Ncert Solution Class 12 Chemistry Chapter 7 Questions And Answers, Notes & Numericals PDF

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Ncert Solution Class 12 Chemistry Chapter 7 Questions And Answers, Notes & Numericals PDF, Ncert Solution Class 12 Chemistry Chapter 7 Questions And Answers, short notes of p-block elements class 12 pdf


Ncert Solution Class 12 Chemistry Chapter 7

Class12th 
Chapter No07
ProvidingQuestions And Answers, Notes & Numericals PDF
Chapter NameThe p-Block Elements
BoardCBSE
Book NCERT
SubjectChemistry
Medium English
Study MaterialsFree VVI Study Materials are Available
Download PDF class 12 Chemistry Ncert Solutions Chapter 7 pdf

Chapter 7 | The p-Block Elements


1. In the long form of a Periodic Table, the p-block consists of group 13 to group 18 elements. Because p-orbitals of valence shell are involved in bond formation, they are known as p-block elements,

2. Gradation in properties 

(a) Ionization energy increases across a period and decreases down the group. 

(b) Electronegativity increases across a period and decreases down the group. 

(c) Atomic radius decreases in a period and increases down the group. of

(d) P, O,S and halogens are very reactive while other elements p-block are less reactive.

(e) Reducing character decreases and oxidizing character increases across a period from group 13 to 17. 3. Nitrogen does not have d-orbitals, so it does not form pentahalides while phosphorus forms the same.

4. Nitrous oxide or nitrogen (I) oxide (NO) is known as laughing gas. 5. Phosphoric acid (H, PO,) is tribasic while phosphorous add (H, PO3) is dibasic in nature.

6. Oxygen exhibits a -2 oxidation state in most of its compounds. Other members of the group exhibit several oxidation states i.e. -2, +2, +4, +6 etc. 7. SO, acts as a reducing agent as well as oxidizing agent.

8. Bleaching action of chlorine is permanent while that of SO, is temporary. 9. Sulphuric acid is manufactured by the contact process. It is a good

dehydrating agent as well as an oxidizing agent. 

10. First member of each group shows strange behaviour from the rest of the elements.

11. Interhalogen compounds are generally more reactive than individual halogens. 

12. Group 18 consists of unreactive elements, called noble gases, having ns2, np valence shell configuration. Helium has a 1s configuration because it has only K shell.

13. Radon is obtained as the decay product of radium. 

14. Inert or noble gases are present in very small amounts (~1%) in the atmosphere. They have various applications. 

15. The first compound of xenon, Xe’ [Pt XI was prepared by Neil Bartlett in 1962.


class 12 chemistry chapter 7 Notes & Intext Questions


Question 1. Why are pentahalides more covalent than trihalides? Covalent character increases with an increase in oxidation state.

Solution. The higher the positive oxidation state of the central atom, the more will be its polarising power, which in turn increases the covalent character of the bond formed between the central atom and the halogen atom. In pentahalides, the central atom is in a +5 oxidation state while in trihalides, it is in a +3 oxidation state. Therefore, pentahalides are more covalent than trihalides.

Question 2. Why is BiH, the strongest reducing agent amongst all the hydrides of group 15 elements?

Reducing agents have a tendency to lose hydrogen atoms and the lesser the bond dissociation enthalpy of the E-H bond more is the reducing character of hydrides.

Solution. Among the 15 group elements, the size of the Bi atom is the largest and hence, the Bi-H bond length is the largest or Bi-H bond dissociation energy is the lowest. That’s why the Bi-H bond dissociates more readily than the other hydrides of the group, and hence, BiH is the strongest reducing agent.

Question 3. Why is N₂ less reactive at room temperature? 

Solution. Dinitrogen is inert or less reactive because the bond enthalpy of the N N bond is very high. 

Question 4. Mention the conditions required to maximize the yield of ammonia.

Write the balance thermochemical equation of the manufacture of ammonia by Haber’s process. Mention temperature, and pressure according to Le Chatelier’s principle, catalyst and promotor.

Solution. Ammonia is produced by Haber’s process as

N₂ (g) + 3H2(g)→→Fe-Mo. 700K, 200 atm

→2NH3(g) ∆H°=-46.1 kJ mol-¹

The yield of ammonia is favoured by high pressure according to Le-Chatelier’s principle. Other conditions, that favour the production of ammonia are as follows:

(i) Temperature – approximately 700 K 

(ii) Pressure-200 atm or 200 x 10’5 Pa

(iii) Catalyst-Iron oxide

(iv) Promotor – Molybdenum or K₂O and Al₂O3 

Question 5. How does ammonia react with a solution of Cu2+p

Copper ion forms a complex with ammonia in solution.

Solution. When ammonia (aqueous solution is ammonium hydroxide) reacts with a solution of Cu2+, a deep blue solution is obtained due to the formation of tetraamine copper (II) ion. 

Cu2+ (aq) + 4NH4OH(aq) →→→ [Cu(NH3)4]²+4H₂0 [Tetraamine copper (II) ion (Deep blue solution)]

Question 6. What is the covalence of nitrogen in N₂O? To find covalence draw the structure of NO, and count the number of bonds formed by each N atom, which is equal to its covalence,

Solution. Structural formula of N₂O5

Since the N atom has 4 shared electron pairs; the valence of N is 4.



Question 7. The bond angle in PH is higher than that in PH3. Why? Find the number of bond pairs and lone pairs. Presence of lone pairs decreases the bond angle.

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Solution. In both PH and PH,, the phosphorus atom is sp³ hybridized. In PH, all the four orbitals are bonded whereas in PH,, there is a lone pair of electrons too. Due to lone pair-bonded pair repulsion in PH, the bond angle is less than 109.5°.

Question 8. What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO₂? 

Solution. White phosphorus dissolves in boiling NaOH in an inert atmosphere of CO, producing phosphine (PH) gas and sodium hypophosphite (NaH PO₂).

P4 (White phosphorus) + 3NaOH(Sodium hydroxide) +3H₂O→→→Heat/CO₂ atmosphere→→→3NaH PO₂(Sodium hypophosphite)+ PH3 ↑(Phosphine) 

Question 9. What happens when PCI, is heated? Draw the structure of PCL to find which bonds are weaker as only these bonds break on heating.

Solution. In PCI,, there are 5P-Cl bonds, out of which three are equatorial (longer) and two are axial (shorter). When PCI, is heated strongly, two less stable axial bonds break and phosphorus trichloride (PCI) is formed.

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Phosphorus pentachloride →Strong heating→→ Phosphorus trichloride + Cl₂ 

If PCI5 is warmed slightly, it sublimes only without decomposing.

Question 10. List the important sources of sulphur.

Solution. Occurrence or sources of sulphur In the earth’s crust, the percentage of sulphur is only 0.03 to 0.1%. In a combined state, it occurs (i) In the form of sulphates, e.g., gypsum (CaSO, 2H, O), Epsom salt

(MgSO, 7H,O), baryte (BaSO4). 

(ii) In the form of sulphides, e.g., galena (PbS), zinc blende (ZnS), copper pyrites (CuFeS2).

In volcanoes, traces of sulphur occur as H.S. In organic materials such as eggs, proteins, garlic, onion, mustard, hair and wool, sulphur is present in trace amounts,

Question 11. Write the order of thermal stability of the hydrides of group 16 elements.

Thermal stability bond dissociation energy,

Adiss H (H-E)/kJ mol¹ ∞1/ Size of central atom

(where, E = 16 group elements)

Solution. The thermal stability of hydrides of group 16 elements is directly proportional to the bond dissociation enthalpy of the H-E bond. On moving down the group, bond dissociation energy decreases because bond length increases. Thus, the order of bond dissociation energy is

H₂O>H₂S>H₂Se>H₂Te>H₂Po

This is also the order of thermal stability.

Question 12. Why is H2O a liquid and H₂S a gas? Discuss different intermolecular interactions as these interactions are stronger in liquid as compared to gas.

Solution. The difference in electronegativity values of O(3.5) and H(2.1) is more than the difference between the electronegativity values of H(2.1) and S (2.5) i.e., the O-H bond is more polar than the S-H bond. That is why H-bonding is present among water molecules but absent in H S. Thus, strong intermolecular interactions cause water to exist as a liquid but due to weak van der Waals’ forces HS exists as a gas.

Question 13. Which of the following does not react with oxygen directly? Zn, Ti, Pt, Fe. Only noble metals do not react directly with oxygen, so select the noble metal.

Solution. Platinum (Pt).

Question 14. Complete the following reactions.

(i) C₂H4 + 0₂ →→→

(ii) 4 Al+3o₂→→→

(1) Hydrocarbons (compounds of carbon and hydrogen) on complete combustion give CO, and H₂O. (1) Metals react with

Solution. (1) 

C₂H4+ 30₂ →→→2CO₂+ 2H2O 

(ii) 4Al + 30₂→→→ 2A1₂03 

Question 15. Why does O, act as a powerful oxidising agent? Consider the decomposition of ozone

Solution. Because ozone liberates nascent oxygen very easily.

Ozone

O3→→∆→o₂+[o]Nascent oxygen

Question 16. How is O; estimated quantitatively? 

Solution. When ozone reacts with an excess of potassium iodide solution buffered with a borate buffer (pH 9.2), iodine is liberated which can be titrated against a standard solution of sodium thiosulphate. In this way, O can be estimated quantitatively.

Question 17. What happens when sulphur dioxide is passed through an aqueous solution of Fe (III) salt?

Fe+ acts as an oxidising agent. It oxidises SO, into SO and itself gets reduced to Fe²* Solution. When SO, is passed through an aqueous solution of Fe (III), i.e., ferric salt, it is reduced to Fe (II) i.e., ferrous salt. Here, SO, acts as a reducing agent.

2Fe³+ SO₂ +2H,O→→→→ 2Fe²+ + SO²/4+4H+

Question 18. Comment on the nature of two S-O bonds formed in SO₂ molecules. Are the two S-O bonds in this molecule equal?

Solution. The two S-O bonds in SO₂ molecules are covalent in nature. These are equal with bond length = 143 pm. The resonating structures of SO₂ are as follows

SO₂ is a resonance hybrid of these two canonical forms. 

Question 19. How is the presence of SO₂ detected? 

Solution. SO₂ has a pungent odour. Two tests to detect the presence of SO₂, are as follows: 

1. SO₂, decolourises acidified KMnO, solution.

5SO₂+2MnO4 + 2H2→→→5S0-²/4+4H + 2Mn2

2. SO, changes the colour of acidified potassium dichromate solution from orange to green. 

3 SO₂+Cr₂O-²/7+2H→→→→3SO-²4+2C3+ H₂O

Question 20. Mention three areas in which H₂SO, plays an important role.

Solution. Uses of sulphuric acid : 

(i) It is used in the manufacture of pigments, paints and dyestuff intermediate.

(ii) It is used in petroleum refining.

(iii) It is also used in the fertiliser industry.

Question 21. Write the conditions to maximise the yield of H₂SO, by contact process. Write the balanced thermochemical equation (equation showing heat change) for the manufacture of H₂SO, by contact process and define the condition which increases the yield of HSO, on the basis of Le-Chatelier’s principle.

Solution. The key step in the manufacture of H-SO is the catalytic oxidation of SO, to produce SO, in the presence of V₂O

The reaction is exothermic, reversible and the forward reaction results in a decrease in volume. Thus according to Le-Chatelier’s principle,

the forward reaction should be favoured by low temperature and high pressure. But the temperature should not be very low otherwise the rate of reaction will become very slow.

(a) It forms only one oxo-acid while other halogens from a number of (b) Hydrogen fluoride (HF) is a liquid (b.p. 293 K) due to strong

Question 22. The sea is the greatest source of some halogens. Comment

Solution. Sea water contains chlorides, bromides and iodides of sodium, potassium, magnesium and calcium. NaCl is the most abundant (2.5% by mass) among all of these. 

Certain forms of marine life contain iodine in their systems; Various seaweeds, for example, contain up to 0.5% of iodine and chile saltpetre contains up to 0.2% of sodium iodate. NaCl and carnallite, KCl MgCl, 6HO etc are present in the deposits of dried-up seas.

Question 23. Give the reason for the bleaching action of Cl₂. 

Solution. The bleaching action of chlorine is due to its oxidising property. When chlorine reacts with water, it gives nascent oxygen which decolourises the coloured substance.

Cl₂ + H₂O→→→2HCI + [O]

Coloured substance + [O]→→→ Colourless substance

The bleaching action of chlorine creates permanent effects. It bleaches

the vegetable or organic matter in the presence of moisture.

Question 24. Name some poisonous gases which can be prepared from chlorine gas.

Solution. 

(i) Phosgene (COCI₂) 

(ii) Tear gas (CCI3NO₂) 

(iii) Mustard gas (CICH2₂CH₂SCH₂CH₂CI)

Question 25. Why is ICL more reactive than I?

Solution. The I-Cl bond in ICI is weaker than the I-I bond in 12. That is why ICI is more reactive than 1₂,

Note In all interhalogen compounds, the X-X bond is weaker than X-X or X-X’bond. So, these compounds are more reactive than the individual halogens.

Question 26. Why is helium used in diving apparatus?

Consider the solubility of helium in blood.

Solution. Helium is used in diving apparatus due to its very low solubility in blood. 

Question 27. Balance the following equation :

Solution.

XeF6+H₂0→→→ XeO₂F₂+ HF. 

XeF6 + 2H₂O→→→Xeo₂F₂+4HF

Question 28. Why has it been difficult to study the chemistry of radon? 

Solution. Radon is a radioactive element with a very short half-life of 3.82 days. That is why the study of the chemistry of radon is a difficult task.


Ncert Solution Class 12 Chemistry Chapter 7 Exercise


Question 1. Discuss the general characteristic of group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionisation enthalpy and electronegativity.

Solution. General characteristics of group 15 elements.

(1) Electronic configuration Valence shell configuration of elements of group 15 is ns np. the s-orbital of these elements is completely configured extra stable.

filled and p-orbital is half-filled, making their electronic

Nitrogen (?N)= [He] 2s22p³

Phosphorus (15P)-[Ne] 3s² 3p3

Arsenic (33 As)=[Ar] 3d104s²4p3

Antimony(51 Sb)= [Kr] 4d105s²5p3

Bismuth (83B)-[Xe] 44-14 5d106s²6p3

(ii) Oxidation states Common oxidation states exhibited by elements of this group are -3,+3 and +5. The tendency to exhibit a -3 oxidation state decreases down the group due to an increase in size and metallic character. Bi (the last member of the group) hardly forms any compound of a -3 oxidation state. The stability of the +5 oxidation state decreases down the group,

whereas that of the +3 oxidation state increases down the group (due to the inert pair effect). N reacts with oxygen showing+1+2+3+4 and + 5 oxidations states. P shows + 1 and + 4 oxidation states in some oxo-acids.

(iii) Atomic size Covalent and ionic radii (in a particular state) increase down the group. From N to P, the covalent radius increases considerably. From As to Bi, there is only a small increase in covalent radius. The reason is the presence of completely filled d and/or f orbitals in heavier members. 

(iv) Ionization enthalpy ionisation enthalpy decreases down the group due to a gradual increase in atomic size. I.E., group 15 elements is greater than that of group 14 elements and group 16 elements in the corresponding periods. The order of successive ionisation enthalpies as expected are AH, <A, H₂<A, H.

(v) Electronegativity The electronegativity values decrease down the group with increasing atomic size. 

Question 2. Why phosphorus? Consider bond dissociation enthalpy.

Solution. Nitrogen exists as a diatomic molecule (N=N). Due to the presence of a triple bond between the two N-atoms, the bond dissociation energy is large (941.4 kJ mol¹). Thus, nitrogen is inert and unreactive in its elemental state. In contrast, white or yellow phosphorus exists as a triatomic molecule (P). 

Since, the P-P single bond is much weaker (213 kJ mol¹) than N=N triple bond (941.4 kJ mol¹) therefore, phosphorus is much more reactive than nitrogen.

Question 3. Discuss the trends in the chemical reactivity of group 15 elements. Solution. Trends in chemical reactivity of group 15 elements.

1. Oxidation state Same as in Question no,1

2. Reactivity towards hydrogen These elements form covalent hydrides with the formula EH,. [where, E = N, P, As, Sb]

(a) Thermal stability of hydrides decreases down the group. 

(b) Reducing the character of hydrides increases the group. 

(c) Basic character of hydrides decreases down the group.

(d) Boiling point of NH is greater than PH, due to H-bonding. Boiling points of other hydrides increase from PH, onwards. 

3. Reactivity towards halogens All group 15 elements ie., N, P, As, Sb, Bi form

(a) Halides of general formula EX3. Except for NBr, and NI,, all are stable with a pyramidal shape, These get hydrolyzed by water. 

(b) Halides of general formula EX,. These are formed by P, As and Sb but not by N. These halides are trigonal bipyramidal with sp3d hybridization.

4. Reactivity towards oxygen. These elements form oxides of the formula E₂O, and E, O,. However, nitrogen because of its tendency to form рл-рл multiple bonds, forms a range of oxides having oxidation states +1 to + 5 (e.g., NO, NO, N2O3, NO, NO, N₂Os) EO, are more acidic than E, O, Acidic character of oxides decreases down the group.

Question 4. Illustrate how copper metal can give different products on reaction with HNO Solution. 

The products of the reaction of copper with HNO, depend on the concentration of HNO, used

6) Copper metal reacts with dilute HNO, to form nitrogen (II) oxide (NO)

3Cu+8HNO3(dilute)→→→3Cu(NO3)₂+2NO+4H₂O

(H) The copper metal reacts with conc. HNO, to form nitrogen (IV) oxide

or nitrogen dioxide (NO) Cu+4HNO, (conc) Cu(NO3)2+2NO₂+2H₂O

Question 5. Give the resonating structures of NO, and N205 Draw all the possible structures in which the octet of all the atoms is complete and which differ only in the position of atoms. NO, is an odd electron molecule. 

KSV P2 CHM APP E01 090 S01

Question 6. The HNH angle value is higher than HPH, HASH and HSbH angles. Why?

Repulsive forces are responsible for the bond angle

Solution. The central atom E (where, E = N, P, As, Sb, Bi) in all the given hydrides is sp hybridised. However, its electronegativity decreases and atomic size increases on moving down the group. Therefore, there is a gradual H decrease in the force of repulsion between the shared electron pairs around the central atom. Thus, the bond angle decreases as we move down the group.

Question 7. Why does RP=O exist but RN-O does not (R = alky group)? The difference in the formation (and not formation) of the dл – pr bond.

Solution. Nitrogen cannot form dx-px multiple bonds with oxygen because d-orbitals do not exist in N-atoms. Thus, the covalency of N cannot go beyond 4. But in R, NO, the valency of N should be 5. So, these compounds do not exist. Whereas d-orbitals are present in phosphorus. So, P can form dr-pл multiple bonds and hence, can expand its covalency beyond 4. Therefore, P forms R, P=O in which the covalency of P is 5.

Question 8. Explain why NH3 is basic while BiH, is only feebly basic. The higher the availability of lone pair for donation, the more will be the basic character.

Solution. The atomic size of nitrogen (atomic radius = 70 pm) is less than that of bismuth (atomic radius = 148 pm). As a result, the electron density on the nitrogen atom is greater than that on the bismuth atom.

 It means that the electron-releasing tendency of ammonia is more than BiH, Therefore, NH, is basic while BiH is only feebly basic. Note Atomic size will be considered while comparing the basicity of NH; with other hydrides of group 15 elements. 

Question 9. Nitrogen exists as a diatomic molecule and phosphorus as P Why?

Solution. Due to its small size and high electronegativity, nitrogen atoms form pл – рr multiple bonds with themselves (triple bond); so it exists as a discrete diatomic molecule in elemental form. But phosphorus atoms have large size and less electronegativity so it forms a single bond instead of a p-pa multiple bond. Therefore, phosphorus exists as a Molecule in its elemental form.

Question 13. Write the main difference between the properties of white Solution. Difference between white phosphorus and red phosphorus.

White (yellow) phosphorusRed phosphorus
1It is a soft waxy solid and can be easily cut with a knife.it is a hard and crystalline solid.
2It has a garlic-like odour.It is odourless.
3White phosphorus is poisonous.Red phosphorus is non-poisonous.
4It is highly reactive.It is less reactive..
5P molecules are held by weak van der Waals forces.P molecules are held by covalent bonds in polymeric structures.
6White phosphorus shows phosphorescence.It does not show phosphorescence.
7Burns easily in Cl2 forming PCI3 and PCI3 Combines with Cl, only on heating. 
8The ignition temperature is low (303 K), so burns easily in the air.Ignition temperature is high (543 K), so does not catch fire easily.

Question 10. Why does nitrogen show catenation properties less than phosphorus?

Consider the bond strength of N-N and P-P bonds.

Solution. The single N-N bond is weaker than the single PP bond due to the high interelectronic repulsion of the non-bonding electrons in N₂, owing to the small bond length. Therefore, the catenation property is weaker in nitrogen as compared to phosphorus.

Question 11. Give the disproportionation reaction of H3PO3.

Those redox reactions in which the same element undergoes oxidation and reduction, are called disproportionation reactions.

Solution. Orthophosphoric acid (or phosphorus acid) disproportionates to produce orthophosphoric acid (or phosphoric and phosphine on heating.

Question 12. Can PCI act as an oxidizing as well as a reducing agent? Justify.

A substance can behave as a reducing agent if the central atom can Increase its oxidation number. In the same way, a substance can behave as an oxidizing agent if the central atom can decrease its own oxidation number.

Solution. In PCI,, the oxidation number of P is + 5 i.e., maximum, which cannot be increased further. So, PCI, cannot behave as a reducing agent. But it can decrease its oxidation number from + 5 to + 3, so it can behave as an oxidizing agent. e.g.,

2 Ag+PCI5 →→→2AgCl + PCI3

Sn+2PC5, — SnC!4 +2PCI3

Question 13. 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. 

Solution. Electronic configuration The electronic configuration of the above-mentioned elements (group 16) is given below.

All these elements have similar valence shell configurations ns2 np. hence their position in group 16 with each other is justified. Oxidation state Oxidation states are mentioned in the above table. 

As we observe that in all elements of group 16 oxidation state 2 is common. S, Se and Te also exhibit + 4 and + 6 oxidation states, and polonium shows +4 oxidation states. 

Oxygen does not exhibit + 4 and + 6 oxidation states due to the absence of d-orbitals. Therefore, their position in group 16 is justified. 

Hydride formation All these elements form hydrides of the general formula H₂E. e.g., H₂O, HS,H, Se, H₂Te and H.Po. All the hydrides with the exception of H₂O, possess reducing properties, have acidic nature and are thermally not stable. On the basis of the above-mentioned similarities, these elements are correctly placed in group 16.

Question 14. Why is oxygen a gas but sulphur a solid?

Consider the bond strength and size of the atom Solution. The O-O bond is weaker than the S-S bond due to greater interelectronic repulsions in small-sized oxygen atoms, moreover being smaller and highly electronegative oxygen forms px -рx multiple bonds. 

So, it exists as O, molecules which are held together by weak van der Waals’ force. 

Thus, oxygen exists as a gas at room temperature. Sulphur has less tendency to form pr-pr multiple bonds, moreover, it has large atomic size, less electronegativity and forms strong S-S single bonds, that’s why it shows more catenation properties and exists as S molecules having puckered ring structure. Hence, sulphur is a solid at room temperature.

Question 15. Knowing the electron gain enthalpy values for 0→→→→→0 and 0-02 as -141 and 702 kJ mol respectively, how can you account for the formation of a large number of oxides having O² species and not O”?

Strong electrostatic attraction forces in the crystal lattice, /e, high lattice energy.

Solution. Given,

O+e →→→O ∆eg H = -141 kJ mol¹

O+2e→→→02 ∆egH = +702 kJ mol-1

As it is obvious from the given data, the formation of a divalent anion (02) needs more energy as compared to a monovalent anion (O) where energy is actually released. Yet the number of oxides with a -2 state of oxygen (e.g., Na2O, KO etc) is more. The reason is that their crystal lattice is more stable due to stronger electrostatic forces of attraction involving divalent oxygen than the oxides in which oxygen is monovalent in nature.

Question 16. Which aerosols deplete ozone?

Solution. Chlorofluorocarbons or freons, oxides sulphur, etc deplete ozone as they react with it. of nitrogen and

Question 17. Describe the manufacture of H₂SO, by contact process.

 Solution. Manufacture of sulphuric acid (contact process) It involves the following steps:

(i) Production of sulphur dioxide is carried out by burning powdered sulphur or roasting of sulphur-rich ores.

(ii) Oxidation of sulphur dioxide to sulphur trioxide

(iii) Conversion of SO, into H, SO, 

(iv) Dilution of oleum and formation of H₂SO, 

Question 18. How is SO, an air pollutant? Solution. Harmful effects of SO₂

(i) Sulphur dioxide is an irritating gas to the respiratory tract. At a concentration of even 5 ppm, it causes throat and eye irritation. This results in coughing, tears and redness in the eyes. It causes breathlessness and affects the voice box also. 

(ii) At low concentrations, it is harmful to plants too. It slows down

the formation of chlorophyll resulting in injured leaves and loss of green colour (chlorosis). 

(iii) SO reacts with moisture (or rainwater) present in the air to form sulphurous acid causing acid rain. It damages the marble of buildings and causes diseases in animals, humans and plants. 

CaCO3 + H₂SO4→→ CaSO, +H₂O + CO₂

Question 19. Why are halogens strong oxidizing agents

Solution. Halogens have a strong electron-accepting tendency due to low bond dissociation enthalpy, high electronegativity and large negative electron gain enthalpy.

X+e →→→→X

So, they easily oxidised various substances and are good oxidising agents.

Question 20. Explain why fluorine forms only one oxo-acid, HOF.

Solution. In fluorine d-orbitals are absent so it shows only +1 oxidation state in oxo-acid, but not +3, +5 or +7. That’s why unlike other members (halogens) fluorine only forms one oxoacid HOF and not others like HOFO, HOFO, and HOFO,

Question 21. Explain why in spite of nearly the same electronegativity, nitrogen forms hydrogen bonding while chlorine does not. 

Solution. The electronegativity of both nitrogen (N), as well as chlorine (Cl), is 3.0. But only nitrogen forms hydrogen bonding, not chlorine. The reason is that the atomic size of N (atomic radius = 70 pm) is less as compared to chlorine (atomic radius =99 pm). Therefore, N can cause greater polarization of the N-H bond than Cl in the case of the Cl-H bond. Hence, the N atom is involved in hydrogen bonding and not chlorine.

Question 22. Write two uses of CIO₂.

Solution. Uses of CIO₂

(i) It is used as a bleaching agent in the paper and textile industry.

(ii) It is used as a germicide for disinfecting water. Question 27. Why are halogens coloured?

Solution. Halogens are coloured due to the absorption of radiations in the visible region. It results in the excitation of valence electrons to a higher energy level. As different halogens absorb different quanta of radiation, they display different colours. Different colours of halogens are:

Fluorine →→→Yellow 

Chlorine →→→Greenish,Yellow

Bromine →→→Iodine

Red →→→Violet

Question 23. Write the reactions of F2 and Cl₂ with water. Solution.

(i) Fluorine reacts with water to produce oxygen and ozone.

2 F₂(g) + 2H2O(1)4HF (aq) + O2(g) 

3F₂(g) + 3H2O(1)— 6HF (aq) + O(g)

Question 24. Why do noble gases have comparatively large atomic sizes?

Solution. Noble gases have comparatively large atomic sizes because they have van der Waals’ radii only which are expected to have larger magnitudes whereas other members of a period have either covalent or metallic radii which are less in magnitude.

Question 25. List the uses of neon and argon gases. Solution. 

Uses of Neon

1. Neon is used in discharge tubes and fluorescent bulbs for advertisement display purposes.

2. Neon bulbs are used in botanical gardens and in greenhouses.

3. Neon is used in voltage regulators and indicators.

Uses of Argon

1. Argon is used for filling electric bulbs.

2. It is used to provide an inert atmosphere in high-temperature metallurgical operations (or welding of metals or alloys). 3. Argon is used in rectifiers and in radio valves.


Selected NCERT Exemplar Problems | Short Answer Type


Question 1. In the preparation of H₂SO, by contact process, why is SO3, not absorbed directly in water to form H₂SO,? 

Solution. Because the reaction is highly exothermic, an acid mist is formed and thus, the reaction becomes difficult to handle.

Question 2. Write the structure of pyrophosphoric acid.

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Question 4. PH forms bubbles when passed slowly in water but NH3 dissolves. Explain why?

Solution. NH forms an H-bond with water so it is soluble but PH does not form H-bond with water so it remains as gas and forms bubbles in the water.

Question 5. In PCl5, phosphorus is in a sp hybridised state but all its five bonds are not equivalent. Justify your answer with reason. 

Solution. It has trigonal bipyramidal geometry, in which two Cl atoms occupy axial positions while three occupy equatorial positions.

Question 6. Why is nitric oxide paramagnetic in a gaseous state but the solid obtained on cooling it, is diamagnetic? 

Solution. In the gaseous state, NO exists as a monomer with one unpaired electron, but in the solid state, it dimerises to N₂O, so no unpaired electron is left. Therefore, NO, is paramagnetic in a gaseous state but diamagnetic in a solid state.

Question 7. Give one reason to explain why CIF exists but FCl3 does.

Solution. Because fluorine is more electronegative as compared to chlorine and has a smaller size. Thus, one large Cl atom can accommodate three smaller F atoms but the reverse is not valid.

Question 8. Out of H2O and H2S, which one has a higher bond angle? Why? 

Solution. The bond angle of H₂O (H-O-H=104.5°) is larger than that of HS (H-S-H = 92°) because oxygen is more electronegative than sulphur therefore, the bond pair electron of the O-H bond will be closer to oxygen and there will be more bond pair-bond pair repulsion between bond pairs of two O-H bonds.

Question 9. SF is known but SCI is not. Why? 

Solution. The fluorine atom is smaller in size so six F ions can surround a sulphur atom. The case is not so with chlorine atoms due to their large size. So, SF is known but SCI is not known due to interionic repulsion between larger Cl ions.


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