Tamil Nadu 10th Standard Science Book Solution | Lesson 8 – PERIODIC CLASSIFICATION OF ELEMENTS

Lesson.8 PERIODIC CLASSIFICATION OF ELEMENTS

PERIODIC CLASSIFICATION OF ELEMENTS - Book Back Answer

Lesson 8 > PERIODIC CLASSIFICATION OF ELEMENTS

Formulae
Metallic radiusDistance between the nuclei of a adjacent metal atoms / 2
Covalent radiusDistance between the nuclei of two covalently bonded atoms of the same element ina molecule / 2

I. Choose the best answer.

1. The number of periods and groups in the periodic table are __________.

  1. 6,16
  2. 7,17
  3. 8,18
  4. 7,18

Ans; 7,18

2. The basis of modern periodic law is __________.

  1. atomic number
  2. atomic mass
  3. isotopic mass
  4. number of neutrons

Ans; atomic number

3. __________ group contains the member of halogen family.

  1. 17th
  2. 15th
  3. 18th
  4. 16th

Ans;- 17th

4. __________ is a relative periodic property.

  1. Atomic radii
  2. Ionic radii
  3. Electron affinity
  4. Electronegativity

Ans; Electronegativity

5. Chemical formula of rust is __________.

  1. FeO.xH2O
  2. FeO4.xH2O
  3. Fe2O3.xH2O
  4. FeO

Ans; Fe2O3.xH2O

6. In the aluminothermic process the role of Al is __________.

  1. oxidizing agent
  2. reducing agent
  3. hydrogenating agent
  4. sulphurising agent

Ans; reducing agent

7. The process of coating the surface of the metal with a thin layer of zinc is called __________.

  1. painting
  2. thinning
  3. galvanization
  4. electroplating

Ans; galvanization

8. Which of the following have inert gases 2 electrons in the outermost shell.

  1. He
  2. Ne
  3. Ar
  4. Kr

Ans; He

9. Neon shows zero electron affinity due to __________.

  1. Stable arrangement of neutrons
  2. Stable configuration of electrons
  3. Reduced size
  4. Increased density

Ans; Stable configuration of electrons

10. __________ is an important metal to form amalgam.

  1. Ag
  2. Hg
  3. Mg
  4. Al

Ans; 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 __________.

Ans; Ionic

2. __________ is the longest period in the periodical table.

Ans; 6, 7

3. __________ forms the basis of modern periodic table.

Ans; Atomic Number

4. If the distance between two Cl atoms in Cl2 molecule is 1.98Å, then the radius of Cl atom is __________.

Ans; 0.99 A

5. Among the given species A–, A+, and A, the smallest one in size is __________.

Ans; A+

6. The scientist who propounded the modern periodic law is __________.

Ans; Henry Moseley

7. Across the period, ionic radii __________ (increases,decreases).

Ans; decreases

8. __________ and __________ are called inner transition elements.

Ans ; Lanthanides, Actinides

9. The chief ore of Aluminium is __________.

Ans; Bauxite

10. The chemical name of rust is __________.

Ans; hydrated ferric hydroxide

III. Match the following

  1. Galvanization – Noble gas elements
  2. Calcination – Coating with Zn
  3. Redox reaction – Silver–tin amalgam
  4. Dental filling – Alumino thermic process
  5. Group 18 elements – Heating in the absence of air

Ans ; 1 – b, 2 – e, 3 – d, 4 – c, 5 – a

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

1. Moseley’s periodic table is based on atomic mass. ( False )

  • Mosley’s periodic table is based on atomic numbers.

2. Ionic radius increases across the period from left to right. ( False )

  • Ionic radius decreases across the period from left to right.

3. All ores are minerals, but all minerals cannot be called ores; ( True )

4. Al wires are used as electric cables due to their silvery-white color. ( False )

  • Aluminum wires are used as electric cables due to their good conductor of heat and electricity.

5. An alloy is a heterogeneous mixture of metals. ( False )

  • An alloy is a homogeneous mixture of metals.

V. Assertion and Reason

Answer the following questions using the data given below:

  1. A and R are correct, R explains the A.
  2. A is correct, R is wrong.
  3. A is wrong, R is correct.
  4. A and R are correct, R doesn’t explain A.

1. Assertion: The nature of the bond in the HF molecule is ionic.
Reason: The electronegativity difference between H and F is 1.9.

  • Ans : (a) 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 : (c) A is wrong, R is correct

3. Assertion: An uncleaned copper vessel is covered with a greenish layer.
Reason: Copper is not attacked by alkali.

  • Ans : (a) A and R are correct, R explains the A

VI. Short answer questions:

1. Aisa reddish-brown metal, which combines with O, at < 1370 K gives B, a black colored compound. At a temperature > 1370 K, A gives C which is red in color. Find A, B, and C with reaction.

  • A- reddish-brown metal – Copper
  • When copper is heated at < 1370 K in the presence of oxygen, copper forms black color Copper II oxide (CuO).
  • 2Cu+ O2 —  below 1370K  —→  2CuO (copper II oxide)
  • When copper is heated at > 13`70 K in the presence of oxygen, copper forms red color Copper I oxide (Cu2O)
  • 4Cu+ O2 —  below 1370K  —→  2CuO (copper I oxide)
  • A – copper (Cu)
  • B-copper II oxide (CuO) – Black colored
  • C – copper-I-oxide (Cu2O) – Red coloured

2. Aisa silvery-white metal. A-combines with O2 to form B at 800°C, the alloy of is used in making the aircraft. Find A.and B.

  • 4Al + 302 → 2Al203 (Aluminium oxide)
  • A is Aluminium (Al)
  • B is Aluminium oxide (Al203)

3. What is rust? Give the equation for the formation of rust.

  • Rust is the formation of scaling reddish-brown hydrated ferric oxide on the surface of iron-containing materials.
  • This compound is known as rust and the phenomenon of formation of rust is known as rust.
  • 4Fe + 302 + xH2O > 2Fe203.xH2O

(Rust)

(Hydrated ferric oxide)

4. State two conditions necessary for rusting of iron.

Conditions necessary for rusting of iron.

  • Iron is exposed to moist air.
  • Presence of water droplets in the atmosphere.
  • Presence of Oxygen.

VII. Long answer questions:

1. a) State the reason for the addition of caustic alkali to bauxite ore during purification of bauxite.
b) Along with cryolite and alumina, another substance is added to the electrolyte.
Addition of caustic alkali to bauxite ore:

  • Bauxite ore is finely ground and heated under pressure with a solution of concentrated caustic soda solution at 150°C to obtain sodium metal aluminate.
  • On diluting sodium meta aluminate with water, a precipitate of aluminum hydroxide is formed.
  • The precipitate is filtered, washed, dried, and ignited at 1000°C to get alumina.
  • 2Al(OH)3 — 1OOoC —→ Al2O3+ 3H20

b)

  • Fluorspar.
  • It lowers the fusion temperature of electrolytes.

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. H,SO, forms C and D along with water. D is a gaseous compound. Find A, B, C, and D.

  • Copper gets covered with a green layer of basic copper carbonate in the presence of CO2, and moisture.
  • 2Cu+02+CO2+H20 > CuCO3.Cu(OH)2
  • Copper reacts with dil H,SO, to from copper sulphate and SO,
  • Cu+2H2SO4 > CuSO4 + SO2 ↑ + 2H2O
  • So, Ais — Copper (Cu)

Bis – CuCO3 Cu(OH), Basic copper carbonate.

Cis ~ CuSO4 (copper sulphate)

D is — SO2, (Sulphur oxide)

3. Explain smelting process.

Smelting (in a Blast Furnace):

  • The charge consisting of roasted ore, coke and limestone in the ratio 8:4:1 is smelted in a blast furnace by introducing it through the cup and cone arrangement at the top.
  • There are three important regions in the furnace.

a) The Lower Region (Combustion Zone):

  • The temperature is at 1500°C.
  • In this region, coke burns with oxygen to form CO2 when the charge comes in contact with a hot blast of air.
  • It is an exothermic reaction since heat is liberated.

1500°C

C + O2 ———————-→ CO2 + Heat

Δ

b) The Middle Region (Fusion Zone):

  • ‘The temperature prevails at 1000°C.
  • In this region, CO2 is reduced to CO.

1000°C

C + O2 ———————-→ 2CO – Heat

Δ

  • Limestone decomposes to calcium oxide and CO2,

  CaCO3 ——————–→ CaO + CO2 – Heat

Δ

  • These two reactions are endothermic due to the absorption of heat.
  • Calcium oxide combines with silica to form calcium silicate slag.

CaO + SiO2 —> CaSiO3

c) The Upper Region (Reduction Zone):

  • The temperature prevails at 400°C.
  • In this region, carbon monoxide reduces ferric oxide to form a fairly pure spongy iron.

1000°C

Fe2O3+ 3CO ———————-→ 2Fe + 3CO2

  • The molten iron is collected at the bottom of the furnace after removing the slag.
  • The iron thus formed is called pig iron.
  • It is remelted and cast into different moulds.
  • This iron is called cast iron.

VIII. Higher Order Thinking Skill (HOTS)

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

  • When steam is passed over red hot aluminium, Aluminium oxide and Hydrogen is produced.

2Al+       3H2O       →        Al2O3         +3H2
(Steam)                (Aluminium oxide)

  • When Aluminium react with strong caustic alkalis forming aluminate.

2Al+ 2NaOH + 2H2O       →          2Na AlO2          +3H2
(Sodium meta Aluminium)

  • A is Aluminium (Al)
  • B is Aluminium oxide (Al2O3)
  • C is Sodium meta aluminate (Na AlO2)

2. Name the acid that renders aluminium passive. Why?

  • Dilute or concentrated nitric acid renders aluminum passive
  • It does not attack aluminium but it forms oxide film on its surface.

3. a) Identify the bond between H and F in HF molecule.
b) What property forms the basis of identification?
c) How does the property vary in periods and in groups?

a) Tonic bond.
b) Electronegativity property.
c)

  • Along the period from left to right in the periodic table the electronegativity increases, because of the increase in the nuclear charge which in turn attracts the electrons more strongly.
  • On moving down a group the electronegativity of the element decreases because of the increased number of energy levels.

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