p-Block Elements and their Compounds Terminal Questions with Answer NIOS
p-Block Elements and their Compounds Terminal Questions with Answer NIOS
National Institute of Open Schooling (NIOS)
Sr. Secondary, Module-6
Lession-19 and 20: p-Block Elements and their Compounds Terminal Questions with Answer
Terminal QUESTIONS with Answer
p-Block Elements and their Compounds-I Terminal Questions with Answer
1. Why boric acid is not a protonic acid?
Answer:
B(OH)
3 + 2H
2O → B(OH)
4− + H
3O
+
Boric acid does not contain hydrogen ions so not a protonic acid but they can accept electrons from OH
− ion. Therefore, it is a Lewis acid. So, we can say that Boric acid is not a protic acid.
2. Discuss the structure of boric acid.
Answer:
Each boric acid molecule contains boron-oxygen single bonds. The boron atom occupies the central position and is linked to three hydroxide groups. Boric acid is sp
2 hybridized. The overall molecular geometry of boric acid is trigonal planar. The structure of boric acid is illustrated below-
3. Draw the Lewis structure of CO and CO2 molecule.
Answer:
4. Why does BF3 acts as Lewis acid?
Answer:
BF3 is an electron deficient species and can accept the lone pair of electrons. That's why it acts like Lewis acid.
5. What is catenation? Why does carbon shows catenation but silicon does not?
Answer:
Catenation is the ability of an atom to form bonds with other atoms of the same element. Smaller size and high electronegativity of carbon makes it to form pπ-pπ multiple bonds with itself or with other elements of the similar size but silicon due to comparatively larger size and smaller electronegativity can not form pπ-pπ multiple bonds. That's why carbon shows catenation property to large extent than silicon.
6. Compare the structure of CO2 and SO2.
Answer:
Carbon dioxide is a linear molecule while sulfur dioxide is a bent molecule. Both molecules contain polar bonds (see bond dipoles on the Lewis structures below), but carbon dioxide is a nonpolar molecule while sulfur dioxide is a polar molecule.
7. Describe briefly the Haber process for manufacture of ammonia.
Answer:
Haber process involves the direct combination of nitrogen and hydrogen in 1:3 ratio by volume at very high temperature(400 – 450oC) and pressure(150-200atm). Generally iron is used as a catalyst and molybdenum as a promoter.
N2 + 3H2 ⇌ 2NH3
The reaction is exothermic because it progresses in a forward direction with a significant decrease in volume. High pressure would favour the formation of ammonia, according to Le Chatelier's principle.
8. Why is graphite a conductor of electricity but diamond is not?
Answer:
In graphite, each carbon atom is linked to three carbon atoms via a carbon-carbon covalent bond. The fourth electron is delocalized so it is mobile and can conduct electricity. However, diamonds have no free electrons because each carbon is four other carbon atoms covalently bonded in a tetrahedral manner. That's why graphite is a good conductor of electricity but not diamond.
9. Explain different types of silicons.
Answer:
Silicones are polymeric compounds.The polymer chain containing alternatively linked silicon and oxygen atoms. Alkyl and aryl groups are attached to the polymer ackbone through covalent bonds to the silicon atoms.
Followings are the type of silicones-
(i) Linear silicones: They are obtained by the hydrolysis and subsequent condensation of dailkyl or diaryl silicon chlorides.
(a) Silicone rubbers: These silicones are bridged together by methylene or similar groups.
(b) Silicone resins: They are obtained by blending silicones with organic resins such as acrylic esters.
(ii) Cyclic silicones: These are obtained by the hydrolysis of
R2SiCl2
(iii) Cross linked silicones: They are obtained by hydroysis of RSiCl3
10. Explain ortho and pyro silicates.
Answer:
Ortho silicates: The simplest silicates which contain discrete [SiO4]4- tetrahedral units are called ortho silicates or neso silicates. Examples: Phenacite – Be2SiO4 (Be2+) ions are tetrahedrally surrounded by O2- ions). Olivine – (Fe/Mg)2 SiO4 (Fe2+ and Mg2+ cations are octahedrally surrounded by O2- ions).
Pyro silicates: Silicates which contain [Si2O7]6- ions are called pyro silicates (or) Soro silicates They are formed by joining two [SiO4]4- tetrahedral units by sharing one oxygen atom at one comer, (one oxygen is removed while, joining).
Example: Thortveitite – SC2Si2O7.
p-Block Elements and their Compounds-II Terminal Questions with Answer
1. Which one of the following oxides can react with an acid as well as with an alkali : SO2, CaO, ZnO, MgO?
Answer:
ZnO can react with acids as well as bases to forms salts and behaves like amphoteric.
ZnO + H2SO4 = ZnSO4 + H2O
ZnO + 2NaOH = Na2ZnO2 + H2O
2. Write two oxides which do not react with either acids or alkalies. Which type of oxides are they?
Answer:
Oxides which does not react with either acids or alkalies is called neutral oxide. Examples are- carbon monooxide, nitrous oxide etc.
3. Is ozone an allotrope of oxygen? Which is more soluble in water oxygen or ozone?
Answer:
Yes, ozone is the allotrope of oxygen. Ozone is almost ten times more soluble in water than oxygen gas because it is a polar molecule and thus is soluble in polar solvents.
4. What is the state of hybridization of the central oxygen atom in O3 molecule?
Answer:
Ozone generally forms a V-shaped molecule where the central oxygen atom is sp2 hybridized.
5. Why does ozone tail mercury?
Answer:
Being a strong oxidising agent due to presence of nascent oxygen, Mercury on shaking with ozone is oxidised to mercurous oxide (Hg2O). Therefore, mercury loses its meniscus and sticks to glass. This is called tailing of mercury.
O3 + 2Hg → Hg2O + O2
6. Which property of ozone makes it useful for bleaching?
Answer:
The oxidizing property of ozone makes it a very useful ingredient for bleaching purpose where ozone is oxidized to nascent oxygen which is actually responsible for the beaching action.
Colored Substance + O3 = Colorless substance + O2
7. Write the conditions which are required to convert SO2 to SO3 in the manufacture of sulphuric acid by contact process.
Answer:
In contact process, the conversion of sulphur dioxide to sulphur trioxide is done by taking a 1:1 ratio of SO2 and O2 at 400oC−450oC and 1-2 atm pressure in presence of Vanadium pentaoxide.
SO2 + O2 ---V2O5---> 2SO3
8. What is oleum?
Answer:
During the manufacturing of sulphuric acid by contact process, SO3 is absorbed by concentrated sulphuric acid and oleum is formed which is basically a dense and corrosive liquid and commonly called fuming Sulphuric acid.
SO3 + H2SO4 → H2S2O7
9. Why is sulphuric acid known as king of chemicals?
Answer:
Sulphuric acid is known as king of chemicals because it is a strong acid , highly reactive and have highly corrosive property.
10. Write down the structure of oxo acids of sulphur.
Answer:
11. Fluorine never acts as a central atom in any interhalogen compounds. Why?
Answer:
In interhalogen compounds, the central atom must be the more electropositive and bigger element. Fluorine has the smallest size and is the most electronegative element. Fluorine exhibits only one oxidation state that is -1 and therefore can not be the central atom in interhalogen compounds.
12. Draw the structure of BrF4−.
Answer:
13. Arrange the hydrogen halides in order of their acid strength in aqueous solution.
Answer:
The acidic strength of hydrogen halides increases in the following order-
HF < HCl < HBr < HI
14. Why is F2O not known as fluorine oxide? Instead it is known as oxygen fluoride, OF2.
Answer:
F is most electronegative element and oxygen is less electronegative element. Therefore, compounds of oxygen with fluorine is known as oxygen fluorides and not fluorine oxide.
15. Which is the strongest acid among the oxoacids of chlorine and why?
Answer:
Perchloric acid is the most strongest acid among the oxoacids of chlorine because it possess the highest number of oxygen atoms.
HOCl < HO2Cl < HO3Cl < HOCl
16. What happens when XeF4 reacts with SbF5? Write the complete equation for the reaction.
Answer:
When XeF4 reacts with strong Lewis acid SbF5 they forms compound.
XeF4 + SbF5 → [XeF3]+[SbF6]-
NIOS Sr. Secondary Chemistry Intext Questions of All Chapters with Answer
