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

Answer:

Nitrogen belongs to second period (n=2) and has only s and p-orbitals. It does not have d-orbitals in its valence shell and therefore, it cannot extend its octet. That is why nitrogen does not form pentahalides.

Answer:

Ammonia exists as associated molecule due to its tendency to form hydrogen bonding. Therefore, it has high boiling point. Unlike NH₃, phosphine (PH₃) molecules are not associated through hydrogen bonding in liquid state. This is because of low electronegativity of P than N. As a result, the boiling point of PH3 is lower than that of NH₃.

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Thermal decomposition of sodium azide gives nitrogen gas.

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Nitrogen atom in NH₃ has one lone pair of electrons which is available for donation. Therefore, it acts as a Lewis base.

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PH₃ reacts with acids like HI to form phosphonium iodide, PH₄I.

This shows that PH₃ is basic in nature. This basic nature of PH3 is due to the presence of lone pair on phosphorus atom and therefore, it acts as a Lewis base.

Answer:

PCl₃ gets hydrolysed in the presence of moisture and gives fumes of HCl.

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PCl5 has trigonal bipyramidal structure in which there are three P—Cl equatorial bonds and two P—Cl axial bonds. The two axial bonds are being repelled by three bond pairs at 90$°$ while the three equatorial bonds are being repelled by two bond pairs at 90$°$. Therefore, axial bonds are repelled more by bond pairs than equatorial bonds and hence are larger (219 pm) than equatorial bonds (204 pm).

Answer:

H₃PO₂ has one P==O, one P—OH and two
P—H bonds as

Since two H atoms are bonded directly to P atom which impart reducing character to the acid.

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The elements of group 15 have extra stable half filled electronic configuration (ns²np³) and therefore, larger amount of energy is required to remove electrons. But the electronic configurations of group 16 elements (ns² np⁴) do not have extra stability and therefore, energy required to remove the electron is less as compared to group 15 elements.

Answer:

In H₂Te, the size of central Te is more than that of S in H₂S and therefore, the distance between the central atom and hydrogen, Te—H is more than that of S—H. As a result of large bond length, the bond dissociation enthalpy of, Te—H is less than that of S—H and bond cleavage of Te—H bond is easy. Therefore, H₂Te is more acidic than H₂S.

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1. It forms hydrogen fluoride.

   

2. Sulphuric acid is formed.

   

Answer:

The halogens have the smallest size in their respective periods and therefore, high effective nuclear charge. Moreover, they have only one electron less than the stable noble gas configuration (ns2np6).
Therefore, they have strong tendency to accept one electron to acquire noble gas electronic configurations and hence have maximum negative electron gain enthalpy in their respective periods.

Answer:

It is because of:

1. low bond dissociation enthalpy of F–F bond.
2. high hydration enthalpy of F ^– ion.

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Fluorine is most electronegative element and cannot exhibit any positive oxidation states. On the other hand, the other halogens are less electronegative and therefore, can exhibit positive oxidation states. They also have vacant d-orbitals and hence can expand their octets and show +1, +3, +5 and +7 oxidation states also.

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This is a disproportionation reaction because the oxidation state of Cl is changed from zero (in Cl₂) to –1 (in NaCl) and +5 (in NaClO₃).

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HCl reacts with finely divided iron and produces H₂ gas

Liberation of hydrogen prevents the formation of ferric chloride.

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It involves sp3d hybridisation of the central halogen atom and the molecule has trigonal bipyramidal geometry with two positions occupied by lone pairs. Its structure is termed as T-shaped.

[Three bonds with F and two positions occupied by lone pairs]

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The elements present in group 18 have completely filled valence shell (except He 1s²). Therefore, they have neither any tendency to lose nor to gain electrons. However, they react with a few elements only under certain conditions. Therefore, they are known as noble gases.

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Noble gases are monoatomic gases and are held together by weak van der Waals forces (dispersion forces). Therefore, they are liquefied at very low temperatures. Hence they have low boiling points.

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The hydrolysis reaction of XeF₆ is

The oxidation states of all the elements in the products remain the same as it was in the reacting state. Hence, it is not a redox reaction.

Answer:

In pentahalides, the oxidation state is more (+5) than in trihalides (+3). As a result of higher positive oxidation state of central atom, they have larger polarizing power and can polarise the halide ion(X^–) to a greater extent than in the corresponding trihalide. Since larger the polarisation, larger is the covalent character, therefore, pentahalides are more covalent than trihalides of P, As, Sb and Bi.

Answer:

Among the hydrides of group 15, BiH3 is least stable because Bi has largest size in the 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.

Answer:

In molecular nitrogen, there is a triple bond between two nitrogen atoms (N $\equiv$$\equiv$ N) and it is non-polar in character. Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol^–1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

Answer:

Ammonia is formed according to the reaction:

The conditions for maximum yield of ammonia are:

1. Low temperature of the order of about 700 K
2. High pressure of 200 × 10⁵ Pa (about 200 atm).
3. Presence of catalyst such as iron oxide with small amount of K₂O and Al₂O₃

Answer:

N₂O₅ has the structure.

The covalence of nitrogen in the above structure is four because it has four shared pair of electrons.

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Phosphine is formed.

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On heating, PCl₅ first sublimes and then decomposes on strong heating:

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H₃PO₄ contains three P–OH bonds and therefore, its basicity is three.

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On heating, H₃PO3 disproportionates to give orthophosphoric acid and phosphine.

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Sulphur occurs in the combined form as sulphide ores and sulphate ores. The common sulphide ores in which sulphur occurs are galena (PbS), zinc blende (ZnS), copper pyrites (CuFeS₂) and sulphate ores are gypsum (CaSO₄.2H₂O), epsom salt (MgSO₄.7H₂O) and baryte (BaSO₄). Traces of sulphur occurs as hydrogen sulphide in volcano. Organic materials such as eggs, proteins, onion, garlic, mustard, hair and wool also contain sulphur.

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The thermal stability of the hydrides of group 16 elements decreases from H₂O to H₂Te as: H₂O > H₂S > H₂Se > H₂Te

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Due to high electronegativity of oxygen and its small size, there are strong hydrogen bonding in water. As a result, the molecules exist as associated and is liquid at room temperature. But there is negligible hydrogen bonding in H₂S because of low electronegativity of S.

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Zn, Ti, Pt, Fe

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Ozone acts as a powerful oxidising agent because it has higher energy content and decomposes readily to give atomic oxygen as:

Therefore, ozone can oxidise a number of non-metals and other compounds. For example,

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Ozone oxidises potassium iodide to iodine as:

The liberated iodine may be titrated against a standard solution of sodium thiosulphate.

Thus, to estimate O3 quantitatively, ozone is allowed to react with known amount of excess potassium iodide solution buffered with a borate buffer (pH = 9.2). The liberated I2 is titrated against Na2S2O3 solution using starch as an indicator. From this amount of ozone can be calculated.

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Fe(III) salt is reduced to Fe(II) salt.

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Sulphur dioxide can be detected by the following tests:

1. It has a pungent characteristic smell.
2. It decolourises acidified potassium permanganate solution.
3. It turns acidified potassium dichromate solution green.
4. It turns blue litmus red.

Answer:

1. manufacture of fertilizers (e.g., ammonium sulphate, superphosphate).
2. petroleum refining
3. metallurgical applications (e.g., cleansing metals before enamelling, electroplating, galvanising, etc.).

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1. Low temperature (optimum temperature 720 K)
2. High pressure (optimum pressure 2 bar)
3. Presence of catalyst (V₂O₅ catalyst).

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Answer:

F₂, is stronger oxidising agent than Cl₂. This can be explained on the basis of bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy. The process of oxidising behaviour may be expressed as:

The overall tendency for the change (i.e., oxidising behaviour) depends upon the net effect of three steps. As energy is required to dissociate or convert molecular halogen into atomic halogen, the enthalpy change for this step is positive. On the other hand, energy is released in steps (II) and (III), therefore, enthalpy change for these steps is negative. Now although fluorine has less negative electron gain enthalpy, yet it is stronger oxidising agent because of low enthalpy of dissociation and very high enthalpy of hydration. In other words, large amount of energy released in step (III) and lesser amount of energy required in step (I) overweigh the smaller energy released in step (II) for fluorine. As a result, the ∆H overall is more negative for F₂ than for Cl₂. Hence, F₂ is stronger oxidising agent than Cl₂.

Answer:

1. Since fluorine is most electronegative element, it shows only a negative oxidation 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.
2. Maximum covalency of fluorine is one because it cannot expand its valency shell beyond octet because there are no d-orbitals in the valence shell. On the other hand, other elements can exercise covalencies upto 7 because of availability of vacant d-orbitals.

Answer:

Sea is an important source of some halogens. Sea water contains chlorides, bromides and iodides of sodium, potassium, magnesium and calcium. But it is mainly sodium chloride (2.5% by mass). The deposits of dried up sea contains sodium chloride, carnallite (KCl. MgCl₂.6H₂O). Certain forms of marine life also contain iodine in their systems. For example, various sea weeds contain upto 0.5% of iodine and chile salt petre contains upto 0.2% of sodium iodate.

Answer:

Bleaching action of chlorine is due to its oxidation. In the presence of moisture, chlorine gives nascent oxygen

Because of nascent oxygen, it bleaches coloured substance as:

It bleaches vegetables or organic matter. The bleaching action of chlorine is permanent.

Answer:

1. Phosgene (COCl₂)
2. Tear gas (CCl₃.NO₂)

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Interhalogens are more reactive compounds than their constituting halogens because of weaker X–Y bonds than X–X and Y–Y bonds. I–Cl bond is weaker than Cl–Cl bond and therefore, more reactive than I₂.

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Helium is used as a diluent for oxygen in diving apparatus because of its low solubility in blood.

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Radon is radioactive and has very short half life period. Therefore, it is difficult to study the chemistry of radon.

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(i) Electronic configuration: The elements of group 15 have 5 electrons in the valence shell, two in s and three in p subshell. The general electronic configuration of this group may be expressed as ns²np³. The value of n increases from 2 onwards. For example,

Oxidation states. The elements of group 15 exhibit various oxidation states from –3 to +5.

(a) These elements have five electrons in the valence shell (ns²np³) and therefore, require three more electrons to acquire the nearest noble gas configuration. Nitrogen being the smallest and most electronegative element of the group forms N³– (nitride) ion and shows an oxidation state of –3 in nitrides of some highly electropositive metals such as Mg₃N₂, Ca₃N₂, etc. The other elements of this group form covalent compounds even with metals and show an oxidation state of –3 with metals. For example, calcium phosphide (Ca₃P₂), sodium arsenide (Na₃As), zinc antimonide (Zn₃Sb₂), magnesium bismuthide (Mg3Bi2).

(b) The elements of group 15 also exhibit positive oxidation states of +3 and +5. However, on moving down the group, the stability of +5 oxidation state decreases while that of +3 oxidation state increases due to inert pair effect.

For example, +5 oxidation state of Bi is less stable than in Sb.

Nitrogen can exist in various oxidation states from –3 to +5 in its hydrides, oxides and oxoacids as shown below:

Phosphorus also exhibits +1 and +4 oxidation states in some oxoacids.

(iii) Atomic size. On moving down the group, the atomic radii increase due to increase in number of shells because of addition of a new principal shell in each succeeding element. However, from As to Bi only a small increase in covalent radius is observed. This is due to the presence of completely filled d and or f-orbitals in the heavier members.

(iv) Ionisation enthalpies. These elements have higher ionisation enthalpies than the corresponding members of group 14 elements. This is due to greater nuclear charge, small size and stable configuration of the atoms of group 15 elements. The electronic configuration of atoms of group 15 are half filled, np_x¹, np_y¹, np_z¹ and are stable. Therefore, they have high ionisation enthalpies. On going down the group, the ionisation enthalpies decrease. The decrease in ionisation enthalpy, as we move down the group, is due to increase in atomic size and screening effect which overweigh the effect of increased nuclear charge.

(v) Electronegativity. On going down the group, the electronegativity value decreases. The decrease in electronegativity on going down the group is due to increase in size of the atoms and shielding effect of inner electron shells on going down the group.

Answer:

Nitrogen has a unique ability to form p$\mathrm{\pi }$–p$\mathrm{\pi }$ multiple bonds with itself and with other elements having small size and high electronegativity (e.g., C, O). Therefore, it has triple bond between two nitrogen atoms (N N) and is non-polar. Due to triple bond it has very high bond enthalpy (941.4kJ) and therefore, it does not react with other elements under normal conditions and is very unreactive. On the other hand, phosphorus forms single bond (P–P) and is reactive in comparison to nitrogen.

Answer:

The important trends in chemical reactivity of group 15 elements are: 1. Reactivity towards hydrogen. All the elements of group 15 form gaseous trihydrides of the formula EH₃ (where E = N, P, As, Sb or Bi) such as:
NH₃ PH₃ AsH₃ SbH₃ BiH₃
Ammonia Phosphine Arsine Stibine Bismuthine
All these hydrides are covalent in nature and have pyramidal structure and these involve sp3 hybridization of the central atom and one of the tetrahedral position is occupied by a lone pair.
These hydrides show regular gradation in properties as given below:

1. All these hydrides have one lone pair of electrons on their central atom, therefore, they act as Lewis bases.
2. Thermal stability of the hydrides of group 15 elements decreases as we go down the group as :
   NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃
3. The reducing character of the hydrides of group 15 elements increases from NH₃ to BiH₃. Thus, increasing order of reducing character is as follows :
   NH₃ < PH₃ < AsH₃ < SbH₃ < BiH₃
4. Ammonia has a higher boiling point than phosphine and then the boiling point increases down the group because of increase in size.

2. Reactivity towards oxygen. The elements of group 15 combine with oxygen directly or indirectly to form two types of oxides, E₂O₃ (trioxides) and E2O5 (pentaoxides). However, nitrogen forms a number of oxides with oxidation states ranging from +1 to +5 such as N₂O, NO, N₂O₃, N₂O₄, and N₂O₅ and which have no analogues of P, As, Sb or Bi.

3. Reactivity towards halogens. The elements of group 15 form two series of halides of the type MX₃ (trihalides) and MX₅ (pentahalides). The trihalides are formed by all the elements while pentahalides are formed by all the elements except nitrogen. Nitrogen cannot form pentahalides due to the absence of vacant d-orbitals in its outermost shell.

4. Reactivity towards metals. All the elements of group 15 combine with metals to form their binary compounds in which the elements show – 3 oxidation state.

Answer:

Because of high electronegativity and small size of nitrogen, ammonia forms hydrogen bonds. On the other hand, P has low electronegativity and large size and hence cannot form hydrogen bonds.

Answer:

In the laboratory, nitrogen is prepared by treating an aqueous solution of ammonium chloride with sodium nitrite.

It may be noted that small amounts of NO and HNO₃ are also formed in this reaction as impurities. These are removed by passing the gas through aqueous sulphuric acid containing potassium dichromate.

Answer:

Ammonia can be manufactured by Haber’s process which involves the reaction:

This is a reversible exothermic reaction. The favourable conditions for high yield of ammonia can be understood by applying Le Chatelier’s principle.

1. Low temperature. Since the forward reaction is exothermic, therefore, low temperature will favour the formation of ammonia. However, an optimum temperature of about 700 K is necessary.
2. High pressure. High pressure of the order of 200 atmospheres or 200 × 10⁵ Pa is required to favour the forward reaction.
3. Presence of catalyst. The use of catalyst such as iron oxide containing a small amounts of molybdenum or potassium oxide (K₂O) and aluminium oxide (Al₂O₃) as promoter, increases the rate of attainment of equilibrium of ammonia. Details of the process

The plant required for the manufacture of ammonia is shown in Fig. In this method a mixture of N₂ and H₂ in the molar ratio of 1 : 3 is compressed to about 200 atmosphere pressure. The compressed gases are then cooled and passed through soda lime tower to free them from moisture and carbon dioxide. Then these are fed into catalyst chamber packed with iron oxide with small amount of K₂O and Al₂O₃ or molybdenum. The chamber is heated electrically to a temperature of 700 K when the two gases combine to form ammonia. The reaction being exothermic, the heat evolved maintains the desired temperature and further electrical heating is not required.

The gases which escape from the chamber contain about 15–20% ammonia and the remaining are unreacted N₂ and H₂. These are passed through condensing pipes where ammonia gets liquefied and is collected in the receiver. The unreacted gases are pumped back to the compression pump where they are mixed with fresh gaseous mixture.

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Concentrated nitric acid is a strong oxidising agent and reacts with metals. The products of oxidation depend upon the concentration of the acid, temperature and the nature of the material undergoing oxidation.
For example, copper reacts with HNO₃ giving different products as:
Conc. HNO3 gives copper nitrate and nitrogen dioxide.

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R₃N==O does not exist because nitrogen cannot have covalency more than four. Moreover, R₃P==O exists because phosphorus can extend its covalency more than 4 as well as it can form d$\mathrm{\pi }$-p$\mathrm{\pi }$ bond whereas nitrogen cannot form d$\mathrm{\pi }$-p$\mathrm{\pi }$ bond.

Answer:

Both N and Bi have a lone pair of electrons in NH₃ and BiH₃ respectively. They can donate the electron pair and therefore behave as Lewis base. In NH₃, N has small size and the lone pair is concentrated on a small region and electron density on it is maximum. Consequently, it has greater electron releasing tendency.

But the size of Bi is large and the electron density of the lone pair is less. As a result, it has lesser tendency to donate electron pair. Hence, NH₃ is basic while BiH3 is only feebly basic.

Answer:

The size of nitrogen atom is quite small. As a result, two nitrogen atoms can be linked to each other by three covalent bonds as : N $\equiv$$\equiv$ N : in order to complete the octets of both the nitrogen atoms However, phosphorus being large in size has less tendency to form three bonds.

Therefore, P atom completes its octet by sharing its valence electrons with three other P atoms. As a result, it exists as P₄ molecule.

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The main differences between white phosphorus and red phosphorus are:

Answer:

Nitrogen has little tendency for catenation because N—N single bond is weak. This is because nitrogen has small size and the lone pairs on two nitrogen atoms repel each other. On the other hand, phosphorus is comparatively large in size so that lone pairs on P atoms donot repel to the same extent. As a result P—P bond is stronger than N—N bond. Therefore, P has a tendency for catenation because of high bond enthalpy of P—P bond.

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Phosphorus can show maximum oxidation state of +5 in its compounds. In PCl₅, its oxidation state is +5. Since it cannot increase its oxidation state beyond +5, it cannot act as a reducing agent. However, it can act as an oxidising agent by undergoing decrease in its oxidation state from +5 to +3. For example, it oxidises silver to AgCl, Sn to SnCl₄ etc.

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All these elements O, S, Se, Te and Po have six electrons in their valence shell and have the general configuration ns²np⁴.
All these elements show minimum oxidation state of – 2. Except oxygen, S, Se, Te and Po show positive oxidation states such as + 2, + 4 and + 6. Oxygen cannot show higher oxidation states because of the absence of d-orbitals in its valence shell.
All these elements form hydrides of the general formula EH₂ i.e., H₂O, H₂S, H₂Se, H₂Te and H₂Po. These hydrides have similar properties.
On the basis of formation of hydrides of the general formula EH₂, electronic configuration and oxidation states, these elements are justified to be placed in the same group, Group 16 of the periodic table.

Answer:

Due to small size and high electronegativity, oxygen atom forms pπ-pπ double bond, O = O. The intermolecular forces in oxygen are weak van der Waals’ forces and therefore, oxygen exists as a gas.

On the other hand, sulphur does not form stable pπ-pπ bonds and do not exists as S₂. It is linked by single bonds 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.

Answer:

The second electron enthalpy of oxygen for the formation of O^2– is positive while first electron gain enthalpy of oxygen for the formation of O^– is negative. This means that if electron gain enthalpy is the only factor involved for the formation of divalent ions, we would except that oxygen would prefer to form O^– ions rather than O^2– ions. Actually, a large number of oxides have O^2– species and not O^–. This is because

1. divalent O^2– has the stable noble gas configuration.
2. In the solid state, large amount of energy known as lattice enthalpy is released to form divalent O^2– ions than monovalent O^– ions. It is the greater lattice enthalpy of O^2– ion which compensates for the high energy required to remove the second electron. This is responsible for greater stability of O^2– ion as compared to O^– ion.

Answer:

Manufacture of Sulphuric Acid Sulphuric acid can be manufactured by Contact process. The process involves the following steps:

1. Preparation of sulphur dioxide. Sulphur dioxide is prepared by burning sulphur or iron pyrites in excess of air.
2. Preparation of R455 dioxide. Sulphur dioxide is prepared by burning sulphur or iron pyrites in excess of air.

   

3. Oxidation of sulphur dioxide into sulphur trioxide. Sulphur dioxide is catalytically oxidised to sulphur trioxide with atmospheric oxygen in the presence of a catalyst (V₂O₅). The reaction is reversible as well as exothermic in nature.

   

4. Absorption of sulphur trioxide into 98% sulphuric acid to form oleum. Sulphur trioxide is absorbed in about 98% H₂SO₄ to form oleum or fuming sulphuric acid.

   

5. Dilution of oleum with water. Oleum is then diluted with required quantity of water to get sulphuric acid of any desired concentration.

   

   Conditions favouring the maximum yield of sulphur trioxide.

The key step in the manufacture of sulphuric acid is the catalytic oxidation of SO₂ with O₂ to give SO₃. This is a reversible and exothermic process.

The conditions for the maximum yield of sulphur trioxide are derived by using Le Chatelier’s principle as follows:

1. Low temperature. The forward reaction is exothermic and therefore, low temperature favours the oxidation of sulphur dioxide. However, it is essential to have minimum temperature of 720 K, called optimum temperature, to get the maximum yield of the product.
2. High pressure. Since the volume of the gaseous products is less than that of the gaseous reactants, high pressure should favour the oxidation of sulphur dioxide. But a very high pressure may cause the corrosion of the vessel in which oxidation is carried. Therefore, a pressure of 2 to 3 bar is sufficient for the oxidation.
3. Use of catalyst. A catalyst increases the speed of reaction. Platinised asbestos was used as catalyst but now has been replaced by vanadium pentoxide (V₂O₅). It is comparatively cheap and is not poisoned by the impurities.
4. Purity of gases. The gases must be purified before subjecting them to oxidation in the presence of catalyst.

Description of the plant. The plant employed for the Contact process has been shown in Fig. 1.

1. Sulphur burners. Sulphur or iron pyrites are burnt in excess of air to form sulphur dioxide.

   

2. Purification unit. The gaseous mixture coming out of sulphur burners is generally impure. The gases are purified as follows:
   1. Dust chamber. Steam is introduced to remove dust particles.
   2. Coolers. The hot gases are cooled to about 373 K by passing them through cooling pipes.
   3. Scrubber. Gases are introduced into a washing tower (packed with quartz) also known as scrubber which dissolves mist and any other soluble impurities.
   4. Drying tower. A spray of conc. H₂SO₄ is used for drying of gases.
   5. Arsenic purifier. This is a small chamber fitted with shelves containing gelatinous ferric hydroxide Fe(OH)3. The impurities of arsenic oxide present in the gases are absorbed by ferric hydroxide.
3. Testing box. The gases coming out of purification unit are tested in this box with the help of a strong beam of light. If some impurities are present, they will scatter light and the path will become visible. In case the gases are impure, they are passed through the purifying unit again.
4. Contact chamber or converter. The pure gases are then, heated to about 723–823 K in a pre-heater. These are then introduced in the contact chamber. It is a cylindrical iron chamber fitted with iron pipes. Each pipe is packed with the catalyst consisting of either platinized asbestos or V₂O₅. In this chamber, sulphur dioxide is oxidised to sulphur trioxide.

   

   As the forward reaction is exothermic, the pre-heating of the incoming gases is stopped once the oxidation reaction has started. The heat produced in the reaction is sufficient to maintain the temperature of the reaction.
5. Absorption tower. It is a cylindrical tower packed with acid proof flint. Sulphur trioxide escaping from the converter is led to the bottom of the tower while concentrated sulphuric acid (98%) is sprayed from the top. Sulphur trioxide gets absorbed by sulphuric acid to form oleum or fuming sulphuric acid.

   

   Oleum is then diluted with calculated amount of water to get acid of desired concentration.

   

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.

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Halogens have strong tendency to accept electrons and therefore, act as strong oxidising agents. Their oxidising power decreases from F₂ to I₂

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Due to small size and high electronegativity, fluorine cannot act as central atom in higher oxoacids.

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Nitrogen has smaller size than chlorine and the smaller size of nitrogen favours hydrogen bonding.

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1. A bleaching agent for paper pulp and textiles.
2. In water treatment.

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All the halogens are coloured. This is due to absorption of radiations in the visible region which results in the 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 (high energy) and therefore, appears pale yellow. On the other hand, iodine needs very less excitation energy and absorbs yellow light of low energy. Thus, it appears dark violet. Similarly, we can explain the greenish yellow colour of chlorine and reddish brown colour of iodine.

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Fluorine reacts vigorously with water giving oxygen and ozone.

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(i) Oxidation of hydrogen chloride gas by atmospheric oxygen in the presence of CuCl₂ catalyst at 723 K (Deacon’s process).

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In 1962, N. Bartlett noticed that platinum hexafluoride, PtF₆ is a powerful oxidising agent which combines with molecular oxygen to form ionic compound, dioxygenyl hexafluoroplatinate (V),

This indicates that PtF6 has oxidised O₂ to O₂⁺. Now, oxygen and xenon have some similarities:

1. The first ionisation energy of xenon gas (1170kJ mol^–1) is fairly close to that of oxygen (1166 kJ mol^–1).
2. The molecular diameter of oxygen and atomic radius of xenon are similar (4Å).

This prompted Bartlett to carry out the reaction between Xe and PtF₆.

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ClO^– is isoelectronic with ClF. Yes, it is a Lewis base.

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Answer:

1. I₂ < F₂ < Br₂ < Cl₂
2. HF < HCl < HBr < HI
3. BiH₃ < SbH₃ < AsH₃ < PH₃ < NH₃

Answer:

(b) NeF₂.

Answer:

(a) XeF₄ square pyramidal structure
(b) XeF₂ linear structure
(c) XeO₃ pyramidal structure
Structures
(a) XeF₄ has square planar structure as shown in Fig. It involves sp3d2 hybridisation of xenon which adopts octahedral geometry in which two positions are occupied by lone pairs.
(b) XeF₂ has linear structure as shown in Fig. It involves sp³d hybridisation of xenon which adopts trigonal bipyramidal geometry in which three positions are occupied by lone pairs. Because of the symmetrical arrangement of three lone pairs (each at an angle of 120ºC), the net repulsions on the Xe—F bond pairs is zero. Thus XeF2 has linear geometry.
(c) XeO₃ has pyramidal structure as shown in Fig. It involves sp³ hybridisation of xenon.

Answer:

In case of noble gases, the atomic radii correspond to van der Waal’s radii, which are always large.

Answer:

Neon is used

1. for filling discharge tubes for optical decorations and advertisements.
2. in safety devices for protecting electrical instruments like voltmeters, relays, rectifiers, etc.
3. for filling sodium vapour lamps.
4. in becon light as safety signal for air navigators because its light has fog penetration power.

Argon is used

1. for filling electric bulbs because of its inert nature.
2. to provide an inert atmosphere in high temperature metallurgical processes (arc welding of metals or alloys).
3. in gas chromatography.
4. It is used to protect metal surfaces from oxidation during the welding of steel.