Samacheer Kalvi 10th Science Solutions Chapter 8 Pdf
Tamilnadu Board Samacheer Kalvi 10th Science Solutions Chapter 8: Tamilnadu State Board Solution Class 10 Science Chapter 8 – Periodic Classification of Elements.
Samacheer Kalvi 10th Science Solutions Chapter 8: Overview
Board |
Samacheer Kalvi |
Class |
10 |
Subject |
Science |
Chapter |
8 |
Chapter Name |
Periodic Classification of Elements |
Samacheer Kalvi 10th Science Solutions Chapter 8
I.) Choose the best answer.
1.) The number of periods and groups in the periodic table are______.
a) 6, 16
b) 7, 17
c) 8, 18
d) 7, 18
Ans:d) 7, 18
Because, there are 7 periods and 18 groups in the periodic table.
2.) The basis of modern periodic law is______.
a) atomic number
b) Atomic mass
c) Isotopic mass
d) Number of neutrons
Ans:a) atomic number
Because, modern periodic law is based on atomic number.
3.) ____ group contains the member of halogen family.
a) 17th
b) 15th
c) 18th
d) 16th
Ans:a) 17th
4.) ____ is a relative periodic property
a) Atomic radii
b) Ionic radii
c) Electron affinity
d) Electronegativity
Ans:d) Electronegativity
5.) Chemical formula of rust is ________.
a) FeO.xH2O
b) FeO4.xH2O
c) Fe2O3.xH2O
d) FeO
Ans:c) Fe2O3.xH2O
6.) In the alumino thermic process the role of Al is _____.
a) Oxidizing agent
b) Reducing agent
c) Hydrogenating agent
d) Sulphurising agent
Ans:b) Reducing agent
Because, Al is the powerful reducing agent.
7.) The process of coating the surface of metal with a thin layer of zinc is called______.
a) Painting
b) Thinning
c) Galvanization
d) Electroplating
Ans:c) Galvanization
8.) Which of the following have inert gases 2 electrons in the outermost shell.
a) He
b) Ne
c) Ar
d) Kr
Ans:a) He
Because, He is the inert gas element or zero group element with stable electronic configuration with atomic number 2.
9.) Neon shows zero electron affinity due to _____.
a) Stable arrangement of neutrons
b) Stable configuration of electrons
c) Reduced size
d) Increased density
Ans:b) Stable configuration of electrons
Because, Neon is the zero group element with all shells are completely filled i.e. having stable electronic configuration.
10.) ______ is an important metal to form amalgam.
a) Ag
b) Hg
c) Mg
d) Al
Ans: b) Hg
Because, amalgam is the alloy of mercury with another metal.
II.) Fill in the blanks
1.) If the electronegativity difference between two bonded atoms in a molecule is greater than 1.7, the nature of bonding is ______
Ans: If the electronegativity difference between two bonded atoms in a molecule is greater than 1.7, the nature of bonding isionic bonding.
2.) ________ is the longest period in the periodical table.
Ans: Sixth periodis the longest period in the periodical table.
3.) ________ forms the basis of modern periodic table.
Ans: Atomic numberforms the basis of modern periodic table
4.) If the distance between two Cl atoms in Cl2 molecule is 1.98Å, then the radius of Cl atom is ________.
Ans:If the distance between two Cl atoms in Cl2 molecule is 1.98Å, then the radius of Cl atom is 0.99A0.
5.) Among the given species A–, A+, and A, the smallest one in size is ________.
Ans: Among the given species A–, A+, and A, the smallest one in size is A+.
6.) The scientist who propounded the modern periodic law is ________.
Ans: The scientist who propounded the modern periodic law is Henry Moseley.
7.) Across the period, ionic radii ________ (increases, decreases).
Ans: Across the period, ionic radii decreases.
8.) ________ and ________ are called inner transition elements.
Ans: Lanthanides and Actinidesare called inner transition elements.
9.) The chief ore of Aluminium is ________.
Ans:The chief ore of Aluminium is bauxite.
10.) The chemical name of rust is ________.
Ans:The chemical name of rust is Hydrated Ferric Oxide.
III.) Match the following.
Ans:
1) Galvanization: Coating with zinc
2) Calcination: Heating in the absence of air
3) Redox reaction:Alumino thermic process
4) Dental filling: Silver-tin amalgam
5) Group 18 elements: Noble gas elements
IV.) True or False: (If false give the correct statement)
1.) Moseley’s periodic table is based on atomic mass
Ans: False
Correct statement: Moseley’s periodic table is based on atomic number.
2.) Ionic radius increases across the period from left to right.
Ans: False
Correct statement: Ionic radius decreases across the period from left to right.
3.) All ores are minerals; but all minerals cannot be called as ores;
Ans: True
4.) Al wires are used as electric cables due to their silvery white colour.
Ans: False
Correct statement: Al wires are used as electric cables because they are good conductors of electricity.
5.) An alloy is a heterogenous mixture of metals.
Ans: False
Correct statement: An alloy is a homogeneous mixture of metals.
V.) Assertion and Reason answer the following questions using the data given below:
i) A and R are correct, R explains the A.
ii) A is correct, R is wrong.
iii) A is wrong, R is correct.
iv) A and R are correct, R doesn’t explains A.
1.) Assertion : The nature of bond in HF molecule is ionic
Reason: The electronegativity difference between H and F is 1.9
Ans: i) A and R are correct, R explains the A.
2.) Assertion : Magnesium is used to protect steel from rusting
Reason: Magnesium is more reactive than iron
Ans:i) A and R are correct, R explains the A.
3.) Assertion: An uncleaned copper vessel is covered with greenish layer.
Reason: copper is not attacked by alkali
Ans:iv) A and R are correct, R doesn’t explains A.
VI. Short answer questions
1) A is a reddish brown metal, which combines with O2 at < 1370 K gives B, a black coloured compound. At a temperature > 1370 K, A gives C which is red in colour. Find A, B and C with reaction.
Ans:
Here, A is the Copper which on heating at different temperatures in the presence of oxygen gives two forms of oxides CuO and Cu2O.
Thus,
At below 1370K, 2Cu + O2 → 2CuO(Copper II Oxide – Black)
At above 1370K, 4Cu + O2 → 2Cu2O (Copper I Oxide – Red)
Hence, here A→ Copper Cu
B→ Copper II Oxide – Black
C→ Copper I Oxide – Red
2) A is a silvery white metal. A combines with O2 to form B at 800oC, the alloy of A is used in making the aircraft. Find A and B
Ans:
Here, A is the aluminium which on heating at 8000C burns very brightly and forms its oxide.
Thus, at 8000C: 4Al + 3O2 → 2Al2O3 (Aluminium Oxide)
And the alloy of aluminium Duralumin is used in making aircraft.
Here, A→ Aluminium
B→ Aluminium oxide
3) What is rust? Give the equation for formation of rust.
Ans:
When iron is exposed to moist air, it forms a layer of brown hydrated ferric oxide on its surface which is called as rust and the process of formation of rust is called as rusting.
Following is the equation for formation of rust:
4Fe + 3O2 + xH2O → 2Fe2O3•xH2O (Rust)
4) State two conditions necessary for rusting of iron.
Ans:
When iron is exposed to moist air, it forms a layer of brown hydrated ferric oxide on its surface which is called as rust and the process of formation of rust is called as rusting.
Following is the equation for formation of rust:
4Fe + 3O2 + xH2O → 2Fe2O3•xH2O (Rust)
Thus, presence of oxygen and water is most essential for rusting process of iron.
Impurities like water vapour, acids, salts and carbon dioxide in iron may hasten the rust.
VII.) Long answer questions
1)
Ans:
a) State the reason for addition of caustic alkali to bauxite ore during purification of bauxite.
Because, naturally bauxite is not soluble in normal solvents, hence by adding the caustic alkali to bauxite helps in extracting the aluminium.
Bauxite ore is finely ground and heated under pressure with the solution of concentrated caustic soda solution at 1500C to form sodium meta aluminate.
Al2O3 + 2NaOH → 2NaAlO2 + H2O
Bauxite Sodium meta aluminate
b.) Along with cryolite and alumina, another substance is added to the electrolyte mixture. Name the substance and give one reason for the addition.
Along with cryolite and alumina, another substance is added to the electrolyte mixture which is called as Fluorspar. By adding Fluorspar the fusion temperature of the electrolyte will be lowered.
2) The electronic configuration of metal A is 2, 8,18,1. The metal A when exposed to air and moisture forms B a green layered compound. A with con. H2SO4 forms C and D along with water. D is a gaseous compound. Find A, B, C and D.
Ans:
Given electronic configuration is 2, 8, 18, 1 and it is the electronic configuration of copper with atomic number Z=29.
When copper is exposed to air and moisture forms copper carbonate, a green layered compound which is shown in below reaction.
2Cu + O2 + CO2 + H2O → CuCO3•Cu(OH)2
Copper Copper carbonate
Also, copper reacting with conc. H2SO4 gives copper sulphate and sulphur dioxide.
Cu + conc. H2SO4→ CuSO4 + SO2↑ + 2H2O
Copper Copper sulphate sulphur dioxide
Thus, here A →Copper Cu
B →Copper Carbonate CaCO3
C →Copper sulphate CuSO4
D →Sulphur dioxide SO2
3) Explain smelting process.
Ans:
Smelting:
The charge consisting of roasted ore, coke and limestone in the ratio 8:4:1 is smelted in blast furnace through the hopper arrangement at the top. There are three important regions in the furnace which are explained as below:
a.) The lower region: Combustion zone
In this region, coke burns with oxygen and forms CO2 when the charge come in contact with the hot blast air, at 15000C.
C + O2 →CO2 + Heat
As here heat is liberated, it is an exothermic reaction.
b.) The middle region: Fusion zone
In this region CO2 is reduced to CO at 10000C.
CO2 + C → 2CO – heat
And limestone decomposes to calcium oxide and CO2, at 10000C.
CaCO3 → CaO + CO2 – Heat
As in this both reactions, heat is absorbed hence this are the endothermic reactions.
And also, calcium oxide combines with silica and forms calcium silicate slag.
CaO + SiO2 → CaSiO3
c.) The upper region: Reduction zone
In this region, the carbon monoxide reduces ferric oxide to form a fairly pure spongy iron, at temperature 4000C.
Fe2O3 + 3CO → 2Fe + 3CO2↑
After removing the slag, molten iron is collected at the bottom.
And the iron formed is called as pig iron.
After remelting it is cast into different moulds and it is called as cast iron.
VIII.) HOT questions
1) Metal A belongs to period 3 and group 13. A in red hot condition reacts with steam to form B. A with strong alkali forms C. Find A, B and C with reactions
Ans:
Here, A is the Al metal which belongs to period 3 and group 13.
In red hot condition, Al reacts with steam and forms aluminium oxide.
2Al + 3H2O → Al2O3 + 3H2↑
Aluminium Steam Aluminium Oxide
Also, Al reacts with strong alkali to form sodium meta aluminate.
2Al + 2NaOH + 2H2O → 2NaAlO2 + 3H2↑
Thus, here A →Aluminium Al
B →Aluminium Oxide Al2O3
C→ Sodium meta aluminate NaAlO2
2) Name the acid that renders aluminium passive. Why?
Ans:
Dilute or concentrated HNO3 does not attacks on Al, but it renders Al passive because of the formation of oxide film on its surface.
3)
Ans:
Identify the bond between H and F in HF molecule.
The bond formed between H and F in HF molecule is ionic bond.
What property forms the basis of identification?
The property of electronegativity forms the basis of identification.
How does the property vary in periods and in groups?
In a period, from left to right the nuclear charge increases due to which the electrostatic force of attraction between electrons and nucleus increases and hence electronegativity also increases.
But in group, from top to bottom, the energy levels are adding continuously due to which the electronegativity also decreases.