NCERT Solution Class 11th Chemistry Chapter – 11 The p-Block Elements
| Textbook | NCERT |
| class | Class – 11th |
| Subject | Chemistry |
| Chapter | Chapter – 11 |
| Chapter Name | The p-Block Elements |
| Category | Class 11th Chemistry Question & Answer |
| Medium | English |
| Source | last doubt |
NCERT Solution Class 11th Chemistry Chapter – 11 The p-Block Elements
?Chapter – 11?
✍The p-Block Elements✍
?Question & Answer?
Q 1. Discuss the pattern of variation in the oxidation states of
(i) B to Tl
(ii) C to Pb
Answer: (i) 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 aluminium, other elements of this group exhibit both +1 and +3 oxidation states. Boron and aluminium show oxidation state of +3 due to 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 becomes 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.
| Group 13 element | Oxidation state |
| B | +3 |
| Al | +3 |
| Ga, In, Ti | +1, +3 |
(ii) 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. Upon moving down the group, the +2 oxidation state becomes increasingly common and the relatively higher oxidation states become less stable. This is because of the inert pair effect. Si and C mostly exhibit the +4 state. Although Sn, Ge and Pb show the +4 and +2 states, the stability of higher oxidation states decrease while moving down the group.
| Group 14 element | Oxidation state |
| C | +4 |
| Si | +4 |
| Ge, Sn, Pb | +2, +4 |
Q 2. How can you explain the higher stability of BCl3 as compared to TlCl3?
Answer: Thallium and boron belong 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 a more stable +1 state.
Q 3. Why does boron trifluoride behave as a Lewis acid?
Answer: 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 mean 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 4. Consider the compounds, BCl3 and CCl4. How will they behave with water? Justify.
Answer: Since it is a Lewis acid, BCl3 readily undergoes hydrolysis to form boric acid. BCl_{3}+3H_{2}O\rightarrow 3HCl+B(OH)_{3}On the other hand, CCl4 does not undergo hydrolysis readily. Since carbon does possess any vacant orbitals, it cannot accept any electrons from water in order to form an intermediate. When CCl4 and water are mixed, separate layers are formed. CCl_{4}+H_{2}O\rightarrow No Reaction
Q 5. Is boric acid a protic acid? Explain.
Answer: 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 a Lewis acid by accepting an electron pair from the OH– ion.
Q 6. Explain what happens when boric acid is heated
Answer: Upon heating orthoboric acid at a temperature of 370 K or above, it is converted into metaboric acid and, upon further heating, yields boric oxide (B2O3).
Q 7. Describe the shapes of BF3 and BH4–. Assign the hybridisation of boron in these species
Answer: (i) BF3
Boron tends to form monomeric covalent halides because of its small size and high electronegativity. These halides of boron usually have a planar triangular geometry, which can be explained by the overlap of three sp2 hybridized orbitals of boron with the sp3 orbitals of halogen atoms. It can be noted that boron is sp2 hybridized in BF3.
(ii) BH4–
BH4– has a tetrahedral structure which is accounted for by the sp3 hybridized orbitals.
Q 8. Write reactions to justify amphoteric nature of aluminium
Answer: Amphoteric substances are substances that exhibit both acidic and basic qualities. Since aluminium dissolves in both acids and bases, it is said to have an amphoteric nature.
(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 9. What are electron deficient compounds? Are BCl3 and SiCl4 electron deficient species? Explain.
Answer: Electron-deficient compounds are chemical compounds with incomplete octets that tend to acquire 1 or more electrons to complete their octet configurations.
BCl3
BCl3 is a good example of an electron-deficient compound. Since boron has three valence electrons, it forms 3 single bonds with chlorine, giving the boron atom a total of 6 electrons in the outermost shell. However, it is still short of 2 electrons to complete its octet.
SiCl4
Silicon has 4 valence electrons (its electronic configuration is 2s2 2p2). Its octet is completed after forming four single bonds with four different chlorine atoms. Therefore, SiCl4 is not an electron-deficient compound.
Q 10. Write the resonance structures of CO_{3}^{2-} and HCO_{3}^{-}.
Answer: For CO_{3}^{2-}
There are only 2 resonating structures for the bicarbonate ion.
For HCO_{3}^{-}
Q 11. What is the state of hybridisation of carbon in (a) CO32– (b) diamond (c) graphite?
Answer: The state of hybridization of carbon in –
(a)CO_{3}^{2-}C in CO_{3}^{2-} is sp2 hybridized and is bonded to 3 oxygen atoms.
(b) Diamond – Each diamond carbon is hybridized by sp3 and is bound to 4 additional carbon atoms.
(c) Graphite – Each carbon atom in graphite is hybridized with sp2 and is bound to 3 other atoms of carbon.
Q 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 sp3 hybridised and bonded to four other carbon atoms through a sigma bond. | In graphite, each atom is sp2 hybridised and is bonded to three other carbon atoms through a sigma bond. The fourth electron forms a pi bond. |
| It has a rigid covalent bond network which is difficult to break. | It is quite soft and its layers can be separated easily. |
| The C-C bond length in diamond is 154 pm. | The C-C bond length in graphite is 141.5 pm. |
| It acts as an electric insulator. | It is a good conductor of electricity. |
Q 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.
Answer: Lead is a member of group 14 of the periodic table. The two oxidation statuses this group shows are + 2 and + 4. The + 2 oxidation state becomes more stable when moving down the group, and the oxidation status + 4 becomes less stable. This is due to the inert pair effect. So, PbCl4 is significantly less stable than PbCl2. However, PbCl4 formation occurs when the chlorine gas is bubbled through a PlCl2 saturated solution.
- Due to the inert pair effect, the higher oxidation state becomes unstable when moving down group IV.
- Lead is not known for forming PbI4. In nature, Pb (+ 4) is oxidizing, and I reduce in nature. A Pb(IV) and iodide ion combination are not stable. Iodide ion reduces sharply in nature. Pb(IV) oxidizes I– to I2, and reduces itself to Pb(II).
PbI_{4}\rightarrow PbI_{2}+I_{2}
Q 14. Suggest reasons why the B–F bond lengths in BF3 (130 pm) and BF4–(143 pm) differ.
Answer: In BF3, the length of the B–F bond is shorter than that of the B–F bond in BF4–. BF3 is a species that has an electron deficiency. The fluorine and boron atoms undergo pn–pn back-bonding with a vacant p-orbital on boron to remove this deficiency. This endows the B–F bond with a double bond character.
This double-bond character causes a shortening of the bond length by BF3 (130 pm). However, a change in hybridization from sp2 (in BF3) to sp3 (in BF4–) occurs when BF3 coordinates with the fluoride ion. Boron now forms 4 sigma bonds and the character of the double-bond is lost. This represents the length of a 143 pm B-F bond in BF4– ion.
Q 15. If B–Cl bond has a dipole moment, explain why BCl3 molecule has zero dipole moment.
Answer: The B – Cl bond is naturally polar because of the difference in the electronegativities of Cl and B. Yet the molecule of BCl3 is non-polar. That is because the shape of BCl3 is trigonal planar. The molecule is symmetrical. So the respective B – Cl bond dipole moments cancel each other, creating a zero dipole moment.
Q 16. Aluminium 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.
Answer: Hydrogen fluoride is a covalent compound with a very strong intermolecular bonding to hydrogen. Thus, it does not provide ions and does not dissolve aluminium fluoride in it. Sodium fluoride is an ionic compound, and Alf dissolves when added to the mixture. This is because of the Free F– available. The reaction involved in the process is:
AlF_{3}+3NaF\rightarrow Na_{3}[AlF_{6}]
When boron trifluoride is added to the solution, aluminium fluoride becomes precipitated out of solution. This is because the boron’s tendency to form complexes is much greater than that of aluminium. So, when boron trifluoride is added to the solution, B replaces Al of the complexes by the following reaction:
Na_{3}[AlF_{6}]+3BF_{3}\rightarrow 3Na[BF_{4}]+AlF_{3}
Q 17. Suggest a reason as to why CO is poisonous.
Answer: Given its ability to form a complex with haemoglobin, carbon monoxide is highly poisonous. The former impedes binding with oxygen by Hb. Consequently, a person dies due to suffocation on not receiving oxygen The complex CO – Hb is more stable than the complex O2–Hb. The CO – Hb complex is found to be roughly 300 times more stable than the O2–Hb complex.
Q 18. How is the excessive content of CO2 responsible for global warming?
Answer: Carbon dioxide is a gas that is necessary for our survival. The increased CO2 content in the environment, however, poses a serious threat. An increase in fossil fuel combustion, calcareous decomposition, and a decrease in the number of trees has resulted in higher carbon dioxide levels. Carbon dioxide has the property of trapping the heat that sun rays provide. The higher the carbon dioxide level, the greater the amount of heat that is trapped. This leads to an increase in atmospheric temperature which causes global warming.
Q 19. Explain the structures of diborane and boric acid.
Answer: (a) Diborane
B2H6 is a compound that lacks an electron. B2H6 only has 12 electrons – 6 e– of 6 H atoms and 3 e– of 2 B atoms each. So none of the boron atoms has any electrons left after being mixed with 3 H atoms. X-ray diffraction studies showed the diborane structure as:
Two boron and four-terminal atoms of hydrogen (Ht) lie in one direction, while the other two bridging atoms of hydrogen (Hb) are on the surface perpendicular to the plane of boron atoms. Once, of the two atoms of hydrogen bridging, one atom of H lies above the plane, and the other below the plane. The terminal bonds are regular two-centre two-electron (2c – 2e– ) bonds, while the two bridgings (B–H–B) bonds are three-centre two-electron (3c – 2e– ) bonds.
(b) Boric acid
Boric acid is structured in a layered form. Each planar unit BO3 is connected by atoms H to each other. The H atoms form a covalent bond with a BO3 unit whereas another BO3 unit forms a hydrogen bond. The dotted lines, in the given figure, represent hydrogen bonds.
Q 11.20
What happens when
(a) Borax is heated strongly
(b) Boric acid is added to water
(c) Aluminum is treated with dilute NaOH
(d) BF3 is reacted with ammonia?
Answer: (a) Borax is heated strongly
Borax undergoes various transformations when heated. It is losing molecules and swells of water at first. Then it becomes a translucent liquid, which solidifies to form a glass-like material called a borax bead.
Na2B4O7.10H2O → Na2B4O7 → 2NaBO2 + B2O3
Borax Sodium Metaborate Boric anhydride
(b) Boric acid is added to water
When boric acid is added to water, it accepts electrons from –OH ion.
B(OH)_{3}+2H_{2}O\rightarrow [B(OH)_{4}]^{-}+H_{3}O^{+}
(c) Aluminum is treated with dilute NaOH
Aluminum reacts to sodium tetrahydroxoaluminate(III) by forming dilute NaOH. In this process, hydrogen gas is liberated.
2Al_{(s)}+2NaOH_{(aq)}+6H_{2}O_{(l)}\rightarrow 2Na^{+}[Al(OH)_{4}]^{-}_{(aq)}+3H_{2(g)}
(d) 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.
BF_{3}+NH_{3}\rightarrow [H_{3}N\rightarrow BF_{3}]
Q 11.21
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: (a) Silicon is heated with methyl chloride at high temperature in the presence of copper
A class of organosilicon polymers called methyl-substituted chlorosilane MeSiCl3, Me2SiCl2, Me3SiCl, and Me4Si) are formed when silicon reacts with methyl chloride in the presence of copper (catalyst) and at a temperature of about 537 K.
(b) Silicon dioxide is treated with hydrogen fluoride.
When the hydrogen fluoride (HF) heats silicon dioxide (SiO2), it forms silicon tetrafluoride (SiF4). The Si – O bond is usually a strong bond, avoiding any attack by halogens and most acids even at high temperatures. It is being attacked by HF though.
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}
(c) 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)}
(d) Hydrated alumina is treated with aqueous NaOH solution
When sodium hydroxide is added to hydrated alumina, the former dissolves in the latter due to the sodium meta-aluminate formation.
Al_{2}O_{3}.2H_{2}O+2NaOH\rightarrow 2NaAlO_{2}+3H_{2}O
Q 22. Give reasons –
(i) Conc.HNO3 can be transported in an aluminium container.
(ii)A mixture of dilute NaOH and aluminium pieces is used to open drain.
(iii) Graphite is used as the lubricant.
(iv) Diamond is used as an abrasive
(v) Aluminum alloys are used to make aircraft body.
(vi) Aluminum utensils should not be kept in water overnight.
(vii) Aluminum wire is used to make transmission cables.
Answer: (i) Conc.HNO3 can be transported in an aluminium container.
As it reacts with aluminium to form a thin protective oxide layer on the aluminium surface, concentrated HNO3 can be stored and transported in aluminium containers.
(ii) A mixture of dilute NaOH and aluminium pieces is used to open the drain.
Sodium hydroxide and aluminium react to form sodium tetra hydro aluminate (III) and hydrogen gas.
2Al+2NaOH+6H_{2}O\rightarrow 2Na^{+}[Al(OH)_{4}]^{-}+3H_{2}
(iii) Graphite is used as the lubricant.
Graphite has a layered structure, and the forces of weak van der Waals bind different layers of graphite together. These layers may slide one over another. Graphite is slippery and soft. Graphite can, therefore, be utilized as a lubricant.
(iv) Diamond is used as an abrasive
Carbon is sp3 hybridized in Diamond. With the help of strong covalent bonds, each carbon atom is bound to four other carbon atoms. These covalent bonds exist all over the surface, giving it a very rigid 3-D structure. This extended covalent bonding is very hard to break and for this reason, diamond is the hardest known substance. It is therefore used as an abrasive and for cutting instruments.
(v) Aluminium alloys are used to make aircraft body.
Aluminium alloys such as duralumin (aluminium–copper alloys) is light, tough and resistant to corrosion and hence is used to make aircraft body.
(vi) Aluminium utensils should not be kept in water overnight.
The oxygen in water reacts to create a thin layer of aluminium oxide with aluminium. This layer prevents further reaction from aluminium. Nevertheless, some amount of aluminium oxide can dissolve in water if the water is held in an aluminium vessel for long periods of time. Water should not be stored overnight in aluminium vessels because aluminium ions are harmful.
(vii) Aluminium wire is used to make transmission cables.
Gold, copper, and aluminium are amongst the best electricity conductors. Silver is a costly metal and very delicate gold links. Copper is pretty expensive, and heavy too. Aluminium wires are good conductors of electricity and also it is cheap metal easily available, light in weight and also very good ductility. So it is used to make transmission cables.
Q 23. Explain why is there a phenomenal decrease in ionization enthalpy from carbon to silicon?
Answer: Carbon ionizing enthalpy (the first element in group 14) is very high (1086 kJ / mol). That is expected because of its small size. However, there is a sharp decrease in enthalpy (786 kJ) when moving down the group to silicon. This is due to a considerable increase in the atomic sizes of elements when the group is moving down.
Q 24. How would you explain the lower atomic radius of Ga as compared to Al?
Answer:
| Atomic radius (in pm) | |
| Aluminium | 143 |
| Gallium | 135 |
Although Ga has more than one shell than Al, it is smaller in size than Al. This is because the 3d-electrons have poor shielding effect. The shielding effect of d-electrons is very poor, and the effective nuclear charge experienced in gallium by the valence electrons is far more than in Al.
Q 25. What are allotropes? Sketch the structure of two allotropes of carbon namely diamond and graphite. What is the impact of structure on the physical properties of two allotropes?
Answer: Allotropy is the presence of an element in more than one form, having different physical properties but the same chemical properties.
Diamond’s solid 3-D structure makes it a really strong material. Diamond is probably one of the most difficult naturally occurring substances. It is used as an abrasive and as a tool for cutting.
It has hybridized carbon sp2, structured in layer form. These layers are held together by the forces of weak van der Walls. These layers may slide over each other, making graphite slippery and soft. Consequently, it is used as a lubricant.
Q 26. (a) Classify the following oxides as neutral, acidic, basic or amphoteric.
CO, B2O3, SiO2, CO2, Al2O3, PbO2, Tl2O3
(B) Write suitable equations to show their nature.
Answer: → CO = Neutral
→ B2O3 = 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
→ 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
→ 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
→ 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
→ 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}
→ Tl2O3 = 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
Q 27. In some of the reactions, thallium resembles aluminium, whereas in others it resembles with the group I metals. Support this statement by giving some evidence.
Answer: Thallium is a group 13 element. For this group, the most common oxidation state is + 3. Heavier members of this group, however, also display the state of oxidation + 1. This is due to the inert pair effect. Aluminium shows the state of oxidation + 3 and alkali metals show the state of oxidation + 1. Thallium demonstrates the oxidation state of both. Hence, it is similar to both aluminium and alkali metals. Thallium forms compounds such as TlCl3 and Tl2O3, like aluminium. It also forms compounds like alkali metals such as Tl2O and TlCl.
Q 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.
Answer: The given metal X gives sodium hydroxide to a white precipitate, and the precipitate dissolves exceeding sodium hydroxide. X must, therefore, be made of aluminium. The obtained white precipitate (compound A) is hydroxide from aluminium. Sodium tetrahydroxy aluminate(III) is the compound B formed when an excess of base is added.
Now, when dilute hydrochloric acid is added to aluminium hydroxide, aluminium chloride (compound C) is obtained.
As well, it gives compound D when compound A is heated strongly. Uses this compound to extract metal X. Extracts aluminium metal from alumina. Therefore, alumina must be compound D.
Q 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, s-block electrons decrease their tendency to participate in chemical bonding. This effect is called the inert pair effect. The electronic configuration for group 13 elements is ns2 np1, and its group valence is + 3. However, the oxidation state of + 1 becomes more stable when moving down the group. This is because the ns2 electrons are poorly shielded by the d- and f- electrons. The ns2 electrons are held tightly by the nucleus as a result of poor shielding, and thus they can not engage in chemical bonding.
(b) Allotropy
Allotropy is the presence of an element in more than one form, having different physical properties but the same chemical properties. The different forms of an element are called allotropes. Carbon occurs in three allotropical forms, for example, diamond, graphite and fullerenes.
(c) Catenation
The atoms of some elements (such as carbon) may connect with each other via strong covalent bonds to form long chains or branches. This characteristic is known as catenation. It is most common in carbon and important in Si and S.
Q 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.
Answer: The salt given to litmus is alkaline. X is, therefore, a salt with a strong base, and a weak acid. Once X is heated excessively, it also swells to form material Y. Hence X has to be borax. When heated, borax loses water and swells in order to form sodium metaborate. When heating is continued, the forming of a glassy material Y solidifies. Thus, Y must be a mixture of metaborate sodium and boric anhydride.
When concentrated acid is added to borax, white crystals of orthoboric acid (Z) are formed.
Q 31. Write balanced equations for –
(i)BF3 + LiH →
(ii) B2H6 + H2O →
(iii)NaH + B2H6 →
(iv) H3BO3 →
(v) Al + NaOH →
(vi) B2H6 + NH3 →
Answer:
Q 32. Give one method for industrial preparation and one for laboratory preparation of CO and CO2 each.
Answer: Carbon dioxide
CO2 can be prepared in the laboratory through the action of dilute hydrochloric acid on calcium carbonate. Their reaction is as follows-
CO2 is commercially prepared by heating limestone. The reaction involved is as follows-
Carbon 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 –
The CO is prepared commercially by passing steam over hot coke –
Q 33. An aqueous solution of borax is
(a) Neutral (b) Amphoteric
(c) Basic (d) Acidic
Answer: (b) Borax is a potent base salt (NaOH) and a weak acid (H3BO3). Therefore, it is a basic fundamental in nature.
Q 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: (b) Boric acid is polymeric due to the hydrogen bonds. The dotted lines in the given figure represent hydrogen bonds.
Q 35. The type of hybridization of boron in diborane is
(a) sp (b)sp2 (c) sp3 (d) dsp2
Answer: (c) Boron in diborane is sp3 hybridized.
Q 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.
Q 37.
Elements of group 14
(b) exhibit oxidation state of +4 only
(a) exhibit oxidation state of +2 and +4
(c) form M2– and M4+ ion
(d) form M2+ and M4+ ions
Answer: (b) Group 14 elements possess 4 valence electrons. Consequently, group oxidation status is + 4. However, the lower oxidation state becomes increasingly stable due to the inert pair effect, and the higher oxidation state becomes less stable.
Therefore, this group exhibits +4 and +2 oxidation states.
| Group 14 element | Oxidation state |
| C | +4 |
| Si | +4 |
| Ge, Sn, Pb | +2, +4 |
Q 38. If the starting material for the manufacture of silicones is RSiCl3, write the structure of the product formed.
Answer:
