P Block Elements (Gp 13 & Gp 14)

Q1 Discuss the pattern of variation in the oxidation states of  (i) B to Ti and (ii) C to Pb.

Ans. (i) B shows +3 oxidation state. Other group 13 elements s show +1 and +3 oxidation state. On moving down the group 13, the stability of +1 oxidation state increases and the stability of +3 oxidation state decreases due to inert pair effect.

B and Al show +3 oxidation state, Ga and In show +1 and +3 oxidation states and Tl mainly shows +1 oxidation state.

(ii) Group 14 elements show +2 and +4 oxidation states. The stability of +2 oxidation state increases and the stability of +4 oxidation state decreases due to inert pair effect. C and Si mainly show +4 oxidation state, Ge and Sn show +2 and +4 oxidation state and Pb mainly shows +2 oxidation state.

Q2 How can you explain higher stability of BCl3 as compared to TlCl3?

Ans. B shows +3 oxidation state and can form stable BCl3 . Thallium shows +1 and +3 oxidation state. Due to inert pair effect, +1 oxidation state is more stable than +3 oxidation state. Hence, BCl3  is stable but TlCl3 not.

Q3 Why does boron triflouride behave as a Lewis acid ?

Ans. B atom in BF3 has 6 valence electrons. To complete octet, it requires 2 electrons. It readily accepts an electron pair from Lewis bases such as fluoride ion, ethoxide ion and behaves as Lewis acid.

BF3 +F− →BF4−

Q4 Consider the compounds BCl3 and CCl4. How will they behave with water? Justify.

Ans. In BCl3, B atom is electron deficient and accepts an electron pair from water. It is hydrolyzed to boric acid. 

BCl3 +3H2O→H3BO3+3HCl

On the other hand, CCl4 is an electron precise molecule. C has completed its octet and cannot expand its valency shell due to absence of d orbitals. Hence, it can neither donate nor accept electrons. Hence, CCl4 is not hydrolyzed.

Q5 Is boric acid a protic acid? Explain.

Ans. Boric acid does not ionize in water to give proton. Hence, it is not a protic acid. It is a Lewis acid as it accepts electrons from hydroxide ions.

B(OH)3 + 2H2O  →  [B(OH4)]- + H3O+ 

therefore, boric acid is not protic acid.

Q6 Explain what happens when boric acid is heated .

Ans. 

Q7 Describe the shapes of BF3 and BH4–. Assign the hybridisation of boron in these species.

Q 8 Write reactions to justify amphoteric nature of aluminium.

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

Ans. In electron deficient compounds, the central atom either does not have eight electrons in the valence shell or it has eight electrons but can expand its valence beyond 4 due to presence of vacant d orbitals. BCl3  has 6 valence electrons on B atom and accepts pair of electrons from ammonia to complete its octet. Hence, it is an electron deficient compound.

In SiCl4, the central Si atom has 8 electrons but it can expand its valency beyond 4 due to presence of vacant d orbitals. Hence, it is also an electron deficient compound.

Q10 Write the resonance structures of CO32-  and HCO3- .

Ans. 

Q 11 What is the state of hybridisation of carbon in (a) CO32-  (b) diamond (c) graphite?

Ans. (a) In carbonate ion, the carbon atom is sp2   hybridized and forms pi bond.

(b) In diamond, carbon atom is sp3  hybridized and has tetrahedral geometry.

(c) In graphite, carbon atom is sp2 hybridized.

Q 12 Explain the difference in properties of diamond and graphite on the basis of their structures.

Ans. 

Q13 Rationalise 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. 

Q14 Suggest reasons why the B–F bond lengths in BF3 (130 pm) and BF4– (143 pm) differ.

Ans.  In BF3 molecule, B−F bond has partial double bond due to presence of pπ−pπ backbonding between B and F, which removes the electron deficienty of the molecule. Due to this, B−F bond length is reduced to 130 pm. When BF3 changes to BF4− , the hybridization changes from sp2  to sp3 .The double bond character changes to single bond and the B−F bond length becomes 143 pm.

Q15 If B–Cl bond has a dipole moment, explain why BCl3 molecule has zero dipole moment.

Ans. B-Cl bond is polar due to electronegativity difference between B and Cl.  In BCl3 , the central B atom undergoes sp2 hybridization which results in plane triangular geometry. The molecule has symmetry and the individual bond dipoles cancel each other. Hence, the molecule has zero dipole moment.

Q16 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. 

Anhydrous HF is a covalent compound and forms strong intramolecular H bonds. It does not give fluoride ions and cannot dissolve AlF3 in it. KF is an ionic compound and contains fluoride ions. It combines with AlF3 to form the soluble complex. 

AlF3  +3NaF→Na3 [AlF6]

B has small size and high electronegativity. It has mauch higher tendency for complex formation than Al. Hence, when BF3 is added to above solution, AlF3 is precipitated. 

Na3[AlF6]+3BF3  →3NaBF4  +AlF3

Q17 Suggest a reason as to why CO is poisonous.

Ans. Carbon monoxide is highly-poisonous because of its ability to form a complex with haemoglobin. The CO–Hb complex is more stable than the O2–Hb complex. The former prevents Hb from binding with oxygen. Thus, a person dies because of suffocation on not receiving oxygen. It is found that the CO–Hb complex is about 300 times more stable than the O2–Hb complex.

Q18 How is excessive content of CO2 responsible for global warming ? 

Ans. During combustion reactions, CO2 is formed. During photosynthesis, plants utilize CO2 and release O2. Due to excessive combustion, the concentration of CO2 increases beyond certain limit and some CO2 remains unutilized which absorbs heat irradiated by earth. Some of the heat is dissipated back in the atmosphere and the remaining is redirected back towards earth surface. This increases the temperature and is known as green house effect which results in global warming.  

Q19 Explain structures of diborane and boric acid.

Ans. 

Diborane B2H6: 

Boric Acid (H3BO3) : 

Q20 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 is heated strongly

(b) Boric acid is added to water

(c) Aluminium is treated with dil NaOH

(d) BF3 is reacted with ammonia 

Q21 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) 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

(i) CO is heated with ZnO because CO acts as a reducing agent, when CO reacts with ZnO, it reduces ZnO to Zn.

ZnO + CO ----→ Zn + CO2

(d)Alumina dissolves in aqueous NaOH solution to form sodium meta-aluminate.

Q22 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) Aluminium utensils should not be kept in water overnight.

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

Ans. (i) HNO3 reacts can be stored and transported in aluminium containers since it reacts with aluminium to form thick protective oxide layer which makes aluminium passive.

(ii) Dilute NaOH reacts with aluminium pieces to form sodium tetrahydroxo aluminate III and hydrogen gas. The pressure of hydrogen gas so produced is used to open jammed drain.

(iii) The layers of graphite are held together by weak forces between the free electrons of either layers. Since this force can be broken easily, the layers slide on one another. Due to this slippery property, graphite is used as a lubricant.

(iv) These covalent bonds are present throughout the surface giving it a very rigid 3-D structure. It is very difficult to break this extended covalent bonding and for this reason diamond is the hardest substance known. Thus it is used as an abrasive and for cutting tools.

(v) Aluminium has a high tensile strength and is very light in weight. It can also be alloyed with various metals such as Cu, Mn, Mg, Si, and Zn. It is very malleable and ductile. Therefore, it is used in making aircraft bodies.

(vi) Aluminium does not react with pure water but when salts are present in water they remove the oxide layer on the surface making aluminium reactive. Then aluminium reacts with water slowly.

(vii) Aluminium wires are good conductors of electricity and also it is cheap metal easily available, light in weight and also very good ductility. All these properties of aluminium make it better to make transmission cables.

Q23 Explain why is there a phenomenal decrease in ionization enthalpy from carbon to silicon?

Ans. Due to increase in atomic size and screening effect, the force of attraction of the nucleus for the valence electron decreases considerably in Si as compared to C. As a result , there is a phenomenal decreases in ionization enthalpy from carbon to silicon .

Q24 How would you explain the lower atomic radius of Ga as compared to Al ?

Ans. This is because Ga has 3 d electrons, which has poor shielding effect. So, the effective nuclear charge on the outermost electrons is higher than Al, as a result of which the atomic radii decreases and is lower than Al.

Q 25 What are allotropes ? Sketch the structure of two allotropes of carbon namely diamond and graphite. What is the impact of structure on physical and chemical properties of two allotropes?

Ans. Allotropes: - The phenomenon of existence of an element in two or more forms which have different physical properties but identical chemical properties is called allotropy and the different forms are called allotropes. Carbon exists in two allotropic forms: - 1. Crystalline 2. Amorphous 

1. Crystalline allotropic forms of carbon: - Three allotropes of carbon having well defined crystal structures are: a. Diamond, b. Graphite and c. Fullerenes.

a. Diamond: 

- It occurs in nature and can be artificially prepared

- Carbon is sp3 hybridised

- Each carbon is tetrahedrally linked to four neighbouring carbon atoms through four strong C-C sp3 - sp3 σ bonds

- Diamond is the purest form of carbon.

b. Graphite: 

- each carbon is sp2 hybridised. 

- each carbon is linked to three other carbon atoms forming hexagonal rings

- Graphite is a 2D sheet like (layered) structure consisting of a no. of benzene rings fused together

- Layers are held together by weak van der Waal's forces of attraction

- Layers are at a distance of 340pm. As two successive layers are held together by weak forces of attraction, one layer can slip over the other

- This makes graphite soft and a good lubricating agent

Q26 (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) The neutral oxide is CO.

The acidic oxides are B2O3,SiO2 and CO2.

The basic oxide is Tl2O3.

The amphoteric oxides are Al2O3 and PbO.

Q27 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 is a group 13 element with common oxidation states +1 and +3.

Due to inert pair effect +1 oxidation state in Tl is more stable than +3 oxidation state. Al shows +3 oxidation state and alkali metals shows +1 oxidation state. Thus, thallium resembles both Al, and alkali metals. Like Al, Tl forms compounds such as TlCl3 and Tl2O3 and like alkali metals, Tl forms compounds such as TlCl and Tl2O. 

Q28 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. 

Q 29 What do you understand by  (a). inert pair effect (b). allotropy (c) catenation?

Ans. 

a. Inert Pair Effect: - heaviour elements of p block show less oxidation state by 2 and this is due to electrons in inner s orbitals come closer to the nucleus and hence these are not available for donation. Hence this pair of electrons of s orbitals becomes inert, so this phenomenon is known as inert pair effect. 

b. Allotropy: - The phenomenon of existence of an element in two or more forms which have different physical properties but identical chemical properties is called allotropy and the different forms are called allotropes. Carbon exists in two allotropic forms: - 1. Crystalline 2. Amorphous.

c. Catenation: - Ability of like atoms to link with one another through covalent bonds. This is due to smaller size and higher electronegativity of carbon atom and unique strength of carbon-carbon bonds. The property of catenation mainly depends upon the strength of element-element bond. Carbon fits into that category which is required for catenation viz. size order, electronegativity, ionization enthalpy.

Q30 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. 

Q31 Write balanced equations for:

(i) BF3 + LiH →

(ii) B2H6 + H2O →

(iii) NaH + B2H6 →

(iv) H3BO3 →∆

(v) Al + NaOH →

(vi) B2H6 + NH3 →

Ans. 

(iv)

(v) 

(vi) When diborane reacts with ammonia, an addition product that is, an adduct is formed when further decomposes on heating at 473K to form a volatile compound called borazine or borazole. -Borazine is isostructural to benzene and hence, it is known as inorganic benzene

Q32 Give one method for industrial preparation and one for laboratory preparation of CO and CO2 each.

Q 33 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). It is, therefore, basic in nature.

Q34 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) Boric acid is polymeric due to the presence of hydrogen bonds. The polymeric structure is not due to the acidic nature, monobasic nature or the geometry.

Q35 The type of hybridisation of boron in diborane is (a) sp (b) sp2 (c) sp3 (d) dsp2 

Ans. (c) Boron in diborane is sp3 hybridised.

Q36 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.

Q37 Elements of group 14

(a) exhibit oxidation state of +4 only

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

(c) form M2– and M4+ ions

(d) form M2+ and M4+ ions

Ans.  (b)The elements of group 14 have 4 valence electrons. Therefore, the oxidation state of group is +4. However, as a result of the inert pair effect, the lower oxidation state becomes more stable and the higher oxidation state becomes less stable. Therefore, this group exhibits +4 and +2 oxidation states.

Q38 If the starting material for the manufacture of silicones is RSiCl3, write the structure of the product formed. 

Ans. Hydrolysis of trichlorosiane leads to the formation of cross linked silicaned.