🧪 Chemistry · Chapter 8 · Samacheer Kalvi Grade 10

Samacheer Class 10 Science - Periodic Classification of Elements

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Samacheer Kalvi Class 10 Science Chapter 8 Periodic Classification of Elements book back questions and answers — all MCQ, short answer and long answer with complete solutions. This chapter covers Döbereiner's triads, Newlands' law of octaves, Mendeleev's periodic table, modern periodic law, periods and groups and periodicity of properties as per the TN SSLC Grade 10 Chemistry syllabus.

Periodic Classification of Elements — key concepts & quick answers

What is the modern periodic law?
The physical and chemical properties of elements are periodic functions of their atomic numbers.
What are groups and periods in the periodic table?
The vertical columns are called groups (18 in all) and the horizontal rows are called periods (7 in all). Elements in the same group have similar chemical properties.
How does atomic radius vary across a period and down a group?
Atomic radius decreases across a period (left to right) as the nuclear charge increases, and increases down a group as new electron shells are added.
What is metallurgy?
Metallurgy is the science and process of extracting metals from their ores and refining them for use.
What is an alloy? Give an example.
An alloy is a homogeneous mixture of two or more metals, or of a metal with a non-metal. For example, brass is an alloy of copper and zinc.
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Multiple Choice Questions Fill in the Blanks True or False Match the Following Assertion & Reasoning Short Answer Questions Numerical Problems Long Answer Questions
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MCQI. Multiple Choice Questions1 mark each
Q.1 The number of periods and groups in the periodic table are______.
✓ Answer: (D) 7,18
Q.2 The basis of modern periodic law is______.
✓ Answer: (A) atomic number
Q.3 _____ group contains the member of halogen family.
✓ Answer: (A) 17th
Q.4 _____ is a relative periodic property
✓ Answer: (A) atomic radii
Q.5 Chemical formula of rust is ________.
✓ Answer: (C) Fe2O3.xH2O
Q.6 In the alumino thermic process the role of Al is _____.
✓ Answer: (B) reducing agent
Q.7 The process of coating the surface of metal with a thin layer of zinc is called______.
✓ Answer: (C) galvanization
Q.8 Which of the following have inert gases 2 electrons in the outermost shell.
✓ Answer: (A) He
Q.9 Neon shows zero electron affinity due to _____.
✓ Answer: (B) stable configuration of electrons
Q.10 ______ is an important metal to form amalgam.
✓ Answer: (B) Hg
FillII. Fill in the Blanks1 mark each
#Statement (Answer in bold)
1If the electronegativity difference between two bonded atoms in a molecule is greater than 1.7, the nature of bonding is ionic.
26th and 7th are the longest periods in the periodic table; each has 32 elements.
3Atomic number forms the basis of modern periodic table.
4If the distance between two Cl atoms in Cl2 molecule is 1.98 Å, then the radius of Cl atom is 0.99 Å.
5Among the given species A-, A+ and A, the smallest one in size is A+.
6The scientist who propounded the modern periodic law is Henry Moseley.
7Across the period, ionic radii decreases.
8Lanthanides and actinides are called inner transition elements.
9The chief ore of aluminium is bauxite.
10The chemical name of rust is hydrated ferric oxide.
MatchIII. Match the Following1 mark each
Column AColumn B
Dental filling : silversilver amalgam (dental amalgam)
T/FIV. True or False1 mark each
#StatementAnswerCorrection (if False)
1Moseley's periodic table is based on atomic mass.FalseMoseley's periodic table is based on atomic number.
2Ionic radius increases across the period from left to right.FalseIonic radius decreases across the period from left to right.
3All ores are minerals; but all minerals cannot be called ores.True
4Al wires are used as electric cables due to their silvery white colour.FalseAl wires are used in electric cables because they are good conductors.
5An alloy is a heterogenous mixture of metals.FalseAn alloy is a homogeneous mixture of metals.
A&RV. Assertion & Reasoning2 marks each
Q.1 Assertion : The nature of bond in HF molecule is ionic
✓ Answer
Reason: The electronegativity difference between H and F is 1.9.
Answer: iv. Assertion is false, but Reason is true. HF has a polar covalent bond, not an ionic bond.
Q.2 Assertion : Magnesium is used to protect steel from rusting
✓ Answer
Reason: Magnesium is more reactive than iron.
Answer: i. Both Assertion and Reason are true, and Reason correctly explains the Assertion. Magnesium acts as a sacrificial metal and protects steel from rusting.
Q.3 Assertion : An uncleaned copper vessel is covered with greenish layer.
✓ Answer
Reason: Copper is not attacked by alkali.
Answer: ii. Both Assertion and Reason are true, but Reason does not explain the Assertion. The green layer on copper is basic copper carbonate formed by reaction with air, moisture and carbon dioxide.
ShortVI. Short Answer Questions2 marks each
Q.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.
✓ Answer
A is copper (Cu), the reddish brown metal.
At a temperature below 1370 K, copper forms black copper(II) oxide:
2Cu + O2 -> 2CuO
So, B is copper(II) oxide (CuO).
At a temperature above 1370 K, copper forms red copper(I) oxide:
4Cu + O2 -> 2Cu2O
So, C is copper(I) oxide (Cu2O).
Q.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
✓ Answer
A is aluminium (Al), a silvery white metal. At about 800°C it combines with oxygen to form aluminium oxide:
4Al + 3O2 -> 2Al2O3
So, B is aluminium oxide (Al2O3). Aluminium alloys are used in making aircraft because they are light and strong.
Q.3 What is rust? Give the equation for formation of rust.
✓ Answer
When iron is exposed to moist air, it forms a layer of brown hydrated ferric oxide on its surface. This compound is known as rust and the phenomenon is known as rusting.
4Fe + 3O2 + xH2O -> 2Fe2O3.xH2O (rust)
Q.4 State two conditions necessary for rusting of iron.
✓ Answer

Rusting, the common term for the corrosion of iron and its alloys, is an electrochemical process that requires specific environmental conditions to occur. The two primary conditions necessary for the rusting of iron are the presence of oxygen and the presence of water or moisture. Oxygen acts as an oxidizing agent, while water serves as an electrolyte, facilitating the movement of ions and completing the electrochemical circuit. In the absence of either oxygen or moisture, iron will not rust. For example, iron kept in dry air or in vacuum-sealed containers does not rust, nor does iron submerged in oil or other non-aqueous liquids.

NumericalVII. Numerical Problems3 marks each
Q.1 a) State the reason for addition of caustic alkali to bauxite ore during purification of bauxite. b) Along with cryolite and alumina, another substance is added to the electrolyte mixture. Name the substance and give one reason for the addition.
✓ Answer

a) During the purification of bauxite ore, caustic soda solution (sodium hydroxide, NaOH) is added to dissolve the alumina (Al₂O₃) present in the ore. Alumina reacts with the hot, concentrated caustic soda solution to form soluble sodium meta-aluminate (Na[Al(OH)₄] or NaAlO₂), while the impurities like iron oxides, silica, and titanium dioxide, which are insoluble in alkali, remain undissolved. This selective dissolution allows for the separation of alumina from its ore. b) Along with cryolite (Na₃AlF₆) and alumina (Al₂O₃), another substance, fluorspar (CaF₂), is added to the electrolyte mixture in the electrolytic refining of aluminum. The primary reason for adding fluorspar is to lower the fusion temperature of the mixture. Pure alumina has a very high melting point (over 2000°C), making electrolysis impractical and energy-intensive. The addition of cryolite and fluorspar significantly lowers the melting point of the electrolyte to around 900-1000°C, reducing energy consumption and increasing the fluidity of the molten bath, which improves the efficiency of the electrolysis process.

Q.2 The electronic configuration of metal A is 2,8,18,1.
✓ Answer
The metal A is copper (Cu). When copper is exposed to air and moisture, it forms a green layer of basic copper carbonate:
2Cu + O2 + CO2 + H2O -> CuCO3.Cu(OH)2
So, B is basic copper carbonate.
With concentrated sulphuric acid, copper forms copper sulphate, sulphur dioxide and water:
Cu + 2H2SO4 -> CuSO4 + SO2 + 2H2O
So, C is copper sulphate (CuSO4) and D is sulphur dioxide (SO2).
Q.3 Explain smelting process.
✓ Answer

Smelting is a pyrometallurgical process used to extract a metal from its ore by heating it strongly in the presence of a reducing agent, typically in a furnace. The primary objective of smelting is to reduce the metallic oxide ore to its molten metallic state. During this process, a flux, which is a chemical agent like limestone or silica, is often added. The flux reacts with the unwanted impurities in the ore (such as silica, alumina, or basic oxides) to form a molten slag. This slag has a lower melting point than the metal and is immiscible with it, allowing it to be easily separated from the molten metal by decantation or tapping. Smelting is a crucial step in the extraction of many important metals, including iron, copper, lead, and tin.

HOTIX. Higher Order Thinking3 marks each
Q.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
✓ Answer
A is aluminium (Al), because it belongs to period 3 and group 13.
In red hot condition, aluminium reacts with steam to form aluminium oxide:
2Al + 3H2O -> Al2O3 + 3H2
So, B is aluminium oxide (Al2O3).
With strong alkali, aluminium forms sodium meta aluminate:
2Al + 2NaOH + 2H2O -> 2NaAlO2 + 3H2
So, C is sodium meta aluminate (NaAlO2).
Q.2 Name the acid that renders aluminium passive. Why?
✓ Answer

Nitric acid renders aluminium passive. This phenomenon, known as passivation, occurs because nitric acid oxidizes the surface of the aluminium metal. A thin, adherent, and protective layer of aluminium oxide (Al2O3) is formed on the aluminium's surface. This oxide film acts as a barrier, preventing further reaction between the underlying aluminium metal and the nitric acid. Therefore, the aluminium is rendered passive and does not react with the acid.

Q.3 a) Identify the bond between H and F in HF molecule.
✓ Answer

a) The bond between H and F in the HF molecule is a polar covalent bond. b) This identification is based on the significant difference in electronegativity between hydrogen and fluorine. Fluorine is highly electronegative, while hydrogen is less electronegative. This unequal sharing of electrons creates a partial positive charge on the hydrogen atom and a partial negative charge on the fluorine atom, resulting in a polar bond. c) Electronegativity generally increases from left to right across a period in the periodic table due to an increasing nuclear charge and a decrease in atomic radius. Conversely, electronegativity decreases as one moves down a group because the atomic radius increases and the outermost electrons are farther from the nucleus, experiencing less attraction.

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