MP Board Class 12th Chemistry Solutions Chapter 7 The p-Block Elements

MP Board Class 12th Chemistry Solutions Chapter 7 The p-Block Elements

The p-Block Elements NCERT Intext Exercises

Question 1.
Why are.pentahalides more covalent than trihalides ?
Answer:
Higher the positive oxidation state of central atom, it will have more polarising power which, in turn will increase the covalent character of the bond formed between the central atom and the other atom. Hence pentahalides are more covalent than trihalide.

Question 2.
Why is BiH₃ the strongest reducing agent amongst all the hydrides of Group 15 elements ?
Answer:
Among the hydrides of Group-15, BiH₃ is the least stable because Bi has largest size in group and has least tendency to form covalent bond with small hydrogen atom. Therefore, it can readily lost H atom and has strongest tendency to act as reducing agent.

Question 3.
Why is N₂ less reactive at room temperature?
Answer:
In N₂ molecule, there is a strong pπ-pπ overlap that results in the formation of triple bond between N atoms (N ≡ N). As a consequence, the bond dissociation energy of N₂ is very high and N₂ is less reactive at room temperature.

Question 4.
Mention the conditions required to maximise the yield of ammonia.
Answer:
Ammonia is prepared by the Haber’s process.

In accordance with Le-chatelier’s principle, to maximize the yield, a high pressure 200 atm is used. To increase the rate of the reaction, a temperature of around 700 K is used and iron oxide mixed with some K₂O and Al₂O₃ is used as a catalyst. Sometimes, Mo is also used as a promoter to increase the efficiency of the Fe catalyst.

Question 5.
How does ammonia reacts with a solution of Cu²⁺?
Answer:
Ammonia reacts with a solution of Cu⁺² ions of solution to form deep blue coloured complex, tetra’amine copper (II) ion.

Question 6.
What is the covalence of nitrogen in N₂O₅ ?
Answer:
Covalency depends upon the number of shared pairs of,electrons. Now in N₂O₅, each nitrogen atom has four shared pairs of electrons as shown :

Therefore, the covalency of N in N₂O₅ is 4.

Question 7.
Bond angle in PH₄⁺ is higher than that in PH₃. Why ?
Answer:
PH₄⁺ is a regular tetrahedral structure, thus bond angle is 109°28′.
PH₃ molecule has a pyramidal structure. The bond angle is 93°. Hence, bond angle in PH₄⁺ ion is higher.

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

Question 9.
What happens when PCl₅ is heated ?
Answer:
PCl₅ has three equatorial and two axial bonds. Since axial bonds are weaker than equatorial bonds, therefore, when PCl₅ is heated, the less stable axial bond breaks to form PCl₃.

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

Question 11.
What is the basicity of H₃PO₄?
Answer:
H₃PO₄ contains three P-OH bonds and therefore, its basicity is three.

Question 12.
What happens when H₃PO₃ is heated ?
Answer:
On heating, H₃PO₃ disproportionates to give orthophosphoric acid and phosphine.

Question 13.
List the important sources of sulphur.
Answer:
Sulphates and sulphides are important sources of sulphur.
Sulphates : Gypsum (CaSO₄.2H₂O), Epsom salt (MgSO₄.7H₂O), Baryte (BaSO₄).
Sulphides : Galena (PbS), Zinc blende (ZnS), Copper pyrites (CuFeS₂), Iron pyrites (FeS₂) etc.
Organic materials such as eggs, proteins, garlic, onion, mustard, hair and wool also contain sulphur in small quantities.

Question 14.
Write the order of thermal stability of the hydrides of Group-16 elements.
Answer:
As the size of element increases down the group, the E-H bond dissociation energy decreases and hence E-H bond breaks more easily. Thus, the thermal stability of the hydrides of Group 16 elements decreases down the group, i.e., H₂O > H₂S > H₂Se > H₂Te > H₂PO.

Question 15.
Why is H₂O a liquid and H₂S a gas ?
Answer:
Due to high electronegativity of oxygen and its small size, there are strong H-bonding in water. As a result, the molecules exist as associated and is liquid at room temperature. But there is no H-bonding in H₂S because of low electronegativity of ‘S’.

Question 16.
Which of the following does not react with oxygen directly ?
Zn, Ti, Pt, Fe
Answer:
Being noble metal, Pt does not react with oxygen directly. Zn, Ti and Fe are active metals so they react with oxygen.

Question 17.
Complete the following reactions :
(i) C₂H₄ + O₂ →
(ii) 4Al + 3O₂ →
Answer:

Question 18.
Why does O₃ acts as a powerful oxidizing agent ?
Answer:
O₃ is an endothermic compound. On heating, it readily decompose to give dioxygen and nascent oxygen.

Since it liberates nascent oxygen easily. It acts as powerful oxidising agent.

Question 19.
How is O₃ estimated quantitatively ?
Answer:
Ozone (O₃) reacts with an excess of potassium iodide (KI) solution buffered with a borate buffer (pH = 9-2) and liberate (I₂).
2I^– + H₂O_(l) + O_3(g) → 2OH^–_(aq) + I_2(s) + O_2(g)
The liberated I₂ is titrated against a standard solution of sodium thiosulphate. At the end point, I₂ combines with the starch to give the deep blue starch-iodine complex.

This method is also used for detecting ozone in polluted air.

Question 20.
What happens when sulphur dioxide is passed through an aqueous solution of Fe(III) salt ?
Answer:
Sulphur dioxide (a reducing agent) reduces Fe (III) ions to Fe (II) ions as per the following reaction:

Question 21.
Comment on the nature of two S-0 bonds formed in SO₂ molecule. Are the two S-O bonds in this molecule equal ?
Answer:
Both the S-0 bonds are covalent and have equal strength due to resonating structures.

Question 22.
How is the presence of SO₂ detected ?
Answer:
The presence of SO₂ detected by following ways :
(a) SO₂ (reducing agent) decolourises acidified potassium permanganate solution.

(b) SO₂ turns orange coloured acidified K₂Cr₂O₇ solution green due to reduction of Cr₂O₇2- to Cr⁺³ ions.

Question 23.
Mention three areas in which H₂SO₄ plays an important role.
Answer:
Uses of H₂SO₄ :
(i) Manufacture of fertilizers (e.g. ammonium sulphate, calcium superphosphate)
(ii) Petroleum refining
(iii) Metallurgical applications (e.g. cleansing metals before enamelling electroplating, galvanising etc.)

Question 24.
Write the conditions to maximise the yield of H₂SO₄ by Contact process.
Answer:

According to Le-chatelier’s principle following are the requirements for maximum yield.
(i) Low temperature as the conversion of SO₂ to SO₃ is exothermic (720K).
(ii) High pressure (2 atm).
(iii) A catalyst (Pt or V₂O₅) to increase the rate of reaction.

Question 25.
Why is K_a2 << K_a1 for H₂SO₄ in water ?
Answer:
H₂SO₄ is a dibasic acid, it ionizes in two stages and hence has two dissociation constants.

K_a2 is less than K_a, because the negatively charged HSO^–₄ ion has much less tendency to denote a proton to H₂O as compared to neutral H₂SO₄ due to electrostatic reason.

Question 26.
Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidizing power of F₂ and Cl₂.
Answer:
Oxidising power is a combined effect of bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy. Comparing F₂ and Cl₂ with the given parameters :

From the data given above, it is clear that bond dissociation enthalpy and electron gain enthalpy are higher for chlorine but hydration enthalpy is much higher for fluorine. It compensates the effect of other two. As a result, the ∆H overall is more negative for F₂ than for Cl₂. Hence, F₂ is stronger oxidizing agent than Cl₂.

Question 27.
Give two examples to show the anomalous behaviour of fluorine.
Answer:
(i) Two anomalous behaviour of fluorine are : (i) Since fluorine is most elec-tronegative element, it shows only a negative oxidition state of-1. It does not show any positive oxidation state. On the other hand, the other halogens show positive oxidation states also such as +1, +3, +5, +6 and +7.
(ii) Maximum co valency of fluorine is one because it can not expand its valence shell beyond octet because there are no d-orbitals in the valence shell. On the other hand, other elements can exercise covalencies up to 7 because of availability of vacant d-orbitals.

Question 28.
Sea is the greatest source of some halogens. Comment.
Answer:
Sea water contains chlorides, bromides and iodides of sodium, potassium, magnesium and calcium, but mainly sodium chloride (2-5% by mass). Dried up sea-weeds contain sodium chloride and carnallite, KCl. MgCl₂. 6H₂O. Certain sea-weeds contain up to 0-5% of iodine as sodium iodide and chile saltpetre (NaNO₃) contains up to 0-2% of sodium iodate. Thus, sea is the greatest source of halogen.

Question 29.
Give the reason for bleaching action of Cl₂.
Answer:
Bleaching action of chlorine is due to its oxidation. In the presence of moisture, chlorine gives nascent oxygen.
Cl₂ + H₂O → 2HCl + [0]
Because of nascent oxygen, it bleaches colouring substance as
Colouring substance +[0] → Colourless substance.
It bleaches vegetables or organic matter. The bleaching action of chlorine is permanent.

Question 30.
Name two poisonous gases which can be prepared from chlorine gas.
Answer:
(i) Phosgene (COCl₂)
(ii) Tear gas (CCl₃.NO₂)

Question 31.
Why is I-Cl more reactive than I₂ ?
Answer:
I-Cl bond is weaker than I-I bond, so ICl is more reactive than I₂.

Question 32.
Why is Helium used in diving apparatus ?
Answer:
Helium is used as a diluent for oxygen is modern diving apparatus because of its very low solubility in blood.

Question 33.
Balance the following equation :
XeF₆ + H₂O → XeO₂F₂ + HF
Answer:
XeF₆ + 2H₂O → XeO₂F₂ + 4HF.

Question 34.
Why has it been difficult to study the chemistry of Radon ?
Answer:
Radon is a radioactive with very short half life of 3-82 day’s that’s why the study of Chemistry of Radon is difficult.

The p-Block Elements NCERT TextBook 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.
Answer:
The group-15 of the periodic table contains five elements, namely nitrogen (N), phosphorus (P), arsenic (As), antimony (Sb) and bismuth (Bi). These elements are called ‘Pnicogens’ and their compounds as ‘Pniconides’. The name is derived from the Greek word ‘Pnicogens’ meaning suffocation.

Its general characteristics are given below :

1. Electronic configuration : these elements have five valency electrons therefore, their general valence shell electronic configuration will be ns² np³, where, n = 2 to 6. The penultimate shell has 2 electrons in case of nitrogen, 8 electrons in phosphorus and 18 electrons in other elements.

In accordance with the Hund’s rule of maximum multiplicity the three np electrons are distributed as np¹_x, np¹_y, np¹_z. Thus, the p-orbitals are half filled and hence are stable and do not show much reactivity.

2. Oxidation state: The valency shell electronic configuration ns² np³ of these elements suggests that these elements can show an oxidation states of 3, +3 and +5.

Nitrogen and phosphorus the first two members show negative oxidation state of -3 in their compound. Since, their electronegativities are high and atomic size are small. They form anions such as nitride ion (Mg₃N₂) and phosphide ion (Mg₃P₂) with highly positive elements. But, this tendency to show negative oxidation state decreases down the group due to the decrease in electronegativity and increase in size of the elements.

All these elements show positive oxidation state when they combine with more electronegative elements. They show +3 and +5 oxidation states in their compounds. However, the stability of +3 oxidation state increases and that of +5 oxidation state decreases down the group on account of inert pair effect (in ability of ns electrons to participate in bonding). This is noticed maximum in Bi due to very large nuclear charge. Thus, a molecule of BiCl₃ can exist and not of BiCl₅.

3. Atomic size: The atomic radii (covalent) and ionic radii (in same oxidation state) of 15^th group members are smaller as compared to 14^th group elements.
Reason : As we move left to right in a period nuclear charge increases and the new electrons are added in the same shell. Due to that effective nuclear charge increases (Z-effective). This results in decrease in atomic and ionic radii.
On going down the group, the covalent and ionic radii (in a same oxidation state) increases with the increase in atomic number. This increase is very large from N to P. However, from As to Bi only a small increase is observed.
Reason: On moving down the group increase in size can be explained on the basis of successive addition of new shell due to that Z-effective decreases. But, after ‘P’ increase in size is comparatively very small, because of insufficient shielding effect of 3d-electrons (Z- effect increases). Similarly, increase in size from Sb to Bi is small due to insufficient screening by 4/-electrons.

4. Ionization enthalpy : (a) The elements of group 15^th element (nitrogen family) have sufficiently high ionization enthalpy which are more than the corresponding elements of group 14th element (carbon family).
(b) Down the group, the value of ionization enthalpies decreases regularly.

5. Electronegativity : As compared to group 14^th element, group 15th elements are more electronegative. The electronegativity value decreases on moving down the group from N to Bi. Nitrogen is the most electronegative element of the family.

Question 2.
Why does the reactivity of nitrogen differ from phosphorus ?
Answer:
Molecular nitrogen exist as a diatomic molecule having a triple bond between the two nitrogen atoms, N = N (due to it stability to form pπ-pπ multiple bonding). The bond dissociation energy is very high (941-4 kJ mol^-1). Thus, under ordinary conditions, nitrogen behaves as an inert gas. On the other hand, white and yellow phosphorus exists as a triatomic molecule (P₄) having single bonds. The dissociation energy of P-P bond is low (213 kJ mol^-1). Thus, phosphorus is much more reactive than nitrogen.

Question 3.
Discuss the trends in chemical reactivity of Group 15^th elements.
Answer:
Chemical reactivity: The elements of group 15 show different reactivity. Nitro¬gen, inspite of its greater electronegativity value of 3, is chemically inert.
The chemical inertness of nitrogen is due to the presence of a triple bond between two nitrogen atoms (N ≡ N) which is very strong and requires high dissociation energy (941.4 kJ mol^-1).

Phosphorus exists in P₄ tetrahedral molecules, each p-p-p angle is 60° instead of 109°28 ‘as expected from sp³ hybridization. Thus P₄ molecule has a highly strained cage like structure and this makes white phosphorus highly reactive, (on the other hand red phosphorus has opened up network structure and not discrete P₄ molecule hence less reactive than white phosphorus). As, Sb and Bi are also not very reactive.

Question 4.
Why does NH₃ form hydrogen bond but PH₃ does not ?
Answer:
The N – H bond in ammonia is quite polar as nitrogen is highly electronegative in nature. As a result, NH₃ molecules are linked by intermolecular hydrogen bonding.

On the other hand, P – H bond is non-polar as both P and H have same electronegativ¬ity. Hence in phosphine no hydrogen bonding is present.

Question 5.
How is nitrogen prepared in the laboratory ? Write the chemical equations of the reactions involved.
Answer:
(i) In the laboratory, dinitrogen is prepared by heating an aqueous solution of ammonium chloride with sodium nitrite.
NH₄C_l(ag) + NaNO_2(aq) → N_2(g) + 2H₂O_(l) + NaCl_(aq)
(ii) Preparation by thermal decomposition of Ammonium dichromate.

(iii) Net N₂ is obtained from sodium or Barium azide.

Question 6.
How is ammonia manufactured industrially ?
Answer:
Theory : One volume of nitrogen and three volume of hydrogen reacts together to form ammonia. It is an exothermic process. Formation of ammonia is accompanied by decrease in volume because four volume of reactants react to form two volumes of product. Thus, according to Le-chatelier’s principle high concentration of N₂ and H₂ low temperature and high pressure is favourable for the formation of ammonia.
Chemical reaction:

Method: Pure nitrogen and pure and dry hydrogen are taken in ratio 1:3. The above mixture is passed in chamber containing ferric oxide (Fe₂O₃) as catalyst and molybdenum as promoter at 400-500C under the high pressure of 200 atmospheric pressure. The gas obtained after reaction contains 15-40% is cooled by condensers and liquid ammonia is obtained.

Precautions : (i) N₂ and H₂ should be pure and dry because impurities damage catalytic activities and
(ii) Temperature and pressure should be controlled.

Question 7.
Illustrate how copper metal can give different products on reaction with HNO₃.
Answer:
On heating with dil.HNO₃, Copper gives copper nitrate and nitric oxide.

With cone. HNO₃, instead of NO, NO₂ is evolved.

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

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].
Answer:
According to VSEPR theory, the lone pair-bond pair repulsion is larger than bond pair-bond pair repulsion. As a result, the tetrahedral shape is distorted and the bond angle becomes less than tetrahedral angle (109°28′). Electronegativity of N is maximum in group 15, so bond pairs of electrons lie much closer to N in NH3. The force of repulsion between the adjacent bond pairs are also high so the HNH bond angle value is higher in NH₃ than HPH, HAsH and HSbH angles.

Question 10.
Why does R₃P = O exist but R₃N = O does not (R = alkyl group)?
Answer:
Nitrogen does not have d-orbitals in its valence shell. Therefore, it can not extend its covalency to five by forming dπ-dπ bonding. As a result, the molecule of R₃N = O does not exist. However, phosphorus has vacant rf-orbitals in the valence shell and can form dx-dxbonding. Thus, a molecule like R₃P = O can exist.

Question 11.
Explain why NH₃ is basic while BiH₃ is only feebly basic.
Answer:
Atomic size of N(70pm) is much smaller than that of Bi (148 pm). Therefore, electron density on the N-atom is much higher than that on Bi-atom. Consequently, the tendency of N in NH₃ to donate the pair of electrons is much higher than that of Bi in BiH₃. Thus, NH₃ is basic while BiH₃ is only feebly basic.

Question 12.
Nitrogen exists as diatomic molecule and phosphorus as P₄ Why?
Answer:
Nitrogen because of its small size and high electronegativity is capable of forming pπ-pπ multiple bonding. Therefore, it exists as diatomic molecule with one σ and two K bonds (N ≡ N). Phosphorus, on the other hand, has a longer size and lower electronegativity and thus is not capable to form pπ-pπ multiple bonds. It prefers to form single bonds hence, it exists as tetrahedral P₄ molecules.

Question 13.
Write main differences between the properties of white phosphorus and red phosphorus.
Answer:
Comparison of properties of White and Red phosphorus :

Question 14.
Why does nitrogen show catenation properties less than phosphorus?
Answer:
The catenation properties depends upon the strength of the element-element bond. The N-N bond strength is much weaker (due to repulsion of lone pairs on nitrogen because of its small size) than the P-P bond strength, therefore, nitrogen shows catenation less than phosphorus.

Question 15.
Give the disproportionation reaction of H₃PO₃.
Answer:

Question 16.
Can PCl₅ acts as an oxidising as well as a reducing agent ? Justify.
Answer:
Phosphorus has maximum oxidation state of +5 in PCl₅. It can not increase its oxidation state further and thus it can not act as reducing agent. However, PCl₅ can act as an oxidising agent as it can itself reduce from +5 to +3 oxidation state. For example, PCl₅ oxidises Ag and Sn in the following reactions :

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.
Answer:
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 (ns²np⁴), they should display an oxidation state of -2. However, only oxygen predominatntly 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 from 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.

Question 18.
Why is dioxygen a gas but sulphur a solid?
Answer:
The oxygen atom has tendency to form pπ-pπ multiple bonding due to its small size and high electronegativity. As a result, oxygen exist as diatomic molecule. These mol¬ecules are held together by weak van der waal’s forces of attraction which can be easily overcome by collisions of the molecules at room temperature. Therefore, O₂ is the gas at room temperature.

Sulphur, on the other hand, because of its bigger size and lower electronegativity, does not form pπ-pπ multiple bonds. Instea it prefers to form S-S single bond and form polyatomic complex molecules having eight atoms per molecule (S₈) and have puckered ring structure. Therefore, S atoms are strongly held together and it exists as a 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 a large number of oxides having O^2- species and not O^– ?
(Hint: Consider lattice energy factor in the formation of compounds).
Answer:
Although the formation of O^2- anions requires more energy in comparison to the formation of O^– anion (actually energy is released). Yet in large number of oxides, oxygen is divalent in nature. This is due to the fact that lattice energies of the oxides having O^2- anions are very high on account of greater magnitude of electrostatic force of attraction.

Question 20.
Which aerosols deplete ozone ?
Answer:
Freon (CCl₂F₂) (Chlorofluoro carbon).

Question 21.
Describe the manufacture of H₂SO₄ by contact process.
Answer:
Manufacture of sulphuric acid : Sulphuric acid is prepared on large scale by contact process. The basic raw material used is either sulphur or iron pyrites.
Principle of contact process : When pure and dry SO₂ mixed with air is passed over V₂O₅ catalyst it gets oxidised to SO₃ which is absorbed by H₂SO₄ to form oleum i.e., H₂S₂O₇.
2SO₂ + O₂ → 2SO₃ + 45.2 cal
SO₃ + H₂SO₄ → H₂S₂O₇

Details of the process are given below :
(a) Sulphur or pyrite burners : SO₂ is produced by burning sulphur or roasting pyrites in pyrite burners.

(b) Purification unit: This unit is an assembly of various parts like.
(i) Dust chamber: SO₂ gas produced is passed through dust chamber through which steam is sprayed from top. Dust particles absorb moisture, become heavy and settle down.
(ii) Cooling pipes : Now, the gaseous mixture is passed through cooling pipes to lower the temperature to 100°C.
(iii) Washing tower or scrubbers : It is filled with quartz and showers of cold water continue from top. Dust particles as well as soluble impurities settle down and are removed.
(iv) Drying tower : It is a high tower full of quartz pieces. Cone. H₂SO₄ is sprayed from the top. Gases enter the tower from bottom and get dried in contact with cone. H₂SO₄.
(v) Arsenic purifier : It is filled with shelves containing freshly prepared ferric hydroxide which absorb the impurities of arsenic purified and dried.

(c) Testing box : Before sending the gaseous mixture to contact chamber it is tested here. A strong beam of light is thrown into the testing box, if the gaseous mixture contains dust or any other particles they become visible, hence gaseous mixture is repurified. The completely pure gas is then sent to the contact chamber.

(d) Preheater : Pure gas free from dust particle is heated in preheater to 450°C and then passed into contact tower, where SO₂ gets oxidized to SO₃. This reaction is exothermic due to which temperature of contact tower raises to 450°C. After this the pure gases are pumped directly into the contact tower.

(e) Contact tower : It consists of a big iron container containing several pipes filled with vanadium pentoxide or platinized asbestos or any other suitable catalyst. Temperature is maintained at 450°C. Pure and dry SO₂ reacts with air in these pipes to form sulphur trioxide.

On account of exothermic nature of reaction, sufficient heat is available.

(f) Absorption tower: The sulphur trioxide is then passed into a tower in which cone, sulphuric acid flows down in the form of spray. Sulphuric acid absorbs SO₃ and becomes more concentrated. This acid, due to excess of SO₃ produces a fog in the tower. The acid obtained is called fuming sulphuric acid or oleum.

Oleum is mixed with calculated quantity of water to form acid of a particular strength.
H₂S₂O₇ + H₂O → 2H₂SO₄
It may be noted that sulphur trioxide is not directly absorbed in water to form sulphuric acid because it forms a dense fog of sulphuric acid which does not condense easily.
Note : For detail refer to your NCERT Text-Book.

Question 22.
How is SO₂ an air pollutant?
Answer:
SO₂ dissolves in rain water and produces acid rain. The acid rain contains sulphuric acid

In addition to H₂SO₄, acid rain also contains HNO₃.

Question 23.
Why are halogens strong oxidising agents?
Answer:
Halogens are strong oxidising agents due to their low bond dissociation enthalpy, high electronegativity and large negative electron gain enthalpy. Halogens have a strong tendency to accept electrons and thus get reduced.
X₂ + 2e^– → 2X^–
As a result, halogen acts as strong oxidising agents. Their oxidising power however decreases from F₂ to I₂ as is evident from their electrode potentials:

Question 24.
Explain why fluorine forms only one oxyacid, HOF ?
Answer:
Due to absence of d-orbital, fluorine does not shqw +3, +5 and + 7 oxidation state, while other halogen do show. Thus, it does not form oxoacids like HOFO, HOFO₂, HOFO₃. It shows only +1 oxidation state and form HOF only.

Question 25.
Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not.
Answer:
Oxygen has smaller size than chlorine, smaller size of oxygen favours hydrogen bonding.

Question 26.
Write two uses of ClO₂.
Answer:
Two uses of ClO₂ :
(i) It is a powerful oxidising and chlorinating agent.
(ii) It is an excellent bleaching agent for wood pulp, flour for making white bread.

Question 27.
Why are halogens coloured ?
Answer:
Halogen absorb part of light in the visible region which causes excitation of outer electrons to higher energy levels. By absorbing different quanta of radiations, they display different colours. Fluorine atom is the smallest and the force of attraction between the nucleus and the outer electrons is very large. As a result, it requires large excitation energy and absorbs violet light and therefore appear yellow (complementary colour). As different colours are absorbed by different halogens, they display different complementary colour also.

Question 28.
Write the reactions of F₂ and CI₂ with water.
Answer:
F₂ being strong oxidising agent oxidises H₂O to O₂ or O₃.

Cl₂, on the other hand, reacts with H₂O to form hydrochloric acid and hypochlorous acid.

Question 29.
How can you prepare Cl₂ from HCl and HCl from Cl₂? Write reactions only.
Answer:

Question 30.
What inspired N. Bartlett for carrying out reaction between Xe and PtF₆?
Answer:
Neil Bartlett observed that PtF6 reacts with O2 to form an ionic solid O+2 PtF6
O_2(g) + PtF_6(g) → O⁺₂[PtF₆]^–
In this reaction, O₂ gets oxidised to O⁺₂ by PtF₆.
Since the first ionisation energy of xenon is fairly close to that of oxygen. Bartlett thought that PtF₆ should also oxidise Xe to Xe⁺. This prompted Bartlett to carry out the reaction between Xe and PtF₆ and formed the first noble gas compound.

Question 31.
What are the oxidation states of phosphorus in the following :
(i) H₃PO₃
(ii) PCl₃
(iii) Ca₃P₂
(iv) Na₃PO₄
(v) POF₃ ?
Answer:
(i) H₃PO₃ = 3 × (+1) + 3 × (-2) + X = 0 or X = +3
(ii) PCl₃ = X + 3(-1) = 0 or X = +3
(iii) Ca₃P₂ = 3 × (+2) + 2 X = 0 or X = -3
(iv) Na₃PO₄ = 3 × (+1) + X + 4 × (-2) = 0 or X = +5
(v) POF₃ = X (-2) + 3 × (-1) = 0 or X = +5.

Question 32.
Write balanced equations for the following :
(i) NaCI is heated with sulphuric acid in the presence of MnO₂.
(ii) Chlorine gas is passed into a solution of Nal in water.
Answer:
(i) Cl₂ is produced
4NaCl + MnO₂ + 4H₂SO₄ → MnCl₂ + 4NaHSO₄ + 2H₂O + Cl₂
(ii) Cl₂ being an oxidising agent oxidises Nal to I₂.
Cl_2(g) + 2NaI_(aq) → 2NaCl_(aq) + I_2(s)

Question 33.
How are xenon fluorides XeF₂, XeF₄ and XeF₆ obtained ?
Answer:
Xenon Fluorides : Three fluorides of xenon are important. These are xenon difluoride (XeF₂), xenon tetrafluoride (XeF₄) and xenon hexafluoride (XeF₆).The fluorides of xenon can be prepared by the direct combination between xenon and fluorine under different conditions.

Question 34.
With what neutral molecule is CIO^– isoelectronic? Is that molecule a Lewis base ?
Answer:
OF₂ and ClF are isoelectronic to CIO^–, out of which ClF is a Lewis base.

Question 35.
How are XeO₃ and XeOF₄ prepared ?
Answer:
Hydrolysis of XeF₄ and XeF₆ with water forms XeO₃.
6XeF₄ + 12H₂O → 4Xe + 2XeO₃ + 24HF + 3O₂
XeF₆ + 3H₂O → XeO₃ + 6HF
Partial hydrolysis of XeF₆ gives XeOF₄.
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.
Answer:
(i) I₂ < F₂ < Br₂ < Cl₂

Here F₂ has very less bond dissociation energy inspite of having small size. It is due to repulsion be¬tween the lone pair of electrons on small size F-atoms.
(ii) HF < HCl < HBr < HI
It depends upon their bond dissociation enthalpy, which decreases from H-F to H-I as the size of atom increases from F to I.
(iii) BiH₃ < SbH₃ < AsH₃ < PH₃ < NH₃.
They are lewis bases due to presence of lone pair of electrons on central atom of hydrides. The availability of lone pair of electrons is maximum in nitrogen owing to its small size therefore, NH3 is maximum basic and as the size of central atom increases availability decreases.

Question 37.
Which one of the following does not exist ?
(i) XeOF₄
(ii) NeF₂
(iii) XeF₂
(iv) XeF₆
Answer:
NeF₂ does not exist as Ne does not have J-orbitals in valence shell and possesses high ionisation enthalpies. Thus, there is no scope of promotion of electrons i.e. formation of half filled orbitals. Hence, the formation of NeF₂ is not possible.

Question 38.
Give the formula and describe the structure of a noble gas species which is isostructural with:
(i) ICl₄^–
(ii) IBr₂^–
(iii) BrO₃^–.
Answer:
(i) ICl₄^– : sp³d² hybridisation, square planar shape.
XeF₄ is an isostructural species.

(ii) IBr₂^– : sp³d hybridisation, Linear shape.
XeF₂ is an isostructural species.

(iii) BrO₃^–: sp³ hybridisation, tetrahedral geometry (Pyramidal shape)
XeO₃ is an isostructural species.

Question 39.
Why do noble gases have comparatively large atomic sizes?
Answer:
In case of noble gases, the atomic radii corresponds to van der Waal’s radii, which are always large.

Question 40.
List the uses of neon and argon gases.
Answer:
Uses of Neon : (i) Neon lights are used for commercial advertisements. It consists of a long tube fitted with electrodes at both ends. On filling the tube with neon gas and passing electric discharge of about 1000 volt potential, a bright red light is produced. Different colours can be obtained by mixing neon with other gases. For producing blue or green light neon is mixed with mercury vapours.

Uses of Argon :
(i) Like helium, it is also used to create inert atmosphere in welding of aluminium ahd stainless steel.
(ii) It is filled in electric bulbs along with 25% nitrogen.
(iii) It is also used in radio valves.
(iv) Argon alone or its mixture with neon is used in tubes for producing lights of different colours.

The p-Block Elements Other Important Questions and Answers

The p-Block Elements Objective Type Questions

Question 1.
(A) Choose the correct answer :

Question 1.
In which compound oxygen exhibits +2 oxidation state :
(a) H₂O
(b) Na₂O
(c) OF₂
(d) MgO.
Answer:
(c) OF₂

Question 2.
Reddish brown gas is formed, when nitric oxide oxidizes by air. This gas is :
(a) Na₂O₂
(b) Na₂O₄
(c) NO₂
(d) N₂O₃.
Answer:
(c) NO₂

Question 3.
H₃PO₃ is :
(a) Dibasic acid
(b) Monobasic acid
(c) Tribasic acid
(d) Tetrabasic acid
Answer:
(c) Tribasic acid

Question 4.
Which one of the following is a typical metal:
(a) P
(b) As
(c) Sb
(d) Bi.
Answer:
(d) Bi.

Question 5.
Oxide which shows paramagnetic character :
(a) N₂O₄
(b) NO₄
(C) P₄O₆
(d) N₂O₅.
Answer:
(b) NO₄

Question 6.
Ammonia can be dried by :
(a) H₂SO₄
(b) P₂O₅
(c) Anhydrous CaCl₂
(d) CaO.
Answer:
(d) CaO.

Question 7.
Nitric acid change iodine into :
(a) Iodic acid
(b) Hydroiodic acid
(c) Iodine pentaoxide
(d) Iodine nitrate.
Answer:
(a) Iodic acid

Question 8.
NH₃ is a Lewis base which forms complex salt with cations. Following cation does not form complex salt with NH₃ :
(a) Ag⁺
(b) Cu²⁺
(c) Cd²⁺
(d) Pb²⁺
Answer:
(d) Pb²⁺

Question 9.
Ammonia is soluble in:
(a) Hg₂Cl₂
(b) PbCl₂
(c) Agl
(d) Cu(OH)₂.
Answer:
(d) Cu(OH)₂.

Question 10.
How much water molecules are required to change one molecule of P₂O₅ into orthophosphoric acid:
(a) 2
(b) 3
(c) 4
(d) 5.
Answer:
(b) 3

Question 11.
Bleaching action of SO₂ is due to :
(a) Reduction
(b) Oxidation
(c) Hydrolysis
(d) Acidic nature.
Answer:
(a) Reduction

Question 12.
What happens when SO₂ is passed through acidic K₂Cr₂O₇ :
(a) Solution turns blue
(b) Solution turns colourless
(c) SO₂ reduced
(d) Green chromic sulphate is formed.
Answer:
(d) Green chromic sulphate is formed.

Question 13.
P₂O₃ forms which of the following acid :
(a) H₄P₂O₇
(b) H₃PO₄
(C) H₃PO₃
(d) HPO₃
Answer:
(C) H₃PO₃

Question 14.
Most acidic halide is :
(a) PCl₅
(b) SbCl₃
(c) BrCl₃
(d) CCl₄
Answer:
(a) PCl₅

Question 15.
The catalyst used in manufacturing of H₂SO₄ is :
(a) Al₂O₃
(b) CrO₃
(c) V₂O₅
(d) MnO₂
Answer:
(c) V₂O₅

Question 16.
The hydrolysis of phosphorus trihalides give :
(a) One monobasic acid and one dibasic acid
(b) One monobasic acid and one tribasic acid
(c) One monobasic acid and a salt
(d) Two dibasic acids.
Answer:
(a) One monobasic acid and one dibasic acid

Question 17.
In the following reaction :
P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂
(a) P oxidises
(b) P oxidises and reduces
(c) P reduces
(d) Na oxidises.
Answer:
(b) P oxidises and reduces

Question 18.
Laughing gas is :
(a) NO
(b) N₂O
(C) N₂O3
(d) N₂O5.
Answer:
(b) N₂O

Question 19.
White phosphorus does not contain :
(a) 6 P-P single bond
(b) 4 P-P single bond
(c) 4 lone pair of electron
(d) Bond angle of P-P-P is 60°.
Answer:
(b) 4 P-P single bond

Question 20.
On heating NH₄Cl and NaNO₂ solution gives :
(a) N₂O
(b) N₂
(C) NO₂
(d) NH₃.
Answer:
(b) N₂

Question 21.
Formula of metaphosphoric acid is :
(a) H₃PO₄
(b) HPO₃
(c) H₃PO₃
(d) H₂PO₃.
Answer:
(b) HPO₃

Question 22.
Gas which cannot be collected on water :
(a) Na
(b) O₂
(c) SO₃
(d) PH3.
Answer:
(c) SO₃

(B) Choose the correct answer :

Question 1.
Chlorine shows bleaching property in the presence of:
(a) Dry air
(b) Moisture
(c) Sunlight
(d) Pure oxygen.
Answer:
(b) Moisture

Question 2.
Which of the following element forms least number of compounds :
(a) He
(b) Ar
(c) Kr
(d) Xe.
Answer:
(a) He

Question 3.
Which is used for bright advertisement display :
(a) Xe
(b) Ar
(c) Ne
(d) He.
Answer:
(c) Ne

Question 4.
Monazite is a source of:
(a) Ne
(b) Ar
(c) Kr
(d) He.
Answer:
(d) He.

Question 5.
Which of the following is not obtained by the direct reactions of the constituents:
(a) XeF₂
(b) XeF₄
(c) XeO₃
(d) XeF₆.
Answer:
(a) XeF₂

Question 6.
Which of the following halides are least stable and whose existence is suspicious:
(a) CI₄
(b) GeI₄
(c) SnI₄
(d) PbI₄
Answer:
(d) PbI₄

Question 7.
Which of the following is the strongest acid:
(a) HBr
(b) HCl
(c) HF
(d) HI.
Answer:
(d) HI.

Question 8.
Most electronegative element is:
(a) F
(b) Cl
(c) Br
(d) I.
Answer:
(a) F

Question 9.
Which halogen is solid at room temperature:
(a) Cl₂
(b) I₂
(c) Br₂
(d) F₂
Answer:
(d) F₂

Question 10.
Which has maximum electron affinity:
(a) F
(b) Cl
(c) Br
(d) I.
Answer:
(b) Cl

Question 11.
Electronic configuration of valence shell of halogen ¡s:
(a) s² p⁵
(b) s²p³
(c) s²p⁶
(d) s²p⁴.
Answer:
(a) s² p⁵

Question 12.
Which element is most alkaline:
(a) F
(b) Cl
(c) Br
(d) I.
Answer:
(d) I.

Question 13.
The strongest reducing agent is:
(a) F^–
(b) Br^–
(c) I^–
(d) I.
Answer:
(c) I^–

Question 14.
Which of the following is weakest acid:
(a) HF
(b) HCI
(c) HBr
(d) HI.
Answer:
(a) HF

Question 15.
Oxidizing property ¡s highest of:
(a) I₂
(b) Br₂
(c) F₂
(d) Cl₂.
Answer:
(c) F₂

Question 16.
Which Noble gas does not have octet complete:
(a) Helium
(b) Neon
(c) Argon
(d) Krypton.
Answer:
(a) Helium

Question 17.
Which Halogen sublimates:
(a) Chlorine
(b) Bromine
(c) Iodine
(d) Fluorine.
Answer:
(c) Iodine

Question 18.
Which of the noble gas is most soluble in water:
(a) He
(b) Ar
(c) Ne
(d) Xe.
Answer:
(d) Xe.

Question 19.
I₂ easily dissolve in KI solution to form :
(a) I
(b) KI₂
(C) KI
(d) KI₃.
Answer:
(d) KI₃.

Question 20.
Gas mixed in oxygen which is used in respiration for asthma patients :
(a) N₂
(b) Cl₂
(c) He
(d) Ne.
Answer:
(c) He

Question 21.
Deacon’s process is used for manufacture of:
(a) Bleaching powder
(b) Chlorine
(c) Nitric acid
(d) Sulphuric acid.
Answer:
(b) Chlorine

Question 22.
Sea grass is the source of industrial manufacture of:
(a) Chlorine
(b) Bromine
(c) Iodine
(d) Fluorine.
Answer:
(c) Iodine

Question 23.
Which gas is more useful to be filled in electric bulb :
(a) He
(b) Ne
(c) Ar
(d) Kr.
Answer:
(c) Ar

Question 2.
(A) Fill in the blanks :

1. N₂O is a …………………. oxide.
2. Formula of Caro’s acid is ………………….
3. Concentrated nitric acid has brown colour due to the dissolution of …………………. gas.
4. Out of nitrogen oxides …………………. and …………………. are paramagnetic.
5. Pyrophosphoric acid is …………………. basic acid.
6. H₂S gas cannot be dried by cone. H₂SO₄ because H₂S …………………. it.
7. Fuming sulphuric acid dissolves in SO₃ to form ………………….
8. Oxidation state of S in H₂S₂O₈ (Marshall gas) is ………………….
9. NH₃ gives white fumes of …………………. when it combines with HCl.
10. Elements of 16^th group are known as ………………….
11. …………………. is used as a refrigerant.
Answers:
1. Neutral
2. H₂SO₅
3. NO₂
4. NO, NO₂
5. Tetra
6. reduces
7. Oleum
8. +6,9. NH₄Cl
10. Chalcogen
11. Liquid NH₃.

(B) Fill in the blanks :

1. ………………….. has the highest electron affinity.
2. In the presence of moisture, chlorine acts as a ………………….. agent.
3. Bleaching powder is also called as …………………..
4. At ordinary temperature bromine is a …………………..
5. Inter halogen compound AX₅ has ………………….. structure.
6. Chlorine was discovered by …………………..
7. Neil Bertlett forms first noble compound which is …………………..
8. Element with highest electron affinity is …………………..
9. In oxy acid halogen present in ………………….. hybride state.
10. Gas which is light due to which it is filled in aircraft tyres is …………………..
11. ………………….. inert gas is mostly used in advertisements.
12. Elements of Group 17^th are commonly known as …………………..
13. ………………….. is a radioactive inert gas.
Answer:
1. Chlorine
2. Bleaching
3. Calcium chlorohypochloride
4. Liquid
5. Square pyramidal
6. Scheele
7. Xe [PtF₆]
8. Chlorine
9. sp³
10. Helium
11. Ne (Neon)
12. Halogen
13. Radon.

Question 3.
Match the following :
I.

Answers:

1. (d)
2. (c)
3. (a)
4. (b)
5. (f)
6. (g)
7. (e)
8. (i)
9. (j)
10. (h)

II.

Answers:

1. (e)
2. (c)
3. (b)
4. (a)
5. (d).

III.

Answers:

1. (e)
2. (d)
3. (b)
4. (f)
5. (c)
6. (a).

Question 4.
(A) Answer in one word / sentence :
1. Who protects earth from ultraviolet rays ?
2. Give the chemical name of oil of vitriol or king of chemical.
3. Which gas is used for refrigeration ?
4. At which temperature water has maximum density ?
5. Give the name of compound formed by the dissolution of SO₃ in sulphuric acid.
6. Write name of an antichlor.
7. Which gas is used for bleaching of oil and elephant teeth ?
8. Ammonium salts react with Nessler’s reagent to give which coloured precipitate ?
9. On moving down from N to Bi, +3 oxidation state of Bi becomes more stable than +5. Why ?
10. State the percentage of N₂ by volume in nature.
Answers:
1. Ozone layer
2. H₂SO₄
3. NH₃
4. 4°C
5. Oleum
6. SO₂
7. Ozone
8. Brown
9. Due to Inert pair effect
10. 80%.

(B) Answer in one word/sentence :

1. Give the name of radioactive halogen.
2. Which noble gas is used in bulb with nitrogen ?
3. Write the name of noble gas which is used in treatment of cancer.
4. Write the formula of Carnalite.
5. Which noble gas is maximum available in the atmosphere ?
6. What is the shape of XeF₆ ?
7. Write one use of fluorine.
8. Which type of hybridization is present in XeO₃?
9. Write the oxidation state of F.
10. Write only equation for the preparation of chlorine in the lab.
11. What is the shape of AX₃ type of Interhalogen compound ?
12. Write the correct order of strength of halogen acid.
13. Which noble gases do not form compounds ?
14. With which noble gas does F form compounds ?
15. Sea weeds are the source of which halogen ?
Answer:
1. Astatine
2. Ar
3. Rn
4. KCl.MgCl₂.6H₂O
5. Argon
6. Distorted octahedron
7. In preparing fluorocarbon, which is used in refrigeration
8. sp³
9. -1
10. MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂
11. T- shape
12. HF < HCl < HBr < HI
13. He, Ne and Ar
14. Xe
15. Iodine.

The p-Block Elements Very Short Answer Type Questions

Question 1.
Why is dinitrogen (N₂) less reactive at room temperature ?
Answer:
Due to high bond enthalpy of N ≡ N bond, dinitrogen is very less reactive at room temperature.

Question 2.
Why is concentrated sulphuric acid a high boiling point oily liquid ?
Answer:
Due to hydrogen bond between H₂SO₄ molecules, it is a high boiling point oily liquid.

Question 3.
Name two poisonous gases which are prepared from chlorine.
Answer:
(i) Phosgene (COCl₂)
(ii) Tear gas (CCl₃NO₂).

Question 4.
Write the order of acidic strength of HCIO, HBrO and HIO.
Answer:
Strength of hypohalous acids decreases from HCIO to HIO.
HCIO > HBrO > HIO.

Question 5.
What are clathrate compounds ?
Answer:
In the cavity or space of crystal lattice of a compound small sized elements like noble gases get trapped within. Such compounds are called clathrate compounds.
Example : Kr₃ (β-quinol).

Question 6.
Electronegativity of F atom is more than I atom still acidic strength of HF is less than HI. Explain.
Answer:
Due to small size of F atom, bond dissociation energy of HF bond is much higher than HI bond, in which size of I atom is large.

Question 7.
Which noble gases can form chemical compound ?
Answer:
Kr and Xe can form compounds in extremely specific conditions.

Question 8.
Fluorine is strong oxidizing agent as compared to chlorine. Why ?
Answer:
Because of following points, fluorine is strong oxidizing agent:

1. The size of F atom is smaller than Cl atom.
2. Electronegativity of fluorine is higher than Cl atom.
3. Dissociation energy of fluorine is less than that of chlorine.
4. E° value of fluorine is more than chlorine.

Question 9.
F₂O is not considered to be the oxide of fluorine. Why ?
Answer:
Fluorine is the most electronegative element. Its electronegativity is higher than oxygen. In naming a compound less electronegative elements are first and then more electronegative elements are written. Therefore, F₂O or OF₂ is known as oxygen difluoride and not fluorine oxide.

Question 10.
Interhalogen compounds are more reactive than halogen. Why ?
Answer:
X — Y bonds of interhalogen compounds are more polar due to difference in the electronegativities of halogen and are generally weaker than X — Y bonds of pure halogen. Therefore, interhalogen compounds are more reactive than halogens.

Question 11.
Helium and Neon do not form compounds with Fluorine. Why ?
Answer:
Due to absence of cf-orbitals in the valence shell of He and Ne, their electrons cannot get excited into higher energy o’-sub-shells. Therefore, He and Ne do not form compounds with fluorine.

The p-Block Elements Short Answer Type Questions

Question 1.
H₂O is liquid, while H₂S is a gas at ordinary temperature. Explain.
Answer:
In H₂O, oxygen is electronegative due to this intermolecular hydrogen bonding is found among H₂O molecule and exists as liquid state.
On the other hand, sulphur is less electronegative due to this intermolecular hydrogen bonding is not found in H₂S and it exists as gas

Question 2.
Write formula and structures of five oxy acids of phosphorus.
Answer:
Five oxy acids of phosphorous are :
(i) H₃PO₂ (Hypophosphorous acid)
(ii) H₃PO₃ (Phosphorous acid)
(iii) H₄P₂O₅ (Pyrophosphorous acid)
(iv) H₄P₂O₆ (Hypophosphoric acid)
(v) H₄P₂O₇ (Pyrophosphoric acid).
Structures of oxy acids of phosphorus are :

Question 3.
Behaviour of oxygen is different from other elements of the group. Why ?
Answer:
Reason for abnormal behaviour of oxygen :

1. Electronegativity is maximum.
2. Ionisation energy is high.
3. No any vacant d-orbitals present in valence shell.
4. Atomic size is small.
5. It forms strong hydrogen bonds.

Question 4.
What is the reason that oxygen is a gas whereas sulphur is a solid ?
Answer:
Oxygen forms diatomic O₂ molecule. Different molecules of oxygen are tied with weak van der Waals forces, so oxygen is gas at normal temperature.

On the other hand, sulphur forms a complex molecular structure of 8 sulphur atoms. Molecular mass of S₈ is very much and it is solid.

Question 5.
Water is neutral, whereas H₂S is a weak acid. Why ?
Answer:
Due to the presence of strong H-bond in water, its molecules mutually associate due to which magnitude of its dissociation constant (k_a) decreases. Therefore, water is neutral. Whereas in H₂S due to lower electronegativity of S, hydrogen bond cannot be formed and size of sulphur atom is larger than oxygen atom due to which release of hydrogen in the form of proton is helpful. Thus, H₂S is a weak acid.

Question 6.
In the laboratory ammonia is dried by quick lime, why ?
Answer:
In the laboratory ammonium chloride is heated with caustic soda solution or with milk of lime in a hard, round bottom flask. Ammonia evolved is directly passed over quick lime and is collected in gas jars by downward displacement of air.

Aqueous NH₃ is dried by passing it over quick lime (CaO). H₂SO₄, CaCl₂ and P₂O₅, etc. cannot be used for drying NH₃ because it reacts with them.

Question 7.
Write the difference between the bleaching action of SO₂ and Cl₂.
Or
Bleaching of flowers by Cl₂ is permanent while bleaching by SO₂ is temporary, why ?
Answer:
Bleaching action of SO₂: Bleaching by SO₂ is due to the process of reduction in the presence of moisture.
SO₂ + 2H₂O → H₂SO₄ + 2[H] (nascent)
The nascent hydrogen liberated in the reaction is responsible for bleaching flower but when the flower comes in contact with air. It gets oxidized and become coloured.

Bleaching action of Cl₂: Bleaching by Cl₂ is due to the process of oxidation in the presence of moisture.
Cl₂ + H₂O → 2HCl + [O] (nascent)
Coloured substance + [O] → Colourless
The oxygen liberated in the reaction is responsible for bleaching, colour bleached by Cl₂ is permanent and original colour cannot be restored on exposure to air.

Question 8.
Write formula and structure of five Oxy acids of Sulphur.
Answer:
Oxy acids of Sulphur:

Question 9.
Write any four functions of Glover’s tower in lead chamber process for manufacturing of H₂SO₄.
Answer:
Functions of Glover’s tower :
(1) Oxide of nitrogen in nitrated sulphuric acid obtained by Glover’s tower, released
2NO + 2HSO₄ + H₂O → NO + NO₂ + 2H₂SO₄

(2) By the temperature of gases, the water present in lead chamber evaporates and acid concentration increases upto 78%.

(3) In this tower SO₂, NO₂ and water vapour react to form H₂SO₄.
SO₂ + NO₂ + H₂O → H₂SO₄ + NO

(4) Stones of Glover’s tower cool down and maintain the temperature 60-80°C.

Question 10.
What is Aqua regia ? Write its uses.
Answer:
Aqua regia : Mixture of 1 part of cone. HNO₃ and 3 part of cone. HCl.
Uses : It dissolves Au and Pt.

Question 11.
Contact process of manufacturing of H₂SO₄ is supposed to be more ad¬vance than lead chamber process. Why ?
Answer:
(i) The acid obtained by contact process is pure while the acid obtained by lead chamber process is impure.
(ii) The plant for lead chamber process requires large area while plant used in contact process occupies comparatively less area.
(iii) The catalyst used in contact process is platinised asbestos while in lead chamber process catalyst is gaseous (NO₂ gas) and to maintain its flow regular is tedious.
(iv) In contact process concentrated sulphuric acid is obtained while in lead chamber process dilute acid is obtained.
(v) In contact process running of the plant is less costly while in lead chamber process running of the plant is costly.

Question 12.
The boiling point of NH₃ is more than PH₃. Why ?
Answer:
Ammonia molecules are associated together by hydrogen bond due to presence of N atom which is more electronegative than P atom of PH₃ in which there is no hydrogen bond that is why boiling point of NH₃ is high.

Question 13.
Draw labelled diagram of laboratory method of phosphine and give chemical equation.
Answer:
Phosphine is prepared in laboratory by heating white phosphorus with NaOH in an inert atmosphere of CO₂ and oil gas. Phosphine produced is highly inflammable due to the presence of phosphorus dehydride as impurity. Vapour of phosphine form vortex noing of smoke in contact with air. Alcoholic KOH can be used in place of NaOH.

Question 14.
Nitrogen differs from its group 15 and show diagonal relationship with sulphur of group 16. Clarify the cause of similarity and difference.
Answer:
Nitrogen shows dissimilarities with other elements of that group due to following factors :

1. Smaller atomic size.
2. High electronegativity
3. Tendency to form multiple bonds
4. Non-availability of d-subshells.

Nitrogen resembles sulphur of group 16 in the following properties :

1. Both are non-metal
2. Both are bad conductor of electricity
3. Both are electronegative element
4. Both form covalent compound
5. Both form stable and covalent hydride
6. Oxide of both elements are soluble in water and form oxy acids
7. Hydride of both elements dissolve in water.

Question 15.
Write the structure of Pyrophosphoric acid.
Answer:

Oxidation number of phosphorus in pyrophosphoric acid is +5. It is tetrabasic acid. Pyro word is used for such acids which is obtained by the loss of one water molecule by heating two molecules. Its chemical formula is H₄P₂O₇(P₂O₅.2H₂O).

Question 16.
Oxygen exhibit -2 to +2 oxidation state while the other member of this group show +2, +4 and +6 oxidation state. Why ?
Answer:
The oxidation state of oxygen is -2, but in H₂O₂, O₂, O₂F₂ and OF₂ etc., the oxidation number of oxygen is -1, 0, +1 and +2. Oxygen is divalent because it contains two unpaired electrons. Its electronic configuration is 1s²,2s², 2p²_x, 2p¹_y 2p¹_z Therefore, there is no vacant nd orbital in oxygen. Thus, it does not have more than 2 valency whereas other element of this group contains vacant nd orbital. The electron from ns and np orbital can jump to nd orbital and gives rise to +2, +4 and +6 oxidation state.

Question 17.
Explain that molecular formula of oxygen is O₂ whereas that of sulphur is Sg.
Answer:
Size of oxygen atom is small, thus it possesses the ability to form stable double bond with itself. Thus, its molecular formula is O₂ where as due to large size of sulphur atom it does not form multiple bond, therefore it does not exist in the form of S₂. Along with it due to high S-S bond energy it has catenation property more than oxygen. Therefore, sulphur exists in the form of S₈ in which each sulphur atom is linked with other sulphur atoms by single covalent bonds and form puckered ring like structure.

Question 18.
Write name, formula and oxidation state of oxides of nitrogen.
Answer:
Five oxides of Nitrogen are :
(i) Nitrous oxide (N₂O): Oxidation state +1.

Question 19.
Explain Siemen-Halskey ozonizer process for the manufacture of ozone and draw labelled diagram.
Answer:

For commercial production of ozone Siemen-Halskey ozonizer is used. It consists of an iron box fitted with 6 to 8 glass cylinders. Each cylinder is 3 ft high and 10″ in width. The box is divided into three com-partments. Cold water flows through the central compartment for cooling. Each cylinder is fitted with an aluminium rod which are resting on insulating glass slab in the lower compartment. There is annular space between rod and sides of cylinder. A current of air flows through this annular space. The aluminium rods are subjected to high potential of 8,000 to 10,000 volts. The air which enters at bot¬tom escapes at the top and contains about 5% of ozone. If pure oxygen is used in place of oxygen then ozonized oxygen contains about 15% of ozone.

Question 20.
H₂S is a stronger reducing agent than H₂O. Why ? Give any three reasons.
Answer:
H₂S is a stronger reducing agent than H₂O because H₂S can easily releases its hydrogen not H₂O due to the following reasons :
(1) In H₂O, H-bond is present which require extra energy to get released.
(2) Size of S is larger than oxygen.
(3) Electronegativity of oxygen is more than sulphur.

Question 21.
Why is sulphurous acid a reducing agent ?
Answer:
Because on sulphur atom of H₂SO₃ a non-bonding electron is still present. By losing this electron pair sulphur atom of H₂SO₃ can achieve higher oxidation state. Therefore, H₂SO₃ act as a reducing agent.

Question 22.
What is general electronic configuration of noble gases ? Write electronic configuration of noble gases.
Answer:
The general electronic configuration of noble gases is ns²np⁶. Where, n is the valence shell’s serial number.

Electronic configuration of noble gases are following :
Helium (He) → 1s²
Neon (Ne) → 1s² 2s² 2p⁶
Argon (Ar) → 1s² 2s² 2p⁶ 3s² 3p⁶
Krypton (Kr) → 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶
Xenon (Xe) → 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶
Radon (Rn) → 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶.

Question 23.
Explain with reason :
(a) HF is liquid while other hydrides of halogens are .gases at normal temperature.
(b) Fluorine does not form polyhalide.
Answer:
(a) The electronegativity of fluorine is maximum than other halogens. Therefore, HF molecule is conjugated by H-bond and the boiling point of HF is more to other halogen acids therefore HF is liquid while halides of other halogens are gases at room temperature.
H – F …. H – F …. H – F …. H – F
(b) The electronic configuration of fluorine is 1s², 2s², 2p⁵. Due to absence of d-orbital in its valency shell it does not show higher oxidation state and so it does not made polyhalide.

Question 24.
Explain, why :
(a) Noble gases are monoatomic ?
(b) Atomic radii of noble gases are largest ?
(c) Ionisation energy of noble gases are highest ?
Answer:
(a) There is no any unpaired electron in electronic configuration of noble gases. So these do not form chemical bonds and are monoatomic.
(b) Outermost shell of noble gases are completely filled so atomic radii of noble gases are maximum.
(c) All electrons in outermost orbit of noble gases are paired and energy required to make them unpaired is very high. Thus, noble gases have maximum ionisation energy in respective periods.

Question 25.
What do you mean by available chlorine ? Explain with reaction.
Answer:
When bleaching powder is treated with excess of dilute acid or carbon dioxide, the whole of chlorine present in the molecule is liberated. The amount of chlorine thus set free is called available chlorine.
CaOCl₂ + 2HCl → CaCl₂ + H₂O + Cl₂
CaOCl₂ + H₂SO₄ → CaSO₄ + H₂O + Cl₂
CaOCl₂ + CO₂ → CaCO₃ + Cl₂
Amount of available chlorine in a good sample is 35-38%.

Question 26.
Write any two uses of helium gas.
Answer:
Uses of Helium : (i) It is used in filling air ships and balloons for weather study.
(ii) A mixture of 80% helium and 20% oxygen is used by divers for respiration.

Question 27.
Explain the structures of XeF₂ and XeF₄.
Answer:
Structure of XeF₂: In XeF₂ molecule Xe atom is in sp³ d hybrid state hence, its structure is trigonal bipyramidal but its geometry is linear.


Structures of XeF₄ : XeF₄ formed under second excited state of Xe where two of sporbitals are unpaired and 4 unpaired electron develop. Being sp³ d² hybridization involved, the structure is octahedral but due to presence of two lone pair of electron geometry of molecules get distorted and becomes square planar.

Question 28.
Fluorine always exhibit -1 oxidation state. Why ?
Answer:
Fluorine is the most electronegative element of periodic table. It always exhibit -1 oxidation state. It has no d-subshell in the outermost orbit. So it does not show any excited state.

Question 29.
Write the name, formula and oxidation number of any two oxy acids of chlorine.
Or,
Write formula structure and oxidation states of any two oxy acids of chlorine.
Answer:
Oxy acids of Chlorine :

(i) Structure of HCIO : Central atom chlorine in CIO ion of HCIO is sp³ hybridized, due to which four hybrid orbitals are formed. Three of it contain lone electron pair and the fourth undergo overlapping with the orbital of oxygen forming sigma bond. The H+ geometry is linear.

(ii) Structure of HClO₃

(iii) Structure of HClO₄ : Its central atom Cl is sp³ hybridized. Thus, ClO₄ ion is tetrahedral.

Question 30.
Explain bleaching action of bleaching powder.
Answer:
The cloth to be bleached is dipped in a solution of bleaching powder in water in a vat. From this vat, the cloth passes into a very dilute solution of hydrochloric acid. Here the cloth is bleached due to the liberation of chlorine by the action of acid on the bleaching agent. Some chlorine remains sticking to the fibre and is likely to damage it. In order to remove the traces of chlorine, the cloth is passed through a vat containing antichlor like sodium thiosulphate.

The cloth is then carried to another vessel where it is washed with excess of water. The washed cloth is then pressed with rollers, dried and ironed in the ironing cylinders and then wrapped in the form of a roll.
CaOCl₂ + H₂SO₄ → CaSO₄ + H₂O + Cl₂

Question 31.
Noble gases are inert. Why ?
Answer:
Noble gases are inert due to following reasons :

1. Noble gases are inert because octet of these gases are complete. It is the most stable state and do not have unpaired electrons.
2. Ionisation energy of noble gases are high and electronegativity and electron affinity are zero. Therefore, these gases neither accept electron nor release electron.

Thus, noble gases are inert in nature.

Question 32.
Write any three anomalous behaviour of fluorine from other member of its group.
Answer:
Anomalous behaviour of fluorine are :

1. Highest electronegativity of fluorine than other halogen. So, it can form F- ion and can displace other element.
2. Due to smaller size, it forms strong covalent bond with other atom.
3. Bond energy of fluorine is very low of about 158 kJ mol^-1. Hence, it requires very less activation energy.

Question 33.
Xenon is a noble gas, then also it forms compound. Why ? Give structure of any two compounds.
Answer:
In 1962 Neil Bartlett observed that oxygen reacts with PtF₆ and form O₂[PtF₆]. He thought that the first ionization energy of oxygen and xenon is nearly same, on this basis he treated Xe and PtF₆ and got the first compound Xe+[PtF₆].

By direct action of Xe on PtF₆, first compound of inert gases Xe⁺[PtF₆]^– was obtained. It is an orange-yellow crystalline solid.

Question 34.
Fluorine is strong oxidizing agent as compared to chlorine. Why ?
Answer:
Because of following points, fluorine is strong oxidizing agent:

1. The size of F atom is smaller than Cl atom.
2. Electronegativity of fluorine is higher than Cl atom.
3. Dissociation energy of fluorine is less than that of chlorine.
4. E° value of fluorine is more than chlorine.

Question 35.
Describe the laboratory preparation of chlorine.
Answer:
Laboratory preparation : HCl and MnO₂ is taken in a round bottom flask. Cone. HCl is added from thistle funnel. The flask is heated slowly so that green yellow colour of Cl₂ gas evolved.
MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂

Question 36.
Write any five physical properties of noble gases.
Answer:
Five physical properties of noble gases :

1. Atomic radii: Ionic radii of noble gases are corresponding to van der Waals’ radii. On moving downwards in the group, van der Waals’ radii increases.
2. Ionisation energy : Due to stable electronic configuration, ionisation energy is very much. Value of ionisation energy decreases with increase in atomic number.
3. Electron affinity : Due to stable electronic configuration ns²np⁶, noble gases do not have tendency to accept electron. Thus, electron affinity of these nearly zero.
4. m. p. and b.p.: Due to weak attractive forces present in atoms m.p. and b.p. of these gases are low. Value of m.p. and b.p. increases on moving downward in the group.
5. Liquification : Noble gases are not liquefied easily due to weak van der Waals’ forces.

Question 37.
Describe properties of hydrides of halogens under the following heads :
(i) Physical state
(ii) Thermal stability
(iii) Reducing property
(iv) Acidic strength
(v) Nature of bond.
Answer:
Properties of hydrides of hydrogen :
(i) Physical state: Hydrogen fluoride is a low boiling (292K) liquid while HC1, HBr and HI are gases. HF is liquid due to H-bonding.

(ii) Thermal stability : The order of thermal stability is
HF > HCl > HBr > HI
On moving down the group thermal stability decreases due to decrease in stability of bond with increase in size of halogen atom.

(iii) Reducing property : Reducing property increases down the group because stability of H – X bond decreases down the group.
HF < HCl < HBr < HI
HF does not have reducing property. HCl is weak reducing agent, HBr is strong reducing agent while HI is strongest reducing agent.

(iv) Acidic strength : Halogen hydrides are covalent compound in gaseous state. They ionize in gaseous medium and behave like acids. Their acidic strength is related to
stability of H – X bond strength. The decreasing order of acidic strength is
HF > HCI > HBr > HI
Dissociation energy of HF is highest hence, it does not ionises in aqueous solution.
Therefore, it is the weakest acid.

(v) Nature of bond: All halides are of covalent nature, but some possess ionic character also. Inonic character decreases from HF to HI.

Question 38.
Fluorine shows only -1 oxidation state while other halogen shows +3, +5 and +7 oxidation state in addition to -1 and +1, why ?
Answer:
F shows only +1 and -1 oxidation state, while other halogen i.e Cl, Br and I show +3, +5, +7 oxidation state in addition to +1, -1 because Cl, Br, and I halogen atom have vacant d-orbital on excited state they exhibit +3, +5 and +7 unpaired in d-orbital.

Therefore, except F other halogen show +3, +5 and +7 oxidation state in addition to -1 and +1.

Question 39.
Elements of group 17 (halogens) are coloured. Explain, why ?
Answer:
All elements of halogen group are coloured. Colour of these elements become deep with increase in atomic number.
Fluorine – Yellow
Bromine – Brown or deep red
Chlorine – Green-Yellow
Iodine – Violet.
Colour of halogens is due to absorption of visible light by molecules. Due to absorption of light outermost electrons excited to higher orbitals and elements are seen coloured. Being smaller atomic size, fluorine absorbs violet radiation of higher energy and are seen light yellow coloured while iodine atom having larger atomic size absorbs yellow radiation of lower energy and are seen violet coloured. For larger atomic size, radiation of less energy are required because electrons are far from the nucleus.

Question 40.
Give reason :
(a) Iodine exhibits some metallic property.
(b) Halogens are strongly oxidizing, Why ?
Answer:
(a) Ionization energy of iodine is small so it gives out one electron from its valence shell. In some reaction Iodine gives I⁺ ion so it has some metallic property.
I → I⁺ + e^–
(b) Halogens are only one electron short to complete their octet and attaining stable structure. So halogens have strong tendency to accept electron. Thus, halogens are strong oxidising agents because of accepting electrons in reduction, oxidising power of halogens decreases from fluorine to iodine.
F₂ > Cl₂ > Br₂ > I₂

Question 41.
Fluorine is more active than other halogen. Explain, in three points.
Answer:
Fluorine is more active than other halogen because of:

1. Bond energy of fluorine is minimum (158 kJ/mol), so F₂ molecules require less activation energy for reaction.
2. Size of fluorine atom is smaller than other halogen atoms, so the covalent bond formed with other atoms is stronger.
3. Electronegativity of fluorine atom is highest and it forms F- ion easily. Due to more electronegativity it substituted other elements from their compounds.

The p-Block Elements Long Answer Type Questions

Question 1.
Give chemical reaction of ozone with
(i) K₂MnO₄
(ii) I₂
(iii) Ag₂O
(iv) CH₂ = CH₂ and
(v) PbS.
Answer:
(i) With K₂MnO₄ : Potassium permangnate is formed.
2K₂MnO₄ + H₂O + O₃ → 2KMnO₄ + 2KOH + O₂

(ii) With I₂ : Iodic acid is formed.
I₂ + H₂O + 5O₃ → 2HIO₃ + 5O₂

(iii) With Ag₂O : Silver is formed.
Ag₂O + O₃ → 2 Ag + 2O₂

(iv) With CH₂ = CH₂: Ethene ozonide is formed.

(v) With PbS : PbSO₄ is formed.
PbS + 4O₃ → PbSO₄ + 4O₂

Question 2.
Explain Ostwald’s process of manufacture of nitric acid drawing labelled diagram.
Answer:

Principle of Ostwald’s process: The mixture of ammonia and air when passed over platinum gauze catalyst at 750 – 90OC, ammonia gets oxidized to pitric oxide.

The reaction is exothermic and the heat liberated maintains the temperature of the catalyst. The nitric oxide is then oxidised to nitrogen dioxide by air which is cooled to 50”C and absorbed by water to produce nitric acid.

Question 3.
In which properties nitrogen exhibit difference with the other elements of that group and why ?
Answer:
Nitrogen shows dissimilarities with other elements of that group due to following factors :

1. Smaller atomic size
2. High electronegativity
3. Tendency to form multiple ,bonds
4. Non-availability of J-subshells.

Differences in properties :

1. Nitrogen is less reactive gas while other members are reactive solids
2. Nitrogen is diatomic while molecules of other elements are tetra-atomic (P₄, AS₄, Sb₄)
3. Nitrogen does not form complex compound while other elements of this group form complex compound due to presence of vacant d-orbitals
4. Various oxides of nitrogen (N₂O, NO, N₂O₃, N₂O₄ , N₂O₅) are known while other members do not form so many oxides
5. NH₃ is a liquid with high b.p. while other hydrides are gases
6. Nitrogen shows catenation capability while it is absent in other elements
7. Nitrogen does not show allotropism while other elements show allotropism
8. Nitrogen does not form complex ions while other elements form complex
   ions e.g., PF^–₄,SbI^–₆ etc.

Question 4.
Describe Brodie’s Ozonizer with diagram.
Answer:

It is U-shaped glass tube, consists Pt wire with dil. H₂SO₄. The whole apparatus is put into glass utensil, outer utensil also contain H₂SO₄ and Pt electrode. Dry O₂ is i passed through U-tube under the influence of high electric discharge oxygen gets converted to ozonised oxygen, the outcoming gas contains 20% of ozone.

Question 5.
Explain hydrides of nitrogen family on the following points :
1. Name and formula
2. Basic nature
3. Reducing property
4. Bond angle
5. Melting and boiling point.
Answer:

Basic nature: Basic nature decreases from NH₃ to BiH₃. Basic nature of these hydrides is due to the presence of lone pair of electron in the central atom.
Reducing nature: Reducing property increases in moving downwards from NH₃ to BiH₃. This is due to the decreasing stability of hydrides.
Bond angle: On going down the group the bond angle decreases due to decreasing electronegativity of central atom :

Melting point and boiling point: There is no regular trend of M.P. and B.P. of 15^th group elements. M.P. increase from N to As whereas again decreases in Sb and Bi. B.P. increases continuously from N to Bi.

Question 6.
Explain hydrides of oxygen family on the following points :
1. Name and formula
2. Thermal stability
3. Reducing property
4. Acidic property
5. Covalent Character
Answer:

Thermal stability : Stability decrease with increase in molecular mass.

Reducing nature : The reducing power of hydrides of oxygen family increase in thermal stability.
Acidic nature : Water is neutral in nature while acidic nature increases from H₂S to H₂Te. This is due to increase in bond length in moving down the group which leads to increase hydrogen releasing tendency.
Covalent Character: Hydrogen has one electron in its outermost shell by which it can form covalent bond with the other elements of oxygen family and complete its outermost shell.

Question 7.
Explain manufacture of chlorine under the following heads :
(i) Labelled diagram of Nelson cell
(ii) Principle
(iii) Deacon’s process.
Answer:
(i) Nelson cell : In Nelson cell, graphite anode is suspended in a perforated cylindrical steel cathode provided with asbestos. Brine solution is taken in steel cathode. On electrolysis, chlorine gas is liberated at anode and is drawn Brine off through the outlet at the top. Na reacts with H₂O at bottom to form NaOH and H₂. Hydrogen escapes through the exit.

(ii) Principle: Nelson method or By electrolysis : Chlorine is manufactured by electrolysis of NaCl solution, chlorine is obtained as by-product.

(iii) Deacon’s process : Previously chlorine was manufactured by Deacon’s process. In this process, HCl gas is heated with oxygen at 450°C in presence of CuCl₂ catalyst.

Question 8.
Why elements of group 17 are called halogen ? Explain the following properties of halogen:
(i) Oxidation state
(ii) Electronegativity
(iii) Oxidizing property
(iv) Formation of bond with other element.
Answer:
Halogen means salt producer. The first four members of this group found in sea water. Therefore, the elements belonging to 17th group are called halogen.

Properties of halogen:
(i) Oxidation state : Common oxidation state of halogen is -1. Except fluorine the other halogen atom exhibit + 3, + 5, + 7 oxidation state.
F = -1
Cl = -1, +1, + 3, + 5, + 7
Br = -1, +1, +3, +5 .
I = -1, +1, +3, + 5, + 7.
Halogen have electronic configuration ns²np⁵ in their outermost energy level. So they have strong tendency to get or to share one electron for completion of octet, so they show -1 oxidation state when they combine with less electronegative elements.On combining with more electronegative elements they show +1 oxidation state. Oxidation state in HF, HC1 and HI is -1 and in ClF, BrF, IF, HClO, HBrO and HIO is + 1.

(ii) Electronegativity: The halogen have very high values of electronegativity. Fluorine is the most electronegative element with the electronegativity value 4.

(iii) Oxidizing property : Halogen are strong oxidizing agent due to high electron affinity, they have strong tendency to accept electron.
Decreasing order of oxidizing power is given as F₂ > Cl₂ > Br₂ > I₂.

(iv) Formation of bond with other element: Halogen form ionic bond with elec-tropositive element with the increase in the size of halogen atom, the tendency to form ionic bond decreases, e.g., AlF₃ is ionic compound but AlCl₃ is covalent compound. It forms covalent bond with non-metal.

Question 9.
Explain bleaching powder under the following points :
(i) Methods of preparation
(ii) Properties
(iii) Uses.
Answer:
(i) Methods of preparation: Bleaching powder is manufactured by the action of chlorine on dry slaked lime.

There are two plants for the manufacture of bleaching powder.

(a) Hasenclever plant : The plant consists of a number of horizontal cylinders provided with rotating shafts carrying blades.

Procedure: Slaked lime is added in the hopper and moved forward from one cylinder to the next by the revolving blades till it falls down. At the same time, dilute chlorine is passed up.

The formation of bleaching powder is based on the principle of counter currents as slaked lime and chlorine move in opposite direction to ensure complete conversion. Bleaching powder formed is collected in the receiver below.

(b) Modern method (Bachmann’s plant): Description of the plant: The plant consists of a vertical tower made of cast iron. It is provided with inlets for chlorine and hot air slightly above the base, a hopper at the top and an exit for the unused chlorine and air just below the top. There are a number of horizontal shelves at regular heights inside the tower.

Each shelf is provided with rotating rakes.

Dry slaked lime is fed into the chlorinating tower through the hopper at the top with the help of a suitable compressed air pumping arrangement. Slaked lime thus added, moves down with the help of rotating rakes. It meets the upgoing chlorine on the way and is con¬verted into bleaching powder which collects in the container, placed at the bottom. A current of hot air is passed to drive away unreacted chlorine.

(ii) Properties of bleaching powder:
(a) It is a pale yellow powder having strong smell of Cl₂.
(b) It is soluble in cold water.
(c) In presence of a little cobalt chloride, it decomposes to liberate oxygen. On long standing it undergoes slow auto-oxidation and is converted into a mixture of calcium chlorate and calcium chloride.
(d) Reaction with insufficient acid: Bleaching powder is treated with small amount of dilute acid, it liberates hypochlorous acid which fumitus oxygen (nascent) and thus acts as oxidizing and bleaching agent.

(e) With excess of acid : Bleaching powder reacts with excess of acid and liberates Cl₂. The amount of Cl₂ thus liberated is called “Available chlorine”.

(f) Reduction: Bleaching powder decomposes in presence of cobalt chloride liberating oxygen.

On keeping for long time it undergoes self oxidation forming calcium chlorate and calcium chloride.

(iii) Uses: (a) It is used as a disinfectant and germicide for the sterilization of drinking water.
(b) For the manufacture of chloroform.
(c) In textile and paper factories, for bleaching cotton, linen and wood pulp.
(d) For rendering wool unshrinkable.

Question 10.
What are interhalogen compounds ? How many types are they ? Give one example of each type with structure.
Or,
Explain hybridization in ABS and AB7 type of interhalogen compounds and draw their structures.
Answer:
Because of difference in the electronegativity value of halogen member it is possible to combine to form compound. When two different halogen combine together to form binary compound, then it is known as interhalogen compound.

There are four types of interhalogen compound, which has general formula XY_n, where n = 1, 3, 5 or 7.

Structure of interhalogen compound :
1. AB Type : Like ClF, BrCl, IBr, ICl.
Their geometry is linear. In the ground state electronic configuration of chlorine it is clearly observed that there is one unpaired electron which forms covalent bond with other halogen atom.

2. AB₃ Type : It is T-shaped. Like ClF₃ molecule. Its central atom X shows sp³d hybridization.

3. AB₅ (IF₅, BrF₅) Type : These type of compounds show sp³d² hybridization. Their structure is square pyramidal in which there is lone pair of electron at one place. Like in the figure formation of IF₅ molecule is shown.

4. AB₇ (IF₇) Type : Its geometry is pentagonal bipyramidal which is formed by sp³d² hybridization. Hybridization in IF7 molecule is sp³d³.

Question 11.
Write the uses of Noble gases.
Answer:
Uses of Noble Gases :
Uses of Helium : (i) It is used in filling air ships and balloons for weather study as it is light and non-inflammable.
(ii) Helium is used in research work for maintaining very low temperatures.
(iii) It is used in producing inert atmosphere, for food preservation and as filler in electric transformer.
(iv) Sea divers use mixture of helium and oxygen for deep sea diving. This is due to the fact that under high pressure helium is less soluble in blood as compared to nitrogen. If for breathing in deep sea diving air is used then more nitrogen will dissolve in blood under pressure and when the diver comes back at the surface of the sea, the solubility of nitrogen in blood decreases due to decrease in pressure and changes into nitrogen bubbles. This causes severe pain in diver’s body. It may also cause blisters and vomiting.

Uses of Neon : (i) Neon lights are used for commercial advertisements. It consists of a long tube fitted with electrodes at both ends. On filling the tube with neon gas and passing electric discharge of about 1000 volt potential, a bright red light is produced. Different colours can be obtained by mixing neon with other gases. For producing blue or green light neon is mixed with mercury vapours.

Uses of Argon : (i) Like helium, it is also used to create inert atmosphere in welding of aluminium ahd stainless steel.
(ii) It is filled in electric bulbs along with 25% nitrogen.
(iii) It is also used in radio valves.
(iv) Argon alone or its mixture with neon is used in tubes for producing lights of different colours.

Uses of Krypton and Xenon : (i) They are also used in electric bulbs in place of argon, but they are costly.
(ii) Both are used in flash photography.
(iii) Krypton is mainly used in producing fluorescence, incandescence and electric lights.
(iv) Krypton lamp is used as indicator at aeroplane terminals for landing of the aeroplane.

Uses of Radon: (i) In treatment of malignant growths (cancer) and non-healing wounds. Also, used in researches associated with radioactivity.
(ii) Used as a substitute for X-rays in radiology.
(iii) Radon is soluble in fatty compounds and recently a radon containing ointment has been manufactured for radiotherapy.

MP Board Class 12th Chemistry Solutions