This means that SO₂ molecule exhibits two resonating structures as shown below.

Question 22.
How is presence of SO₂ detected ?
Ans.
Presence of SO₂ is detected by bringing a paper dipped in acidified potassium dichromate near the gas. If the paper turns green, it shows the presence of SO₂ gas.

Question 23.
Mention three areas in which H₂SO₄ plays an important role.
Answer:
(i) It is used in the manufacture of fertilizers such as (NH₄)² SO₄ , calcium superphosphate.
(ii)It is used as an electrolyte in storage batteries.
(iii)It is used in petroleum refining, detergent industry and in the manufacture of paints, pigments and dyes.

Question 24.
Write the conditions to maximise the yield of sulphuric acid by Contact process.
Answer:
Catalytic oxidation of sulphur dioxide into sulphur trioxide. Sulphur dixocide is oxidised to sulphur trioxide with air in the presence of V₂O₅ or platinised asbestos acting as the catalyst

Question 25.
Why is ({ K }_{ { a }_{ 2 } }) < ({ K }_{ { a }_{ 1 } }) for H₂SO₄ in water ?
Answer:
H₂SO₄ is a very strong acid in water largely because of its first ionisation to H₃O⁺ and HSO₄^– The ionisation of HSO₄^– to H₃O⁺ and SO₄^2- is very very small. That is why, ({ K }_{ { a }_{ 2 } }) < ({ K }_{ { a }_{ 1 } }).

Question 26.
Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F₂ and Cl₂.
Answer:
Fluorine is a better oxidising agent than chlorine because E°_F2/F- is higher than E°_Cl2/Cl- It is mainly due to low bond dissociation energy, high hydration energy and lower electron gain enthalpy, non-availability of d-orbitals in valence shell, that results in higher reduction potential of F₂ than chlorine.

Question 27.
Give two examples to show the anomalous behaviour of fluorine.
Answer:
The anomalous behaviour of fluorine, the first member of the halogen family as compared to the rest of the members is due to its very small size, very high electronegativity and absence of vacant d-orbitals in the valence shell. It is supported by the following points.
(i) Fluorine shows only negative oxidation state of -1 in its compounds. The other members exhibit both positive and negative oxidation states.
(ii) Fluorine forms hexafluoride with sulphur (SF₆). No other member of the family forms hexahalide with sulphur.

Question 28.
Sea is the greatest source of some halogens. Comment.
Answer:
The name halogen is a Greek Word meaning lsea salt forming’. Sea is a major source of a members of halogens particularly chlorine, bromine and iodine and they exist as the soluble salts of sodium, potassium, calcium, magnesium etc. The deposits of dried up sea water contain sodium chloride and carnallite (KClMgCl₂.6H₂O). Sea weeds contain nearly 0-5 percent of iodine. Similarly Chile saltpeter contains about 0-2% of sodium iodate (NaIO₃).

Question 29.
Give reason for the bleaching action of Cl₂.
Answer:
Bleaching by chlorine occurs in the presence of moisture. In fact, it releases nascent oxygen on reacting with HO which carries bleaching. Since the reaction cannot be reversed, the bleaching by chlorine is permanent.
Chlorine bleaches by oxidation Cl₂ + H₂O → HCl + HOCl → HCl + [O]
The nascent oxygen reacts with dye to make it colourless.

Question 30.
Name two poisonous gases which can be prepared from chlorine gas.
Answer:
Two poisonous gases are phosgene and mustard gas.

Question 31.
Why is ICl more reactive than I₂ ? (C.B.S.E. Outside Delhi 2012)
Answer:
The reactivity of ICl is due to its polar nature (I—Cl). Iodine (I₂) being non-polar is comparatively less reactive chemically.

Question 32.
Why is helium used in diving apparatus?
Answer:
A mixture of helium and oxygen does not cause pain due to very low solubility of helium in blood as compared to nitrogen.

Question 33.
Balance the equation : XeF₆ + H₂O → XeO₂F₂ +HF (H.P. Board 2013)
Answer:

Question 34.
Why has it been difficult to study the chemistry of radon ?
Answer:
Radon (Rn) is a radioactive element with very short half period of 3-82 days. Therefore, it becomes quite difficult to study the details about the chemistry of the element.

NCERT EXERCISE

Question 1.
Discuss the general characteristics of group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionisation enthalpy and electronegativity.
Answer:
The valence shel 1 electronic configuration of group 15 elements is ns²np³. Due to half- filled p-orbitals, these elements have extra stability associated with them.

The common oxidation states of these elements are – 3, +3 and +5. The tendency to exhibit -3 oxidation state decreases down the group. The stability of +5 state decreases and that of +3 state increases down the group due to inert pair effect.

The size of group 15 elements increases down the group. There is a considerable increase in covalent radius from N to P. However, from As to Bi only a small increase in covalent radius is observed. This is due to presence of completely filled d and or f orbitals in heavier members.

Down the group, ionisation enthalpy decreases due to increase in atomic size. Due to stable half-filled configuration, they have much greater value than that of group 14 elements.

The electronegativity value, in general, decreases down the group with increasing atomic size. However, amongst the heavier elements, the difference is not that much pronounced.

Question 2.
Why does the reactivity of nitrogen differ from phosphorus?
Answer:
N₂ exist as a diatomic molecule containing triple bond.batoeen two N-atoms. Due to the presence —of triple bond betweep the two N-atoms, the bond dissociation energy is large (941 .4 kJ mol^-1 ). As aresult of this N₂ is inert and unreactive whereas, phosphorus exists as a tetratomic molecule, containg P – P single bond. Due to the presence of single bond, the bond dissociation energy is weaker (213 kJmol^-1 ) than N a N triple bond (941 .4 kJ mol^-1 ) and moreover due to presence of angular strain in P₄ tetrahedra. As a result of this, phosphorus is much more reactive than nitrogen.

Question 3.
Discuss the trends in chemical reactivity of group 15 elements.
Answer:
(a) Hydrides: The hybrides are covalent with pyramidal structures  and the central atom is sp³ hybrilised. The presence of the lone pair of electrons on the central atom distorts the geometry of the molecules and the bond angle less tham that of a regular tetrahedron.

(b) Halides: Elements of group 15 form two types of halides viz. trihalides and pentahalides. The halides are predominantly basic (Lewis bases) in nature and have lone pair of electrons (central atom is sp3 hybridized). The pentahalides are thermally less stable than the trihalides.

(c) Oxides: All the elements of this group form two types of oxides ie., M₂O₃ and M₂O₅ and are called trioxides and pentoxides

Question 4.
Why does NH₃ form hydrogen bonding while PH₃ does not ?
Answer:
The N—H bond in ammonia is quite polar on account of the electronegativity difference of N (3·0) and H (2·1). On the contrary, P—H bond in phosphine is almost non-polar because both P and H atoms have almost same electronegativity (21). Due to polarity, intermolecular hydrogen bonding is present in the molecules of ammonia but not in those of phosphine.

Question 5.
How is nitrogen prepared in the laboratory ? Write the chemical equations of the reactions involved.
Answer:
Laboratory preparation: Dinitrogen is prepared in the laboratory by heating a solution containing an equivalent amount of sodium nitrite and ammonium chloride.

Question 6.
How is ammonia manufactured industrially ?
Answer:
Ammonia is prepared on commercial scale by Haber’s Process from dinitrogen and dihydrogen by the following chemical reaction
N₂ + 3H₂ ⇌ 2NH₃; ∆_fH° = – 46.1 kJ mon^-1
Dihydrogen needed for the commercial preparation of ammonia is obtained by the electrolysis of water while dinitrogen is obtained from the liquefied air as a fractional distillation. The two gases are purified and also dried.
These are compressed to about 200 atmosphere pressure and are then led into the catalyst chamber packed with the catalyst and the promoter.

Question 7.
Illustrate how copper metal gives different products on reaction with HNO₃.
Answer:
Anhydrous nitric acid is colourless, fuming and pungrnt smelling liquid. However, it acquires a yellowish brown colour in the presence of sun light. HNO₃ decomposes to give NO₂ gas which dissolves and imparts it yellowish brown colour.

On strong heating, the acid decomposes to give NO₂ and O₂

Question 8.
Give the resonating structures of NO₂ and N₂O₅.
Answer:

Question 9.
The HNH angle value is higher than those of HPH, HAsH and HSbH angles ; why ?
Answer:


The central atom (E) in all the hydrides is sp³ hybridised. However, its electronegativity decreases and atomic size increases down the group. As a result, there is a gradual decrease in the force of repulsion in the shared electron pairs around the central atom. This leads to decrease in the bond angle.

Question 10.
Why does R₃P = O exist but R₃N = O does not (R = alkyl group)?
Answer:
Nitrogen does not contains d-orbitals. As a result, it cannot expand its covalency beyond four and cannot form pπ – dπ multiple bonds. In constrast, P contains the d-orbitals, and can expand its covalency beyond 4 and can form pπ-dπ multiple bonds. Hence R₃P = O exist but R₃N = O does not.

Question 11.
Explain why is NH₃ basic while PH₃ is feebly basic in nature. (C.B.S.E. Outside Delhi 2008, 2009, Jharkhand Board 2009)
Answer:
Both NH₃ and PH₃ are Lewis bases due to the presence of lone electron pair on the central atom. However, NH₃ is more basic than PH₃. The atomic size of nitrogen (Atomic radius = 70 pm) is less than that of phosphorus (Atomic radius = 110 pm) As a result, electron density on the nitrogen atom is more than on phosphorus. This means that electron releasing tendency of ammonia is also more and is therefore, a stronger base than phosphine.

Question 12.
Nitrogen exists as diatomic molecule (N₂) while phosphorus as tetra-atomic molecule (P₄). Why ?
Answer:
Nitrogen is diatomic gaseous molecule at ordinary temperature due to its ability to form pπ – pπ multiple bonds. The molecule has one σ and two π – bonds. Phosphorus exists as discrete tetratomic tetrahedral molecules as these are not capable of forming multiple bonds due to repulsion between non-bonded electrons of the inner core.

Question 13.
Write the main difference between the properties of white and red phosphorus. (C.B.S.E. Delhi 2012)
Answer:

Question 14.
Why does nitrogen show catenation properties less than phosphorus?
Answer:
The extent of catenation depends upon the strength of the element – element bond. The N – N bond strength (159 kJ mol^-1 ) is weaker than P – P bond strength (213 kJ mol^-1 ). Thus, nitrogen shows less catenation properties than phosphorus.

Question 15.
Give one disproportionation reaction of phosphorus acid (H₃PO₃).
Answer:
Upon heating to about 573 K, phosphorus acid undergoes disproportionation as follows :

Question 16.
Can PCl₅ act as oxidising as well as reducing agent ? Justify.
Answer:
In general, the molecules of a substance can behave as a reducing agent if the central atom is in a position to increase its oxidation number. Similarly, they can act as an oxidising agent if the central atom is in a position to decrease its oxidation number. Now, the maximum oxidation state or oxidation number of phosphorus (P) is +5. It cannot increase the same but at the sametime can decrease its oxidation number. In PCl₅, oxidation number of P is already +5. It therefore, cannot act as a reducing agent. However it can behave as an oxidising agent in certain reactions in which its oxidation number decreases. For example,

Question 17.
Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation states and hydride formation.
Answer:
The members of the oxygen family are placed in group 16 of p-block. Their inclusion in the same group is justified on the basis of following characteristics.
1. Electronic configuration. Members of the family has ns2p4 configuration. Their group (10 + 6) is 16.
2. Oxidation states. With exception of oxygen which exhibits -2 oxidation state in its compounds (OF₂ and H₂O₂ are exceptions), rest of the members of the family show variable oxidation states (-2, +2, +4, +6) in their compounds.
3. Hydride formation. All the members of the family form covalent hydrides (MH₂) in which the central atom is sp³ hybridised. These have angular structures and their characteristics show regular gredation.
FeS + H₂SO₄ (dil.) → FeSO₄ + H₂S
Na₂Se + H₂SO₄ (dil.) → Na₂SO₄ + H₂Se

Question 18.
Why is dioxygen a gas while sulphur is a solid ? (C.B.S.E. Delhi 2013)
Answer:
Oxygen atom has the tendency to form multiple bonds (pπ – pπ interaction) with other oxygen atom on account of small size while this tendency is missing in sulphur atom. The bond energy of oxygen-oxygen double bond (0 = 0) is quite large (about three times that of oxygen-oxygen single bond, O – O = 34.9 kcal mol^-1) while sulphur-sulphur double bond (S = S) is not so large (less than double of sulphur-sulphur single bond, S – S = 63.8 kcal mob1). As a resul —O—O—O— chains are less stable as compared to O = O molecule while —S—S—S— chains are more stable than S = S molecule. Therefore, at room temperature, while oxygen exists as a diatomic gas molecule, sulphur exists as S₈ solid.

Question 19.
Knowing the electron gain enthalpy values for O → O^– and O → O^2- as – 141 kJ mol^-1 and 702 kJ mol^-1 respectively, how can you account for the formation of a large number of oxides having O^2- species and not O^– ?
Answer:
According to available data :
O + e^– → O^–; ∆_(eg) H = – 141 kJ mol^-1
O + 2e^– → O^2-; ∆_(eg) H = + 702 kJ mol ^-1
Although the formation of divalent anion (O^2-) needs more energy as compared to monovalent anion (O^–) where energy is actually released, still in large number of oxides (e.g Na₂O, K₂O, CaO etc.) oxygen is divalent in nature. This is on account of a more stable crystal lattice because of greater magnitude of electrostatic forces of attraction involving divalent oxygen than the oxides in which oxygen is monovalent in nature.

Question 20.
Which aerosols deplete ozone ?
Answer:
Aerosols or chlorofluorocarbons (CFC’s) such as freon (CCl₂F₂) deplete ozone layer by supplying chlorine free radicals (Cl) which convert ozone into oxygen.

Question 21.
Describe the manufacture of H₂SO₄ by Contact process.
Answer:
Sulphuric acid is manufactured by the Contact process which involves three steps :

1. burning of sulphur or sulphide ores in air to generate SO₂.
2. conversion of SO₂ to SO₃ by the reaction with oxygen in the presr nee of a catalyst (V₂O₅), and
3. absorption of SO₃ in H₂SO₄ to give oleum (H₂S₂O₇).

A flow diagram for the manufacture of sulphuric acid is shown in the figure. The SO₂ produced is purified by removing dust and other impurities such as arsenic compounds. The key step in the manufacture of H₂SO₄ is the catalytic oxidation of SO₂ with O₂ to give SO₃ in the presence of V₂O₅ (catalyst).

The reaction is exothermic, reversible and the forward reaction leads to a decrease in volume. Therefore, low temperature and high pressure are the favourable conditions for maximum yield. But the temperature should not be very low otherwise rate of reaction will become slow.

In practice, the plant is operated at a pressure of 2 bar and a temperature of 720 K. The SO₃ gas from the catalytic converter is absorbed in concentrated H₂SO₄ to produce oleum. Dilution of oleum with water give H₂SO₄ of the desired concentration. In the industry, two steps are carried out simultaneously to make the process a continuous one and also to reduce the cost.

Question 22.
How is SO₂ an air pollutant?
Answer:
(i) Sulphur dioxide released in the atmosphere during the combustion of fuels combines with H20 molecules and oxygen present to form sulphuric acid. The acid being poisonous in
SO₂ + 1/2O₂ + H₂O → H₂SO₄
nature causes pollution. It causes the corrosion of statues and monuments made from marble (CaC03) due to the formation of sulphate.
CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂
(ii) Sulphur dioxide adversely affects respiratory tract due to its poisonous as well as irritating nature. It causes throat infection as well as irritation in the eyes.
(iii) Even very low concentration of the gas (0·03 ppm) has a very damaging effect on plants and vegetation. This is called chlorosis. It slows down the formation of chlorophyll and the leaves slowly wither.

Question 23.
Why are halogens strong oxidising agents?
Answer:
The halogens are strong oxidising agents due to
low bond dissociation enthalpy, high electronegativity and large negative electron gain enthalpy.

Question 24.
Explain why does fluorine form only one oxoacid (HOF).
Answer:
The members of the halogen family with the exception of fluorine show variable oxidation states due to the availability of rf-orbitals for the bond formation. They form a number of oxoacids such as HOX, HOXO, HOXO₂ and HOXO₃. However, fluorine which is highly electronegative and has no rf-orbitals, forms only one oxoacid (HOF) in which its oxidation state is +1.

Question 25.
Explain why inspite of nearly the same electronegativity, nitrogen is involved in hydrogen bonding while chlorine is not.
Answer:
Both nitrogen (N) and chlorine (Cl) have electronegativity of 3·0. However, only nitrogen is involved in the hydrogen bonds (e.gNH₃) and not chlorine. This is an account of smaller atomic size of nitrogen (atomic radius = 70 pm) as compared to chlorine (atomic radius = 99 pm). Therefore, N can cause greater polarisation of N—H bond than Cl in case of Cl—H bond. Consequently, N atom is involved in hydrogen bonding and not chlorine.

Question 26.
Write two uses of ClO₂.
Answer:
ClO₂ is a strong oxidising agent. Therefore,
(i) It acts as bleaching agent for paper pulp in paper industry and in textile industry.
(ii) It acts as germicide for disinfecting water.

Question 27.
Why are halogens coloured ?
Answer:
All the halogens are coloured in nature. The colour deepens with the increase in atomic number of the element from fluorine to iodine.

The cause of the colour is due to absorption of energy from the vivible light by the molecules fot the excitation of outer non -bonded electrons to higher energy levels. The excitation energy depends upon the size of the atom. Fluorine has the smallest size and the force of attraction between the nucleus and electrons is very large.

Question 28.
Write the reactions of F₂ and Cl₂ with water.
Answer:

Question 29.
How can you prepare Cl₂ from HCl and HCl from Cl₂ ? Write chemical equations only.
Answer:
(i) HCl can be oxidised to chlorine with the help of a number of oxidising agents like MnO₂, KMnO₄, K₂Cr₂O₇ etc.
Mn0O₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O
(ii) Cl₂ can be reduced to HCl by reacting with H₂ in the presence of sunlight. The gas on passing through water dissolves to form hydrochloric acid

Question 30.
What inspired N. Bartlett for carrying out the reaction between Xe and PtF₆ ?
Answer:
The X-ray study of the compound has shown it be a crystalline solid consisting of O₂⁺ ans (PtF₆]^– ions. In this reaction, PtF₆ has oxidised O₂ to O₂⁺ ion. Bartlett through than PtF₆ should Xe to xe⁺ since first ionisation enthalpy of xenon (1176 kJ mol^-1) is quite close to that of O₂ (1180 kJ mol^-1).

Question 31.
What is the oxidation state of phosphorus in the following ?
(a) H₃PO₃
(b) PCl₃
(c) Ca₃P₂
(d) Na₃PO₄
(e) POF₃
Answer:

Question 32.
Write balanced equations for the following :
(i) NaCl is heated with sulphuric acid in the presence of MnO₂
(iii) Chlorine gas is passed into a solution of Nal in water.
Answer:

Question 33.
How are xenon fluorides XeF₂, XeF₄ and XeF₆ obtained ?
Answer:
Xenon forms three binary fluorides, XeF₂, XeF₄ and XeF₆ by the direct reaction of elements under appropriate experimental conditions.

Question 34.
With what neutral molecule is CIO^– isoelectronic? Is that molecule a Lewis base?
Answer:
CIO^– has 17 + 8 +1 = 26 electrons.Also, OF₂ has (8 + 2 x 9) = 26 electrons, and ClF has (17 + 9) = 26 electrons.Out of these, ClF can act as Lewis base. The chlorine atom has three lone pair of electrons which it donates to form compounds like ClF₃, ClF₅ and ClF₇

Question 35.
How are XeO₃ and XeOF₄ prepared ?
Answer:
Preparation of XeO₃. By complete hydrolysis of XeF₆.
XeF₆ + 3H₂O → XeO₃+ 6HF.
Preparation ofXeOF₄. By partial hydrolysis of XeF₆.
XeF₆ + H₂O → XeOF₄+ 2HF.

Question 36.
Arrange the following in the order of property indicated for each set :
(i) F₂, Cl₂, Br₂,I₂ – increasing bond dissociation enthalpy.
(ii) HF, HCl, HBr, HI – increasing acid strength.
(iii) NH₃, PH₃, AsH₃, SbH₃, BiH₃ – increasing base strength. (Pb. Board 2009, C.B.S. Sample Paper 2017)
Answer:
(i) I₂ (151-1 kJmol^-1) < F₂ (158-8 kJ mol^-1) < Br₂ (192-8 kJ mol^-1) < Cl₂ (242-6 kJ mol^-1)
(ii) HF < HCl < HBr < HI
(iii) BiH₃ < SbH₃ < ASH₃ < PH₃ < NH₃.

Question 37.
Which one of the following does not exist ?
(i) XeOF₄
(ii) NeF₂
(iii) XeF₂
(iv) XeF₆.
Answer:
NeF₂ does not exist because the element Ne(Z = 10) with 1s²2s²2p⁶ does not have vacant 2d orbitals. As such there is no scope of any electron promotion even by a highly electronegative element.

Question 38.
Give the formula and describe the structure of a noble gas species which is isostructural with :
(i) (IC{ l }_{ 4 }^{ – })
(ii) (IB{ r }_{ 2 }^{ – })
(iii) (Br{ O }_{ 3 }^{ – })
Answer:

(i) Structure of (IC{ l }_{ 4 }^{ – }). The central I atom has in all 8 electrons (7 valence electrons +1 due to negative charge). Out of these, it shares 4 electrons with four atoms of Cl and the remaining four electrons constitute two lone pairs. In all, there are six pairs. The structure of the ion must be octahedral or distorted square planar in order to minimise the forces of repulsion among the two electrons pairs. (IC{ l }_{ 4 }^{ – }) has (7 + 4 x 7 + 1) = 36 valence electrons and it iso-electronic and iso-structural with XeF₄ (8 + 4 x 7) which has also 36 valence electrons.

(ii) Structure of (IB{ r }_{ 2 }^{ – }). In (IB{ r }_{ 2 }^{ – }) ion, the central I atom has 8 valence electrons (7 + 1). Out of these, it shares 2 electrons with two atoms of Br and the remaining 6 electrons constitute three lone pairs. In all, there are five pairs. The structure of the ion must be trigonal bipyramidal or linear in order to minimise the force of repulsion among the three lone electrons pairs. IBrf has (7 + 2 x 7 + 1) = 22 valence electrons and is isoelectronic as well as iso-structural with noble gas species XeF₂ which has also 22 (8 + 2 x 7) electrons.

(iii) Structure of (Br{ O }_{ 3 }^{ – }) ion. In (Br{ O }_{ 3 }^{ – }) ion, the central Br atom has 8 valence electrons (7 + 1). Out of these, it shares 4 with two atoms of O forming Br=0 bonds. Out of the remaining four electrons, 2 are donated to the third O atom and account for its negative charge. The remaining 2 electrons constitute one lone pair. In order to minimise the force of repulsion, the structure of (Br{ O }_{ 3 }^{ – }) ion must be pyramidal. (Br{ O }_{ 3 }^{ – }) ion has (7 + 3 x 6 + 1) = 26 valence electrons and is isoelectronic as well as iso-structural with noble gas species Xe03 which has also 26 (8 + 3 x 6) electrons.

Question 39.
Why do noble gases have comparatively large atomic sizes?
Answer:
This is because noble gases have only van der Waal’s radii while others have covalent radii, van der Waal’s radii are larger than covalent radii.

Question 40.
List the uses of neon and argon gases.
Answer:
Uses of Neon:

• Neon lamps are used in botanical gardens and also in green houses as they are useful in the formation of chlorophyll and thus, stimulate plant growth.
• It is used in filling sodium vapour lamps.
• It is used in safety devices for protecting certain electrical instruments (voltmeters, relays, rectifiers etc.)

Uses of Argon:

• It is used in metal flament electric lamps since it increases the lofe of the tungsten filament by retarding its vapourisation.
• A mixture of argon and mercury vapours is used in fluorescent tubes.
• It is used to create an inert atmosphere for welding and for carrying certain chemical reactions.

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