10th Science - Book Back Answers - Chemistry Unit 8 - English Medium Guides

 

SSLC / 10th - Science - Book Back Answers - Chemistry Unit 8 - English Medium

Tamil Nadu Board 10th Standard  Science - Chemistry Unit 8: Book Back Answers and Solutions

    This post covers the book back answers and solutions for Unit 8 – Chemistry from the Tamil Nadu State Board 10th Standard  Science textbook. These detailed answers have been carefully prepared by our expert teachers at KalviTips.com.

    We have explained each answer in a simple, easy-to-understand format, highlighting important points step by step under the relevant subtopics. Students are advised to read and memorize these subtopics thoroughly. Once you understand the main concepts, you’ll be able to connect other related points with real-life examples and confidently present them in your tests and exams.

    By going through this material, you’ll gain a strong understanding of Chemistry Unit 8 along with the corresponding book back questions and answers (PDF format).

Question Types Covered:

• 1 Mark Questions: Choose the correct answer, Fill in the blanks, Identify the correct statement, Match the following 
• 2 Mark Questions: Answer briefly 
• 3, 4, and 5 Mark Questions: Answer in detail

All answers are presented in a clear and student-friendly manner, focusing on key points to help you score full marks.

All the best, Class 10 students! Prepare well and aim for top scores. Thank you!

Topic: Periodic Classification of Elements

I. Choose the correct 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
Answer Key:
(d) 7, 18

2. The basis of modem periodic law is ____.
(a) atomic number
(b) atomic mass
(c) isotopic mass
(d) number of neutrons.
Answer Key:
(a) atomic number

3. ……….. group contains the member of the halogen family.
(a) 17th
(b) 15th
(c) 18th
(d) 16th
Answer Key:
(a) 17th
 
4. _______ is a relative periodic property.
(a) atomic radii
(b) ionic radii
(c) electron affinity
(d) electronegativity.
Answer Key:
(b) ionic radii

5. Chemical formula of rust is:
(a) Fe0.xH₂O
(b) Fe04.xH₂O
(c) Fe₂O₃. xH₂O
(d) FeO
Answer Key:
(c) Fe₂O₃. xH₂O

6. In the aluminothermic process the role of Al is:
(a) oxidizing agent
(b) reducing agent
(c) hydrogenating agent
(d) sulphurising agent
Answer Key:
(b) reducing agent

7.The process of coating the surface of the metal with a thin layer of zinc is called ____.
(a) painting
(b) thinning
(c) galvanization
(d) electroplating.
Answer Key:
(c) galvanization

8. Which of the following inert gas has electrons in the outermost shell?
(a) He
(b) Ne
(c) Ar
(d) Kr
Answer Key:
(a) He

9. Neon shows zero electron affinity due to ____.
(a) stable arrangement of neutrons
(b) stable configuration of electrons
(c) reduced size
(d) increased density.
Answer Key:
(b) stable configuration of electrons

10. ……….. is an important metal to form amalgam.
(a) Ag
(b) Hg
(c) Mg
(d) Al
Answer Key:
(b) Hg

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 ………..
2. …………. is the longest period in the periodical table.
3. ………… forms the basis of modern periodic table.
4. If the distance between two Cl atoms in Cl₂ molecule is 1.98 Å, then the radius of Cl atom is ………..
5. Among the given species A^- A⁺, and A, the smallest one in size is ……….
6. The scientist who propounded the modern periodic law is …………
7. Across the period, ionic radii ………… (increases,decreases).
8. ……….. and ………… are called inner transition elements.
9. The chief ore of Aluminium is ………….
10. The chemical name of rust is ………….
Answer Key:
1. ionic
2. 6th (sixth) period
3. Atomic number
4. 0.99 Å
5. A⁺
6. Dimitri Mendeleev
7. decreases
8. Lanthanides, Actinides
9. bauxite
10. hydrated ferric oxide

III. Match the following

Column I		Column II
1	Galvanisation	i	Noble gas elements
2	Calcination	ii	Coating with Zn
3	Redox reaction	iii	Silver-tin amalgam
4	Dental filling	iv	Alumino thermic process
5	Group 18 elements	v	Heating in the absence of air

Answer Key:

Column I		Column II
1	Galvanisation	ii	Coating with Zn
2	Calcination	v	Heating in the absence of air
3	Redox reaction	iv	Alumino thermic process
4	Dental filling	iii	Silver-tin amalgam
5	Group 18 elements	i	Noble gas elements

IV.  True or False (If false give the correct statement) 

1. Moseley’s periodic table is based on atomic mass.
2. Ionic radius increases across the period from left to right.
3. All ores are minerals; but all minerals cannot be called as ores.
4. Aluminium wires are used as electric cables due to their silvery white colour.
5. An alloy is a heterogenous mixture of metals.
Answer Key:
1. False – Moseley’s periodic table is based on atomic number.
2. True
3. True
4. False – Aluminium wires are used as electric cables because it is a good conductor of heat and electricity.
5. False – An alloy is an homogeneous mixture of metals.

V.  Assertion & Reasoning

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.
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer Key:
(a) Assertion and Reason are correct, Reason explains the Assertion.

2. Assertion: Magnesium is used to protect steel from rusting.
Reason: Magnesium is more reactive than iron.
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer Key:
(a) Assertion and Reason are correct, Reason explains the Assertion.

3. Assertion: An uncleaned copper vessel is covered with greenish layer. Reason: copper is not attacked by alkali.
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer Key:
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.

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.
Answer Key:
(A) is a reddish brown metal – Copper
 $\mathbf{2}\mathbf{c}\mathbf{u}\mathbf{+}{\mathbf{o}}_{\mathbf{2}}\stackrel{\mathbf{<}\mathbf{1370}\mathbf{k}}{\mathbf{\to }}\mathbf{2}\mathbf{c}\mathbf{u}\mathbf{o}\mathbf{ }\mathbf{\left(}\mathbf{B}\mathbf{\right)}$

                            cupric oxide(Black)

$\mathbf{4}\mathbf{c}\mathbf{u}\mathbf{+}{\mathbf{o}}_{\mathbf{2}}\stackrel{\mathbf{<}\mathbf{1370}\mathbf{k}}{\mathbf{\to }}\mathbf{2}\mathbf{c}{\mathbf{u}}_{\mathbf{2}}\mathbf{o}\mathbf{ }\mathbf{\left(}\mathbf{c}\mathbf{\right)}$

                             cuprous oxide(Red)
A – Copper; B – Cupric oxide; C – Cuprous oxide.

2. A is a silvery white metal. A combines with O₂ to form B at 800°C, the alloy of A is used in making the aircraft. Find A and B.
Answer Key:
A – Silvery white metal – Aluminium
 $\mathbf{4}\mathbf{A}\mathbf{l}\mathbf{+}\mathbf{3}{\mathbf{o}}_{\mathbf{2}}\mathbf{\to }\mathbf{2}\mathbf{A}{\mathbf{l}}_{\mathbf{2}}{\mathbf{o}}_{\mathbf{3}} \left(\mathrm{B}\right)$

                       Aluminium
The alloys of aluminium, Duralumin and Magnalium are used in making the aircraft.
A – Aluminium; B – Aluminium oxide.

3. What is rust? Give the equation for the formation of rust.
Answer Key:
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 of formation of rust is known as rusting.
4Fe + 3O₂ + xH₂O → 2Fe₂O₃. xH₂O (Rust).

4. State two conditions necessary for rusting of iron.
Answer Key:
(i) The presence of water and oxygen is essential for the rusting of iron.
(ii) Impurities in the iron, the presence of water vapour, acids, salts and CO₂ speeds up rusting.

VII. Long answer questions

1. (a) State the reason for addition of caustic alkali to bauxite ore during purification of bauxite.
Answer Key:
Caustic alkali is added to bauxite, to dissolve bauxite ore and obtain a solution of sodium 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.
Answer Key:
CaF₂ (Fluorspar) is added along with cryolite and alumina. It is added to reduce the high melting point of the electrolyte.

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.
Answer Key:
Metal (A) with electronic configuration- 2, 8, 18, 1 is copper.
2Cu + O₂ + CO₂ + H₂O → CuCO₃. Cu(OH)₂(B)
Copper carbonate (Green layer)
$\mathbf{2}\mathbf{c}\mathbf{u}\mathbf{+}{\mathbf{o}}_{\mathbf{2}}\mathbf{+}\mathit{c}{\mathbf{o}}_{\mathbf{2}}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{o}\mathbf{\to }\mathit{c}\mathit{u}\mathit{c}{\mathit{o}}_{\mathbf{3}}\mathbf{.}\mathit{c}\mathit{u}\mathbf{\left(}\mathit{O}\mathit{H}{\mathbf{)}}_{\mathbf{2}}\mathbf{\right(}\mathit{B}\mathbf{)}$

                                copper carbonet (Green layer)

$\mathbf{c}\mathbf{u}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{S}{\mathbf{O}}_{\mathbf{4}}\mathbf{\to }\mathbf{C}\mathbf{u}\mathbf{S}{\mathbf{O}}_{\mathbf{4}}\mathbf{+}\mathbf{S}{\mathbf{O}}_{\mathbf{2}}\mathbf{\uparrow }\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$

                            copper     sulphet

                           sulphet (C) dioxide (D)
(A) – Copper (Cu)
(B) – Copper Carbonate (CuCO₃. Cu(OH)₂)
(C) – Copper Sulphate (CuSO₄)
(D) – Sulphur dioxide (SO₂)

3. Explain the smelting process.
Answer Key:
The roasted ore of copper is mixed with powdered coke and sand and is heated in a blast furnace to obtain matte (Cu₂S + FeS) and slag. The slag is removed as waste.
2 FeS + 3O₂ → 2 FeO + 2 SO₂
2 Cu₂S + 3O₂ → 2 Cu₂O + 2SO₂
Cu₂O + FeS → Cu₂S + FeO
FeO + SiO₂ → FeSiO₂ (slag)

 

 

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.
Answer Key:
Metal A belongs to Period 3 and Group 13. So metal ‘A’ is aluminium.
(A) in red hot condition reacts with steam to form ‘B’.
$\mathbf{2}\mathbf{A}\mathbf{l}\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\to }\mathbf{A}{\mathbf{l}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}$

           steam   alumium 

                         oxide(B)
‘A’ with strong alkali forms ‘C’

$\mathbf{2}\mathbf{A}\mathbf{l}\mathbf{+}\mathbf{2}\mathbf{N}\mathbf{a}\mathbf{O}\mathbf{H}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\to }\mathbf{2}\mathbf{N}\mathbf{a}\mathbf{A}\mathbf{l}{\mathbf{O}}_{\mathbf{2}}\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}\mathbf{\uparrow }$

                                           Sodium meta 

                                           aluminate (C)
(A) – Aluminium
(B) – Aluminium oxide
(C) – Sodium meta aluminate

 

2. Name the acid that renders aluminium passive. Why?
Answer Key:
Dilute or concentrated nitric acid (HNO₃) renders aluminium passive. Because nitric acid does not attack aluminium but it renders aluminium passive due to the formation of an oxide film on its surface.
 

3. (a) Identify the bond between H and F in HF molecule.
Answer Key:
Ionic, because the electronegativity difference is more than 1.7.
(b) What property forms the basis of identification?
Answer Key:
Electronegativity.
(c) How does the property vary in periods and in groups?
Answer Key:
In a period, from left to right the electronegativity increases because of the increase in the nuclear charge.
In a Group, from top to bottom, the electronegativity decreases because of the increase in size of the elements.