Class 11th Chemistry - The p-block Elements Case Study Questions and Answers 2022 - 2023
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The p-block Elements Case Study Questions With Answer Key
11th Standard CBSE
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Reg.No. :
Chemistry
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Group 13 and 14 are p-block elements, consisting of metals, non-metals and metalloids. These elements show variable oxidation st~tes, lower oxidation state becomes more stable, when we go down the group. The combined effect of size and availability of d-orbitals influence their ability to form \(\pi\) - bonds, Lighter elements form \(p \pi-p \pi\) bonds where as heavier elements form \(d \pi-p \pi \text { or } d \pi-d \pi\) bonds.
Boron forms electron deficient compounds. AI exhibits +3 oxidation state. Tl+ is more stable than TI3+due to inert pair effect. Carbon is typical non-metal of group 14, shows a property of catenation and forms large number of compounds. It also shows allotropy, diamond, graphite and fullerene are crystalline allotropes of carbon. Group 14 elements show +4 and +2 oxidation states. Pb2+ is more stable than Pb4+.CO and CO2 are oxides of carbon. CO2 is acidic and CO has lone pair, therefore, forms metal carbonyls. CO is deadly poisonous. CO2 is greenhouse gas. Silica, silicates and silicones are important classes of compounds of silicon.
(a) Why is Pb4+ good oxidising agent?
(b) Why is CO deadly poisonous?
(c) Why are silicones used in implants?
(d) What is use of zeolites?
(e) Why is fullerene purest allotrope of carbons?
(f) What is structure of \(\mathrm{SiO}_{4}^{4}\) ?
(g) Why is NaOH not stored in glass bottles?(a) -
Boron is element of Group 13. It is metalloid. Borax, orthoboric acid and diborane are important compounds of boron. Borax is used to prepared heat resistant pyrex glass. It is also used for cleaning metals like gold. It is also used in borax bead test to detect coloured salts in lab.
Orthoboric acid is a weak acid used as antiseptic in eye lotion. It is solid having extensive H-bonding. It acts as Lewis acid.
Diborane is a colourless, highly toxic gas. it catches fire spontaneously upon exposure to air. It burns in oxygen releasing enormous amount of energy.
(a) Why is boric acid considered as weak Lewis acid?
(b) What happens when boric acid is heated strongly?
(c) What happens when sodium hydride reacts with B2H6?
(d) What happens when sodium borohydride is reacted with I2?
(e) Give one use of NaBH4.(a) -
Study the table of atomic and physical properties of group 14 elements and answer the questions that follow based on table and related studies concepts.
Atomic and Physical Properties of Group 14 ElementsProperty Element Carbon
CSilicon
SiGermanium
GeTin
SnLead
PbAtomic number 6 14 32 50 82 Atomic mass (g mol-1) 12.01 28.09 72.60 118.71 207.2 Electronic configuration [He]2s2 2p2 [Ne]3s 3p2 [Ar]3d104s24p2 [Kr]4d105s2 Sp2 [Xe]4f145d106s26p2 Covalent radius/pma 77 118 122 140 146 Ionic radius M4+/pmb - 40 53 69 78 Ionic radius M2+/pmb - - 73 118 119 Ionization
enthalpy/kJ
mol-1\(\Delta_{t} \mathbf{H}_{1}\)
\(\Delta_{t} \mathbf{H}_{2}\)
\(\Delta_{t} \mathbf{H}_{3}\)
\(\Delta_{t} \mathbf{H}_{4}\)1086
2352
4620
6220786
1577
3228
4354761 708
1411
2942
3929715
1450
3081
40821537 3300 4409 Electronegativityc 2.5 1.8 1.8 1.8 1.9 Densityd/g cm3 3.51e 2.34 5.32 7.26f 11.34 Melting point/K 4373 1693 1218 505 600 Boiling point/K - 3550 3123 2896 2024 Electrical resistivity/ ohm em (293 K) 1014 -1016 50 50 10-5 2 x 10-5 (a) Why does carbon not form C4+ ions?
(b) Why is density of lead higher than Sn?
(c) Why is melting point of TiD lowest in group 14?
(d) Which is better conductor lead or Tin and why?
(e) Why does \(\mathrm{CF}_{6}^{2}\) Dot formed but \(\mathrm{SiF}_{6}^{2-}\)exists?
(f) Why is CO2 gas whereas Si02 solid?
(g) What is carborundum? Give its use.(a)
Case Study
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Answers
The p-block Elements Case Study Questions With Answer Key Answer Keys
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(a) It is due to inert pair effect in Pb2+ which is more stable than Pb4+. Pb 4+ gains 2 electrons to form Pb2+.
(b) It reacts with haemoglobin to form carboxy haemoglobin which does not act as oxygen carries.
(c) These are Biocompatible.
(d) They act as water softeners
(e) It has soccer ball-like structure and does not have edges, therefore, impurities can't enter.
(f) It is tetrahedral, sp3 hybridised.
(g) It reacts with SiO2 present in glass to form sodium silicate.
2NaOH + SiO2 \(\longrightarrow\) Na2SiO3 + H2O -
(a) Because it is not able to release H+ ions on its own. It receives OH- from water molecule to complete its octet and in turn releases H+ ions, hence acts as weak Lewis acid.
B(OH)3 + 2HOH \(\longrightarrow\) \(\left[\mathrm{B}(\mathrm{OH})_{4}\right]^{\ominus}\) H3P+
(b) B2 O3 is formed.
\(\mathrm{H}_{3} \mathrm{BO}_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{HBO}_{2} \longrightarrow \mathrm{B}_{2} \mathrm{O}_{3}\)
Boric acid Metaboric acid Boron trioxide
(c) Sodium borohydride is form
2NaH + B2H \(\longrightarrow\) 2NaBH4
Sodium Sodium
hydride borohydride
(d) B2H6 ,NaI and H2 are formed.
2NaBH4 + I \(\longrightarrow\) 2Nal + B2H2 + H2
(e) It acts as reducing agent. -
(a) It is because its fourth ionisation enthalpy is very high.
(b) It is because Pb has more atomic mass per unit volume than tin.
(c) It is due to weak metallic bonds.
(d) Sn is better conductor because it has lower resistivity.
(e) It is because 'C' does not have d-orbitals, can't expand its valency beyond 4.
Si has vacant d-orbitals and can expand its valency to 6.
(f) CO2 exist as discrete molecule, has weak van der Waals' forces of attraction whereas SiO2 is 3D covalent solid.
(g) Silicon carbide is called carborundum. It is used as abrasive.
Case Study
