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1. Discuss the pattern of variation in the oxidation states of B to Tl and (ii) C to Pb.
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Solution:
The electronic configurations of these elements are ns2p1. Therefore these elements they should
show +1 and +3 oxidation states. But in real practice, boron shows only +3 oxidation state while
other elements show +1 as well as +3 oxidation states. The +1 oxidation state becomes more and
more stable on moving down the group from B to Tl .
The electronic configurations of these elements are ns2p2 .Of these elements indicate that these
elements can lose or gain 4 electrons to attain inert gas configuration. It means, theoretically they
can form M4+ or M4- ions. However, due to high ionization energies, these elements do not form
stable M4+ ions. Moreover, due to low electron affinity, these elements do not exhibit much tendency
to form M4- ions by gaining four electrons, although C4- ions in thought to be present in Be2C and
Al4C3. Hence these elements, particularly C and Si, form mainly covalent bonds in which these
elements exhibit oxidation state of +4. The last of these elements (Ge,Sn and Pb) have a tendency
to form M4+as well as M2+ ions. The +2 oxidation state arises due to inert pair effect which becomes
more pronounced in the heavier elements.
2. How can you explain higher stability of BCl3 as compared to TlCl3 ?
Solution:
The reason for the decrease in the stability of +3 oxidation state is the increase in the inert pair
effects from Ga to Tl. Be and Al do not show inert pair effect, hence exist in +3 oxidation state, but
on moving from Ga to Tl inert pair effect will increase, hence out ns2 np1 only p1 will take part in
bond formation in Tl and ns2 pair will be inert. Therefore, +3 oxidation state in TI will not be stable.
Hence it explains higher stability of BCl3 as compared to TICl3.
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3. Why does boron triflouride behave as a Lewis acid ?
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Solution:
BF3 has six electrons in the outermost orbit in boron atom and thus it can accept a pair of electrons
from a donor atom like N,O,P or S to complete its octet. Hence boron fluorides behave as Lewis
acids.
BCl3+ 3H2O →€B(OH)3 + 3HCl.
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Solution:
Boron trichloride gives boric acid when treated with water.
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4. Consider the compounds, BCl3 and CCl4. How will they behave with
water ? Justify.
Carbon tetrachloride gives phosgene with superheated steam in presence of iron or copper.
CCl4 + H2O → COCl2 + 2HCl
(steam) Phosgene
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5. Is boric acid a protic acid? Explain.
Solution:
Boric acid is not a protonic acid but acts as a Lewis acid by accepting electrons from a hydroxyl ion.
B(OH)3 +2HOH →€[B(OH)4]- + H3O+
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6. Explain what happens when boric acid is heated.
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Solution:
On heating, boric acid it gives metaboric acid at 1000C, tetraboric acid at 1600 C and boron trioxide
at red heat.
7. Describe the shapes of BF3 and BH4 –. Assign the hybridisation of boron in these
species.
Solution:
BF3= ½[3=3-0+0] =6/2=3 =sp2
BF3 has a trigonal planar structure.
BH-4 has tetrahedral structure.
8. Write reactions to justify amphoteric nature of aluminium.
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Reaction with base
2Al + 2NaOH + 2H2O → 2NaAlO2 +3H2
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Solution:
Reaction with acid
2Al +6HCl →€2AlCl3 +3H2
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BH4- =1/2[3+4-0+1] =8/2 =sp2
The above reactions proves the amphoteric nature of aluminium.
9. What are electron deficient compounds? Are BCl3 and SiCl4 electron deficient species?
Explain.
Solution:
2
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Electron deficient compounds are the compounds in which the central atom in their molecules has
the need or tendency to take up electron pairs. The electron deficient compounds are also called
Lewis acids.
Yes, Both BCl3 and SiCl4 are electron deficient species. Boron atom has a vacant 2p orbital. Si atom
has vacant 3d orbitals. Both these atoms can take up electron pairs from electron donor species.
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10. What is the state of hybridisation of carbon in
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(a) CO32-
(b) diamond
(c) graphite?
Solution:
(a) In CO32- ion the carbon is in its sp2 hybridisation state. It forms π bonds.
(b) In diamond carbon is in its sp3 hybridized state.
(c) In graphite carbon is in sp2 hybridized state.
11. Explain the difference in properties of diamond and graphite on the basis of their
structures.
Solution:
.in
12. 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.
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6. It has is a bad conductor electricity.
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5. It is very hard and have high melting point.
Graphite
1. In graphite, each carbon atom undergoes
sp2 hybridisation and is covalently bonded to
three other carbon atoms.
2. The fourth electron in each carbon atom
results in the formation of πbonds. Thus, tt has
hexagonal rings in two dimensions.
3. The C- covalent distance in rings is 142 pm
indicating strong bonding. These arrays of rinds
form layers.
4. The layers in graphite are separated by a
distance of 340pm. The large distance between
these layers indicates that only weak van der
waals forces hold these layers together.
5. Graphite is soft and the layer's move over one
another. Therefore it is used as a lubricant.
6. It is good conductor of electricity.
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Diamond
1. Each carbon atom is sp3 hybridised and is
bonded to four other carbon atoms by single
covalent bonds.
2. Diamond has a network structure of a very
large numbers of carbon atoms bonded to each
other.
3. Each carbon atom lies at the tetrahedron and
other four carbon atoms are present at the
corners of the tetrahedron.
4. C-C bond length is equal to 154 pm. Therefore,
there is a three dimensional network of strong
covalent bonds in diamond.
Solution:
PbCl2 + Cl2 → PbCl4.
Due to inert pair effect, lead shows oxidation state of +2. Hence, Lead(II) chloride is more stable
3
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than PbCl4.
Lead is known not to form the iodide Pbl4, due to the fact that Pb4+ ion a strong oxidising agent
while Br- and I- ions are is strong reducing agents. Thus Pb4+ ion cannot survive in presence of Bror I- ion and is reduce to Pb2+ ion.
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13. Suggest reasons why the B–F bond lengths in BF3 (130 pm) and BF4 -(143 pm) differ.
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Solution:
BF3 acquires a partial double bond character due to the pπ -pπ back bonding. Therefore the B-F bond
length becomes lesser than the value expected for a single covalent B-F bond.
14. If B–Cl bond has a dipole moment, explain why BCl3 molecule has zero dipole
moment.
Solution:
Molecules with symmetrical and linear geometries have zero dipole moments because being a vector
in nature, the dipoles of different bonds cancel with another.
Hence BCl3 has zero dipole moment.
15. Aluminium forms [AlF6]3- where as BF6 is not formed. Why?
Solution:
Aluminium can expand its octet to from bonds with six fluoride ions due to the presence of vacant d
orbitals, whereas Boron cannot since they do not have d-orbitals.
16. Suggest a reason as to why CO is poisonous.
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Solution:
CO combines with haemoglobin, the oxygen- carrier of the blood, to from a stable compound
carboxyhameoglobin. With the result the oxygen transportation is disturbed and thus tissues do not
get necessary oxygen and ultimately death occurs.
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17. How is excessive content of CO2 responsible for global warming?
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Solution:
The carbon dioxide absorbs longer wave- length infra- red rays reflected by the earth. A blanket of
carbon dioxide gas in the atmosphere traps the infra-red rays in the atmosphere. Since infra- red
rays have a heating effect, so atmosphere gets heated up producing the green house effect. Thus,
the role of carbon dioxide gas in the green house effect is to trap the infra – red heat rays and
prevent them from leaving the atmosphere. Thus excess content of CO2 is responsible for global
warming.
18. Explain structures of diborane and boric acid.
Solution:
Structure of Diborane
The four terminal hydrogen atoms and the two boron atoms lie in the same plane. Above and below
this plane, there are two bridging hydrogen atoms shown with the help of dotted lines. Each boron
4
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atom forms four bonds even though it has only three electrons. The terminal B-H bonds are regular
covalent bonds but the bridge B-H bonds are different.
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Structure of boric acid
Planar BO3 units are linked to one another through H- atoms. The H atoms constitute covalent bond
with one unit and hydrogen bond with the other unit.
19. What happens when
(a) Borax is heated strongly,
(b) Boric acid is added to water,
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(c) Aluminium is treated with dilute NaOH, BF3 is reacted with ammonia ?
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Solution:
(a) On heating borax loses water and swells into a white mass, which on further heating melts to
form a transparent glassy solid called borax glass and borax bead
Na2B4O7 + 10H2O
2NaBO2 + B2O3
Sodium meta borate
(b) Boric acid is slightly soluble in cold water and is readily soluble in hot water
.in
heat
Na2B4O7
on
heat
Na2B4O7. 10H2O
2Al + 6NaOH → 2Na3AlO3 + H2
Sodium aluminate
Aluminium dissolves in dilute NaOH and hydrogen is liberated.
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20. Explain the following reactions
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(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; Hydrated alumina is treated with aqueous NaOH solution.
Solution:
Dimethyl dichloro silane is formed
Cu
2CH3Cl + Si
(CH3)2SiCl2
dimethyl dichloro silane
(a) SiO2 + 4HF
(b) ZnO + CO
SiF4 +2H2O
Zn + CO2
(c) Al2O3.2H2O + 2NaOH
2NaAlO2 +3H2O.
21. Give reasons :
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(i) Conc. HNO3 can be transported in aluminium container.
(ii) A mixture of dilute NaOH and aluminium pieces is used to open drain.
(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.
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(iv) Diamond is used as an abrasive.
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(iii) Graphite is used as lubricant.
Solution:
(i) Aluminium becomes passive on coming in contact with conc. HNO3 due to the coating of
aluminium oxide and thus conc. HNO3 caan be transported in an aluminium container.
(ii) Aluminium dissolves in dil NaOH with the evolution of H2. This hydrogen helps to open drains.
2Al + 2NaOH + 2 H2O → 2NaAlO2 +3H2
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(iii) Graphite has sp2 hybridized carbon with a layer structure. Due to wide separation and weak
inter- layer bonds, the two adjacent layers can easily slide over each other. This makes graphite act
as a lubricant.
(iv) Diamond is the hardest substance known and thus used as abrasive and for cutting glass.
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(v) Aluminium alloys are used to make aircraft body because they are light and yet strong and also
resist corrosion.
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(vi) Aluminium utensils should not be kept in water overnight because aluminium is readily corroded
by water.
(vii) Aluminium is cheaply available on a weight –to weight basis, the electrical conductivity of
aluminium is twice that of copper. Hence, aluminium wire is used to make transmission cables.
22. Explain why is there a phenomenal decrease in ionization enthalpy from carbon to
silicon?
Solution:
Silicon (Atomic radius =118 pm) has large atomic size as compared to that of carbon (atomic radius
–77pm). Therefore, the release of electrons is comparatively easy. From silicon to germanium, the
decrease in ionization enthalpy is not so sharp because atomic radii of the elements increase
gradually.
23. How would you explain the lower atomic radius of Ga as compared to Al?
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Solution:
The decrease in atomic radius of gallium as compared to that of aluminium may be attributed to the
presence of ten elements of the first transition series (Z = 21 to 30) which have electrons in the 3d
orbitals. Since d- orbitals have large size than the p- orbitals, the intervening electrons do not have
sufficient shielding effect to counter the increase in the nuclear charge. Therefore, the effective
nuclear charge in case of Ga is less than the expected value. This decreases its atomic radius which
otherwise is expected to increase.
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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?
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Solution:
Allotropy is the ability of a chemical to exhibit a number of different and physically distinct forms in
its pure elemental state. Carbon, for instance can exist as graphite, diamond and fullerene. Typically,
elements capable of variable coordination numbers and/or oxidation states tend to exhibit greater
numbers of allotropic forms. Another contributing factor is the ability of an element to catenate. .
The term allotropes may also be used to refer to the molecular forms of an element (such as a
diatomic gas), even if there is only one such additional form.
7
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Structure of diamond
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Structure of graphite
In diamond, each carbon atom is sp3 hybridized with C-C bond length 54 pm (1.54 A0). While in
graphite, these are sp2 hybridized with C-C bond length of 142 p, (1.42A0). Due to sp3 hybridization,
the carbon atoms in diamond are closely packed in space with no free electrons available. Thus,
diamond is very hard and is also a poor conductor of electricity. But due to sp2 hybridization, one punhybridized orbital of the carbon atom in graphite is not involved in the bond formation. The
corresponding electron being a mobile electron, will
8
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25. Classify following oxides as neutral, acidic, basic or amphoteric:
CO, B2O3, SiO2, CO2, Al2O3, PbO2, Tl2O3. Write suitable chemical equations to show their
nature.
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Solution:
CO - Neutral
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B2O3 - Acidic
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B2O3 + CuO → Cu(BO2)2
SiO2 - Acidic
SiO2 + 2NaOH → Na2SiO3+H2O
Al2O3 - Amphoteric
Al2O3 +6HCl → 2AlCl3 +3H2O
Al2O3 + 2NaOH → 2NaAlO2 + H2O
PbO2 - Amphoteric
PbO2 + 4HCl → PbCl4 + 2H2O
PbO2 + 2NaOH → Na2PbO3 + H2O
Sodium plumbate
Tl2O3 - Basic
Tl2O3 + 3H2SO4 → Tl2 (SO4)3 + 3H2O
Thallium sulphate
(b) Both of them tarnished in air due to the formation of oxides.
(ii)The fluorides of Al and Tl are ionic and have get tarnished high melting -points.
Resemblance of thallium with group 1 elements:
(a) Tl (OH) is soluble in water giving strong alkaline solution very similar to NaOH.
.in
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(a) Both of them have outer electronic configuration of ns2np1
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Solution:
(i) Similarities of thallium with aluminium
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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.
(b) Like alkali metals, thallium (Tl) forms double salts with aluminium salts e.g.
K2SO4. Al2(SO4)3. 24H2O and
Potash alum.
Tl2SO4. Al2(SO4)3. 24H2O
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Thallus alum.
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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.
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Solution:
X is aluminium
Al3+ + 3OH- → Al(OH)3
(From sodium (A)
hydroxide)
Hence (A) is aluminium hydroxide
Al(OH)3 + NaOH → NaAlO2 + 2H2O
(A)
(B)
Hence B is NaAlO2 (Sodium meta aluminate)
Al(OH)3 + 3HCl → 2AlCl3 + 3H2O
(A)
(C)
Hence C is AlCl3 which is soluble in dilute HCl
2Al(OH)3 → Al2O3 +3H2O
(A)
(D)
Hence (D) is Al2O3 (aluminia) Alumina is used to extract metal aluminium.
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28. What do you understand by (a) inert pair effect (b) allotropy and catenation?
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Solution:
(a) Inert pair effect
On moving down the group in p block elements two electrons present in the s- orbital become inert
and only the electrons in the p orbital take part in chemical combination. This is known as inert pair
effect which is a special feature of p –block elements.
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(b) Allotropy
Allotropy is the existence of an element in two or more different forms in the same chemical state.
The different forms of the element are called allotropes. All the allotropes of an element have similar
chemical properties but different physical properties.
(c) Catenation
Carbon atoms can link with one another by means of covalent bonds to form long chains or rings of
carbon atoms. This property of carbon element due to which its atoms can join with one another to
form a long chain is called Catenation.
29. A certain salt X, gives the following results.
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(i) Its aqueous solution is alkaline to litmus.
(ii) It swells up to a glassy material Y on strong heating.
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(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.
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Solution:
X is Borax, Na2B4O7.
(i) An aqueous solution of Borax is alkaline to litmus because the alkali (NaOH) is stronger than the
acid (H3BO3) formed as a result of hydrolysis.
Na2B4O7 + 7H2O → 2NaOH + 4H3BO3+ (weak acid)
(X)
(strong alkali)
(ii) When borox (X) is heated, it swells up and forms a glassy material
Heat
Na2B4O7
2NaBO2 + B2O3
(X)
glass material (Y)
(iii) When conc. H2SO4 is added to a hot solution of X (Na2B4O7), white crystals of Boric acid (Z)
separates out
Na2B4O7 + H2SO4 + 5H2O → 4H3BO3 + Na2SO4
(X)
(Z)
1000C
(iv) H3BO3
HBO2 + H2O
Metaboric acid
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(iii) 2NaH + B2H6 → 2Na+[BH4](sodium borohydride)
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(ii) B2H6 + 6H2O → 2H3BO3 +6H2O
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Solution:
(i) 8BF3 + 6LiH → B2H6 + 6LiBF4
Diborane Lithium borofluoride
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30. Write balanced equations for:
(i) BF3 + LiH _
(ii) B2H6 + H2O _
(iii) NaH + B2H6 _
(iv) H3BO3 . ._
(v) Al + NaOH _
(vi) B2H6 + NH3 _
(v) 2Al + 6 NaOH → 2Na3AlO3 +3H2
Sodium aluminate
High temp
(vi) 3B2H6 + 6NH3
2B3N3H6 + 12H2
Borazole
11
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31. Give one method for industrial preparation and one for laboratory preparation of CO
and CO2 each.
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Solution:
Carbon monoxide: it is formed by incomplete combustion of carbon and carbon containing fuels.
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2C +O2 → 2CO (Industrial preparation)
Laboratory Method: In the laboratory, carbon monoxide is obtained by dehydration of formic acid
with conc. H2SO4 at 373 K.
Carbon dioxide: By heating carbonates heavy metals, like calcium, magnesium, zinc etc. or
bicarbonates of alkali metals.
CaCO3 → CaO + CO2(g) (Industrial preparation)
Laboratory preparation: By the action of acids on carbonates or bicarbonates
CaCO3 + 2HCl → CaCl2 + H2O + CO2 (laboratory preparation)
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