The p-Block Elements
Book Solutions1
Discuss the pattern of variation in the oxidation states of
(i) B to Tl and (ii) C to Pb.
Answer
(i) B to Tl
The electric configuration of group 13 elements is ns2 np1. Therefore, the most common oxidation state exhibited by them should be +3. However, it is only boron and aluminium which practically show the +3 oxidation state. The remaining elements, i.e., Ga, In, Tl, show both the +1 and +3 oxidation states. On moving down the group, the +1 state becomes more stable. For example, Tl (+1) is more stable than Tl (+3). This is because of the inert pair effect. The two electrons present in the s-shell are strongly attracted by the nucleus and do not participate in bonding. This inert pair effect becomes more and more prominent on moving down the group. Hence, Ga (+1) is unstable, In (+1) is fairly stable, and Tl (+1) is very stable. The stability of the +3 oxidation state decreases on moving down the group.
(ii) C to Pb
The electronic configuration of group 14 elements is ns2 np2. Therefore, the most common oxidation state exhibited by them should be +4. However, the +2 oxidation state becomes more and more common on moving down the group. C and Si mostly show the +4 state. On moving down the group, the higher oxidation state becomes less stable. This is because of the inert pair effect. 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.
2
How can you explain higher stability of BCl3 as compared to TlCl3?
Answer
Boron and thallium belong to group 13 of the periodic table. In this group, the +1 oxidation state becomes more stable on moving down the group. BCl3 is more stable than TlCl3 because the +3 oxidation state of B is more stable than the +3 oxidation state of Tl. In Tl, the +3 state is highly oxidising and it reverts back to the more stable +1 state.
3
Answer
The B atom in BF3 has only 6 electrons in the valence shell and thus needs two more electrons to complete its octet. Therefore, it easily accepts a pair of electrons from nucleophiles such as F–, (C2H5)2O, RCH2OH etc. and thus behaves as a Lewis acid.4
Answer
The B atom in BCl3 has only six electrons in the valence shell and hence is an electron– deficient molecule. It easily accepts a pair of electrons donated by water and hence BCl3 undergoes hydrolysis to form boric acid (H3BO3) and HCl.
BCl3+3H2O→3HCl+B(OH)3
In contrast, C atom in CCl4 has 8 electrons in the valence shell. Therefore, it is an electron– precise molecule and hence neither accepts nor donates a pair of electrons. In other words, it does not accept a pair of electrons from H2O molecule and hence CCl4 does not undergo hydrolysis in water.
CCl4+H2O→No Reaction
5
Answer
Boric acid is a weak monobasic acid which behaves as a Lewis acid. So, it is not a protic acid. It behaves as an acid by accepting a pair of electrons from OH- ion.6
Answer
7
Answer
In BF3, boron is sp2–hybridized and, therefore, BF3 is a planar molecule. On the other hand, in [BH4]– boron is sp3–hybridized and hence [BH4]– is a tetrahedral species.
8
Write reactions to justify amphoteric nature of aluminium.
Answer
9
What are electron deficient compounds? Are BCl3 and SiCl4 electron deficient species? Explain.
Answer
In an electron-deficient compound, the octet of electrons is not complete, i.e., the central metal atom has an incomplete octet. Therefore, it needs electrons to complete its octet.
(i) BCl3
BCl3 is an appropriate example of an electron-deficient compound. B has 3 valence electrons. After forming three covalent bonds with chlorine, the number of electrons around it increases to 6. However, it is still short of two electrons to complete its octet.
(ii) SiCl4
The electronic configuration of silicon is ns2 np2. This indicates that it has four valence electrons. After it forms four covalent bonds with four chlorine atoms, its electron count increases to eight. Thus, SiCl4 is not an electron-deficient compound.
10
Answer
11
What is the state of hybridisation of carbon in
(a) CO32–
(b) diamond
(c) graphite?
Answer
(a) CO32-:
C in CO32- is sp2 hybridised and is bonded to three oxygen atoms.
(b) Diamond:Each carbon in diamond is sp3 hybridised and is bound to four other carbon atoms.
(c) Graphite: Each carbon atom in graphite is sp2 hybridised and is bound to three other carbon atoms.
12
Explain the difference in properties of diamond and graphite on the basis of their structures.
Answer
Diamond | Graphite |
It has a crystalline lattice. | It has a layered structure. |
In diamond, each carbon atom is sp3hybridised and is bonded to four other carbon atoms through a σ bond. | In graphite, each carbon atom is sp2 hybridised and is bonded to three other carbon atoms through a σ bond. The fourth electron forms a π bond. |
It is made up of tetrahedral units. | It has a planar geometry. |
The C–C bond length in diamond is 154 pm. | The C–C bond length in graphite is 141.5 pm. |
It has a rigid covalent bond network which is difficult to break. | It is quite soft and its layers can be separated easily. |
It acts as an electrical insulator. | It is a good conductor of electricity. |
13
Rationalise the given statements and give chemical reactions:
(i) Lead(II) chloride reacts with Cl2 to give PbCl4.
(ii) Lead(IV) chloride is highly unstable towards heat.
(iii) Lead is known not to form an iodide, PbI4.
Answer
(i) Due to inert pair effect, Pb is more stable in + 2 than in + 4 oxidation state. Therefore, lead (II) chloride is more stable than lead (IV) chloride and hence lead (II) chloride does not react with Cl2 to form lead (IV) chloride.
(ii) Due to greater stability of +2 over +4 oxidation state because of inert pair effect, lead (IV) chloride on heating decomposes to give lead(II) chloride and Cl2.
(iii) I– being a powerful reducing agent reduces Pb4+ to Pb2+ in solution. Thus, PbI2 is formed.
14
Suggest reasons why the B–F bond lengths in BF3 (130 pm) and BF4- (143 pm) differ.
Answer
BF3 is a planar molecule in which B is sp2 -hybridized. It has an empty 2p-orbital. F–atom has three lone pairs of electrons in the 2p-orbitals. Because of similar sizes, pπ-pπ back bonding occurs in which a lone pair is transferred from F to B as shown below:
As a result of this back bonding, B–F bond acquires some double bond character. In contrast, in [BF4]– ion, B is sp3 -hybridized and hence does not have an empty p-orbital available to accept the electrons donated by the F atom. Consequently, in [BF4]–, B–F is a purely single bond. Since double bonds are shorter than single bonds. Therefore, the B–F bond length in BF3 is shorter (130 pm) than B–F bond length (143 pm) in [BF4]–.
15
Answer
Due to electronegativity difference between B (E.N. = 2.0) and Cl (E.N. = 3.0), the B–Cl bond is polar and hence has a finite dipole moment. The overall dipole moment of a molecule, however, depends upon its geometry. Now BCl3 is a planar molecule in which the three B–Cl bonds are inclined at an angle of 120°. Therefore, the resultant of two B–Cl bonds is cancelled by equal and opposite dipole moment of the third B–Cl bond as shown below:
As a result, overall dipole moment of BCl3 is zero.
16
Answer
(i) Anhydrous HF is a covalent compound and is strongly H– bonded. Therefore, it does give F– ions hence AlF3 does not dissolve in HF. In contrast, NaF being an ionic compound contains F– ions and hence combines with AlF3 to form the soluble complex.
17
Answer
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.
18
Answer
Carbon dioxide is a very essential gas for our survival. However, an increased content of CO2 in the atmosphere poses a serious threat. An increment in the combustion of fossil fuels, decomposition of limestone, and a decrease in the number of trees has led to greater levels of carbon dioxide. Carbon dioxide has the property of trapping the heat provided by sunrays. Higher the level of carbon dioxide, higher is the amount of heat trapped. This results in an increase in the atmospheric temperature, thereby causing global warming.
19
Answer
In diborane structure 2 boron and 4 terminal hydrogen atoms (Ht) lie in one plane, while the other two bridging hydrogen atom lie in a plane perpendicular to the plane of boron atom. Again, of the two bridging hydrogen atoms, one H-atom lies above the plane and other lies below the plane. The terminal bonds are regular two-centre two-electron (2c – 2e) bonds, while two bridging (B – H – B) bonds are three-centre to-eletron (3c – 2e) bonds.
Structure: Due to sp2 -hybridisation of boron, boric acid is a planar molecule and due to hydrogen bonding between different molecules, boric acid has layer structure as shown below
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?
Answer
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.
Answer
22
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.
Answer
(i) Conc HNO3 reacts with aluminium to form a protective layer of aluminium oxide on its surface which prevents it from further reaction with aluminium.
(ii) NaOH and Al react to form sodium tetrahydroxo aluminate (III) and hydrogen gas. The pressure of the hydrogen gas produced is used to open the blocked drains.
(iii) Graphite has layered structure in which the different layers are held together by weak vander waal's forces and therefore can easily slip over one another. Hence, making graphite a good lubricant.
(iv) Carbon is sp3 hybridised in diamond. Each carbon atom is bonded to four other carbon atoms with the help of strong covalent bond. These covalent bonds are present throughout the surface, giving it a very rigid 3-D structure which makes diamond the hardest known substance on earth. Hence making it suitable to used as an abrasive.
(v) Aluminium alloys such as duralumin is light, tough and resistant to corrosion and hence is used to make aircraft body.
(vi) Al reacts with H2O and dissolved O2 to form a thin film of Aluminium oxide.
A very small amount of Al2O3 may dissolve to give a few ppm of Al3+ ions in the solution. Since Al3+ ions are injurious to health, therefore, drinking water should not be kept in Aluminium utensils overnight.
(vii) Aluminium is a good conductor of electricity. On weight to weight basis, conductivity of Aluminium is twice as Cu. Therefore, it is used in transmission cables.
23
Answer
Ionisation enthalpy of carbon (the first element of group 14) is very high (1086 kJ/mol). This is expected owing to its small size. However, on moving down the group to silicon, there is a sharp decrease in the enthalpy (786 kJ). This is because of an appreciable increase in the atomic sizes of elements on moving down the group.
24
Answer
Due to poor shielding of the valence electrons of Ga by the inner 3d–electrons, the effective nuclear charge of Ga is greater in magnitude than that of Al. As a result, the electrons in gallium experience greater force of attraction by the nucleus than in Al and hence atomic size of Ga (135 pm) is slightly less than that of Al (143 pm).25
Answer
The phenomenon of existence of an element in two or more forms having different physical properties but identical chemical properties are called allotropy and the different forms are called allotropes.
In diamond, carbon is sp3 hybridized. Each carbon is tetrahedrally linked to four neighbouring carbon atoms through four strong C – C, sp3 σ bonds. This network extends in three dimensions and is very rigid. Diamond is the purest form of carbon.
Since diamond exists as a 3-D network solid, it is the hardest substance known with high density and melting point. It is a bad conductor of electricity as all the electrons are firmly held in C–C σ bonds. Because of its high refractive index, diamond can reflect and refract light. It is therefore transparent substance.
Graphite has carbon having sp2 hybridization. Each carbon is thus linked to three other carbon atoms forming hexagonal rings. Thus graphite has a two dimensional sheet like structure. The various sheets are held together by weak van der Waal's forces of attraction. The bond lengths in graphite are smaller than diamond. Since any two successive layers are held together by weak forces of attraction, one layer can slip over another. This makes graphite a good lubricating agent. As only three electrons
26
Answer
(1) CO = Neutral
(2) B2O3 = Acidic
Being acidic, it reacts with bases to form salts. It reacts with NaOH to form sodium metaborate.
(3) SiO2 = Acidic
Being acidic, it reacts with bases to form salts. It reacts with NaOH to form sodium silicate.
(4) CO2 = Acidic
Being acidic, it reacts with bases to form salts. It reacts with NaOH to form sodium carbonate.
(5) Al2O3 = Amphoteric
Amphoteric substances react with both acids and bases. Al2O3 reacts with both NaOH and H2SO4.
(6) PbO2 = Amphoteric
Amphoteric substances react with both acids and bases. PbO2 reacts with both NaOH and H2SO4.
(7) Tl2O3 = Basic
Being basic, it reacts with acids to form salts. It reacts with HCl to form thallium chloride.
27
Answer
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. This happens because of the inert pair effect. 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.
28
Answer
As the metal given in question on treatment with NaOH given a white ppt (A) which dissolves in excess of NaOH to given metal complex (B) therefore the given metal must be Al. The reactions may given as
29
What do you understand by (a) inert pair effect (b) allotropy and (c) catenation?
Answer
(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. In case of group 13 elements, the electronic configuration is ns2 np1and 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
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. For example, carbon exists in three allotropic forms: diamond, graphite, and fullerenes.
(c) Catenation
The atoms of some elements (such as carbon) can link with one another through strong covalent bonds to form long chains or branches. This property is known as catenation. It is most common in carbon and quite significant in Si and S.
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.
Answer
(i) Since the aqueous solution of X is alkaline to litmus, here it must be a salt of strong base and weak acid. Since (X) on strong heating swells up to a glassy material (Y). i.e. giving Borax Bead test thus it is now confirm that the compound X is Borax.
(iii) When conc. H2SO4 is added to a hot solution of X, white crystal of an acid separate out, thus Z must be ortho boric acid.
31
Write balanced equations for:
(i)BF3 + LiH →
(ii) B2H6 + H2O →
(iii)NaH + B2H6 →
(iv) H3BO3 →
(v) Al + NaOH →
(vi) B2H6 + NH3 →
Answer
32
Answer
33
An aqueous solution of borax is
(a) Neutral
(b) Amphoteric
(c) Basic
(d) Acidic
Answer
(c) Borax is a salt of a strong base (NaOH) and a weak acid (H3BO3). It is, therefore, basic in nature.34
Boric acid is polymeric due to
(a) its acidic nature
(b) the presence of hydrogen bonds
(c) its monobasic nature
(d) its geometry
Answer
Boric acid is polymeric due to the presence of H–bonds. Therefore, option (b) is correct.35
The type of hybridisation of boron in diborane is
(a) sp
(b) sp2
(c) sp3
(d) dsp2
Answer
(c) Boron in diborane is sp3 hybridised.36
Thermodynamically the most stable form of carbon is
(a) diamond
(b) graphite
(c) fullerenes
(d) coal
Answer
(b) Graphite is thermodynamically the most stable form of carbon.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
Answer
Due to inert pair effect, elements of group 14 exhibit oxidation states of +2 and +4. Thus, option (b) is correct.Group 14 element | Oxidation state |
C | +4 |
Si | +4 |
Ge, Sn, Pb | +2, +4 |
38
Answer
