NCERT Solutions for Class 11 Chemistry Chapter 11
The p-Block Elements Class 11
Chapter 11 The p-Block Elements Exercise Solutions
Exercise : Solutions of Questions on Page Number : 323
Discuss the pattern of variation in the oxidation states of
Answer :
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.
Group 13 element |
Oxidation state |
B |
+3 |
Al |
+3 |
Ga, In, Tl |
+1, +3 |
The stability of the +3 oxidation state decreases on moving down the group.
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.
Group 14 element |
Oxidation state |
C |
+4 |
Si |
+4 |
Ge, Sn, Pb |
+2, +4 |
How can you explain higher stability of BCl3 as compared to TlCl3?
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. BCl3is more stable than TlCl3because the +3 oxidation state of B is more stable than the
Why does boron trifluoride behave as a Lewis acid?
Consider the compounds, BCl3 and CCl4. How will they behave with water? Justify.
Being a Lewis acid, BCl3 readily undergoes hydrolysis. Boric acid is formed as a result.
CCl4completely resists hydrolysis. Carbon does not have any vacant orbital. Hence, it cannot accept electrons from water to form an intermediate. When CCl4and water are mixed, they form separate layers.
Is boric acid a protic acid? Explain.
Boric acid is not a protic acid. It is a weak monobasic acid, behaving as a Lewis acid.
It behaves as an acid by accepting a pair of electrons from –OH ion.
Explain what happens when boric acid is heated.
4
Describe the shapes of BF3and BH -.
Assign the hybridisation of boron in these species.
(i) BF3
As a result of its small size and high electronegativity, boron tends to form monomeric covalent halides. These halides have a planar triangular geometry. This triangular shape is formed by the overlap of three sp2hybridised orbitals of boron with the sporbitals of three halogen atoms. Boron is sp2hybridised in BF3.
(ii)
4
BH -
|
Write reactions to justify amphoteric nature of aluminium.
What are electron deficient compounds? Are BCl3and SiCl4electron deficient species? Explain.
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
BCl3is 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.
Write the resonance structures of and . Answer :
(b)
What is the state of hybridisation of carbon in (a) (b) diamond (c) graphite?
The state of hybridisation of carbon in:
C in is sp2hybridised and is bonded to three oxygen atoms.
Each carbon in diamond is sp3 hybridised and is bound to four other carbon atoms.
Explain the difference in properties of diamond and graphite on the basis of their structures.
Diamond |
Graphite |
|
It has a crystalline lattice. |
It has a layered structure. |
|
In diamond, each carbon atom is sp3 hybridised and is bonded to four other carbon atoms through a ΓΖ’ bond. |
In graphite, each carbon atom is sp2 hybridised and is bonded to three other carbon atoms through a ΓΖ’ 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. |
|
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Q13 :
”Ά Lead(II) chloride reacts with Cl2to give PbCl4.
Answer :
(a) Lead belongs to group 14 of the periodic table. The two oxidation states displayed by this group is +2 and +4. On moving down the group, the +2 oxidation state becomes more stable and the +4 oxidation state becomes less stable.
This is because of the inert pair effect. Hence, PbCl4is much less stable than PbCl2. However, the formation of PbCl4takes place when chlorine gas is bubbled through a saturated solution of PlCl2.
(b)
(c) Lead is known not to form PbI4. Pb (+4) is oxidising in nature and I-is reducing in nature. A combination of Pb(IV) and iodide ion is not stable. Iodide ion is strongly reducing in nature. Pb(IV) oxidises I–to I2and itself gets reduced to Pb(II).
Suggest reasons why the B–F bond lengths in BF3(130 pm) and (143 pm) differ.
This double-bond character causes the bond length to shorten in BF3(130 pm). However, when BF3coordinates with
If B-Cl bond has a dipole moment, explain why BCl3molecule has zero dipole moment.
Aluminium trifluoride is insoluble in anhydrous HF but dissolves on addition of NaF. Aluminium trifluoride precipitates out of the resulting solution when gaseous BF3is bubbled through. Give reasons.
Hydrogen fluoride (HF) is a covalent compound and has a very strong intermolecular hydrogen-bonding. Thus, it does not provide ions and aluminium fluoride (AlF) does not dissolve in it. Sodium fluoride (NaF) is an ionic compound and when it is added to the mixture, AlF dissolves. This is because of the availability of free F–. The reaction involved in the process is:
When boron trifluoride (BF3) is added to the solution, aluminium fluoride precipitates out of the solution. This happens because the tendency of boron to form complexes is much more than that of aluminium. Therefore, when BF3is added to the solution, B replaces Al from the complexes according to the following reaction:
Suggest a reason as to why CO is poisonous.
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
How is excessive content of CO2 responsible for global warming?
Explain structures of diborane and boric acid.
(a) Diborane
2 boron and 4 terminal hydrogen atoms (Ht) lie in one plane, while the other two bridging hydrogen atoms (Hb) lie in a plane perpendicular to the plane of boron atoms. Again, of the two bridging hydrogen atoms, one H atom lies above the plane and the other lies below the plane. The terminal bonds are regular two-centre two-electron (2c - 2e-) bonds, while the two bridging (B-H-B) bonds are three-centre two-electron (3c - 2e-) bonds.
Boric acid has a layered structure. Each planar BO3unit is linked to one another through H atoms. The H atoms form a covalent bond with a BO3 unit, while a hydrogen bond is formed with another BO3unit. In the given figure, the dotted lines represent hydrogen bonds.
What happens when
(b) Boric acid is added to water,
(d) BF3is reacted with ammonia?
(a)
(b) When boric acid is added to water, it accepts electrons from –OH ion.
(c) Al reacts with dilute NaOH to form sodium tetrahydroxoaluminate(III). Hydrogen gas is liberated in the process.
(d) BF3(a Lewis acid) reacts with NH3(a Lewis base) to form an adduct. This results in a complete octet around B in BF3.
Explain the following reactions
(b) Silicon dioxide is treated with hydrogen fluoride;
(d) Hydrated alumina is treated with aqueous NaOH solution.
(a)
(b) When silicon dioxide (SiO2) is heated with hydrogen fluoride (HF), it forms silicon tetrafluoride (SiF4). Usually, the Si–O bond is a strong bond and it resists any attack by halogens and most acids, even at a high temperature. However, it is attacked by HF.
The SiF4formed in this reaction can further react with HF to form hydrofluorosilicic acid.
(c) When CO reacts with ZnO, it reduces ZnO to Zn. CO acts as a reducing agent.
(d) When hydrated alumina is added to sodium hydroxide, the former dissolves in the latter because of the formation of sodium meta-aluminate.
Give reasons:
(ii) A mixture of dilute NaOH and aluminium pieces is used to open drain.
(iv) Diamond is used as an abrasive.
(vi) Aluminium utensils should not be kept in water overnight.
Answer :
(i) Concentrated HNO3can be stored and transported in aluminium containers as it reacts with aluminium to form a thin protective oxide layer on the aluminium surface. This oxide layer renders aluminium passive.
(ii)
(iii) Graphite has a layered structure and different layers of graphite are bonded to each other by weak van der Waals' forces. These layers can slide over each other. Graphite is soft and slippery. Therefore, graphite can be used as a lubricant.
(iv) In diamond, carbon is sp3hybridised. Each carbon atom is bonded to four other carbon atoms with the help of strong covalent bonds. 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) The oxygen present in water reacts with aluminium to form a thin layer of aluminium oxide. This layer prevents aluminium from further reaction. However, when water is kept in an aluminium vessel for long periods of time, some amount of aluminium oxide may dissolve in water. As aluminium ions are harmful, water should not be stored in aluminium vessels overnight.
(vii)
Explain why is there a phenomenal decrease in ionisation enthalpy from carbon to silicon?
How would you explain the lower atomic radius of Ga as compared to Al?
Atomic radius (in pm) |
|
Aluminium |
143 |
Gallium |
135 |
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?
The rigid 3-D structure of diamond makes it a very hard substance. In fact, diamond is one of the hardest naturally- occurring substances. It is used as an abrasive and for cutting tools.
It has sp2hybridised carbon, arranged in the form of layers. These layers are held together by weak van der Walls' forces. These layers can slide over each other, making graphite soft and slippery. Therefore, it is used as a lubricant.
(a) Classify following oxides as neutral, acidic, basic or amphoteric:
CO, B2O3, SiO2, CO2, Al2O3, PbO2, Tl2O3
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. Al2O3reacts with both NaOH and H2SO4.
(6) PbO2= Amphoteric
Amphoteric substances react with both acids and bases. PbO2reacts 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.
In some of the reactions thallium resembles aluminium, whereas in others it resembles with group I metals. Support this statement by giving some evidences.
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.
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.
The given metal X gives a white precipitate with sodium hydroxide and the precipitate dissolves in excess of sodium hydroxide. Hence, X must be aluminium.
The white precipitate (compound A) obtained is aluminium hydroxide. The compound B formed when an excess of the base is added is sodium tetrahydroxoaluminate(III).
Now, when dilute hydrochloric acid is added to aluminium hydroxide, aluminium chloride (compound C) is obtained.
Also, when compound A is heated strongly, it gives compound D. This compound is used to extract metal X. Aluminium metal is extracted from alumina. Hence, compound D must be alumina.
What do you understand by (a) inert pair effect (b) allotropy and(c) catenation?
(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 np1 and 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 ns2electrons by the d- and f- electrons. As a result of the poor shielding,
the ns2electrons are held tightly by the nucleus and so, they cannot participate in chemical bonding.
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.
A certain salt X, gives the following results.
(ii) It swells up to a glassy material Yon strong heating.
Answer :
The given salt is alkaline to litmus. Therefore, Xis a salt of a strong base and a weak acid. Also, when Xis strongly heated, it swells to form substance Y. Therefore, Xmust be borax.
When concentrated acid is added to borax, white crystals of orthoboric acid (Z) are formed.
Write balanced equations for:
(ii) B2H6 + H2O ΓΆβ€ ’
(iv) H3BO3
(vi) B2H6 + NH3 ΓΆβ€ ’
(iv)
Give one method for industrial preparation and one for laboratory preparation of CO and CO2 each.
Caron dioxide
CO2is commercially prepared by heating limestone. The reaction involved is as follows:
CO is commercially prepared by passing steam over hot coke. The reaction involved is as follows:
An aqueous solution of borax is
(c) basic (d) acidic
Boric acid is polymeric due to
(c) its monobasic nature (d) its geometry
(b) Boric acid is polymeric because of the presence of hydrogen bonds. In the given figure, the dotted lines represent hydrogen bonds.
The type of hybridisation of boron in diborane is
(a) sp (b) sp2(c) sp3 (d) dsp2
Thermodynamically the most stable form of carbon is
(c) fullerenes (d) coal
Elements of group 14
(b) exhibit oxidation state of +2 and +4
(c) form M2-and M4+ion
(d) form M2+and M4+ions
(b)The elements of group 14 have 4 valence electrons. Therefore, the oxidation state of the group is +4. However, as a result of the inert pair effect, the lower oxidation state becomes more and more stable and the higher oxidation state becomes less stable. Therefore, this group exhibits +4 and +2 oxidation states.
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Group 14 element |
Oxidation state |
|
C |
+4 |
||
Si |
+4 |
||
Ge, Sn, Pb |
+2, +4 |
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If the starting material for the manufacture of silicones is RSiCl3, write the structure of the product formed.
(i)