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Practice each question, then open the answer or worked solution.
The reason for the small radius of Ga compared to Al is _______.
(i) poor screening effect of d and f orbitals
(ii) increase in nuclear charge
(iii) presence of higher orbitals
(iv) higher atomic number
Option (i) and (ii) are the answers.
The linear shape of CO2 is due to _________.
(i) sp3 hybridisation of carbon
(ii) sp hybridisation of carbon
(iii) pπ– pπ bonding between carbon and oxygen
(iv) sp2 hybridisation of carbonSolution:
Option (ii) and (iii) are the answers.
Solution coming soon.
Me3SiCl is used during polymerisation of organic silicones because
(i) the chain length of organic silicone polymers can be controlled by adding
Me3SiCl
(ii) Me3SiCl blocks the end terminal of a silicone polymer
(iii) Me3SiCl improves the quality and yield of the polymer
(iv) Me3SiCl acts as a catalyst during polymerisationSolution:
Option (i) and (ii) are the answers.
Solution coming soon.
Which of the following statements are correct?
(i) Fullerenes have dangling bonds
(ii) Fullerenes are cage-like molecules
(iii) Graphite is a thermodynamically most stable allotrope of carbon
(iv) Graphite is slippery and hard and therefore used as a dry lubricant in
Machines
Option (ii) and (iii) are the answers.
Which of the following statements are correct. Answer based on Fig.11.1.
(i) The two bridged hydrogen atoms and the two boron atoms lie in one plane;
(ii) Out of six B–H bonds two bonds can be described in terms of 3 centre 2-electron bonds.
(iii) Out of six B-H bonds, four B-H bonds can be described in terms of 3 centres 2 electron bonds;
(iv) The four-terminal B-H bonds are two centre-two electron regular bonds.
Option (i), (ii) and (iv) are the answers.
Identify the correct resonance structures of carbon dioxide from the ones given
below :
(i) O – C ≡ O
(ii) O = C = O
(iii) –O ≡ C – O+
(iv) –O – C ≡ O+
Option (ii) and (iv) are the answers.
III. Short Answer Type
Draw the structures of BCl3.NH3 and AlCl3 (dimer).
Explain the nature of boric acid as a Lewis acid in water.
Boric acid is a weak monobasic acid and acts as a Lewis acid by accepting electrons from a hydroxyl ion.
B(OH)3 + 2H2O → [B(OH)4]– + H3O+
OH- is accepted by the Boric acid and the formation of hydroxyl ion takes place.
Thus, Boric acid act as a Lewis acid in water.
Draw the structure of boric acid showing hydrogen bonding. Which species is present in water? What is the hybridisation of boron in this species?
When boric acid reacts with water it forms [B(OH)4]– species.
B(OH)3 + 2H2O → [B(OH)4]– + H3O+
The hybridization of Boron in this species is sp3.
Explain why the following compounds behave as Lewis acids?
(i) BCl3
(ii) AlCl3
(i) BCL3 is an electron-deficient compound because it has 6 electrons in its outermost orbital and a vacant p orbital. Hence, it acts as Lewis acid and accepts a lone pair of electrons.
(ii) AlCl3 forms a covalent bond with chlorine by forming three single bonds of chlorine as aluminium has three electrons in its valence shell and act as an electron-deficient compound and act as lewis acid.