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

Question 1:

Though nitrogen exhibits + 5 oxidation state, it does not form pentahalides. Give reason.

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.

Question 2:

PH3 has lower boiling point than NH3. Why?

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.

Question 3:

Write the reaction of thermal decomposition of sodium azide.

Answer:

Thermal decomposition of sodium azide gives nitrogen gas.

Question 4:

Why does ammonia act as a Lewis base?

Answer:

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

Question 5:

Why does NO2 dimerise ? Explain.

Answer:

Question 6:

In what way it can be proved that PH3 is basic in nature.

Answer:

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.

Question 7:

Why does PCl3 fume in moisture ?

Answer:

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

Question 8:

All the five bonds in PCl5 are not equivalent. Justify.

Answer:

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).

Question 9:

How do you account for the reducing behaviour of H3PO2 on the basis of its structure.

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.

Question 10:

Elements of group 16 show lower value of first ionisation enthalpy compared to the corresponding periods of groups 15. Why?

Answer:

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.

Question 11:

Why is H2S less acidic than H2Te ?

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.

Question 12:

Which form of sulphur shows paramagnetic behaviour?

Answer:

Question 13:

What happens when

  1. Concentrated H2SO4 is added to calcium fluoride.
  2. SO3 is passed through water?
Answer:
  1. It forms hydrogen fluoride.

  2. Sulphuric acid is formed.

Question 14:

Halogens have maximum negative electron gain enthalpy in the respective periods of the periodic table. Why ?

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.

Question 15:

Although electron gain enthalpy of fluorine is less negative as compared to chlorine, fluorine is a stronger oxidising agent than chlorine. Why?

Answer:

It is because of:

  1. low bond dissociation enthalpy of F–F bond.
  2. high hydration enthalpy of F ion.
Question 16:

Fluorine exhibits only –1 oxidation state where as other halogens exhibit positive oxidation states also such as +1,+3,+5 and+7. Why?

Answer:

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.

Question 17:

Write the balanced chemical equation for the reaction of Cl2 with hot and concentrated NaOH. Is this reaction a disproportionation reaction ? Justify.

Answer:

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).

Question 18:

HCl when reacts with finely divided iron forms ferrous chloride and not ferric chloride. Why ?

Answer:

HCl reacts with finely divided iron and produces H2 gas

Liberation of hydrogen prevents the formation of ferric chloride.

Question 19:

Discuss the molecular shape of BrF3 on the basis of VSEPR theory.

Answer:

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]

Question 20:

Why are the elements of group 18 known as noble gases?

Answer:

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.

Question 21:

Noble gas have low boiling points. Explain

Answer:

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.

Question 22:

Does the hydrolysis of XeF6 lead to a redox reaction ?

Answer:

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.

Question 23:

Why are pentahalides of P, As, Sb and Bi more covalent than trihalides?

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.

Question 24:

Why is BiH3 the strongest reducing agent amongst all the hydrides of group 15?

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.

Question 25:

Why is N2 less reactive at room temperature?

Answer:

In molecular nitrogen, there is a triple bond between two nitrogen atoms (N 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.

Question 26:

Mention the conditions required to maximise the yield of ammonia.

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
Question 27:

What is the covalence of nitrogen in N2O5?

Answer:

N2O5 has the structure.

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

Question 28:

(a) Bond angle in PH4+ is higher than that in PH3. Why?
(b) What is formed when PH3 reacts with an acid?

Answer:

Question 29:

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

Answer:

Phosphine is formed.

Question 30:

What happens when PCl5 is heated?

Answer:

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

Question 31:

Write a balanced equation for the reaction of PCl5 with water.

Answer:

Question 32:

What is the basicity of H3PO4?

Answer:

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

Question 33:

What happens when H3PO3 is heated?

Answer:

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

Question 34:

List the important sources of sulphur.

Answer:

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.

Question 35:

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

Answer:

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

Question 36:

Why is H2O a liquid and H2S a gas?

Answer:

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.

Question 37:

Which of the following does not react with oxygen directly?

Answer:

Zn, Ti, Pt, Fe

Question 38:

Complete the following reactions:

  1. C2H4 + O2
  2. 4Al + 3O2
Answer:

Question 39:

Why does O3 act as a powerful oxidising agent?

Answer:

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,

Question 40:

How is ozone estimated quantitatively?

Answer:

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.

Question 41:

What happens when sulphur dioxide is passed into aqueous solution of Fe (III) salt?

Answer:

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

Question 42:

Comment on the nature of two S—O bonds formed in SO2 molecule. Are the two S—O bonds in this molecule equal?

Answer:

Question 43:

How is the presence of SO2 detected?

Answer:

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.
Question 44:

Mention three areas in which sulphuric acid plays an important role.

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.).
Question 45:

Write the conditions to maximize the yield of H2SO4 by Contact process.

Answer:
  1. Low temperature (optimum temperature 720 K)
  2. High pressure (optimum pressure 2 bar)
  3. Presence of catalyst (V2O5 catalyst).
Question 46:

Why is Ka2 << Ka1 for H2SO4 in water?

Answer:

Question 47:

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

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.

Question 48:

Give two examples to show the anomalous behaviour of fluorine.

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.
Question 49:

Sea is the greatest source of some halogens. Comment.

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.

Question 50:

Give the reason for bleaching action of chlorine.

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.

Question 51:

Name two poisonous gases which can be prepared from chlorine gas.

Answer:
  1. Phosgene (COCl2)
  2. Tear gas (CCl3.NO2)
Question 52:

Why is ICl more reactive than I2?

Answer:

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.

Question 53:

Why is helium used in diving apparatus?

Answer:

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

Question 54:

Balance the following equation

Answer:

Question 55:

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

Answer:

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

Question 56:

Discuss the general characteristics of group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionization enthalpy and electronegativity.

Answer:

(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, npx1, npy1, npz1 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.

Question 57:

Why does the reactivity of nitrogen differ from phosphorus?

Answer:

Nitrogen has a unique ability to form pπ–pπ 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.

Question 58:

Discuss the trends in chemical reactivity of group 15 elements.

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.

Question 59:

Why does NH3 forms hydrogen bonds but PH3 does not?

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.

Question 60:

How is nitrogen prepared in the laboratory? Write the chemical equation of the reactions involved.

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.

Question 61:

How is ammonia manufactured industrially?

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.

Question 62:

Illustrate how copper metal can give different products on reaction with HNO3.

Answer:

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.

Question 63:

Give the resonating structures of NO2 and N2O5.

Answer:

Question 64:

The HNH angle value is higher than HPH, HAsH and HSbH angles. Why?

Answer:

Question 65:

Why does R3P==O exists but R3N==O does not(R = alkyl group)?

Answer:

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 dπ-pπ bond whereas nitrogen cannot form dπ-pπ bond.

Question 66:

Explain why NH3 is basic while BiH3 is only feebly basic?

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.

Question 67:

Nitrogen exists as diatomic molecule and phosphorus as P4. Why?

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 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.

Question 68:

Write main differences between the properties of white phosphorus and red phosphorus.

Answer:

The main differences between white phosphorus and red phosphorus are:

Question 69:

Why does nitrogen show catenation properties less than phosphorus?

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.

Question 70:

Give the disproportionation reaction of H3PO3.

Answer:

Question 71:

Can PCl5 act as an oxidising as well as a reducing agent? Justify.

Answer:

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.

Question 72:

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:

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.

Question 73:

Why is dioxygen a gas but sulphur a solid?

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.

Question 74:

Knowing the electron gain enthalpy values for OOand OO2– as –141 and 702 kJ mol–1 respectively, how can you account for the formation of a large number of oxides having O2– species and not O?

Answer:

The second electron enthalpy of oxygen for the formation of O2– 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 O2– ions. Actually, a large number of oxides have O2– species and not O. This is because

  1. divalent O2– has the stable noble gas configuration.
  2. In the solid state, large amount of energy known as lattice enthalpy is released to form divalent O2– ions than monovalent O ions. It is the greater lattice enthalpy of O2– ion which compensates for the high energy required to remove the second electron. This is responsible for greater stability of O2– ion as compared to O ion.
Question 75:

Describe the manufacture of sulphuric acid by Contact process.

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.

Question 76:

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 77:

Why are halogens strong oxidising agents?

Answer:

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

Question 78:

Explain why fluorine forms only one oxoacid HOF?

Answer:

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

Question 79:

Explain why inspite of nearly the same electronegativity, nitrogen forms hydrogen bonding while chlorine does not?

Answer:

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

Question 80:

Write two uses of ClO2.

Answer:
  1. A bleaching agent for paper pulp and textiles.
  2. In water treatment.
Question 81:

Why are halogens coloured?

Answer:

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.

Question 82:

Write the reactions of F2 and Cl2 with water.

Answer:

Fluorine reacts vigorously with water giving oxygen and ozone.

Question 83:

How can you prepare Cl2 from HCl and HCl from Cl2. Write reactions only.

Answer:

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

Question 84:

What inspired N. Bartlett for carrying out reaction between Xe and PtF6?

Answer:

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.

Question 85:

What are the oxidation states of phosphorus in the following:
(a) H3PO3 (b) PCl3 (c) Ca3P2 (d) Na3PO4 (e) POF3?

Answer:

Question 86:

Write balanced equations for the following:

  1. NaCl is heated with sulphuric acid in the presence of MnO2.
  2. Chlorine gas is passed into a solution of NaI in water.
Answer:

Question 87:

How are xenon fluorides XeF2, XeF4 and XeF6 obtained?

Answer:

Question 88:

With what neutral molecule is ClO isoelectronic? Is that molecule a Lewis base?

Answer:

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

Question 89:

How are XeO3 and XeOF4 prepared?

Answer:

Question 90:

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

  1. F2, Cl2, Br2, I2 . increasing bond dissociation enthalpy.
  2. HF, HCl, HBr, HI. increasing acid strength.
  3. NH3, PH3, AsH3, SbH3, BiH3. increasing base strength.
Answer:
  1. I2 < F2 < Br2 < Cl2
  2. HF < HCl < HBr < HI
  3. BiH3 < SbH3 < AsH3 < PH3 < NH3
Question 91:

Which one of the following does not exist?
(a) XeOF4 (b) NeF2 (c) XeF2 (d) XeF6
Ans. (b) NeF2.

Answer:

(b) NeF2.

Question 92:

Give the formula and describe the structure of a noble gas species which is isostructural with:
(a) Icl4 (b) Ibr2 (c) BrO3

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.

Question 93:

Why do noble gases have comparatively large atomic sizes?

Answer:

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

Question 94:

List the uses of neon and argon gases.

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.
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