# NCERT Solutions for Class 11 Chemistry Chapter 11: The p-Block Elements

11.1. Discuss the pattern of variation in the oxidation states of (i) B to Tl and (ii) C to Pb.

Ans. (i) B to Tl : Boron (B), aluminium (Al), galium (Ga), indium (In) and thallium (TI) are the elements of group 13 of periodic table. The electronic configuration of group 13 elements is ns2np1. Therefore, the most common oxidation state exhibited by them should be +3 but only boron and aluminium show +3
oxidation state. Ga, In, Tl, show both the +1 and +3 oxidation states. On moving down the group, the +1 state becomes more stable due to inert pair effect. The relative stability of +1 oxidation state progressively increases for heavier elements: Al < Ga< In <Tl. Thus, inert pair effect becomes more and more prominent on moving down the group. Hence, Ga(+1) is unstable, Ln (+1) is fairly stable and TI (+1) is very stable. Also, The stability of the +3 oxidation state decreases on moving down the group.

 Group 13 element Oxidation state B +3 Al +3 Ga,Ln,Tl +1,+3

(ii) C to Pb : Carbon (C), silicon (Si), germanium (Ge), tin (Sn) and lead (Pb) are the elements that belong to group 14 of the periodic table. The group 14 elements have four electrons in outermost shell and its electronic configuration is ns2np2. The common oxidation states exhibited by these elements are +4 and +2. Carbon also exhibits negative oxidation states. However, the +2 oxidation state becomes more and more common on moving down the group. For heavier members, the tendency to show +2 oxidation state increases in the sequence Ge < Sn < Pb. It is due to the inability of ns2 electrons  of valence shell to participate in bonding. Thus, although Ge, Sn, and Pb show both the +2 and +4 states, the stability of the lower oxidation state increases and that of the
higher oxidation state decreases on moving down the group.

 Group 14 element Oxidation state C +4 Si +4 Ge,Sn,Pb +2,+4

$$\underrightarrow{\text{stability\space of }+2\space \text{state\space increases}}\\\text{C\space Si\space Ge\space Sn\space Pb}\\\overrightarrow{\text{stability\space of +2 state\space increases}}$$

11.2. How can you explain higher stability of BCl3 as compared to TlCl3?
Ans. Both boron and thallium belong to group 13 of the periodic table, BCl3 is more stable than TlCl3 because the +3 oxidation state of B is more stable than the +3 oxidation state of TI. This is because in TI, the +3 oxidation state is highly oxidising and it reverts back to the more stable +1 oxidation state.

11.3. Why does boron trifluoride behave as a Lewis acid?

Ans. The Lewis structure of BF3 is :

It is clear from its Lewis structure, that BF3 is an electron deficient compound. The electronic configuration of boron is ns2 np1 . It has three electrons in its valence shell. In the formation of BF3, it forms three covalent bonds with fluorine. The total electrons around boron are six. Hence, boron trifluoride remains electron-deficient and in order to complete its octet, it can accept an electron pair. Hence, it acts as Lewis acid. All such electron-deficient compounds are electron pair acceptors and so are Lewis acids.

11.4. Consider the compounds, BCl3 and CCl4. How will they behave with water? Justify.

Ans. Boron trichloride (BCl3) is an electron deficient compound and acts as a Lewis acid which means it easily accepts a pair of electron from water. Thus, BCl3 undergoes hydrolysis readily and forms boric acid and hydrochloric acid. The reaction can be represented as :

$$\text{BCl}_3 + 3\text{H}_2\text{O} \rightarrow 3\text{HCl} + \text{H}_3\text{BO}_3$$

In case of CCl4 , C has eight electrons in its octet. Thus, its octet is complete and it completely resists hydrolysis. Also, carbon does not have any vacant orbital. It has no tendency to accept electrons donated by water, Hence, when CCl4 and water are mixed, they form separate layers i.e., no reaction occurs.

$$\text{CCl}_4 + \text{H}_2\text{O} \rightarrow \text{No\space reaction}$$

11.5. Is boric acid a protic acid? Explain.
Ans. Boric acid is a weak monobasic acid. It not a protic acid because it does not ionise in water to give a proton. Infact, it act as Lewis acid because it accepts electrons from hydroxide ion.

$$\text{B(OH)}_3 + 2\text{HOH} \rightarrow[\text{B(OH)}_4]^– + \text{H}_3\text{O}^+$$

11.6. Explain what happens when boric acid is heated.
Ans. Boric acid on heating above 170°C gives metaboric acid, which on further heating above 300°C gives tetraboric acid (or boric acid). On further heating gives boron trioxide.

$$\underset{\text{BOric\space acid}} {\text{H}_3\text{BH}_3} \space \underrightarrow{170\degree \text{C}}\space \underset{\text{metaboric\space acid}}{\text{HBO}_2} + \text{H}_2\text{O}\\\underset{\text{BOric\space acid}} {4\text{HBO}_2} \space \underrightarrow{130\degree \text{C}} \space \underset{\text{Tetraboric\space acidor\space Boric\space acid}}{\text{H}_2\text{B}_4\text{O}_7} + \text{H}_2\text{O}\\ \underset{\text{Tetraboric\space acidor\space Boric\space acid}}{\text{H}_2\text{B}_4\text{O}_7} \underrightarrow{\text{Further\space heating}} \underset{\text{Boron\space trioxide}}{2\text{B}_2\text{O}_3}+ \text{H}_2\text{O}$$

11.7. Describe the shapes of BF3 and BH4. Assign the hybridisation of boron in these species.
Ans. (i) BF3 : BF3 is a planar molecule in which B is sp2 hybridised. It has empty 2p orbital. It has a planar triangular geometry. It is formed by the overlap of three sp2hybridised orbitals of Boron with the sp orbitals of three fluorine atoms. Thus, BF3 has planar triangular geometry. The structure can be given as :

(ii) BH4 : Boron-hydride ion (BH4) is formed by the sp3 hybridisation of boron orbitals with hydrogen orbitals. It has tetrahedral geometry. The structure can be given as:

11.8. Write reactions to justify amphoteric nature of aluminium.
Ans. Amphoteric means an element or compound which shows characteristics of both acids and bases. Aluminium is amphoteric in nature. Some reactions which shows the amphoteric nature of aluminium are:

$$\text{(i) Reaction with acids:}\\2\text{Al(s)} + 6\text{HCl(aq)} \rightarrow 2\text{Al}^{3+}(\text{aq}) + 6\text{Cl}^– (\text{aq})+ 3\text{H}_2(\text{g})\\\text{(ii) Reaction with base} :\\2\text{Al(s)} + 2\text{NaOH(aq)} + 6\text{H}_2\text{O(l)} \rightarrow2\text{Na}^+[\text{Al(OH)}_4]^– (\text{aq}) + 3\text{H}_2(\text{g})$$

11.9. What are electron deficient compounds? Are BCl3 and SiCl4 electron deficient species? Explain.
Ans. Compounds that possess very few electrons for a Lewis structure to be written are called electron deficient compounds. In these compounds, octet of central atom is not complete. Therefore, it needs electrons to complete its octet. Thus, they have tendency to accept one or more electron pairs.
(i) BCl3: The electric configuration of boron is ns2 np1 . It has three electrons in its valence shell. After forming three covalent bonds with chlorine, the number of electrons around it increases to 6. Thus, its octet is not complete and it needs two more electrons. Therefore, it is an example of electron deficient compound.
(ii) SiCl4: The electronic configuration of silicon is ns2 np2 . It has four valence electrons. After forming four covalent bonds with four chlorine atoms, its electron count increases to eight. Its octet is complete. Thus, SiCl4 is not an electron-deficient compound.

11.10. Write the resonance structures of CO32– and HCO3.

11.11. What is the state of hybridisation of carbon in
(a) CO3 2– (b) diamond (c) graphite?
Ans. The state of hybridisation of carbon in:
(a) CO 3 2– : C in CO3 2– is sp2 hybridised and is bonded to three oxygen atoms.
(b) Diamond: Each carbon in diamond is bonded to four other carbon atoms and is sp3 hybridised.
(c) Graphite: Each carbon atom in graphite is bonded to three other carbon atoms and is sp2 hybridised.

11.12. Explain the difference in properties of diamond and graphite on the basis of their structures.
Ans. The difference between carbon and graphite is:

 Diamond Graphite It has a crystalline lattice. It has a layered structure. C–C bond length is 154 pm. C–C bond length is 141.5 pm. It occurs as octahedral crystals. It occurs as hexagonal crystals. Each carbon in is sp3 hybridised. Each carbon atom in graphite is sp2 hybridised. it is hardest substance known. It is soft and slippery.

11.13. Rationalise the given statements and give chemical reactions:

(a) Lead(II) chloride reacts with Cl2 to give PbCl4.
(b) Lead(IV) chloride is highly unstable towards heat.
(c) Lead is known not to form an iodide, PbI4.

Ans. (a) Lead can show two oxidation states i.e, +2 and +4. On moving down the group, the +2 oxidation state becomes more stable because of the inert pair effect. Therefore, PbCl4 is much less stable than PbCl2. However, the formation of PbCl4 takes place when chlorine gas is bubbled through a saturated solution of PbCl2.

$$\text{PbCl}_2(\text{s}) + \text{Cl}_2(\text{g}) \rightarrow \text{PbCI}_4(\text{l})$$

(b) On moving down the group 14, the higher oxidation states become unstable because of inert pair effect. Therefore, Lead(IV) chloride is highly unstable and when heated, it reduces to Pb(II).

$$\text{PbCl}_4(\text{l})\xrightarrow{\Delta}\text{PbCl}_2(\text{s})+\text{Cl}_2(\text{g})$$

(c) Pb4+ is highly oxidising in nature because of its tendency to change to Pb2+ (inert-pair effect). Iodide ions, I, on the other hand, have strong reducing power hence PbI4 does not exist. High oxidising power of Pb4+ and high reducing power of Iions always result in the formation of PbI2 and not PbI4.

$$\text{PbI}_4 \rightarrow \text{PbI}_2 + \text{I}_2$$

11.14. Suggest reasons why the B–F bond lengths in BF3 (130 pm) and BF4(143 pm) differ.
Ans. BF3 is a planar molecule in which B is sphybridised. It has empty 2p orbital. Because of similar sizes of B and F, back bonding occurs in which a lone pair is transferred from F to empty p-orbital of B forming pπ-pπ bond. As a result, B – F bond acquire some double bond character. On the other hand, in BF4ion, B is sp3hybridised and forms four B-F single bonds. It does not have vacant p- orbital available to accept the electrons from F atom and hence B-F bond is purely single bond. Since, double is shorter than single bond, therefore B-F bond lengths in BF3 – is shorter than in BF4.

11.15. If B–Cl bond has a dipole moment, explain why BCl3 molecule has zero dipole moment. Ans. Bond moment depends on the difference in the electronegativity values whereas the overall dipole moment of the molecule depends on the geometry of the molecule. There is difference in the electronegativities of B and Cl. Due to this, the B–Cl bond is polar in nature. Thus, it has dipole moment. However, BCl3 molecule is non-polar in nature. Also, since BCl3 is trigonal planar in shape and it is a symmetrical molecule. The respective dipole- moments of the B–Cl bond cancel each other, and net dipole moment of BCl3 molecule is zero.

11.16. Aluminium trifluoride is insoluble in anhydrous HF but dissolves on addition of NaF. Aluminium trifluoride precipitates out of the resulting solution when gaseous BF3 is bubbled through. Give reasons.
Ans. Hydrogen fluoride (HF) is a covalent compound and has very strong hydrogen bonding. Also, it has high dissociation energy and does not dissociate into ions. As a result, aluminium trifluoride (AlF3) does not dissolve in it. On the other hand, Sodium fluoride (NaF) is an ionic compound. It dissociates into its ions and therefore, AlF3 dissolves in it due to availability of ions. The reaction can be given as:

$$\text{AIF}3 + 3\text{NaF} \rightarrow \underset{\text{Sodium hexafluoroaluminate(III)}}{\text{Na}_3[\text{AIF}_6]}$$

When boron trifluoride (BF3) is added to the solution, aluminium trifluoride precipitates out of the solution. This is because complex forming tendency of boron is much more than that of aluminium. Therefore, when BF3 is added to the solution, B replaces Al from the complexes according to the following reaction:

$$\text{Na}_3[\text{AlF}_6] + 3 \text{BF}_3 \rightarrow \underset{\text{Sodium tetrafluoridoborate (III)}}{3 \text{Na}[\text{BF}_4] + \text{AlF}_3}$$

11.17. Suggest a reason as to why CO is poisonous.
Ans. Carbon monoxide is a highly poisonous gas because of its ability to form complex with haemoglobin. Air containing even less than 1% of carbon monoxide, can be fatal, if breathed in for about 10 to 15 minutes. It combines with haemoglobin in the blood and form carboxy-haemoglobin. Carboxy haemoglobin is more
stable than oxy-haemoglobin which is formed by combination of oxygen and haemoglobin.

$$\text{Haemoglobin} +\text {O}_2 \rightarrow \text{Oxy-haemoglobin}$$

This oxy-haemoglobin carries oxygen to various body parts but when carbon monoxide is combined to haemoglobin it stops the normal circulation of oxygen and may cause death due to suffocation. It is found that Co–Hb complex is about 300 times more stable than the O2–Hb complex.

11.18. How is excessive content of CO2 responsible for global warming?
Ans. Carbon dioxide is produced during combustion. It is utilised by plants during photosynthesis and oxygen is released into the atmosphere. As a result of this carbon dioxide cycle, a constant percentage of 21% oxygen is maintained in the atmosphere. However, if the concentration of carbon dioxide increases beyond 0.03% by volume in the atmosphere due to combustion, some of the CO2 remain unutilized. This excess of CO2 absorbs heat radiated by the earth. Some of it is dissipated into the atmosphere while the remaining part is radiated back to earth and other bodies present on the earth. As a result, temperature of the earth and other bodies on earth increases, thereby causing global warming. This is called green house effect and carbon dioxide is called as green house gas. As a result of green house effect, global warming occurs which has serious consequences.

11.19. Explain structures of diborane and boric acid.
Ans. (i) Diborane: B2H6 is an electron-deficient compound. Ithas only 12 electrons – 6 e from 6 H atoms and 3e each from 2B atoms. Thus, after combining with 3H atoms, none of the boron atoms has any electrons left. X-ray diffraction studies have shown the structure of diborane as:

In the above structure, two types of hydrogen bonds are there:

(a) Terminal hydrogen bonds (Ht): In the above structure, the four terminal hydrogen atom and the two boron atoms lie in one plane. Above and below this plane, there are two bridging hydrogen atoms. The four terminal B–H bonds are normal covalent bond which is formed by sharing of a pair of electrons between B and H atoms. Thus, the bonds are quite strong and are known as two centre-two electron bond (2c-2e).

(b) Bridging hydrogen bonds (Hb): The two bridge bonds B------H-----B are quite different from normal covalent bonds. Each bridging hydrogen atom is bonded to two boron atoms by a pair of electrons. There are three centre-two electron bond, (3c-2e) due to which these bonds are quite weak (3c-2e) because of their resemblance to a banana, these centre bonds are called as banana bond.

(ii) Boric acid: Boric acid has a layered structure. Each planar BO3 unit is linked to one another through H atoms. The H atoms form a covalent bond with a BO3 unit, while a hydrogen bond is formed with another BO3 unit. In the given figure, the dotted lines represent hydrogen bonds.

11.20. What happens when
(a) Borax is heated strongly,
(b) Boric acid is added to water,
(c) Aluminium is treated with dilute NaOH,
(d) BF3 is reacted with ammonia?

Ans. (a) When heated, borax undergoes various transitions. It first loses water molecules and swells. Then, it turns into a transparent liquid, solidifying to form a glass-like material called borax bead.

$$\underset{\text{Borax}}{\text{Na}_2\text{B}_4\text{O}_7.10\text{H}_2\text{O}}\xrightarrow{\Delta}\underset{\text{Sodium\space metaborate}}{\text{Na}_2\text{B}_4\text{O}_7}\xrightarrow{\Delta}2\text{NaBO}_2+\underset{\text{Boric\space anhydride}}{\text{B}_2\text{O}_3}$$

b) When boric acid is added to water, it accepts electrons from hydroxyl ion (OH–) ion.

$$\text{B(OH)}_3 + 2\text{HOH} \rightarrow [\text{B(OH)}_4]^– + \text{H}3\text{O}^+$$

(c) Al reacts with dilute NaOH to form sodium tetrahydroxoaluminate(III). Hydrogen gas is liberated in the process.

$$\text{2Al(s)} + 2\text{NaOH(aq)} + 6\text{H}_2\text{O(l)} \rightarrow\underset{\text{Sodium\space tetrahydroxoaluminate(III)}} {2\text{Na}^+[\text{Al(OH)}_4]^–(\text{aq}) + 3\text{H}_2(\text{g})}$$

(d) BF3 is an electron deficient compound and thus behaves like a Lewis acid. It reacts with NH3 (a Lewis base) to form an adduct. This results in a complete octet around B in BF3.

11.21. Explain the following reactions
(a) Silicon is heated with methyl chloride at high temperature in the presence of copper;
(b) Silicon dioxide is treated with hydrogen fluoride;
(c) CO is heated with ZnO;
(d) Hydrated alumina is treated with aqueous NaOH solution.

Ans. (a) When silicon reacts with methyl chloride in the presence of copper (catalyst) and at a temperature of about 537K, a class of organosilicon polymers called methyl-substituted chlorosilanes (MeSiCl3, Me2SiCl2, Me3SiCl, and Me4Si) are formed.

$$2\text{CH}_3\text{Cl}+\text{Si}\xrightarrow[570k]{\text{Cu\space powder}}\underset{\xrightarrow[-2\text{HCl}]{2\text{H}_2\text{O}}(\text{CH}_3)_2\text{Si(OH)}_2}{(\text{CH}_3)_2\text{SiCl}_2}$$

b) Usually, the Si–O bond is a strong bond and it resists any attack by halogens and most acids, even at a high temperatures. However, when silicon dioxide (SiO2) is heated with hydrogen fluoride (HF), it forms silicontetrafluoride (SiF4).

$$\text{SiO}_2+4\text{HF}\rightarrow \text{SiF}_4+2\text{H}_2\text{O}$$

The SiF4 formed in this reaction can further react with HF to form hydrofluorosilicic acid.

$$\text{SiF}_4+2\text{HF}\rightarrow \text{H}_2\text{SiF}_6$$

(c) When CO reacts with ZnO, it reduces ZnO to Zn. CO acts as a reducing agent.

$$\text{ZnO(s)}+\text{CO(g)}\xrightarrow{\Delta} \text{Zn(s)}+ \text{CO}_2(\text{g})$$

(d) When hydrated alumina is added to sodium hydroxide, the former dissolves in the latter because of the formation of sodium meta-aluminate.

$$\text{Al}_2\text{O}_3 \bullet 2\text{H}_2\text{O}+2\text{NaOH} \rightarrow 2\text{NaAlO}_2+3\text{H}_2\text{O}$$

11.22. Give reasons:
(i) Conc. HNO3 can be transported in aluminium container.
(ii) A mixture of dilute NaOH and aluminiumm pieces is used to open drain.
(iii) Graphite is used as lubricant.

(iv) Diamond is used as an abrasive.
(v) Aluminiumalloys areusedtomake aircraft body.
(vi) Aluminium utensils should not be kept in water overnight.
(vii) Aluminium wire is used to make transmission cables.

Ans. (i) Aluminium reacts with conc. nitric acid and becomes passive due to the formation of very thin film of aluminium oxide on its surface which protects it from further action. Therefore, aluminium containers can be used to transport conc. nitric acid.
(ii) Sodium hydroxide and aluminium react to form sodium tetrahydroxoaluminate(III) and hydrogen gas. The pressure of the produced hydrogen gas is used to open blocked drains.

$$2 \text{Al(s)}+2 \text{NaOH(aq)}+6 H_2\text{O(l)}\rightarrow \underset{\text{Sodium\space tetrahydroxoaluminate(III)}}{2\text{na}^+[\text{Al(OH)}_4]^-(\text{aq})+3\text{H}_2(\text{g})}$$

(iii) In Graphite, the carbon atoms are arranged in flat parallel layers as regular hexagons. Each layer is bonded to adjacent layers by weak Van der Waals forces. This allows each layer to slide over the other easily. Due to this type of structure graphite is soft and slippery and can be used as a lubricant.
(iv) Diamond is used as an abrasive because of its structure. Each carbon atom in Diamond is attached to four carbon atoms by strong covalent bonds. Diamond is hardest substance known.
(v) Aluminium alloys are used to make aircraft body because they have high tensile strength and are very light in weight. It is very malleable and ductile and can be alloyed with various metals such as Cu, Mn, Mg, Si and Zn. Therefore, it is used in making aircraft bodies.
(vi) Aluminium reacts with water and dissolved oxygen forms a thin layer of aluminium oxide. A very small amount of aluminium oxide dissolves to give Al3+ ions in solution. Since Al3+ is injurious to health. Therefore, drinking water should not be kept in aluminium utensils overnight.
(vii) Aluminium is a good conductor of electricity and is ductile. Also, it is cheaper than other metals like silver and copper. Therefore, Aluminium wire is used to make transmission cables.

11.23. Explain why is there a phenomenal decrease in ionisation enthalpy from carbon to silicon?

Ans. Ionisation energy decreases down the group as atomic size increases. Ionisation enthalpy of carbon (the first element of group 14) is very high because of its small size. However, on moving down the group to silicon, there is a sharp decrease in the enthalpy because of appreciable increase in the atomic sizes of elements on moving down the group.

11.24. How would you explain the lower atomic radius of Ga as compared to Al?
Ans. Atomic radius of aluminium is 143 pm and that of gallium is 135 pm. Although, Ga has one shell more than Al, its atomic size is lesser than Al. The atomic radius of Ga is less than Al because of the poor shielding effect of the 3d electrons. The shielding effect of d-electrons is very poor and the effective nuclear charge experienced by the valence electrons in gallium is much more than it is in the case of Al.

11.25. What are allotropes? Sketch the structure of two allotropes of carbon namely diamond and graphite. What is the impact of structure on
physical properties of two allotropes?
Ans. Allotropy is the existence of an element in more than one form, having the same chemical properties but different physical properties. The various forms of an element are called allotropes. The two allotropes of carbon are diamond and graphite.

In Diamond, it has a crystalline lattice. In diamond each carbon atom undergoes sp3 hybridisation and linked to four other carbon atoms by using hybridised orbitals in tetrahedral fashion. The C– C bond length is 154 pm. The structure extends in space and produces a rigid three-dimensional network of carbon atoms. The rigid 3-D structure of diamond makes it a very hard substance. Diamond is one of the hardest naturally-occurring substances. It is used as an abrasive and for cutting tools.

Graphite has layered structure. Layers are held by Van der Waals forces. It has sp2 hybridised carbon, arranged in the form of layers. These layers can slide over each other, making graphite soft and slippery. Therefore, it is used as a lubricant.

11.26. (a) Classify following oxides as neutral, acidic, basic or amphoteric: CO, B2O3, SiO2, CO2, Al2O3, PbO2, Tl2O3

(b) Write suitable chemical equations to show their nature.

Ans. (a) CO = Neutral
Acidic: B2O3, SiO2, CO2,
Basic: Tl2O3
Amphoteric: Al2O3, PbO2

(b) B2O3: It is an acidic oxide.Being acidic, it reacts with bases to form salts. It reacts with NaOH to form sodium metaborate.

$$\text{B}_2\text{O}_3+2\text{NaOH}\rightarrow 2\text{NaBO}_2+\text{H}_2\text{O}$$

SiO2 : It is an acidic oxide. It reacts with bases to form salts. It reacts with NaOH to form sodium silicate.

$$\text{SiO}_2+2\text{NaOH}\rightarrow \text{NaSiO}_3+\text{H}_2\text{O}$$

CO2: It is an acidic oxide. It reacts with bases to form salts. It reacts with NaOH to form sodium carbonate.

$$\text{CO}_2+2\text{NaOH}\rightarrow \text{Na}_2\text{CO}_3+\text{H}_2\text{O}$$

Al2O3: It is an amphoteric oxide i.e, react with both acids and bases. Al2O3 reacts with both NaOH and H2SO4.

$$\text{Al}_2\text{O}_3+2\text{NaOH}\rightarrow 2\text{NaAlO}_2+\text{H}_2O\\\text{Al}_2\text{O}_3+3\text{H}_2\text{SO}_4\rightarrow \text{Al}_2(\text{SO}_4)_3+3\text{H}_2\text{O}$$

PbO2 : It is an Amphoteric oxide. PbO2 reacts with both NaOH and H2SO4.

$$\text{PbO}_2+2\text{NaOH} \rightarrow \text{Na}_2\text{PbO}_3+\text{H}_2\text{O}\\2\text{PbO}_2+2\text{H}_2\text{SO}_4 \rightarrow 2\text{PbSO}_4+2\text{H}_2\text{O}+\text{O}_2$$

Tl2O3 : It is a basic oxide. It reacts with acids to form salt and water. It reacts with HCl to form thallium chloride.

$$\text{Tl}_2\text{O}_3+6\text{HCl} \rightarrow 2\text{TlCl}_3+3\text{H}_2\text{O}$$

11.27. In some of the reactions thallium resembles aluminium, whereas in others it resembles with group I metals. Support this statement by
giving some evidences.
Ans. Thallium belongs to group 13 of the periodic table. The most common oxidation state for this group is +3. However, heavier members of this group also display the +1 oxidation state. Aluminium displays the +3 oxidation state and alkali metals display the +1 oxidation state. Thallium displays both the oxidation states. Therefore, it resembles both aluminium and alkali metals. Thallium, like aluminium, forms compounds such as TlCl3 and Tl2O3. It resembles alkali metals
in compounds Tl2O and TlCl. It forms basic oxides like alkali metals.

11.28. When metal X is treated with sodium hydroxide, a white precipitate (A) is obtained, which is soluble in excess of NaOH to give soluble complex (B). Compound (A) is soluble in dilute HCl to form compound (C). The compound (A) when heated strongly gives (D), which is used to extract metal. Identify (X), (A), (B), (C) and (D). Write suitable equations to support theiridentities.

Ans. The given metal X gives a white precipitate with sodium hydroxide and the precipitate dissolves in excess of sodium hydroxide. Hence, X must be aluminium. The white precipitate (compound A) obtained is aluminium hydroxide. The compound B formed when an excess of the base is added is sodium tetrahydroxoaluminate(III).

$$\underset {\text{Aluminium (X)}}{\text{Al}}+\underset{\text{Sodium\space hydroxide}}{3\text{NaOH}} \rightarrow \underset{\text{White ppt.(A)}} {\text{Al(OH)}_3} \downarrow+3\text{Na}^+\\\underset{(\text{A})}{\text{Al(OH)}_3}+\text{NaOH }\rightarrow \underset{\text{Sodium\space tetrahydroxoaluminate(III) (Soluble complex B)}}{\text{Na}^+[\text{Al(OH)}_4]^-}$$

Now, when dilute hydrochloric acid is added to aluminium hydroxide, aluminium chloride (compound C) is obtained.

$$\underset{(\text{A})}{\text{Al(OH)}_3}+3\text{HCl}\rightarrow \underset{\text{Aluminium\space chloride(C)}}{\text{AlCl}_3}+3\text{H}_2\text{O}$$

Also, when compound A is heated strongly, it gives compound D. This compound is used to extract metal X. Aluminium metal is extracted from alumina. Hence, compound D must be alumina.

$$\underset{(\text{A})}{\text{2Al(OH)}_3}\xrightarrow{\Delta}\underset {\text{Alumina\space (D)}}{\text{Al}_2\text{O}_3}+3\text{H}_2\text{O}$$

11.29. What do you understand by (a) inert pair effect (b) allotropy and (c) catenation?
Ans. (a) Inert pair effect: As one moves down the group, the tendency of s-block electrons to participate in chemical bonding decreases. This effect is known as inert pair effect. The inert-pair effect refers to the empirical observation that the heavier elements of groups 13–17 often have oxidation states that are lower by 2 than the maximum predicted for their group. This happens because of the poor shielding of the ns2 electrons by the d- and f- electrons.
For example: In case of group 13 elements,the electronic configuration is ns2 np1 and their group valency is +3.
However, on moving down the group, the +1 oxidation state becomes more stable. This happens because of the poor shielding of the ns2 electrons by the d- and f-electrons. As a result of the poor shielding, the ns2 electrons are held tightly by the nucleus and so, they cannot participate in chemical bonding.
(b) Allotropy: The existence of an element in more than one form, having the same chemical properties but different physical properties is called allotropy. The various forms of an element are called allotropes. For example, carbon exists in three allotropic forms: diamond, graphite, and fullerenes. The allotropic forms of carbon can be represented as :

(c) Catenation: The self linking property of an element by which an atom combines with the other atoms of the same element to form long chains or branches is called catenation. Carbon has the largest tendency of catenation and it is quite significant in Si and S.

11.30. A certain salt X, gives the following results.
(i) Its aqueous solution is alkaline to litmus.
(ii) It swells up to a glassy material Y on strong heating.
(iii) When conc. H2SO4 is added to a hot solution of X, white crystal of an acid Z separates out. Write equations for all the above reactions and identify X, Y and Z.
Ans. The given salt is alkaline to litmus. Therefore, X is a salt of a strong base and a weak acid. Also, when X is strongly heated, it swells to form substance Y. Therefore, X must be borax.

When borax is heated, it loses water and swells to form sodium metaborate. When heating is continued, it solidifies to form a glassy material Y. Hence, Y must be a mixture of sodium metaborate and boric anhydride.

$$\underset{\text{Borax(X)}}{\text{Na}_2\text{B}_4\text{O}_7}+7\text{H}_2\text{O}\xrightarrow{\text{water}}\underset{\text{Sodium\space hydroxide}}{2\text{NaOH}+4\text{H}_3\text{BO}_3}+\underset{\text{Orthoboric\space acid}}{4\text{H}_3\text{BO}_3}\\\text{Na}_2\text{B}_4\text{O}_7.10\text{H}_2\text{O}\xrightarrow{\Delta}\underset{ \text{Sodium\space metaborate}}{\text{Na}_2\text{B}_4\text{O}_7}\xrightarrow{\Delta}\underset{\text{Boric\space anhydride (Glassy\space material)}}{\text{B}_2\text{O}_3}+\text{NaBO}_2$$

When concentrated acid is added to borax, white crystals of orthoboric acid (Z) are formed.

$$\underset{\text{Borax (X)}}{\text{Na}_2\text{B}_4\text{O}_7.10\text{H}_2\text{O}}+\text{H}_2\text{SO}_4\xrightarrow{\Delta}\text{Na}_2\text{SO}]_4+\underset{\text{Orthoboric\space acid (Z)}}{4\text{H}_3\text{BO}_3}+5\text{H}_2\text{O}$$

11.31. Write balanced equations for:

$$(i)BF_3+LiH\rightarrow\\(ii)B_2H_6+H_2O\rightarrow\\(iii)NaH+B_2H_6\rightarrow\\(iv)H_3BO_3\xrightarrow{\Delta}(v)Al+NaOH\rightarrow\\(vi)B_2H_6+NH_3\rightarrow$$

Ans. $$(i)\underset{\text{Boron\space trifluoride}}{2\text{BF}_3}+\underset{\text{Lithium\space hydride}}{6\text{LiH}}\rightarrow \underset{\text{Diborane}}{\text{B}_2\text{H}_6}+\underset{\text{Lithium\space fluoride}}{6\text{LiF}}\\(ii)\underset{\text{Diborane}}{\text{B}_2\text{H}_6}+\underset{\text{Water}}{6\text{H}_2\text{O}}\rightarrow \underset{\text{Orthoboric\space acid}}{2\text{H}_3\text{BO}_3}+\underset{\text{Hydrogen}}{6\text{H}_2}\\(iii)\underset{\text{Diborane}}{\text{B}_2\text{H}_6}+\underset{\text{Sodium\space hydride}}{2\text{NaH}}\xrightarrow{\text{ether}}\underset{\text{Sodium\space borohydride}}{2\text{NaBH}_4}\\(iv)4\text{H}_3\text{BO}_3\overrightarrow{-4\text{H}_2\text{O}}\underset{\text{Metaboric\space acid}}{4\text{HBO}_2}\xrightarrow[-\text{H}_2\text{O}]{410k} \underset{-\text{H}_2\text{O}\underset{\underset{\text{Boron\space trioxide}}{2\text{B}_2\text{O}_3}}{\downarrow} \text{red\space hot}}{\underset{\text{Tetraboric\space acid}}{\text{H}_2\text{B}_4\text{O}_7}}\\(v)2\text{Al}+2\text{NaOH}+6\text{H}_2\text{O}\rightarrow \underset{\text{Sodium\space tetrahydroxoaluminate(III)}}{2\text{Na}^+[\text{Al(OH)}_4]^-(\text{aq})+3\text{H}_3}\\(vi)3\text{B}_2\text{H}_6+6\text{NH}_3\rightarrow 3[\text{BH}_2(\text{NH}_3)_2]^+[\text{BH}_4]^-\rightarrow \underset{\text{Borazene} }{2\text{B}_3\text{N}_3\text{H}_6+l2\text{H}_2}$$

11.32. Give one method for industrial preparation and one for laboratory preparation of CO and CO2 each.
Ans. Carbon dioxide: In the laboratory, CO2 can be prepared by the action of dilute hydrochloric acid on calcium carbonate. The reaction involved is as follows:

$$\text{CaCO}_3(\text{s})+2\text{HCl(aq)}\rightarrow \text{CaCl}_2(\text{aq})+\text{CO}_2(\text{g})+\text{H}_2\text{O(l)CO}_2$$

is commercially prepared by heating limestone. The reaction involved is as follows:

$$\text{CaCO}_3(\text{s})\xrightarrow{\Delta}\text{CaO(s)}+\text{CO}_2\uparrow(\text{g})$$

Carbon Monoxide: In the laboratory, CO is prepared by the dehydration of formic acid with conc. H2SO4, at 373 K. The reaction involved is as follows:

$$\text{HCOOH}\xrightarrow[\text{conc H}_2\text{SO}_4]{373k}\text{H}_2\text{O}+\text{CO}\uparrow$$

Commercially, CO is prepared by passing steam over hot coke. The reaction involved is as follow:

$$\text{C(s)}+\text{H}_2\text{O(g)}\space\underrightarrow{473.1273\text{k}}\space \underset{\text{Synthetic\space gas}}{\text{CO(g)}+\text{H}_2\text{(g)}}$$

11.33. An aqueous solution of borax is :
(a) neutral
(b) amphoteric
(c) basic
(d) acidic

Ans. (c) basic

Explanation:
Borax is a salt of a strong base (NaOH) and a weak acid (H3BO3). It is therefore, basic in nature.

11.34. Boric acid is polymeric due to
(a) its acidic nature
(b) the presence of hydrogen bonds
(c) its monobasic nature
(d) its geometry
Ans. (b) the presence of hydrogen bonds
Explanation:
Boric acid is polymeric because of the presencem of hydrogen bonds.
The figure given below represents the structure of boric acid in which the dotted lines represent hydrogen bonding:

11.35 The type of hybridisation of boron in diborane is :
(a) sp
(b) sp2
(c) sp3
(d) dsp2

Ans. (c) sp3

Explanation:
Each boron atom in diborane is sp3 hybridised. In the structure of diborane four H-atoms, two on the left and two on the right, known as terminal hydrogens, are in different environments from the other two hydrogen atoms which are known as bridging atoms. The two boron atoms and the four terminal hydrogen atoms lie in the same plane while the two boron atoms and the two bridging hydrogen atoms, one above and the other below, lie in a plane perpendicular to this plane.

If we consider the molecule B2H6 (diborane), there are 12 valence electrons at our disposal for chemical bonding (B has 3, and H has 1, so 2×B + 6×H =12). Each terminal B–H bond is a standard vanilla two electron bond, and there are four of these, thus accounting for a total of eight electrons. This leaves a total of four electrons to share between the two bridging H atoms and the two B atoms. Consequently, two B–H–B bridging bonds are formed, each of which consists of two electrons forming what are called three enter-two-electron bonds (i.e., 3 atoms share 2 electrons) – sometimes called ‘banana’ bonds, as they are not linear but curved. Each B atom is, approximately, sp3 hybridized and if we consider just one of the B atoms, two of the four sp3 hybrid orbitals form s-bonds to the terminal H atoms (1s orbitals). That leaves two B sp3 hybrid orbitals, one of which contains an electron, one of which is empty. For each bridge therefore, one sp3 orbital from each of the B atoms combines with the 1s orbital of the bridging H atom to form three new molecular orbitals (MOs) – as always, n atomic orbitals (AO) form
n MOs. One B atom gives its remaining valence electron to one bridge, and the other B atom gives to the other. Each bridge, therefore, has two electrons, which fill our new MO scheme starting with the lowest energy bonding MO.

11.36. Thermodynamically the most stable form of carbon is
(a) diamond
(b) graphite
(c) fullerenes
(d) coal

Ans. (b) graphite
Explanation:
Graphite is thermodynamically the most stable form of carbon. It shows conjugation effect due to the presence of free electron in its structure. Due to this conjugation effect, graphite is thermodynamically most stable form of carbon and its enthalpy of formation is taken as zero. Also, enthalpy of formation of graphite is less than that of diamond, fullerene and coal at room temperature and atmospheric pressure, therefore, it is most stable form of carbon.

11.37. Elements of group 14
(a) exhibit oxidation state of +4 only
(b) exhibit oxidation state of +2 and +4
(c) form M2– and M4+ ion
(d) form M2+ and M4+ ions

Ans. (b) exhibit oxidation state of +2 and +4
Explanation:
The elements of group 14 have 4 valence electrons. Therefore, the oxidation state of the group is +4. However, due to inert pair effect the lower oxidation state becomes more and more stable and the higher oxidation state becomesm less table. Thus, this group also exhibits +2 oxidation states. Hence, elements of group 14 exhibit oxidation state of +2 and +4.

11.38. If the starting material for the manufacture of silicones is RSiCl3, write the structure of the product formed.