Explain the following:
(i) Due to ineffective shielding of valence electrons by the intervening 3d electrons, the effective nuclear charge on Ga is slightly higher than that on A1 and hence the ∆Hi of gallium is slightly higher than that of Al. Boron has three electrons in the valence shell. Because of its small size and high sum of the first three ionisation enthalpies (i.e., ∆
(ii) Boron has three electrons in the valence shell. Because of its small size and high sum of the first three ionisation enthalpies (i.e., ∆iH1 + ∆iH2 + ∆iH3), boron does not lose all its valence electrons to form B3+ ions.
(iii) Al has vacant d-orbitals and hence can expand its coordination number from 4 to 6 and hence forms octahedral [A1F6]3- ion in which Al undergoes sp3d2 hybridisation. In contrast, B does not have d-orbitals. Therefore, it can have a maximum coordination number of 4. Therefore, B forms [BF4]– (in which B is sp3-hybridised) but not [BF6]3-.
(iv) Due to inert pair effect, +2 oxidation state of Pb is more stable than its +4 oxidation state. Consequently, PbX2 in which the oxidation state of Pb is +2 is more stable than PbX4 in which the oxidation state of Pb is +4.
(v) Inert pair effect is less prominent in Sn than in Pb. Therefore, +2 oxidation of Sn is less stable than its +4 oxidation state. In other words, Sn2+ can easily lose two electrons to form Sn4+ and hence Sn2+ acts as a reducing agent.
Sn2+→Sn4+ + 2e
In contrast, the inert pair effect is’ more prominent in Pb than in Sn. Therefore, +2 oxidation state of Pb is more stable than its +4 oxidation state. In other words, Pb4+ can easily lose two electrons to form Pb2+ and hence Pb4+ acts as an oxidising agent.
Pb4+ + 2e–→-Pb2+
(vi) Due to small size, the electron-electron repulsions in the relatively compact 2p-subshell of F are quite strong and hence the incoming electron is not accepted with the same ease as in case of bigger Cl atom where repulsions are comparatively weak. Thus, electron gain enthalpy of chlorine is more negative as compared to that of fluorine.
(vii) Due ta strong inert pair effect, the +3 oxidation state of T1 is less stable than its +1 oxidation state. Since in T1(N03)3, oxidation state of T1 is +3, therefore, it can easily gain two electrons to form T1N03 in which the oxidation state of T1 is +1. Consequently, T1(N03)3 acts as an oxidising agent.
(viii) Property of catenation depends upon the strength of element-element bond which, in turn, depends upon the size of the element. Since the atomic size of carbon is much smaller than that of lead, therefore, carbon-carbon bond strength is much higher than that of lead-lead bond. Due to stronger C-C than Pb-Pb bonds, carbon has a much higher tendency for catenation than lead.
(ix) Unlike other boron halides, BF3 does not hydrolyse completely. Instead, it hydrolyses incompletely to form boric acid and fluoroboric acid. This is because the HF first formed reacts with H3B03.
(x) In graphite, carbon is sp2-hybridised and each carbon is linked to three other carbon atoms by forming hexagonal rings. Each carbon is now left with one unhybridised p-orbital which undergoes sideways overlap to form three p-p double bonds. Thus, graphite has two-dimensional sheet like (layered) structure consisting of a number of benzene rings fused together. Silicon, on the other hand, does not form an analogue of carbon because of the following reason:
Due to bigger size and smaller electronegativity of Si than C, it does not undergo sp2-hybridisation and hence it does not form p-p double bonds needed for graphite like structure. Instead, it prefers to undergo only sp3-hybridisation and hence silicon has diamond like three dimensional network
Boric acid is an acid because its molecule
(a) contains replaceable H+ ion
(b) gives up a proton.
(c)accepts OH–from water releasing proton.
(d) combines with proton from water molecule.
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 cone.H2SO4is added to a hot solution of X, white crystal of an acid Z separates out.
Describe the shapes of BF3 and BH4–. Assign the hybridisation of boron in these species.
Boric acid is polymeric due to
(a) its acidic nature (b) the presence of hydrogen bonds
(c) its monobasic nature (d) its geometry
The reason for small radius of Ga compared to Al is_________ .
(a) poor screening effect of d and f orbitals
(b) increase in nuclear charge
(c) presence of higher orbitals
(d) higher atomic number
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.
Ionisation enthalpy (∆ tH1 kJ mol-1) for the elements of Group 13 follows the order.
(a) B > A1 > Ga > In > T1
(b) B < A1 < Ga< In
(d) B > A1 < Ga > In < T1
Give the chemical reactions as an evidence for each of the following observations.
(i) Tin (II) is a reducing agent whereas lead (II) is not.
(ii) Gallium (I) undergoes disproportionation reaction.
Give the chemical reaction as an evidence for each of the following observations.
(i) Tin (II) is a reducing agent whereas lead (II) is not.
(ii) Gallium (I) undergoes disproportionation reaction.
Which of the following statements are correct?
(a) Fullerenes have dangling bonds.
(b) Fullerenes are cage-like molecules.
(c) Graphite is thermodynamically most stable allotrope of carbon.
(d) Graphite is slippery and hard and therefore used as a dry lubricant in
Carbon and silicon both belong to the group 14, but in spite of the stoichiometric similarity, the dioxides (i.e., carbon dioxide and silicon dioxide) differ in their structures. Comment.
In some of the reactions thallium resembles aluminium, whereas in others it resembles with group 1 metals. Support this statement by giving some evidences.
Explain why is there a phenomenal decrease in ionization enthalpy from carbon to silicon.
What happens when
(i) Quick lime is heated with coke?
(ii) Carbon monoxide reacts with Cl2?
Match the species given in Column I with properties given in Column II.
| Column I | Column II |
| (i) Diborane | (a) Used as a flux for soldering metals |
| (ii) Gallium ‘ | (b) Crystalline form of silica |
| (iii) Borax | (c) Banana bonds |
| (iv) Aluminosilicate | (d) Low melting, high boiling, useful for measuring high temperatures |
| (v) Quartz | (e) Used as catalyst in petrochemical industries |
Match the species given in Column I with the hybridisation given in Column II.
| Column I | Column II |
| (i) Boron in [B(OH)4]" | (a) sp2 |
| (ii) Aluminium in [A1(H20)6]3+ | (b) sp3 |
| (iii) Boron in B2H6 | (c) sp3d2 |
| (iv) Carbon in Buckminsterfullerene | |
| (v) Silicon in SiO44- | |
| (vi) Germanium in [GeCl6]2- |
What are electron deficient compounds? Are BCl3 and SiCl4 electron deficient species? Explain.
Give reasons for the following:
(a) CCl4 is immiscible in water, whereas SiCl4 is easily hydrolysed.
(b) Carbon has a strong tendency for catenation compared to silicon.
Thermodynamically the most stable form of carbon is
(a)diamond (b) graphite (c) fullerenes (d) coal
The linear shape of C02 is due to ______ .
(a) sp3 hybridisation of carbon
(b) sp hybridisation of carbon
(c) pπ-pπ bonding between carbon and oxygen
(d) sp2 hybridisation of carbon
The+1 oxidation state in group 13 and +2 oxidation state in group 14 becomes more and more stable with increasing atomic number. Explain.
Match the species given in Column I with the properties mentioned in Column II.
| Column I | Column II |
| (i) BF4– | (a) Oxidation state of central atom is +4 |
| (ii) A1C13 | (b) Strong oxidising agent |
| (iii) SnO | (c) Lewis acid |
| (iv) Pb02 | (d) Can be further oxidised |
| (e) Tetrahedral shape |
Assertion (A): If aluminium atoms replace a few silicon atoms in three dimensional network of silicon dioxide, the overall structure acquires a negative charge.
Reason (R): Aluminium is trivalent while silicon is tetravalent.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) Both A and R are not correct.
(d) A is not correct but R is correct.
Assertion (A): Silicones are water repelling in nature.
Reason (R): Silicones are organosilicon polymers, which have (-R2SiO-) as repeating unit.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) Both A and R are not correct. ‘
(d) A is not correct but R is correct.
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.
Give reason.
(i) C and Si are always tetravalent but Ge, Sn, Pb show divalency.
(ii) Gallium has higher ionization enthalpy than Al. Explain.
The geometry of a complex species can be understood from the knowledge of type of hybridisation of orbitals of central atom. The hybridisation of orbitals of central atom in [B(OH)4]- and the geometry of the complex are respectively
(a) sp3, tetrahedral
(b) sp3, square planar
(c) sp3d2, octahedral
(d) dsp2, square planar
Me3SiCl is used during polymerisation of organo silicones because
(a) the chain length of organo silicone polymers can be controlled by adding Me3
(b) Me3SiCl blocks the end terminal of silicone polymer.
(c) Me3SiCl improves the quality and yield of the polymer. –
(d) Me3SiCl acts as a catalyst during polymerization.
Assertion (A): Silicones are water repelling in nature.
Reason (R): Silicones are organosilicon polymers, which have (-R2SiO-) as repeating unit.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) Both A and R are not correct. ‘
(d) A is not correct but R is correct.