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Q1. Give a comparative account of the chemistry of carbon and silicon with regard to their property of catenation.

Solution

Carbon shows property of catenation to more extent than silicon due to small size and tendency to form p∏-p∏ multiple bonds with itself; bond length of C-C is more than that of Si-Si. C-C > Si-Si

Q2. Name two minerals of Boron with their chemical formulae.

Solution

Borax (Na₂B₄O₇.10H₂O) and Kernite (Na₂B₄O₇.4H₂O).

Q3. What is an allotrope?

Solution

The various physical forms in which an element can exist are called allotropes of the element.

Q4. What are boranes?

Solution

 

• The hydrides of boron are known by a general name borane.
• These boranes are gaseous substances at room temperature.
• The general formulae of the two important series of boranes are B_nH_n+4 and B_nH_n+6.
• Amongst the well known boron hydrides are: diborane (B₂H₆), tetraborane (B₄H₁₀), pentaborane (B₅H₉) and hexaborane (B₆H₁₀), etc.
• The simplest and the most important hydride of boron is diborane.

Q5. Write reactions to justify amphoteric nature of aluminium.

Solution

Amphoteric nature means aluminium shows both acidic and basic character. Reaction with acids: 2Al_(s) + 6HCl _(aq) → 2Al³⁺ _(aq) + 6Cl^– _(aq) + 3H_{2 (g)} Reaction with Alkalis: 2Al _(s) + 2NaOH _(aq) + 6H₂O _(l) → 2 Na⁺ [Al(OH)₄] ^–_(aq) + 3H_2(g)                                                        Sodium                                                    Tetrahydroxoaluminate (III)  

Q6. Give reasons for the solid carbon dioxide being known as dry ice.

Solution

Since solid CO₂ is a snow like substance which directly changes into the gaseous state without changing into liquid state and hence it is known as dry ice.

Q7. Aluminum can form [AlF₆]⁻ but Boron cannot form [BF₆]⁻. Why?

Solution

 Due to unavailability of vacant d-orbital as in Al, B is unable to expand its octet and thus cannot form [BF₆]⁻  as  [AlF₆]⁻ . 

Q8. The hydrolysis of CCl₄ is not possible while SiCl₄ is easily hydrolysed. Why?

Solution

• CCl₄ doesnot undergo hydrolysis by water because the carbon atom is small and is shielded by larger chlorine atoms.
• Carbon does not have 3d atomic orbitals that water can use to form co-ordinate bonds.
• In SiCl_4, the silicon atom is larger than the carbon atom and also has available 3d atomic orbitals for bonding, thus hydrolysis is possible.

Q9. Which allotrope of carbon has a three-dimensional solid structure?

Solution

Diamond has a three dimensional solid structure.

Q10. Graphite conducts electricity. Explain.

Solution

• Graphite has delocalised Π electrons which are relatively free to move under the influence of electric field.Therefore graphite is a good conductor of electricity.
• However in diamond all the valence electrons are involved in forming carbon-carbon bonds and does not have free electrons and therefore it is a bad conductor of electricity.

Q11. Why does boron triflouride behave as a Lewis acid?

Solution

In BF₃ central atom i.e. boron has incomplete octet. It lacks one electron pair to complete its octet. Since it can accept one electron pair, it acts as Lewis acid because Lewis acids are electron pair acceptor.

Q12. What are allotropes of carbon?

Solution

There are many allotropes of carbon, some are:(a) Amorphous Carbon (b) Diamond (c) Graphite (d) Buckyballs

Q13. Which of the following element exist as liquid in summer among group 13 elements?

Solution

Gallium

Q14. How many groups belong to p-block elements?

Solution

There are six groups of p-block elements in the periodic table.

Q15. What is silica?

Solution

• Silicon dioxide, commonly known as silica, occurs in several crystallographic forms.
• 95% of the earth’s crust is made up of silica and silicates.
• Quartz, cristobalite and tridymite are some of the crystalline forms of silica.
• It is a covalent, three-dimensional network solid in which each silicon atom is covalently bonded in a tetrahedral manner to four oxygen atoms.

• Silica in its normal form is almost non-reactive because of very high Si—O bond enthalpy.

Q16. The boron compounds generally behave as Lewis acid. Explain.

Solution

• As the number of valence electrons in boron is three, it forms compounds which are electron deficient.
• Such electron deficient molecules have tendency to accept a pair of electrons to achieve stable electronic configuration and thus, behave as Lewis acids.

Q17. Is boric acid a protic acid? Explain

Solution

• No , boric acid is not a protic acid.
• Boric acid is a weak monobasic acid.
• It is not a protonic acid but acts as a Lewis acid by accepting electrons from a hydroxyl ion and in turn releases H⁺ ions.

            B(OH)₃ + 2HOH → [B(OH)₄]^– + H₃O⁺  

Q18. What is the use of studying p-Block elements?

Solution

The p-block contains several elements of great social and economic importance as well as chemical interest. Examples include the use of aluminium as a structural material, the importance of silicon and germanium as semiconductors, and the use of sulphur, phosphorus and nitrogen in fertilizers.

Q19. Give the differences in structures of the following pair of compounds CO₂ and SiO₂.

Solution

CO₂ is linear molecule and exists as monomer. It is gas while SiO₂ is solid at room temperature due to three dimensional networks in which each Si atom is covalently bonded to four oxygen atoms tetrahedrally. In CO₂, 'C' is sp hybridized while in SiO₂, 'Si' is sp³ hybridised.

Q20. What are the similarities between graphite and diamond?

Solution

• Both graphite and diamond are forms of carbon. As such, they are said to be allotropes of carbon.
• Both occur naturally. Both are mined for industrial purposes, though larger diamonds are sought and used for other things.
• Both are produced in the earth in geothermal processes.
• Both can be made artificially.
• Both are normally solids and highly stable. And they are both difficult to burn, even in an oxygen environment.

Q21. Why the p-block elements are also known as representative elements?

Solution

Each group in the p-block has its own characteristic properties and hence they are called as representative elements.

Q22. Explain that lead nitrate on heating gives a pale yellow gas which on strong heating turns dark brown.

Solution

Lead nitrate on heating gives dinitrogen tetraoxide, N₂O₄, which is pale yellow gas. But on heating N₂O₄ decomposes into NO₂ which is dark brown in colour.   2Pb(NO₃)₂ → 2PbO+4NO₂+O₂

Q23. Give the type of hybridization involved in BCl₃NH₃.

Solution

sp³hybridization.

Q24. What are silicones?

Solution

• These are organosilicon polymers containing Si-O-Si linkages.
• They are formed by the hydrolysis of alkyl or aryl substituted  chlorosilanes  and their subsequent polymerisation.
• The alkyl or aryl substituted  chlorosilanes are prepared by the reaction of Grignard reagent and silicon tetrachloride.

            RMgCl     + SiCl₄   →    RSiCl₃   + MgCl₂        Grignard reagent            R stands for -CH₃ , -C₂H₅ or -C₆H₅ groups

Q25. Why Boron forms only covalent compounds?

Solution

•  Due to small size of boron, the sum of its first three ionization enthalpies is very high.
• This prevents it to form +3 ions and forces it to form only covalent compounds.

Q26. Why the melting point of diamond is very high?

Solution

• It has a  three dimensional network involving strong C-C bonds which are difficult to break.
• A large amount of energy is required to break down these bonds.
• Therefore , the melting point of diamond is very high.

Q27. Mention some uses of silica.

Solution

Uses of silica:a) Silica is used as a building material. It is used in making of cement, bricks, etc.b) Bricks made from a mixture of powdered sand, clay and lime are used for lining the furnace used in the manufacture of steel.c) It is used as an acid flux in metallurgy.d) Colored crystalline silica are used as gems.e) Few transparent  varieties of quartz are used for the manufacture of lenses, optical instruments.f) Quartz glass used for making special glassware suitable for working with u.v radiation as quartz is transparent.

Q28. Give one reaction for the preparation of orthoboric acid.

Solution

Na₂B₄O₇ + 2HCl + 5H₂O → 2NaCl + 4B(OH)₃                                                               Boric acid

Q29. Give the formula of Borax.

Solution

Na₂B₄O₇.10H₂O.

Q30. Write a brief note on silicon.

Solution

• Silicon is the most common metalloid.
• It is represented as Si and has atomic number 14.
• It is the second most abundant element after oxygen in earth's crust. Pure silicon is used in the semiconductor industry.
• It is used in photovoltaic and electronic applications.

Q31. What makes diamond an excellent abrasive?

Solution

• Diamond has a solid three dimensional structure.
• Each carbon atom is present at the same distance to each of its neighboring carbon atoms.
• In this rigid network atoms cannot move. This makes diamond hard.
• Eventually, diamond is the hardest known natural mineral. Due to its hardness, it is used  for rock drilling and for making abrasives for sharpening hard tools.

Q32. Which is the most stable allotrope of carbon?

Solution

Graphite is the most stable allotrope of carbon.

Q33. What are  p-block elements?

Solution

• The elements belonging to groups 13 to 18 of the periodic table are called p-block elements.
• They are also called as representative elements or normal elements.
• The chemistry of p-block elements plays an important role in our daily life.

Q34. What is inert pair effect?

Solution

• The reluctance of s-electron pair, of heavier elements, to take part in bond formation is known as inert pair effect.
• This is due to poor shielding effect of intervening d and f orbitals, the increased effective nuclear charge holds ns electrons tightly and thereby, restricting their participation in bonding.
• As a result of this, only p-orbital electron may be involved in bonding.

Q35. Classify following oxides as neutral, acidic, basic or amphoteric. B₂O₃, Al₂O₃, Ga₂O₃, In₂O₃, Tl₂O₃.

Solution

Acidic oxide: B₂O₃ Amphoteric oxide: Al₂O₃, Ga₂O₃ Basic oxide: In₂O₃, Tl₂O₃

Q36. What are fullerenes? Explain the structure of C₆₀ molecule.

Solution

• A new family of carbon allotropes consisting of cluster of carbon atoms such as C₃₂ , C₅₀ , C₆₀ , C₇₀ , C₈₄ etc.  are called fullerenes.
• Fullerenes are made by heating of graphite in an electric arc in the presence of inert gases such as helium or argon.
• Among these the allotrope having the molecular formula C₆₀ is important. It is called Buckminster fullerene.
• C₆₀ molecule is a perfect sphere and looks like a soccer ball and is also popularly known as bucky-ball.
• It contains twenty six- membered rings and twelve five membered rings.
• A six membered ring is fused with six or five membered rings but a five membered ring can only fuse with six membered rings.
• All the carbon atoms are equal and they undergo sp² hybridisation.
• Each carbon atom forms three sigma bonds with other three carbon atoms.
• The remaining electron at each carbon is delocalised in molecular orbitals, which in turn give aromatic character to molecule.
• This ball shaped molecule has 60 vertices and each one is occupied by one carbon atom and it also contains both single and double bonds.

     

Q37. Suggest reasons why the B–F bond lengths in BF₃ (130 pm) and BF₄^- (143 pm) differ.

Solution

• In BF₃, B atom is sp² hybridised and has one unhybridised empty 2p atomic orbital.
• This empty 2p orbital overlaps sideways with fully filled 2p atomic orbital of F to form dative pΠ- pΠ bond.
• This reduces the bond length.
• In BF^-₄, B atom is sp³ hybridised, all B-F bond are single covalent bond.

Q38. Give the molecular formula for borax and boric acid.

Solution

The molecular formula for borax is Na₂[B₄O₅(OH)₄].8H₂O and boric acid is H₃BO₃.

Q39. What are electron deficient compounds? Explain with example.

Solution

The molecule in which central atom has lesser electrons than octet are electron deficient. For example in AlCl₃, Al has incomplete octet, having 6 electrons in outermost orbital. Since, it lacks one electron pair hence it is electron deficient compound

Q40. What are the applications of silicon compounds?

Solution

Applications of Silicon Compounds:a. Ferro-silicon and calcium silicate are used as alloying elements in the development of steel or cast iron.b. Silicon carbide possess a diamond like crystalline structure. Due to its hardness it is used as an abrasive. c. CaSiO₃ is used as a component of cement.