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Q1. Generally atomic radii increases down the group, but atomic radius of Ga is less than that of Al. Why?

Solution

The presence of additional 10 d-electrons offer only poor screening effect for the outer electrons from the increased nuclear charge in gallium hence the atomic radius of Ga is less than Al.

Q2. 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.

Q3. 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.

Q4. Give the type of hybridization involved in BCl₃NH₃.

Solution

sp³hybridization.

Q5. 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.

Q6. Give any two uses of  silicones.

Solution

Uses of silicones: Silicone oils are used for high temperature oil baths, high vacuum pumps and low temperature lubricantion. They are used as an insulating materials for electric motors and other electrical appliances.

Q7. Draw the structure of diborane, boron trichloride and boric acid.

Solution

                                Diborane             Boron trichloride                  Boric acid

Q8. If B–Cl bond has a dipole moment, explain why BCl₃ molecule has zero dipole moment.

Solution

B-Cl has a dipole due to the difference in the electronegativity of boron and chlorine atom. The overall dipole of a molecule also depends on the geometry. The geometry of BCl₃ is planar with a bond angle of 120 degree. The resultant dipole of two B-Cl bonds cancels the third one, resulting in net zero dipole.  

Q9. 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)  

Q10. Explain what happens when boric acid is heated.

Solution

Orthoboric acid when heated above 370K forms metaboric acid, HBO₂ which on further heating yields boric oxide, B₂O₃.                   

Q11. Explain structures of diborane.

Solution

In the structure of diborane, the four terminal hydrogen atoms and the two boron atoms lie in one plane. Above and below this plane, there are two bridging hydrogen atoms. The four terminal B-H bonds are regular two centre-two electron bonds while the two bridge (B-H-B) bonds are different and can be described as three centre -two electron bonds or banana bond.  

Q12. Give one reaction for the preparation of orthoboric acid.

Solution

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

Q13. 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

Q14.

Solution

Q15. Give an account of Borax bead test.

Solution

Borax on heating first loses water molecules and forms sodium metaborate which on further heating turns into a transparent liquid, which solidifies into glass like material known as borax bead. Na₂B₄O₇.10H₂ONa₂B₄O₇  2NaBO₂ + B₂O₃                                                                Sodium      Boric                                                              metaborate   anhydride The metaborates of many transition metals have characteristic colours and, therefore, borax bead test can be used to identify them in the laboratory.  

Q16. Give any one method of preparation of silica.

Solution

   

Q17. Which is the most stable allotrope of carbon?

Solution

Graphite is the most stable allotrope of carbon.

Q18. 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.

Q19. What is “inorganic benzene”?

Solution

Borazine, B₃N₃H₆  is known as “inorganic benzene”.

Q20. 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.

Q21. What happens when: (give reactions only) (a) Boron is heated in air (b) Boron reacts with dilute HCl (c) Boron reacts with Chlorine

Solution

(a) Boron reacts with oxygen and nitrogen.       4B_(s) + 3O_2(g)  2B₂O_3(s)       2B_(s) + N_2(g)   2BN_(s)   (b)  B_(s) + Dil. HCl  No reaction   (c) 2B_(s) + 3Cl_2(g)  2BCl_3(s)

Q22. Why graphite is soft but diamond is hard in nature?

Solution

Graphite has layers structure. There are weak van der Waal's forces between the layers. So, the layers can slide over one another. Therefore, graphite is soft. In diamond each carbon atom is the same distance to each of its neighboring carbon atoms. In this rigid network atoms cannot move.

Q23. How is  cross linked polymer of the silicones synthesized?

Solution

Q24. 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]⁻ .