# NCERT Solutions for Class 11 Chemistry Chapter 11 P Block Elements

## NCERT Solutions For Class 11 Chemistry Chapter 11 PDF Free Download

NCERT solutions for class 11 chemistry chapter 11 p-Block Elements help students test their knowledge about p-block elements and learn about the different concepts related to it. The chemistry solutions for chapter 11 we are offering consists of the main questions covering all the topics in the NCERT textbook which is also relevant for class 11 studies in CBSE schools Along with answers to the questions in the textbook, these solutions provide exemplary problems and worksheet that guides you in understanding the topic thoroughly.

The question and explanations have been prepared and solved by our top subject experts as per the latest CBSE syllabus (2018-19). The set of solutions for NCERT class 11 chemistry chapter 11 p-block elements is available in the form of a PDF which is very beneficial as students can freely download it from our website and use it at any other time as resource material.

Class 11 chemistry chapter 11 p-Block Elements is an important topic for the students. One needs to have a good command on this topic in order to excel in the examination. Class 11 chemistry chapter 11 basically deals with the topic of p-block elements. Just to give you a gist, p-block elements are nothing but elements where the last electron enters or is found on the outermost p-subshell.

P-block consists of all the elements between the 13th group and 18th group in the periodic table. Since this chapter covers a lot of topics, students need to learn and be in touch with the subject. At such juncture, students can and should refer to NCERT chemistry solutions for class 11 to retain the concepts in their memory as well as to clear all their doubts.

Along with the structured questions, the NCERT solutions come with explanations to help students learn and understand the different chapter 11 concepts in a more seamless manner. Besides, students who read through or practice these NCERT chemistry solutions will definitely be able to score better marks in exams.

## Subtopics of class 11 chemistry chapter 11 p-Block Elements

1. Group 13 Elements: The Boron Family
1. Electronic Configuration
3. Ionization Enthalpy
4. Electronegativity
5. Physical Properties
6. Chemical Properties
2. Important Trends And Anomalous Properties Of Boron
3. Some Important Compounds Of Boron
1. Borax
2. Orthoboric Acid
3. Diborane, B2h6
4. Uses Of Boron And Aluminium And Their Compounds
5. Group 14 Elements: The Carbon Family
1. Electronic Configuration
3. Ionization Enthalpy
4. Electronegativity
5. Physical Properties
6. Chemical Properties
6. Important Trends And Anomalous Behaviour Of Carbon
7. Allotropes Of Carbon
1. Diamond
2. Graphite
3. Fullerenes
4. Uses Of Carbon
8. Some Important Compounds Of Carbon And Silicon
1. Carbon Monoxide
2. Carbon Dioxide
3. Silicon Dioxide, Si02
4. Silicones
5. Silicates
6. Zeolites.

### Class 11 Chemistry p-Block Elements Important questions

Q 11.1

Explain the variation pattern in oxidation states of

(i) C to Pb

(ii) B to Tl

Ans:

(i) C to Pb

The electronic configuration of group 14 elements is ns2 np2. Hence, the most common oxidation state exhibited by them should be +4. On moving down the group, the +2 oxidation state becomes more and more common and the higher oxidation state becomes less stable because of the inert pair effect.Si and c mostly show the +4 state. Although Sn,Ge and Pb shows both the +4 and +2 states, the stability of higher oxidation states decreases than lower oxidation state on moving down the group.

(ii) B to Tl

Group 13 elements have their electronic configuration of ns2 np1 and the oxidation state exhibited by these elements should be 3.  Apart from these two electrons boron and aluminum, other elements of this group exhibit both +1 and +3 oxidation states. Boron and aluminum show oxidation state of +3. This is because of the inert pair effect. The two electrons, which are present in the S-shell do not participate in bonding as they are strongly attracted by the nucleus. As we move down the group, the inert pair effect become more prominent. Therefore Ga (+1) is unstable and TI (+1) is very stable

On moving down the group, the stability of the +3 oxidation state gets decreased.

Q 11.2

Compare

higher stability of BClto TlCl3  and explain.

Ans:

Thallium and boron belongs to group 13 of the periodic table and +1 oxidation state becomes more stable as we move down the group. Boron is more stable than thallium because +3 state of thallium is highly oxidizing and it reverts back to more stable +1 state.

Q 11.3

Why boron trifluoride behave as lewis acid?

Ans:

The electronic configuration of boron is ns2 np1. It contains 3 electrons in its valence shell. Thus, it can form only 3 covalent bonds which means that there are only 6 electrons around boron and its octet remains incomplete. When 1of the boron’s atom combines with 3 fluorine atoms, its octet (8) remains incomplete. Therefore, boron trifluoride remains electron-deficient and acts as Lewis acid.

Q 11.4

Consider the compounds CCl4, BCland explain their behaviour with water.

Ans:

Being a Lewis acid, BCl3 readily undergoes hydrolysis. Boric acid is formed as a result.

$BCl_{3}+3H_{2}O\rightarrow 3HCl+B(OH)_{3}$

CCl4 completely resists hydrolysis. Carbon does not have any vacant orbital. Therefore, it cannot accept electrons from water to form an intermediate. Separate layers are formed when CCl4 and water are mixed.

$CCl_{4}+H_{2}O\rightarrow No Reaction$

Q 11.5

Is boric acid a protic acid? Explain.

Ans:

Boric acid is a weak monobasic acid which behaves as a Lewis acid. So, it is not a protic acid.

$B(OH)_{3}+2HOH\rightarrow [B(OH)_{4}]^{-}+H_{3}O^{+}$

It behaves as an acid by accepting a pair of electrons from OH ion.

Q 11.6

Explain the reaction while heating boric acid.

Ans:

On heating orthoboric acid at 370 K or above, it changes to metaboric acid and On further heating, this yields boric oxide B2O3.

$H_{3}BO_{3}\rightarrow HBO_{2}\rightarrow B_{2}O_{3}$

Q 11.7

Explain the shapes of BH4and BF3 and assign the hybridisation of boron in these species.

Ans:

(i) BH4

Boron-hydride ion (BH4 ) is formed by the sp3 hybridisation of boron orbitals. Therefore, it is a tetrahedral structure.

(ii)  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 sp2 hybridized orbitals of boron with the sp orbitals of 3 halogen atoms. Boron is sp2 hybridized in BF3.

Q 11.8

Justify amphoteric nature of aluminum with reactions.

Ans:

A substance which displays both characteristics of acids and bases is known as amphoteric. Aluminium gets dissolved in both acids and bases, showing amphoteric behaviour.

(i)

$2Al_{(s)}+6HCl_{(aq)}\rightarrow 2Al^{3+}_{(aq)}+6Cl^{-}_{(aq)}+3H_{2(g)}$

(ii)

$2Al_{(s)}+2NaOH_{(aq)}+6H_{2}O_{(l)}\rightarrow 2Na^{+}[Al(OH)_{4}]^{-}_{(aq)}+3H_{2(g)}$

Q 11.9

What are electron deficient compounds? Are SiCl4and BCl3electron deficient species?

Ans:

In an electron-deficient compound, the octet of electrons is not complete, i.e., the central metal atom has an incomplete octet. Hence, it needs electrons to complete its octet.

(i) SiCl4

The electronic configuration of silicon is ns2 np2 . This indicates that it has 4 valence electrons. After it forms 4 covalent bonds with 4 chlorine atoms, its electron count increases to 8. Thus, SiCl4 is not an electron-deficient compound.

(ii) BCl3

It is an appropriate example of an electron-deficient compound. B has three valence electrons. After forming 3 covalent bonds with chlorine, the number of electrons around it increases to six. However, it is still short of 2 electrons to complete its octet.

Q 11.10

Write the resonance structures of $CO_{3}^{2-}$ and $HCO_{3}^{-}$.

Ans:

(a)$HCO_{3}^{-}$

(b)$CO_{3}^{2-}$

There are only 2 resonating structures for the bicarbonate ion.

Q 11.11

Mention the states of hybridization of carbon in

(a) Graphite

(b)  $CO_{3}^{2-}$

(c) Diamond

Ans:

The state of hybridization of carbon in:

(a)  Graphite

Each carbon atom in graphite is sp2 hybridized and is bound to 3 other carbon atoms.

(b)$CO_{3}^{2-}$

C in $CO_{3}^{2-}$ is sp2 hybridized and is bonded to 3 oxygen atoms.

(c) Diamond

Each carbon in diamond is sp3 hybridized and is bound to 4 other carbon atoms.

Q 11.12

Explain the difference between properties of graphite and diamond on their structure basis.

Ans:

Q 11.13

Rationalize 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:

(a)

Lead belongs to group fourteen 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, PbCl4 is much less stable than PbCl2. However, the formation of PbCl4 takes place when chlorine gas is bubbled through a saturated solution of PlCl2.

(b)

On moving down group IV, the higher oxidation state becomes unstable because of the inert pair effect. Pb(IV) is highly unstable and when heated, it reduces to Pb(II).

(c)

Lead is known not to form PbI4. Pb (+4) is oxidizing 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) oxidizes I to I2 and itself gets reduced to Pb(II).

$PbI_{4}\rightarrow PbI_{2}+I_{2}$

Q 11.14

Suggest reasons why the B–F bond lengths in BF3 (130 pm) and ??4 − (143 pm) differ.

Ans:

The B–F bond length in BF3 is shorter than the B–F bond length in ??4 .                               BF3is an electron deficient species. With a vacant p-orbital on boron, the fluorine and boron atoms undergo pn–pn back-bonding to remove this deficiency. This imparts a double bond character to the B–F bond.

This double-bond character causes the bond length to shorten in BF3 (130 pm). However, when BF3 coordinates with the fluoride ion, a change in hybridization from sp2 (in BF3) to sp3 (in ??4) occurs. Boron now forms 4σ bonds and the double-bond character is lost. This accounts for a B–F bond length of 143 pm in ??4 ion.

Q 11.15

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

Ans:

As a result of the difference in the electronegativities of Cl and B, the B–Cl bond is naturally polar. However, the BCl3 molecule is non-polar. This is because BCl3 is trigonal planar in shape. It is a symmetrical molecule. Hence, the respective dipole moments of the B–Cl bond cancel each other, thereby causing a zero dipole moment.

Q 11.16

Aluminum trifluoride is insoluble in anhydrous HF but dissolves when NaF is added. It precipitates out of the resulting solution when gaseous BF3(boron trifluoride) is bubbled through. Give reasons.

Ans:

Hydrogen fluoride is a covalent compound and has a very strong intermolecular hydrogen-bonding. Thus, it does not provide ions and aluminum fluoride does not dissolve in it. Sodium fluoride 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:

$AlF_{3}+3NaF\rightarrow Na_{3}[AlF_{6}]$

Aluminum fluoride gets precipitated out of the solution when boron trifluoride is added to the solution. This happens because the tendency of boron to form complexes is much more than that of aluminum. Therefore, when boron trifluoride is added to the solution, B replaces Al from the complexes according to the following reaction:

$Na_{3}[AlF_{6}]+3BF_{3}\rightarrow 3Na[BF_{4}]+AlF_{3}$

Q 11.17:

Explain the reason why Carbon Monoxide is poisonous.

Ans:

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

Q 11.18

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

Ans:

Carbon dioxide is an 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 sun rays. 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.

Q 11.19

Explain structures of diborane and boric acid.

Ans:

(a) Diborane

B2H6 is an electron-deficient compound. B2H6 has only 12 electrons – 6 e from 6 H atoms and 3 e each from 2 B atoms. Thus, after combining with 3 H atoms, none of the boron atoms has any electrons left. X-ray diffraction studies have shown the structure of diborane as:

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.

(b)  Boric acid

Boric acid has a layered structure. Each planar BO3 unit 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 BO3 unit. In the given figure, the dotted lines represent hydrogen bonds.

Q 11.20

What happens when

(i) Aluminum is treated with dilute NaOH,

(ii) BF3 is reacted with ammonia,

(iii) Boric acid is added to water,

(iv)Borax is heated strongly.

Ans:

(i) Aluminium is treated with dilute NaOH

Aluminium reacts with dilute NaOH to form sodium tetrahydroxoaluminate(III). Hydrogen gas is liberated in the process.

$2Al_{(s)}+2NaOH_{(aq)}+6H_{2}O_{(l)}\rightarrow 2Na^{+}[Al(OH)_{4}]^{-}_{(aq)}+3H_{2(g)}$

(ii) BF3 is reacted with ammonia

BF3 (a Lewis acid) reacts with NH3 (a Lewis base) to form a product. This results in a complete octet around B in BF3.

$F_{3}B+:NH_{3}\rightarrow F_{3}B\leftarrow :NH_{3}$

(iii) Boric acid is added to water

When boric acid is added to water, it accepts electrons from OH ion.

$B(OH)_{3}+2HOH\rightarrow [B(OH)_{4}]^{-}+H_{3}O^{+}$

(iv) Borax is heated strongly

When heated, borax undergoes various transitions. It first loses water molecules and swells. Then, it turns into a transparent liquid, solidifying to form a glass-like material called borax bead.

Q 11.21

Explain the following reactions

(i) CO is heated with ZnO;

(ii) Silicon is heated with methyl chloride at high temperature in the presence of copper;

(iii) Hydrated alumina is treated with aqueous NaOH solution;

(iv) Silicon dioxide is treated with hydrogen fluoride.

Ans:

(i) CO is heated with ZnO

When CO reacts with ZnO, it reduces ZnO to Zn. CO acts as a reducing agent.

$ZnO_{(s)}+CO_{(g)}\rightarrow Zn_{(s)}+CO{2(g)}$

(ii) Silicon is heated with methyl chloride at high temperature in the presence of copper

When silicon reacts with methyl chloride in the presence of copper (catalyst) and at a temperature of about 537 K, a class of organosilicon polymers called methyl substituted chlorosilane MeSiCl3, Me2SiCl2, Me3SiCl, and Me4Si) are formed.

(iii) Hydrated alumina is treated with aqueous NaOH solution

When hydrated alumina is added to sodium hydroxide, the former dissolves in the latter because of the formation of sodium meta-aluminate.

$Al_{2}O_{3}.2H_{2}O+2NaOH\rightarrow 2NaAlO_{2}+3H_{2}O$

(iv)Silicon dioxide is treated with hydrogen fluoride.

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.

$SiO_{2}+4HF\rightarrow SiF_{4}+2H_{2}O$

The SiF4 formed in this reaction can further react with HF to form hydro-fluorosilicic acid.

$SiF_{4}+2HF\rightarrow H_{2}SiF_{6}$

Q 11.22

Provide reasons:

(i) Diamond is used as an abrasive

(ii)A mixture of dilute NaOH and aluminum pieces is used to open drain.

(iii) Aluminum alloys are used to make aircraft body.

(iv) Conc.HNO3 can be transported in an aluminum container.

(v) Aluminum wire is used to make transmission cables.

(vi) Graphite is used as the lubricant.

(vii) Aluminum utensils should not be kept in water overnight.

Ans:

(i) Diamond is used as an abrasive

In diamond, carbon is sp3 hybridized. Each carbon atom is bonded to 4 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.

(ii)A mixture of dilute NaOH and aluminum pieces is used to open drain.

Sodium hydroxide and aluminum react to form sodium tetra hydroxy aluminate(III) and hydrogen gas. The pressure of the produced hydrogen gas is used to open blocked drains.

$2Al+2NaOH+6H_{2}O\rightarrow 2Na^{+}[Al(OH)_{4}]^{-}+3H_{2}$

(iii) Aluminum alloys are used to make aircraft body.

Aluminum has high tensile strength and it is light weight. It can also be alloyed with various metals such as Si, Mg, Cu, MnandZn. It is very malleable and ductile. Therefore, it is used in making of aircraft bodies.

(iv) Conc.HNO3 can be transported in an aluminum container.

Concentrated HNO3 can be stored and transported in aluminum containers as it reacts with aluminum to form a thin protective oxide layer on the aluminum surface. This oxide layer renders aluminum passive.

(v) Aluminum wire is used to make transmission cables.

Silver, copper, and aluminum are among the best conductors of electricity. Silver is an expensive metal and silver wires are very expensive. Copper is quite expensive and is also very heavy. Aluminum is a very ductile metal. Thus, aluminum is used in making wires for electrical conduction.

(vi) Graphite is used as the lubricant.

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.

(vii) Aluminum utensils should not be kept in water overnight.

The oxygen present in water reacts with aluminum to form a thin layer of aluminum oxide. This layer prevents aluminum from further reaction. However, when water is kept in an aluminum vessel for long periods of time, some amount of aluminum oxide may dissolve in water. As aluminum ions are harmful, water should not be stored in aluminum vessels overnight.

Q 11.23

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

Ans:

Ionization 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.

Q 11.24

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

Ans:

Although Ga has one shell more than Al, its size is lesser than Al. This is because of the poor shielding effect of the 3d-electrons. The shielding effect of d-electrons is very poor and the effective nuclear charge experienced by the valence electrons in gallium is much more than it is in the case of Al.

Q 11.25

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?

Ans:

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.

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 sp2 hybridized 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.

Q 11.26

(A) Classify the following oxides as

Acidic, Neutral, Amphoteric or basic.

→ SiO2

→ Tl2O3

→ CO

→ CO2

→ B2O3

→ PbO2

→ Al2O3

(B) Write suitable chemical equations to show their nature.

Ans:

→ SiO2= Acidic

Being acidic, it reacts with bases to form salts. It reacts with NaOH to form sodium silicate.

$SiO_{2}+2NaOH\rightarrow 2Na_{2}SiO_{3}+H_{2}O$

→ Tl2O= Basic

Being basic, it reacts with acids to form salts. It reacts with HCl to form thallium chloride.

$Tl_{2}O_{3}+6HCL\rightarrow 2TlCl_{3}+3H_{2}O$

→ CO = Neutral

→ CO2 = Acidic

Being acidic, it reacts with bases to form salts. It reacts with NaOH to form sodium carbonate.

$Co_{2}+2NaOH\rightarrow Na_{2}CO_{3}+H_{2}O$

→ B2O= Acidic

Being acidic, it reacts with bases to form salts. It reacts with NaOH to form sodium metaborate.

$B_{2}O_{3}+2NaOH\rightarrow 2NaBO_{2}+H_{2}O$

→ PbO2  = Amphoteric

Amphoteric substances react with both acids and bases. PbO2 reacts with both NaOH and H2SO4.

$PbO_{2}+2NaOH\rightarrow Na_{2}PbO_{3}+H_{2}O$ $2PbO_{2}+2H_{2}SO_{4}\rightarrow 2PbSO_{4}+2H_{2}O+O_{2}$

→ Al2O3= Amphoteric

Amphoteric substances react with both acids and bases. Al2O3 reacts with both NaOH and H2SO4.

$Al_{2}O_{3}+2NaOH\rightarrow NaAlO_{2}$ $Al_{2}O_{3}+3H_{2}SO_{4}\rightarrow Al_{2}(SO_{4})_{3}+3H_{2}O$

Q 11.27

In some of the reactions, thallium resembles aluminum, whereas in others it resembles with the group I metals. Support this statement by giving some evidence.

Ans:

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. Aluminum displays the +3 oxidation state and alkali metals display the +1 oxidation state. Thallium displays both the oxidation states. Therefore, it resembles both aluminum and alkali metals. Thallium, like aluminum, forms compounds such as TlCl3 and Tl2O3. It resembles alkali metals in compounds Tl2O and TlCl.

Q 11.28

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 the metal. Identify (X), (A), (B), (C) and (D). Write suitable equations to support their identities.

Ans:

The given metal X gives a white precipitate with sodium hydroxide and the precipitate dissolves in excess of sodium hydroxide. Hence, X must be aluminum. The white precipitate (compound A) obtained is aluminum hydroxide. The compound B formed when an excess of the base is added is sodium tetrahydroxy aluminate(III)

Now, when dilute hydrochloric acid is added to aluminum hydroxide, aluminum 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.

Q 11.29

What do you understand by

(a) Inert pair effect

(b) Allotropy and

(c) Catenation?

Ans:

(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 the case of group 13 elements, the electronic configuration is ns2 np1 and their group valence 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.

Q 11.30

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, a white crystal of an acid Z separates out.

Write equations for all the above reactions and identify X, Y, and Z.

Ans:

The given salt is alkaline to litmus. Therefore, X is a salt of a strong base and a weak acid. Also, when X is strongly heated, it swells to form substance Y. Therefore, X must be borax. When borax is heated, it loses water and swells to form sodium metaborate. When heating is continued, it solidifies to form a glassy material Y. Hence, Y must be a mixture of sodium metaborate and boric anhydride.

When concentrated acid is added to borax, white crystals of orthoboric acid (Z) are formed.

Q 11.31

Write balanced equations for:

(i)BF3 + LiH →

(ii) B2H6 + H2O →

(iii)NaH + B2H6

(iv) H3BO3

(v) Al + NaOH →

(vi) B2H6 + NH3

Ans:

Q 11.32

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

Ans:

Caron dioxide

In the laboratory, CO2 can be prepared by the action of dilute hydrochloric acid on calcium carbonate. The reaction involved is as follows:

CO2 is commercially prepared by heating limestone. The reaction involved is as follows:

Caron monoxide

In the laboratory, CO is prepared by the dehydration of formic acid with conc. H2SO4, at 373 K. The reaction involved is as follows:

CO is commercially prepared by passing steam over hot coke. The reaction involved is as follows:

Q 11.33

An aqueous solution of borax is

(a) Basic

(b) Amphoteric

(c) Neutral

(d) Acidic

Ans:

a) Borax is a salt of a strong base (NaOH) and a weak acid (H3 BO3). It is, therefore, basic in nature.

Q 11.34

Boric acid is polymeric due to

(a) Its acidic nature

(b) Its geometry

(c) Its monobasic nature

(d) The presence of hydrogen bonds

Ans:

(d) Boric acid is polymeric because of the presence of hydrogen bonds. In the given figure, the dotted lines represent hydrogen bonds.

Q 11.35

The type of hybridization of boron in diborane is

(a) sp

(b)sp3

(c) sp2

(d) dsp3

Ans:

(b) Boron in diborane is sp3 hybridized.

Q 11.36

Thermodynamically the most stable form of carbon is

(a) Diamond

(b) Coal

(c) Fullerenes

(d)  Graphite

Ans:

(d) Graphite is thermodynamically the most stable form of carbon.

Q 11.37

Elements of group 14

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

(b)exhibit oxidation state of +4 only

(c) form M2– and M4+ ion

(d) form M2+ and M4+ ions

Ans:

(a)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.

Q 11.38

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

Ans:

 Also Access NCERT Exemplar for class 11 chemistry Chapter 11 CBSE Notes for class 11 chemistry Chapter 11