NCERT Solutions Class 11 Chemistry Chapter 11

Q.1 Discuss the pattern of variation in the oxidation states of i) B to Tl and ii) C and Pb.

Ans.

i) The elements present in group 13 are B, Al, Ga, In and Tl. Among them B and Al have no d- or f- electrons, so they cannot exhibit inert pair effect. They show +3 oxidation states due to the presence of 3 electrons in the valence shell, out of them two electrons are present in the s-orbital and one electron is present in the p-orbital. Other elements, Ga to Tl have either only d electrons or both d- and f- electrons, hence they show oxidation state of +1 and +3 due to the inert pair effect, this effect is more prominent in Tl. As we move down the group, due to increase in the number of electrons in the d- and f- block electrons, orbitals the inert pair effect becomes more prominent and pronounced. From Ga to Tl, the stability of +1 oxidation state increases (Ga < In <tl), of=”” that=”” while=””> Tl).</tl),>

Thus for Tl, +1 oxidation state is more stable than +3 oxidation state.

ii)The elements present in group 14 are C, Si, Ge, Sn, Pb. C and Si do not have d- or f- electrons, hence they don’t exhibit inert pair effect. They show +4 oxidation states due to the presence of 4 electrons in the valence shell, two in s- and two in p- orbital. But the other elements from Ge to Pb contain d- or f- electrons; hence show +2 and +4 oxidation states due to inert pair effect. When we move down the group, the inert pair effect becomes more prominent and stability of +2 oxidation state increases (Ge<sn

Sn>Pb).</sn

Thus for Pb, +2 oxidation state is more stable than +4 oxidation state.

Q.2 How can you explain higher stability of BCl₃ as compared to TlCl₃?

Ans.

The shielding or screening effect follows the order s>p>d>f, thus s-orbital has strong shielding effect and f-orbital has poor shielding effect. In Tl, 3d, 4d, 5d and 4f electrons show poor shielding effect on 6s electrons. As a result, only 6p¹ electron of Tl participates in bond formation. Thus Tl shows +1 oxidation state rather than +3 oxidation state. TlCl is stable compound but TlCl₃ is unstable.

B doesn’t contain d- or f-electrons, thus can’t show inert pair effect. The three valence electrons ((2s² 2p¹) participate in bond formation and hence it can show +3 oxidation state. Thus BCl₃ is stable compound than TlCl₃.

Q.3 Why does boron triflouride behave as a Lewis acid?

Ans.

In BF₃ molecule, B atom has only 6 electrons in its valence shell and hence is an electron-deficient molecule. To complete its octet, B requires two more electrons, so it easily accepts a pair of electrons from nucleophiles such as F^–, CN^– etc. Thus, BF₃ acts as Lewis acid.

Q.4 Consider the compounds, BCl₃ and CCl₄. How will they behave with water? Justify.

Ans.

BCl₃ is electron-deficient molecule as B has six electrons in the valence shell. To complete its octet, it accepts a pair of electrons donated by water molecule. Thus, BCl₃ undergoes hydrolysis to form boric acid (H₃BO₃) and HCl.

BCl₃ + 3H₂O → H₃BO₃ + 3HCl

In comparison to BCl₃, in CCl₄ molecule, the C atom has 8 electrons in the valence shell. The compound is an electron-precise molecule and hence it can neither accept nor donate a pair of electrons. So CCl₄ does not undergo hydrolysis in water.

Q.5 Is boric acid a protic acid? Explain.

Ans.

No, boric acid is not a protic acid, it is an electron-deficient molecule having incomplete octet. It doesn’t ionize in water to give a proton rather it accepts a lone pair of electrons from the oxygen atom of H₂O molecule to form a hydrated species.

Oxygen atom is electronegative, so +ve charge on oxygen atom pulls the s – electrons of O–H bond towards itself. Thus it releases the proton and B(OH)₃ acts as a weak monobasic Lewis acid. When it reacts with NaOH solution, sodium metaborate is formed.

\text{B}{\left(\text{OH}\right)}_{\text{3}}\text{+ NaOH}\stackrel{}{\to }{\text{Na}}^{\text{+}}{\left[\text{B}{\left(\text{OH}\right)}_{\text{4}}\right]}^{\text{–}}\text{or}{\text{Na}}^{\text{+}}{\text{BO}}_{\text{2}}^{\text{–}}\text{+}{\text{2H}}_{\text{2}}\text{O}

Q.6 Explain what happens when boric acid is heated.

Ans.

Upon heating boric acid loses water in different stages at different temperature. The final product is boron trioxide or boric anhydride.

\underset{\text{Boric acid}}{{\text{H}}_{\text{3}}{\text{BO}}_{\text{3}}}\stackrel{\text{370K}}{\to }\underset{\begin{array}{l}\text{Metaboric}\\ \text{acid}\end{array}}{{\text{HBO}}_{\text{2}}\text{+}}{\text{H}}_{\text{2}}\text{O}

\underset{\begin{array}{l}\text{Meta}\text{boric}\\ \text{acid}\end{array}}{{\text{4HBO}}_{\text{2}}}\underset{{\text{–H}}_{\text{2}}\text{O}}{\overset{\text{410K}}{\to }}\underset{\begin{array}{l}\text{Tetra}\text{boric}\\ \text{acid}\end{array}}{{\text{H}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}}\stackrel{\text{Red}\text{heat}}{\to }\underset{\begin{array}{l}\text{Boron}\\ \text{trioxide}\end{array}}{{\text{2B}}_{\text{2}}{\text{O}}_{\text{3}}\text{+}}{\text{H}}_{\text{2}}\text{O}

Q.7 Describe the shapes of BF₃ and BH₄^–. Assign the hybridization of boron in these species.

Ans.

In BF₃ molecule, boron is sp²–hybridized and therefore, it is a planer molecule. On the other hand, in BH₄^–, boron is sp³–hybridized, so it has tetrahedral structure.

Q.8 Write reactions to justify amphoteric nature of aluminium.

Ans.

Al reacts with both acid and alkali to form dihydrogen.

2Al(s) + 3H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3H₂(g)

2Al(s) + 2NaOH(aq) + 6H₂O(l) → 2Na⁺[Al(OH)₄]^–(aq) + 3H₂(g)

Sod.

tetrahydroxoaluminate(III)

Q.9 What are electron deficient compounds? Are BCl₃ and SiCl₄ electron deficient species? Explain.

Ans.

The compounds in which the central atom does not have 8 electrons in the valence shell or those which have 8 valence electrons but can expand its covalency beyond 4 due to the presence of vacant d-orbitals, are called electron deficient compounds.

(i) In BCl₃, central atom B contains only 6 electrons in valence shell, thus it is an electron deficient compound. NH₃ donates its pair of electrons to BCl₃ molecule and Cl₃B ← NH₃ adduct is formed.

(ii) In SiCl₄, the central Si atom has 8 electrons but it contains vacant d orbital, so it can expand its covalency beyond 4. Thus SiCl₄ is also an electron-deficient compound.

Q.10 What is the state of hybridization of carbon in a) CO₃^2- b) diamond, c) graphite?

Ans.

a) Hybridization of C in CO₃^2– is sp²,

b) Hybridization of C in diamond is sp³,

c) Hybridization of C in graphite is sp².

Q.11 Explain the difference in properties of diamond and graphite on the basis of their structures.

Ans.

Q.12 Rationalize the given statements and give chemical reactions:

• Lead (II) chloride reacts with Cl₂ to give PbCl₄.

• Lead (IV) chloride is highly unstable towards heat.

• Lead is known not to form an iodide, PbI₄.

Ans.

• Lead has d- and f- orbital electrons, it shows inert pair effect. For that reason, +2 oxidation state of Pb is more prominent rather than +4 oxidation state. Thus lead (II) chloride is more stable than lead (IV) chloride. Therefore, PbCl₂ does not react with chlorine to form PbCl₄.

{\text{PbCl}}_{\text{2}}\left(\text{s}\right)\text{+}{\text{Cl}}_{\text{2}}\left(\text{g}\right)\xcancel{\stackrel{}{\to }}{\text{PbCl}}_{\text{4}}\left(\text{l}\right)

• Due to greater stability of +2 oxidation state of Pb over +4 oxidation state (due to inert pair effect); PbCl₂ is more stable than PbCl₄.

{\text{PbCl}}_{\text{4}}\left(\text{l}\right)\stackrel{\text{Δ}}{\to }{\text{PbCl}}_{\text{4}}\left(\text{s}\right)\text{+}{\text{Cl}}_{\text{2}}\left(\text{g}\right)

• In PbI₄, Pb is in +4 oxidation state. But due to inert pair effect, +4 oxidation state is not stable for Pb. Pb–I bond formed initially during the reaction does not release enough energy to unpair the 6s² electrons and excite them to 6p–orbital to have 4 unpaired electrons around Pb atom needed for the formation of PbI₄. Again, due to the oxidizing power of Pb⁴⁺ ion and reducing power of I^– ion, PbI₄ does not exist.

Q.13 Suggest reasons why the B–F bond lengths in BF₃ (130 pm) and BF₄^–(143 pm) differ.

Ans.

In BF₃, boron is sp² hybridized, it has vacant 2p- orbital and F has three lone pair of electrons in 2p-orbitals. Because of similar sizes of 2p-orbitals in F and B, pp – pp back bonding occurs in which a lone pair of electrons is transferred from F to B. Thus B–F bond acquires some partial double bond character.

But in BF₄^– ion, B is sp³– hybridized and does not have any vacant p-orbital available to accept the electrons donated by F atom, so pp–pp back bonding does not occur here. The bonds present in BF₄^– are purely single bonds. Double bonds are shorter in length than single bonds. So, in BF₃ bond length is 130 pm and in BF₄^– ,it is 143pm.

Q.14 If B–Cl bond has a dipole moment, explain why BCl₃ molecule has zero dipole moment.

Ans.

Both B and Cl atoms have electronegativity difference, so B–Cl bond has polar character and has a definite dipole moment. But polarity also depends upon the geometry of the molecule. BCl₃ has planer geometry, the bond angle between the three B–Cl bonds is 120°. Thus the resultant dipole moment of two B–Cl bonds is cancelled by equal and opposite dipole moment of the third B–Cl bond. As a result, overall dipole moment of BCl₃ becomes zero. A clear view of dipole moment of BCl₃ can be depicted by the following picture.

Q.15 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.

HF is a covalent compound in anhydrous form and the molecules are held together by hydrogen bonds. It does not give F^– ion in the medium, so AlF₃ does not dissolve in HF. But NaF is ionic solid, it produces F^– ions which react with AlF₃ to form the soluble complex.

3NaF + AlF₃ → Na₃[AlF₆]

Sod. hexafluoroaluminate (III)

B has much higher tendency to form complexes than Al, due to its smaller size and higher electronegativity. So, when BF₃ is added to the above solution, AlF₃ gets precipitated.

Na₃[AlF₆] + 3BF₃ → 3Na[BF₄] + AlF₃(s)

Sod. Tetrafluroborate (IIII)

Q.16 Suggest a reason as to why CO is poisonous.

Ans.

Haemoglobin present in red blood cells of human body combines with oxygen to form oxyhaemoglobin. It is a reversible reaction; oxyhaemoglobin carries oxygen from lungs to different tissues of body.

\text{Haemoglobin}\text{+}{\text{O}}_{\text{2}}\underset{}{⇌}\text{Oxyhaemoglobin}

When CO and O₂ both are available, haemoglobin preferably combines with CO than O₂. CO combines with haemoglobin irreversibly to form carboxyhaemoglobin, which is more stable than oxyhaemoglobin. The percentage of oxygen in blood gets reduced. Thus the oxygen carrying capacity of haemoglobin is destroyed and human being dies of suffocation.
Haemoglobin + CO → Carboxyhaemoglobin
For this reason, CO is poisonous.

Q.17 How is excessive content of CO₂ responsible for global warming?

Ans.

The normal amount of CO₂ in air is 0.03%. During respiration, CO₂ is evolved and it is utilized by plants during photosynthesis. Thus, CO₂ concentration is balanced in the earth. If concentration of CO₂ increases beyond 0.03% (due to excessive combustion), some of the CO₂ will always remain unutilized. This excess CO₂ absorbs heat radiated by earth and some of it is dissipated into the atmosphere, the remaining part is radiated back to the earth. Thus the temperature of the earth increases gradually. This phenomenon is called greenhouse effect and CO₂ is called green house gas which is a major contribution to the global warming.

Q.18 Explain structures of diborane and boric acid.

Ans.

Structure of diborane:

In diborane (B₂H₆), two types of hydrogen atoms are present-

• The four hydrogen atoms (two on the left and two on the right side of the boron atom) are called the terminal hydrogens, and these 4 H atoms and 2 boron atoms lie in the same plane.
• The remaining two hydrogen atoms, one lying above the plane and another lying below the plane form bridges and are called bridge hydrogens.

The four terminal B-H bonds are normal covalent bonds which are formed by sharing a pair of electrons between B and H atoms. So it is also called two centre electron pair bonds or two centre-two electron bond (2c-2e).

Another two bridge bonds i.e. B—-H—-B are quite different from normal covalent bonds. Each bridge hydrogen atom is bonded with two B atoms by a pair of electrons. So it is called three centre-two electron bonds (3c-2e). The 3c-2e bond resembles to a banana, so it is also called banana bond. This kind of bond is weak.

On the basis of hybridization, explanation of diborane structure:

The electronic configuration of boron atom is as follows:

\begin{array}{l}{\text{5}}^{\text{B}}\left(\text{Ground}\text{state}\right)\text{:}{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}_{\text{x}}^{\text{1}}{\text{2p}}_{\text{y}}^{\text{0}}{\text{2p}}_{\text{z}}^{\text{0}}\\ {\text{5}}^{\text{B}}\left(\text{Excited}\text{state}\right)\text{:}{\text{1s}}^{\text{2}}{\text{2s}}^{\text{1}}{\text{2p}}_{\text{x}}^{\text{1}}{\text{2}}_{\text{y}}^{\text{1}}{\text{2}}_{\text{z}}^{\text{0}}\end{array}

So it undergoes sp³-hybridization. The two half filled hybride orbitals of each B atom overlap with the half filled orbital of hydrogen atoms to form normal covalent bonds. The third half filled hybrid orbital of one boron atom and the vacant hybrid orbital of 2^nd boron atom overlap simultaneously with the half filled orbital of H atom, thus electron cloud spreads over three atoms (2 Boron and 1 H atom). For this reason this bond is called 3 centre-2 electron bonds (3c-2e) or banana bond.

Structure of boric acid: In borate ion (BO₃^3-), the boron atom is sp²-hybridized and has trigonal planer structure. In excited state, boron has three half filled atomic orbitals (2s, 2p_x and 2p_y), these three atomic orbitals undergo sp²-hybridization to give three sp²-hybrid orbitals. Each of these orbital overlaps with 2p-orbitals of O^–, forming B–O^– bonds.

In boric acid, the planer BO₃^3– ions are joined by unsymmetrical hydrogen bonds to give layered structure. The adjacent layers in the crystal of boric acid are held together by weak forces of attraction, so one layer can slide over the other. This property makes boric acid soft and soapy to touch.

Q.19 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) Borax contains 10 molecules of water of crystallization (Na₂B₄O₇.10H₂O). When it is heated, the water molecules get vapourised. When borax is strongly heated, a transparent glassy bead which consists of sodium metaborate (NaBO₂) and boric anhydride is formed.

\underset{\text{Borax}}{{\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}\text{.}}{\text{10H}}_{\text{2}}\text{O}\stackrel{\text{Heat}}{\to }{\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}\text{+}{\text{10H}}_{\text{2}}\text{O}

b) Boric acid acts as weak Lewis acid. It gives proton into the aqueous solution.

c) When Al is treated with aqueous solution of NaOH, sodium aluminate and dihydrogen are formed.

2Al(s) + 2NaOH(aq) + 6H₂O(l) → 2Na⁺[Al(OH)₄]^–(aq) + 3H₂(g)

d) BF₃ is a Lewis acid because it has incomplete octet and NH₃ is a strong base having lone pair of electrons. BF₃ accepts a pair of electrons from NH₃ to form the corresponding adduct.

Q.20 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 is heated with methyl chloride in presence of Cu power (acting as a catalyst), a mixture of mono-, di- , tri-methylchlorosilanes with small amount of tetra-methylsilane is formed.

\underset{\begin{array}{l}\text{Methyl}\\ \text{chloride}\end{array}}{{\text{CH}}_{\text{3}}\text{Cl}\text{+}}\text{Si}\underset{\text{570K}}{\overset{\text{Cu}\text{power}}{\to }}{\text{CH}}_{\text{3}}{\text{SiCl}}_{\text{3}}\text{+}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{SiCl}}_{\text{2}}\text{+}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{SiCl}\text{+}{\left({\text{CH}}_{\text{3}}\right)}_{\text{4}}\text{Si}

b) Silicon dioxide initially dissolves in HF to form silicon tetra-fluoride and it further reacts with remaining HF to form hydrofluorosilicic acid.
SiO₂ + 4HF → SiF₄ + 2H₂O
SiF₄ + 2HF → H₂SiF₆
c) ZnO is reduced by CO to Zn metal.
ZnO + CO → Zn + CO₂
d) Hydrated alumina reacts with NaOH to form sodium meta-aluminate.
Al₂O₃.2H₂O(s) + 2NaOH(aq) + H₂O(l) → 2Na[Al(OH)₄](aq)
Hydrated alumina Sod. tetrahydroxoaluminate (III)
or Bauxite

Q.21 Give reasons:

(i) Conc. HNO₃ can be transported in aluminium container.

(ii) A mixture of dilute NaOH and aluminium pieces is used to open drain.

(iii) Graphite is used as lubricant.

(iv)Diamond is used as an abrasive.

(v) Aluminium alloys are used to make aircraft body.

(vi)Aluminum utensils should not be kept in water overnight.

(vii) Aluminium wire is used to make transmission cables.

Ans.

i) Al reacts with conc. HNO₃ to form a very thin layer of aluminium oxide on its surface which protects it from further reaction.

2Al(s) + 6HNO₃(l) → Al₂O₃(s) + 6NO₂(g) + 3H₂O(l)

Alumina

Thus, Al becomes passive and the Al container can be used for transportation of Conc. HNO₃.

ii) Al reacts with NaOH to form sodium aluminate and dihydrogen gas is evolved. The pressure of the hydrogen gas thus produced can be used to open clogged drains.

2Al(s) + 2NaOH(aq) + 2H₂O(l) → 2NaAlO₂(aq) + 3H₂(g)

iii) Graphite has a layered structure; two successive layers are held together by weak van der Waals forces of attraction. So the layers can slip over one another. For this reason it can act as lubricant.

iv) Diamond exists as three dimensional network structure in which each carbon atom is tetrahedrally bonded with four neighboring carbon atoms. So it is very hard and can be used as abrasive.

v) The following characteristics of aluminium alloys make them useful for making aircraft body:

• Lightweight
• Resistance to corrosion.
• Tough

vi)Al reacts with water and dissolved oxygen to form thin film of aluminium oxide.

2Al(s) + O₂(g) + H₂O(l) → Al₂O₃(s) + H₂(g)

Very small amount of Al₂O₃ dissolves in water to give a few ppm of Al³⁺ ions. Since Al³⁺ ions affect our health, so drinking water should not be kept in Al utensils for overnight.

Vii) Al is light weight and good conductor of electricity, so it is used for making transmission cables and for winding the moving coils of dynamo or motor.

Q.22 Explain why is there a phenomenal decrease in ionization enthalpy from carbon to silicon?

Ans.

The decrease in ionization enthalpy from C to Si is phenomenal due to large increase in atomic size and decrease in screening effect (force of attraction of the nucleus for valance electrons) from C to Si.

Q.23 How would you explain the lower atomic radius of Ga as compared to Al?

Ans.

Ga has electrons in 3d orbital, which shows poor shielding effect on the valence electrons. So the effective nuclear charge of Ga is greater in magnitude as compared to Al. As a result, the valence electrons of Ga experience greater force of attraction by the nucleus than Al, so atomic size of Ga (135pm) is slightly less than Al (143pm).

Q.24 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 two or more different forms which have different physical properties and but identical chemical properties. Different forms of an element are called allotropes. For example, diamond and graphite are the allotropes of carbon.

The impact of structure on physical properties of two allotropes is as follows:

1. Diamond exists in a three dimensional network structure and has high density. Therefore, it is hard. Graphite has two-dimensional sheet like structure and any two consecutive layers are held together by weak van der Waals forces of attraction. Thus one layer can slip over the other and it becomes soft and can be used as lubricant.

2. In diamond, all the electrons are held together by strong sigma bond, as a result, no electrons are available for conducting electricity. So diamond is a poor conductor of electricity. But in graphite, the fourth valance shell electron of each carbon is free to move, therefore, it can conduct the electricity. So graphite is a good conductor of electricity.

3. Due to its high refractive index, diamond can reflect and refract light, so it is transparent. Graphite can’t reflect and refract light, so it is opaque.

Q.25 (a) Classify following oxides as neutral, acidic, basic or amphoteric: CO, B₂O₃, SiO₂, CO₂, Al₂O₃, PbO₂, and Tl₂O₃

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

Ans.

a) Neutral oxide: CO.

Acidic oxide: B₂O₃, SiO₂, CO₂

Basic Oxide: Tl₂O₃

Amphoteric oxide: Al₂O₃, PbO₂

b) Acidic oxides react with alkalis to form salts. The chemical reactions involved are given below:

B₂O₃ + 2NaOH → 2NaBO₂ + H₂O

Boric anhydride Sod. Meta borate

\underset{\text{Silica}}{{\text{SiO}}_{\text{2}}\text{+}}\text{2NaOH}\stackrel{\text{Δ}}{\to }\underset{\text{Sod}\text{.}\text{silicate}}{{\text{Na}}_{\text{2}}{\text{SiO}}_{\text{3}}\text{+}}{\text{H}}_{\text{2}}\text{O}

CO₂ + 2NaOH → Na₂CO₃ + H₂O

Amphoteric oxides react with both acid and base. The chemical reactions involved are as follows:

Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O

Alumina Sod. Meta aluminate

Al₂O₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂O

PbO₂ + 2NaOH → Na₂PbO₃ + H₂O

Lead dioxide Sod. plumbate

2PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O + O₂

Basic oxides dissolve in acids. For example: Tl₂O₃ dissolves in acids.

Tl₂O₃ + 6HCl → 2TlCl₃ + 3H₂O

Q.26 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.

Aluminium shows +3 oxidation state in all compounds and Tl also shows +3 oxidation state in some compounds like TlCl₃, Tl₂O₃, etc.

Group I metals show oxidation state of +1. Due to inert pair effect, Tl also shows +1 oxidation state in the compounds like Tl₂O, TlCl etc. Like group I oxides, Tl₂O is strongly basic in nature.

Q.27 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 their identities.

Ans.

In a schematic diagram, the reaction can be represented as:

X + NaOH → A (white ppt.) + excess NaOH → B (soluble compound)

A + dil HCl → C

\text{A}\stackrel{\text{Δ}}{\to }\text{D}\left(\text{used}\text{to}\text{extract}\text{metals}\right)

X must be Al, ppt. A is Al(OH)₃ and complex B must be sodium tetrahydroxoaluminate(III).

Al + 3NaOH → Al(OH)₃ ↓ + 3Na⁺

(X) Aluminimum

hydroxide ppt.

Al(OH)₃ + NaOH → Na⁺ [Al(OH)₄]^–

(A) (B)

Sod. Tetrahydroxoaluminate (III)

Compound C is AlCl₃. Al(OH)₃ reacts with dil. HCl to give compound C.

Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O

(A) (C)

Upon heating, A decomposes to compound D. D is Al₂O₃ (alumina) which is used to extract metals.

2Al(OH)₂ → Al₂O₃ + 3H₂O

(A) (D)

Q.28 What do you understand by (a) inert pair effect (b) allotropy and (c) catenation?

Ans.

(a) Inert pair effect: As we move down a group, the tendency of s-electrons of the valence shell to participate in bond formation gradually decreases. This reluctance of s- electrons to participate in bond formation is called inert pair effect.

As we move down the group, additional shells are incorporated (having d- or/and f- orbitals) and d- or f- orbitals have poor shielding effect on ns² electrons. Due to increase in effective nuclear charge, the s-electrons become more tightly held (more penetrating) and tend to remain paired, therefore, become more reluctant to participate in bond formation.

(b) Allotropy: The phenomenon of existence of an element in two or more different forms which have different physical properties but identical chemical properties is called allotropy and the different forms are called allotropes. Diamond and graphite are the two allotropes of carbon. The allotropes of sulphur are monoclinic sulphur and rhombic sulphur.

(c) Catenation: Catenation is the property of an element which may be defined as the ability of like atoms to link with one another by covalent bonds. Due to the smaller size and high electronegativity of carbon, it shows this unique property. Since bond energy of C–C single bond is very large, it can form long straight or branched C–C chains or rings of different sizes and shapes. For this catenation property, carbon can form a large number of hydrocarbons, no other atom can show. Other than carbon Si shows the catenation property.

Q.29 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.

i) Since the aqueous solution of the salt (X) turns red litmus paper blue, then it should be salt of strong base and a weak acid.

ii) Salt (X) swells up to glassy material (Y) on strong heating, so (X) is borax and (Y) must be a mixture of sodium meta-borate and boric anhydride.

iii) When conc. H₂SO₄ is added to salt X, white crystals of an acid Z separate out, so Z must be orthoboric acid.

The involved reactions are as follows:

\underset{\text{Borax}\left(\text{x}\right)}{{\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}{\text{.10H}}_{\text{2}}\text{O}}\stackrel{\text{water}}{\to }\underset{\text{Strong}\text{​}\text{base Weak}\text{acid}}{\text{2NaOH}\text{+}{\text{H}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}}{\text{+ 8H}}_{\text{2}}\text{O} \underset{\text{Borax}}{{\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}{\text{.10H}}_{\text{2}}\text{O}}\stackrel{\text{Heat}}{\to }\underset{}{{\text{Na}}_2{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}}{\text{+ 10H}}_{\text{2}}\text{O}

{\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}{\text{.10H}}_{\text{2}}{\text{O + H}}_{\text{2}}{\text{SO}}_{\text{4}}\stackrel{}{\to }\underset{\text{Boric}\text{acid}\left(\text{z}\right)}{{\text{4H}}_{\text{3}}{\text{BO}}_{\text{3}}{\text{+ Na}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{+5H}}_{\text{2}}\text{O}

Q.30  Write balanced equations for:

(i) BF₃ + LiH →

(ii) B₂H₆ + H₂O →

(iii) NaH + B₂H₆ →

\text{(iv)}{\text{H}}_{\text{3}}{\text{BO}}_{\text{3}}\stackrel{\text{Δ}}{\to }

(v) Al + NaOH →

(vi) B₂H₆ + NH₃ →

Ans.

Q.31 Give one method for industrial preparation and one for laboratory preparation of CO and CO₂ each.

Ans.

Industrial preparation of CO:

\text{2C}\left(\text{s}\right)\text{+}{\text{O}}_{\text{2}}\left(\text{g}\right)\stackrel{\text{Limited}\text{air}}{\to }\text{2CO}\left(\text{g}\right)

Laboratory preparation of CO:

\text{HCOOH}\stackrel{{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\to }\text{CO}\text{+}{\text{H}}_{\text{2}}\text{O}

Industrial preparation of CO₂

\text{C}\left(\text{s}\right)\text{+}{\text{O}}_{\text{2}}\left(\text{g}\right)\stackrel{\text{Excess}\text{air}}{\to }{\text{CO}}_{\text{2}}\left(\text{g}\right)

Laboratory preparation of CO₂

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)

Q.32 An aqueous solution of borax is

(a) Neutral (b) Amphoteric

(c) Basic (d) Acidic

Ans.

c) Borax is a salt of a strong base (NaOH) and a weak acid (H₃BO₃), thus it is basic in nature.

Therefore, option (c) is correct.

Q.33 Boric acid is polymeric due to

(a) its acidic nature (b) the presence of hydrogen bonds. (c) its monobasic nature (d) its geometry.

Ans.

Boric acid is polymeric due to the presence of H-bonds. Thus, (b) is the right option.

Q.34 The type of hybridisation of boron in diborane is (a) sp (b) sp² (c) sp³ (d) dsp²

Ans.

The hybridisation of B in diborane is sp³.Therefore, option (c) is correct.

Q.35 Thermodynamically the most stable form of carbon is

(a) Diamond (b) Graphite

(c) Fullerenes (d) Coal

Ans.

Thermodynamically the most stable form of carbon is graphite. Thus, option (b) is the correct one.

Q.36 Elements of group 14

(a) exhibit oxidation state of +4 only

(b) exhibit oxidation states of +2 and +4

(c) form M^2– and M⁴⁺ ions

(d) form M²⁺ and M⁴⁺ ions

Ans.

Group 14 elements show +2 and +4 oxidation states. Due to the inert pair effect, the lower oxidation state becomes more and more stable and the higher oxidation state becomes less stable. Thus option (b) is the correct answer.

Q.37 If the starting material for the manufacture of silicones is RSiCl₃, write the structure of the product formed.

Ans.

Silicones are formed by hydrolysis of alkyltri- chlorosilanes.

Q.38 Write the resonance structures of CO₃^2–and HCO₃^–.

Ans.

Resonance structures of CO₃^2- :

Resonance structures of bicarbonate HCO₃^–