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

NCERT Solutions for Class 12 Chemistry Chapter 7 – The p-Block Elements – FREE PDF Download

NCERT Solutions Class 12 Chemistry Chapter p-Block by CoolGyan is a simplified version of your Chemistry textbook. It is very natural to hear that students get stressed out while studying this subject. This is because all the textbooks come with advanced language and it becomes difficult for the students to understand. For this reason, we always focus on providing study material with simple language. So, to excel in this subject, make sure to opt for our NCERT Solutions of Chapter 7 p-Block elements Class 12.

Revision Notes Class 12 chemistry
NCERT Solutions Class 12 chemistry

Subtopics of Class 12 Chemistry Chapter 7 – The p Block Elements

1. Group 15 Elements
2. Dinitrogen
3. Ammonia
4. Oxides of Nitrogen
5. Nitric Acid
6. Phosphorus — Allotropic Forms
7. Phosphine
8. Phosphorus Halides
9. Oxoacids of Phosphorus
10. Group 16 Elements
11. Dioxygen
12. Simple Oxides
13. Ozone
14. Sulphur — Allotropic Forms
15. Sulphur Dioxide
16. Oxoacids of Sulphur
17. Sulphuric Acid
18. Group 17 Elements
19. Chlorine
20. Hydrogen Chloride
21. Oxoacids of Halogens
22. Interhalogen Compounds
23. Group 18 Elements

Class 12 Chemistry NCERT Solutions (Chapter 7 The p-Block Elements) – Important Questions

Chemistry NCERT Solutions for Class 12  Chapter 7 INTEXT Questions

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

Solution:
N₂ + 3H₂  2NH₃; ∆_fH = -46.1 kj mol^-1
In accordance with Le Chatelier’s principle high pressure and low temperature would favour formation of ammonia. The optimum conditions for the production of ammonia are a pressure of about 200 atm, a temperature of about 700K and use of a catalyst such as iron oxide with small amount of K₂O and Al₂,O₃ as promoters.

Question 5.
How does ammonia react with a solution of Cu²⁺?
Solution:
NH₃ in the form of solution reacts with Cu²⁺ to form a complex with deep blue colour.

Question 6.
What is the covalence of nitrogen in N₂O₅?
Solution:
From the structure of N₂O₅, it is evident that covalence of nitrogen is four.

Question 7.
Bond angle in PH₄⁺ is higher than in PH₃. Why?
Solution:
Both are sp³ hybridised. In PH₄⁺ all the four orbitals are bonded whereas in PH₃ there is a lone pair of electrons on P, which is responsible for lone pair-bond pair repulsion in PH₃ reducing the bond angle to less than 109° 28′.

Question 8.
What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO₂?
Solution:
When white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO₂, phosphine gas is liberated.

Question 9.
What happens when PCl₅ is heated?
Solution:
On heating PCl₅ sublimes and is converted to PCl₃ on stronger heating.

Question 10.
Write a balanced equation for the hydrolytic reaction of PCl₅ in heavy water.
Solution:
PCl₅ + D₂O → POCl₃, + 2DCl

Question 11.
What is the basicity of H₃PO₄?
Solution:
Three P – OH groups are present in the molecule of H₃PO₄. Therefore, its basicity is three.

Question 12.
What happens when H₃PO₃ is heated?
Solution:
On heating phosphorous acid dispropotionates to give orthophosphoric acid and phosphine.
4H₃PO₃ → 3H₃PO₄ + PH₃

Question 13.
List the important sources of sulphur.
Solution:
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 thermal stability of the hydrides of group 16 element.
Solution:
H₂O > H₂S > H₂Se > H₂Te

Question 15.
Why is H₂O a liquid and H₂S a gas?
Solution:
Because of small size and high electronegativity of oxygen, molecules of water are highly associated through hydrogen bonding resulting in its liquid state.

Question 16.
Which of the following does not react with oxygen directly? Zn,Ti, Pt, Fe.
Solution:
Platinum does not react with oxygen directly

Question 17.
Complete the following reactions :

1. C₂H₄ + O₂ →
2. 4Al + 3 O₂ →

Solution:

1. C₂H₄ + 3O₂ → 2CO₂ + 2H₂O
2. 4Al + 3O₂ → 2Al₂O₃

Question 18.
Why does O₃ act as a powerful oxidising agent?
Solution:
Due to the ease with which ozone liberates nascent oxygen atoms, it acts as a powerful oxidising agent.
O₃ → O₂ + O

Question 19.
How is O₃ estimated quantitatively?
Solution:
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 is passed through an aqueous solution of Fe(lll) salt?
Solution:
When sulphur dioxide is passed through an aqueous solution of ferric ions, ferric ions are reduced to ferrous ions.

Question 21.
Comment on the nature of two S — O bonds formed in SO₂ molecule. Are the two S — O bonds in this molecule equal?
Solution:
Both the S—O bonds are covalent and have equal strength due to resonating structures.

Question 22.
How is the presence of SO₂ detected?
Solution:
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.
Solution:

1. In manufacture of fertilisers.
2. In manufacture of pigments, paints and dyestuff intermediates.
3. In detergent industry

Question 24.
Write the conditions to maximise the yield of H₂SO₄ by Contact process.
Solution:
The key step in the manufacture of sulphuric acid is oxidation of SO₂ to SO₃ in presence of V₂O₅ catalyst.

The reaction is exothermic and reversible. Hence, low temperature and high pressure are the favourable conditions for maximum yield of SO₃. In practice a pressure of 2 bar and temperature of 720 K is maintained.

Question 25.
Why is K_a2 « K_a1 for H₂SO₄ in water?
Solution:
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_a2« K_a1.

Question 26.
Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F₂ and Cl₂.
Solution:
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.
Solution:

1. Ionisation enthalpy, electro-negativity and electrode potential are higher for fluorine than the expected trends of other halogen.
2. Fluorine does not show any positive oxidation state except in HOF.

Question 28.
Sea is the greatest source of some halogens. Comment.
Solution:
Sea water contains chlorides, bromides and iodides of sodium, potassium, magnesium and calcium but sodium chloride being the maximum makes sea water saline. Various sea weeds contain upto 0.5% iodine.

Question 29.
Give the reason for bleaching action of Cl₂.
Solution:
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.
Solution:
COCl₂ (phosgene), CCl₃NO₂ (tear gas)

Question 31.
Why is ICI more reactive than l2?
Solution:
In general, interhalogen compounds are more reactive than halogens due to weaker X-X’ bonding than X-X bond. Thus, ICI is more reactive than I₂.

Question 32.
Why is helium used in diving apparatus?
Solution:
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 following equation :
XeF₆ + H₂O → XeO₂F₂ + 4HF
Solution:

Question 34.
Why has it been difficult to study the chemistry of radon?
Solution:
Radon is radioactive with very short half-life which makes the study of chemistry of radon difficult.

NCERT Exercises

Question 1.
Discuss the general characteristics of Group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionisation enthalpy and electronegativity.
Solution:
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?
Solution:
Nitrogen has a unique ability to form pπ – pπ multiple bonds with itself and with other elements having small size and high electronegativity. Consequently, its bond enthalpy is very high and reactivity is less. Another factor which affects the chemistry of N is the absence of d-orbitals in its valence shell. As a result, not only the covalency of N is restricted to four, it can also not form dn – pn bonds. P on the other hand, does not form pn – pn bonds and hence it only forms a P – P single bond, which can easily be broken. Also, phosphorus has vacant d-orbitals and can form dn – dn bond with transition metals to form compounds which can act as ligands.

Question 3.
Discuss the trends in chemical reactivity of group 15 elements.
Solution:
Nitrogen has very low reactivity due to unavailability of vacant d-orbital and high bond dissociation energy of N ≡ N bond.

(a) Hydrides : General formula for hydrides is MH₃, e.g., NH₃, PH₃, ASH₃, SbH₃, BiH₃. All these hydrides are covalent in nature and have pyramidal structure (sp³ hybridized).

(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 bond but PH₃ does not?
Solution:
Hydrogen bond is formed between electronegative atom and hydrogen atom. Nitrogen is an electronegative atom and electronegativity decreases down the group so PH₃ cannot form H – bond.

Question 5.
How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions involved.
Solution:
In the laboratory, dinitrogen is prepared by treating an aqueous solution of ammonium chloride with sodium nitrite.

Small amounts of NO and HNO₃ are also formed in this reaction. These impurities can be removed by passing the gas through aqueous sulphuric acid containing potassium dichromate. It can also be obtained by the thermal decomposition of ammonium dichromate.

Very pure nitrogen can be obtained by the thermal decomposition of sodium or barium azide.

Question 6.
How is ammonia manufactured industrially?
Solution:
Industrially ammonia is manufactured by Haber’s process.

In accordance with Le-Chatelier’s principle, high pressure and low temperature would favour the formation of ammonia. The optimum conditions for production of ammonia are a pressure of 200 × 105 Pa (about 200 atm), a temperature of -700 K and the use of a catalyst such as iron-oxide with small amounts of K₂O and Al₂O₃ to increase the rate of attainment of equilibrium.

Question 7.
Illustrate how copper metal can give different products on reaction with HNO₃.
Solution:
acid is a strong oxidising agent and attacks most metals. The products of oxidation depend upon the concentration of the acid, temperature and the nature of the material undergoing oxidation. Copper with . cold dil. HNO₃ forms nitric oxide (NO).

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

Question 9.
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].
Solution:
The actual bond angles are

The decreased bond angle in other hydrides can be explained by the fact that the sp³ hybridisation becomes less and less distinct with increasing size of the central atom i.e., pure p-orbitals are utilised in M-H bonding.

Question 10.
Why does R₃P = O exist but R₃ N = O does not (R = alkyl group)?
Solution:
In R₃ N = O, covalency required is five. The maximum covalency of nitrogen is four as it does not possess d-orbitals in the valence shell i.e., it cannot extend its valency beyond four. On the other hand, other members have d-orbitals and can utilise these orbitals to show covalency of five or six e.g., R₃P = O, PCl_5, [SbF6]^– etc. These can form dπ – pπ bonds.

Question 11.
Explain why NH₃ is basic while BiH₃ is only feebly basic.
Solution:
The basic character decreases from NH₃ to BiH₃. The basic nature is due to the presence of lone pair of electrons on the central atom. NH₃ is the strongest electron pair donor due to its small size as the electron density of the electron pair is concentrated over a small region. As the size increases the electron density gets diffused over a large region and hence the ability to donate the electron pair (basic nature) decreases.

Question 12.
Nitrogen exists as diatomic molecule and phosphorus as P₄. Why?
Solution:
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 main differences between the properties of white phosphorus and red phosphorus.
Solution:


Structure of white and red phosphorus are given below :

Question 14.
Why does nitrogen show catenation properties less than phosphorus?
Solution:
The single N-N bond is weaker than the single P-P bond because of high interelectronic repulsion of the non-bonding electrons, owing to the small bond length. As a result the catenation tendency is weaker in nitrogen.

Question 15.
Give the disproportionation reaction of H₃PO₃.
Solution:
The acids in +3 oxidation state of phosphorus tend to disproportionate to higher and lower oxidation states, e.g., phosphorous acid on heating disproportionates to give orthophosphoric acid and phosphine.
4H₃PO₃ → 3H₃PO₄ + PH₃

Question 16.
Can PCl₅ act as an oxidising as well as a reducing agent? Justify.
Solution:
The oxidation state of P in PCl₅ is +5. As P has five electrons in its valence shell, it cannot increase its oxidation state beyond +5 by donating electrons, therefore, PCl₅ cannot act as a reducing agent. However, it can decrease its oxidation number from +5 to +3 or some lower value, so, PCl₅ acts as an
oxidising agent. For example, it oxidises Ag to AgCl, Sn to SnCl₄.

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 state and hydride formation.
Solution:

All these elements have same ns²np⁴ (n = 2 to 6) valence shell electronic configuration and hence are justified to be placed in group 16 of the periodic table.

(ii) Oxidation states : They need two more electrons to form dinegative ions by acquiring the nearest noble gas configuration. So, the minimum oxidation state of these elements should be -2. Oxygen predominantly and sulphur to some extent being electronegative show an oxidation state of -2. Since these elements have six electrons in the valence shell, therefore, the maximum oxidation state they can show is, + 6. Other positive oxidation states shown by these elements are +2 and +4. Although, oxygen due to the absence of d-orbitals does not show oxidation states of +4 and +6. Thus, on the basis of minimum and maximum oxidation states, these elements are justified to be placed in the same group i.c., group 16 of the periodic table.

(iii) Formation of hydrides : All the elements complete their respective octets by sharing two of their valence electrons with ls-orbilal of hydrogen to form hydrides of the general formula EH₂ i.c., H₂O, H₂S, H₂Se, H₂Te and H₂Po. Therefore, on the basis of formation of hydrides of the general formula, EH2, these elements are justified to be placed in group 16 of the periodic table.

Question 18.
Why is dioxygen a gas but sulphur a solid?
Solution:
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 and 702 kJ mol^-1 respectively, how can you account for the formation of large number of oxides having O^2- species and not O^– ?
Hint : Consider lattice energy factor in the formation of compound.
Solution:
This can be explained with the help of electronic configuration.

As O^2- has most stable configuration amongst these. So, formation of O^2- is much more easier. In solid state, large amount of energy (lattice enthalpy) is released to form divalent O^2- ions. It is greater lattice enthalpy of O^2- which compensates for the high energy required to remove the second electron.

Question 20.
Which aerosols deplete ozone?
Solution:
CFC – Chlorofluorocarbon (freons) : These compounds commonly known as freons are introduced into the atmosphere from aerosol sprays and refrigerating equipments. They undergo photochemical decomposition and destroy ozone as shown by the following sequence of reactions.

Question 21.
Describe the manufacture of H₂SO₄ by Contact process?
Solution:
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?
Solution:
SO₂ present in atmosphere combines with water to give sulphuric acid.
2SO₂ + O₂ + 2H₂O → 2H₂SO₄
This is called acid rain. Acid rain causes extensive damage to building and statues made of marble, limestone, etc. due to its reaction to give CaSO4.
CaCO₃ + H₂SO₄ → CaSO₄ + CO₂ + H₂O
This way SO₂ acts as pollutant.

Question 23.
Why are halogens strong oxidising agents?
Solution:
General electronic configuration of halogen is

It will easily accept one electron to fulfill its shell. This easily acceptance of electron makes halogens strong oxidising agents.

Question 24.
Explain why fluorine forms only one oxoacid, HOF.
Solution:
Due to high electronegativity and absence of d orbitals, F does not form oxoacids such as HOFO, HOFO₂ and HOFO₃ in which the oxidation state of F is +3, +5 and +7. It just forms one oxoacid, i.e., HOF in which the oxidation state of F is + 1.

Question 25.
Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not.
Solution:
Although O and Cl have about the same electronegativity, yet their atomic size (covalent radii) are much different: O = 66 pm and Cl = 99 pm. Thus, electron density per unit volume of oxygen atom is much higher than that of chlorine atom. Hence, oxygen forms hydrogen bonds while chlorine does not though both have approx, the same electronegativity.

Question 26.
Write two uses of ClO₂.
Solution:
ClO₂ is a powerful oxidising agent. ClO₂ is used as a bleaching agent for paper pulp and textiles and in water treatment.

Question 27.
Why are halogens coloured?
Solution:
All halogens are coloured because of absorption of radiations in visible region which results in the excitation of outer electrons to higher level. By absorbing different quanta of radiation, they display different colours as
F₂ → yellow
Cl₂ → greenish yellow
Br₂ → red
I₂ → violet

Question 28.
Write the reactions of F₂ and Cl₂ with water.
Solution:
Fluorine reacts vigorously with water and oxidises water to oxygen.

Chlorine dissolves in water to form chlorine water. It slowly reacts with the water to form a mixture of hydrochloric acid and hypochlorous acid.

Hypochlorous acid is very unstable. In presence of sunlight, it decomposes to give HCl and nascent oxygen.

This oxygen is responsible for oxidising and bleaching properties of chlorine.

Question 29.
How can you prepare Cl₂ from HCI and HCI from Cl₂? Write reactions only.
Solution:
Preparation of chlorine by Deacon’s process : By oxidation of hydrogen chloride gas by atmospheric oxygen in the presence of CuCl₂ at 723 K

Question 30.
What inspired N. Bartlett for carrying out reaction between Xe and PtF₆?
Solution:

Question 31.
What are the oxidation states of phosphorus in the following :

1. H₃PO₃
2. PCl₃
3. Ca₃P₂
4. Na₃PO₄
5. POF₃?

Solution:

Question 32.
Write balanced equations for the following :

1. NaCI is heated with sulphuric acid in the presence of MnO₂.
2. Chlorine gas is passed into a solution of Nal in water.

Solution:

Question 33.
How are xenon fluorides XeF₂, XeF₄ and XeF₆ obtained?
Solution:
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 ClO^– isoelectronic? Is that molecule a Lewis base?
Solution:
Replace O^– (9 electrons) in ClO^– by F (9 electrons). The resulting neutral molecule is ClF. Since ClF can combine further with F to form ClF₃, so, ClF is a Lewis base.

Question 35.
How are XeO₃ and XeOF₄ prepared?
Solution:

Question 36.
Arrange the following in the order of property indicated for each set :

1. F₂, Cl₂, Br₂, l₂ – increasing bond dissociation enthalpy.
2. HF, HCI, HBr, HI – increasing acid strength.
3. NH₃, PH₃, ASH₃, SbH₃, BiH₃ – increasing base strength.

Solution:

1. l₂ < Br₂ < F₂ < Cl₂
2. HF < HCI < HBr < HI
3. NH₃ > PH₃ > AsH₃ > SbH, > BiH₃

Question 37.
Which one of the following does not exist?
(i) XeOF₄
(ii) NeF₂
(iii) XeF₂
(iv) XeF₆
Solution:
(ii) Amongst all noble gases, only Xe (except KrF₂) forms compounds.

Question 38.
Give the formula and describe the structure of a noble gas species which is isostructural with :

1. lCl^–₄
   
2. IBr^–₂
   
3. Br0^–₃

Solution:


(iii) Structure of BrO^–₃ : The central atom Br has seven electrons. Four of these electrons form two double bonds or coordinate bonds with two oxygen atoms while the fifth electron forms a single bond with O^– ion. The remaining two electrons form one lone pair. Hence, in all there are three bond pairs and one lone pair around Br atom in BrO^–₃ Therefore, according to VSEPR theory, BrO^–₃ s should be pyramidal.

Here, BrO^–₃ has 26(7 + 3 × 6 + 1 = 26) valence electrons. A noble gas species having 26 valence electrons is XeO₃ (8 + 3 × 6 = 26). Thus, like BrO^–₃, XeO₃ is also pyramidal.

Question 39.
Why do noble gases have comparatively large atomic sizes?
Solution:
The atomic radii of noble gases are by far the largest in their respective periods. This is due to the reason that noble gases have only van der Waals radii while others have cova lent radii, van der Waals radii, by definition are larger than covalent radii.

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

• Neon is used in discharge tubes and fluorescent bulbs for advertisement display purposes.
• Glow of different colours ‘neon signs’ can be produced by mixing neon with other gases.
• Neon bulbs and used in botanical gardens and in green houses.

Uses of argon:

• Argon is used mainly to provide an inert atmosphere in high temperature metallurgical processes such as arc welding of metals and alloys. In the laboratory, it is used for handling substance which are air sensitive.
• It is used in filling incandescent and fluorescent lamps where its presence retards the sublimation of the filament and thus increases the life of the lamp.
• It is also used in “neon signs” for obtaining lights of different colours.