10th Science Unit 8 Periodic Classification of Elements Book Back Answers

10th Science Unit 8 Periodic Classification of Elements book back answers for all 42 questions in the 2026-27 Samacheer Kalvi (Tamil Nadu) textbook.

This unit moves on from the early attempts at grouping elements to the modern periodic law, where atomic number decides an element's place. You study the seven periods and the groups, how properties change across a period and down a group, and then the world of metals: minerals and ores, steps in metallurgy, alloys, amalgams and corrosion. The book-back questions mix multiple choice, fill-ups, matching and short answers on periodic trends, extraction methods and rusting.

Answers at a glance28 answers · full steps below

I. Choose the best answer.

  1. Q1d) 7,18
  2. Q2a) atomic number
  3. Q3a) 17th17^{th}
  4. Q4d) electronegativity
  5. Q5c) Fe2O3.xH2O\mathrm{Fe_2O_3.xH_2O}
  6. Q6b) reducing agent
  7. Q7c) galvanization
  8. Q8a) He
  9. Q9b) stable configuration of electrons
  10. Q10b) Hg

II. Fill in the blanks

  1. Q1more ionic (ionic in nature)
  2. Q2Sixth period
  3. Q3Atomic number
  4. Q40.99 Å
  5. Q5A+\mathrm{A^+}
  6. Q6Henry Moseley
  7. Q7decreases
  8. Q8Lanthanides and Actinides
  9. Q9Bauxite (Al2O3⋅2H2O\mathrm{Al_2O_3 \cdot 2H_2O})
  10. Q10Hydrated ferric oxide (Fe2O3⋅xH2O\mathrm{Fe_2O_3 \cdot xH_2O})

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

  1. Q1False – Moseley's periodic table is based on atomic number.
  2. Q2False – Ionic radius decreases across the period from left to right.
  3. Q3True
  4. Q4False – Al wires are used as electric cables because aluminium is a good conductor of electricity and is ductile and light.
  5. Q5False – An alloy is a homogeneous mixture of two or more metals (or metals with certain non-metals).

V. Assertion and Reason

  1. Q1a) A and R are correct, R explains the A.
  2. Q2a) A and R are correct, R explains the A.
  3. Q3d) A and R are correct, R doesn't explains A.

I. Choose the best answer.

Q1. The number of periods and groups in the periodic table are______.

  1. a) 6,16
  2. b) 7,17
  3. c) 8,18
  4. d) 7,18
  1. The horizontal rows are periods and the vertical columns are groups.

  2. The modern periodic table has 7 periods and 18 groups.

Answer

d) 7,18

Q2. The basis of modern periodic law is______.

  1. a) atomic number
  2. b) atomic mass
  3. c) isotopic mass
  4. d) number of neutrons
  1. Moseley discovered atomic number, which gave a better basis for arranging elements.

  2. Modern periodic law: the physical and chemical properties of elements are periodic functions of their atomic numbers.

Answer

a) atomic number

Q3. _____ group contains the member of halogen family.

  1. a) 17th17^{th}
  2. b) 15th15^{th}
  3. c) 18th18^{th}
  4. d) 16th16^{th}
  1. In the family-wise grouping, group 17 elements are called halogens.

  2. Group 15 is nitrogen family, 16 is oxygen family and 18 is noble gases.

Answer

a) 17th17^{th}

Q4. _____ is a relative periodic property

  1. a) atomic radii
  2. b) ionic radii
  3. c) electron affinity
  4. d) electronegativity
  1. Electronegativity is the tendency of an atom to attract the shared pair of electrons towards itself in a covalent bond.

  2. It is measured by comparing atoms on the Pauling scale, so it is a relative property; the others have absolute measured values.

Answer

d) electronegativity

Q5. Chemical formula of rust is ________.

  1. a) FeO.xH2O\mathrm{FeO.xH_2O}
  2. b) FeO4.xH2O\mathrm{FeO_4.xH_2O}
  3. c) Fe2O3.xH2O\mathrm{Fe_2O_3.xH_2O}
  4. d) FeO\mathrm{FeO}
  1. Rusting is the formation of a brown layer when iron is exposed to moist air.

  2. Rust is chemically hydrated ferric oxide, Fe2O3.xH2O\mathrm{Fe_2O_3.xH_2O}.

Answer

c) Fe2O3.xH2O\mathrm{Fe_2O_3.xH_2O}

Q6. In the alumino thermic process the role of Al is _____.

  1. a) oxidizing agent
  2. b) reducing agent
  3. c) hydrogenating agent
  4. d) sulphurising agent
  1. In the aluminothermic process: Fe2O3+2Al→2Fe+Al2O3+Heat\mathrm{Fe_2O_3 + 2Al \rightarrow 2Fe + Al_2O_3 + Heat}.

  2. Aluminium removes oxygen from ferric oxide, so it acts as a powerful reducing agent.

Answer

b) reducing agent

Q7. The process of coating the surface of metal with a thin layer …

The process of coating the surface of metal with a thin layer of zinc is called______.

  1. a) painting
  2. b) thinning
  3. c) galvanization
  4. d) electroplating
  1. Coating zinc on iron sheets by using electric current is galvanization.

  2. It is a surface coating method used to prevent corrosion.

Answer

c) galvanization

Q8. Which of the following have inert gases 2 electrons in the outermost …

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

  1. a) He
  2. b) Ne
  3. c) Ar
  4. d) Kr
  1. Helium has only 2 electrons, both in its single outermost shell.

  2. Ne, Ar and Kr have 8 electrons in their outermost shell.

Answer

a) He

Q9. Neon shows zero electron affinity due to _____.

  1. a) stable arrangement of neutrons
  2. b) stable configuration of electrons
  3. c) reduced size
  4. d) increased density
  1. In noble gases like neon the outer s and p orbitals are completely filled.

  2. This stable electron configuration means no more electrons can be added, so electron affinity is zero.

Answer

b) stable configuration of electrons

Q10. ______ is an important metal to form amalgam.

  1. a) Ag
  2. b) Hg
  3. c) Mg
  4. d) Al
  1. An amalgam is an alloy of mercury (Hg) with another metal.

  2. Example: silver–tin amalgam used for dental filling.

Answer

b) Hg

II. Fill in the blanks

Q1. If the electronegativity difference between two bonded atoms in a molecule is …

If the electronegativity difference between two bonded atoms in a molecule is greater than 1.7, the nature of bonding is ______

  1. Pauling's rule: difference greater than 1.7 means the shared electrons are pulled almost completely to one atom.

  2. So the bond is considered to be more ionic in nature.

Answer

more ionic (ionic in nature)

Q2. ________ is the longest period in the periodical table.

  1. The sixth period covers atomic numbers 55 to 86 (Caesium to Radon) and holds 32 elements — 8 representative, 10 transition and 14 inner transition (lanthanides).

Answer

Sixth period

Q3. ________ forms the basis of modern periodic table.

  1. Moseley's discovery of atomic number led to the modern periodic law: properties are periodic functions of atomic number.

  2. Hence elements are arranged in the increasing order of atomic number.

Answer

Atomic number

Q4. If the distance between two Cl atoms in Cl2\mathrm{Cl_2} molecule is 1.98Å …

If the distance between two Cl atoms in Cl2\mathrm{Cl_2} molecule is 1.98Å, then the radius of Cl atom is ________.

  1. Covalent radius = half the distance between the nuclei of two covalently bonded atoms of the same element.

  2. r=1.982=0.99r = \dfrac{1.98}{2} = 0.99 Å.

Answer

0.99 Å

Q5. Among the given species A−\mathrm{A^-} , A+\mathrm{A^+} , and A, the smallest …

Among the given species A−\mathrm{A^-}, A+\mathrm{A^+}, and A, the smallest one in size is ________.

  1. A cation is formed by loss of an electron, so the remaining electrons are held more strongly by the nucleus.

  2. Hence a cation is smaller than the neutral atom, and the anion is the largest.

Answer

A+\mathrm{A^+}

Q6. The scientist who propounded the modern periodic law is ________.

  1. Henry Moseley (1912) discovered atomic number and modified the periodic law based on it.

Answer

Henry Moseley

Q7. Across the period, ionic radii ________(increases,decreases).

  1. Along a period the shell number is the same but nuclear charge increases, pulling the electron cloud inward.

  2. So, like atomic radius, ionic radius decreases from left to right.

Answer

decreases

Q8. ________ and ________ are called inner transition elements.

  1. The two series placed separately below the table form part of Group 3.

  2. They are the lanthanides and actinides.

Answer

Lanthanides and Actinides

Q9. The chief ore of Aluminium is ________.

  1. Aluminium is extracted profitably from bauxite (Al2O3⋅2H2O\mathrm{Al_2O_3 \cdot 2H_2O}) by Baeyer's process followed by Hall's process.

Answer

Bauxite (Al2O3⋅2H2O\mathrm{Al_2O_3 \cdot 2H_2O})

Q10. The chemical name of rust is ________.

  1. Rust is the reddish brown scaly layer formed on iron in moist air.

  2. It is chemically hydrated ferric oxide, Fe2O3⋅xH2O\mathrm{Fe_2O_3 \cdot xH_2O}.

Answer

Hydrated ferric oxide (Fe2O3⋅xH2O\mathrm{Fe_2O_3 \cdot xH_2O})

III. Match the following

Question 1

1. Galvanisation - 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
  1. Galvanisation is the process of coating zinc on iron sheets using electric current → Coating with Zn.

  2. Calcination is the strong heating of the concentrated ore in the absence (or limited supply) of air → Heating in the absence of air.

  3. In the alumino thermic process Fe2O3+2Al→2Fe+Al2O3+HeatFe_2O_3 + 2Al \rightarrow 2Fe + Al_2O_3 + Heat, aluminium is oxidised and ferric oxide is reduced → Redox reaction.

  4. Silver-tin amalgam, an alloy of mercury, is used for dental filling → Dental filling.

  5. Group 18 elements have stable, completely filled valence shells and are unreactive → Noble gas elements.

Answer

1 - b (Coating with Zn), 2 - e (Heating in the absence of air), 3 - d (Alumino thermic process), 4 - c (Silver-tin amalgam), 5 - a (Noble gas elements)

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

Q1. Moseley's periodic table is based on atomic mass.

  1. Moseley discovered the atomic number and showed that it is the fundamental property of an element.

  2. The modern periodic law states that properties of elements are periodic functions of their atomic numbers, not atomic masses.

Answer

False – Moseley's periodic table is based on atomic number.

Q2. Ionic radius increases across the period from left to right.

  1. Along a period the shell number stays the same but the nuclear charge (number of protons) increases.

  2. The stronger pull shrinks the electron cloud, so atomic and ionic radii decrease from left to right.

Answer

False – Ionic radius decreases across the period from left to right.

Q3. All ores are minerals; but all minerals cannot be called as ores;

  1. A mineral is any compound or mixture of compounds of a metal found in the Earth.

  2. An ore is a mineral from which the metal can be extracted readily and economically, e.g. bauxite is an ore of aluminium while clay is only a mineral.

Answer

True

Q4. Al wires are used as electric cables due to their silvery white …

Al wires are used as electric cables due to their silvery white colour.

  1. Aluminium is used in the electrical cable industry because it is a good conductor of electricity, and it is also ductile and light (low density 2.7).

  2. Silvery white colour is only an appearance; it has nothing to do with conduction of current.

Answer

False – Al wires are used as electric cables because aluminium is a good conductor of electricity and is ductile and light.

Q5. An alloy is a heterogenous mixture of metals.

  1. An alloy is a homogeneous mixture of two or more metals, or of one or more metals with certain non-metallic elements.

  2. Alloys behave like solid solutions, e.g. brass is a solid solution of zinc (solute) in copper (solvent).

Answer

False – An alloy is a homogeneous mixture of two or more metals (or metals with certain non-metals).

V. Assertion and Reason

Q1. Answer the following questions using the data given below …

Answer the following questions using the data given below:

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

  1. a) A and R are correct, R explains the A.
  2. b) A is correct, R is wrong.
  3. c) A is wrong, R is correct.
  4. d) A and R are correct, R doesn't explains A.
  1. Electronegativity values given in the text: H = 2.1 and F = 4.0.

  2. Difference =4.0−2.1=1.9= 4.0 - 2.1 = 1.9, so the Reason is correct.

  3. Rule: if the electronegativity difference is greater than 1.7, the bond is more ionic. Since 1.9 > 1.7, the H–F bond is ionic, so the Assertion is correct.

  4. The large electronegativity difference is exactly the cause of the ionic nature, so the Reason explains the Assertion.

Answer

a) A and R are correct, R explains the A.

Q2. Assertion : Magnesium is used to protect steel from rusting …

Assertion : Magnesium is used to protect steel from rusting Reason : Magnesium is more reactive than iron

  1. a) A and R are correct, R explains the A.
  2. b) A is correct, R is wrong.
  3. c) A is wrong, R is correct.
  4. d) A and R are correct, R doesn't explains A.
  1. Assertion is correct: magnesium is coated/attached to steel as a sacrificial metal in cathodic protection to prevent rusting.

  2. Reason is correct: magnesium is more reactive than iron.

  3. Being more reactive, magnesium corrodes first (acts as anode) and saves the iron, which acts as cathode.

  4. So the Reason correctly explains the Assertion.

Answer

a) A and R are correct, R explains the A.

Q3. Assertion : An uncleaned copper vessel is covered with greenish layer. …

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

  1. a) A and R are correct, R explains the A.
  2. b) A is correct, R is wrong.
  3. c) A is wrong, R is correct.
  4. d) A and R are correct, R doesn't explains A.
  1. Assertion is correct: in the presence of CO2CO_2 and moisture, copper gets covered with a green layer of basic copper carbonate — 2Cu+O2+CO2+H2O→CuCO3.Cu(OH)22Cu + O_2 + CO_2 + H_2O \rightarrow CuCO_3.Cu(OH)_2.

  2. Reason is also correct: copper is not attacked by alkalis.

  3. But the green layer is formed by the action of air, moisture and carbon dioxide, not by any reaction with alkali.

  4. Hence both statements are true, but the Reason does not explain the Assertion.

Answer

d) A and R are correct, R doesn't explains A.

VI. Short answer questions

Q1. A is a reddish brown metal, which combines with O2\mathrm{O_2} at < …

A is a reddish brown metal, which combines with O2\mathrm{O_2} 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.

  1. A is a reddish brown metal with high lustre, high density and high melting point (1356 °C) — so A is copper (Cu).

  2. On heating with oxygen below 1370 K, copper forms black copper(II) oxide, CuO, which is B: 2Cu+O2→below 1370K2CuO2Cu + O_2 \xrightarrow{below\ 1370K} 2CuO (black).

  3. On heating with oxygen above 1370 K, copper forms red copper(I) oxide, Cu2OCu_2O, which is C: 4Cu+O2→above 1370K2Cu2O4Cu + O_2 \xrightarrow{above\ 1370K} 2Cu_2O (red).

  4. Thus A = Cu, B = CuO (copper II oxide, black), C = Cu2OCu_2O (copper I oxide, red).

Answer

A = Copper (Cu), B = CuO (black copper(II) oxide), C = Cu2OCu_2O (red copper(I) oxide).

Q2. A is a silvery white metal. A combines with O2\mathrm{O_2} to form …

A is a silvery white metal. A combines with O2\mathrm{O_2} to form B at 800∘C800^{\circ}\mathrm{C}, the alloy of A is used in making the aircraft. Find A and B

  1. A is a silvery white metal whose alloys (duralumin, magnalium) are used in making aircraft — so A is aluminium (Al).

  2. Aluminium is not affected by dry air, but on heating at 800 °C it burns brightly forming its oxide.

  3. 4Al+3O2→2Al2O34Al + 3O_2 \rightarrow 2Al_2O_3 (aluminium oxide) — so B is Al2O3Al_2O_3.

  4. (At the same temperature it also forms aluminium nitride: 2Al+N2→2AlN2Al + N_2 \rightarrow 2AlN.)

Answer

A = Aluminium (Al) and B = Aluminium oxide (Al2O3Al_2O_3).

Q3. What is rust? Give the equation for formation of rust.

  1. When iron is exposed to moist air, a scaling reddish brown layer is formed on its surface; this is rust.

  2. Rust is chemically hydrated ferric oxide, Fe2O3⋅xH2OFe_2O_3 \cdot xH_2O, and the phenomenon of its formation is called rusting.

  3. Equation: 4Fe+3O2+xH2O→2Fe2O3⋅xH2O4Fe + 3O_2 + xH_2O \rightarrow 2Fe_2O_3 \cdot xH_2O (rust).

  4. Rusting is a type of corrosion, in which iron loses its metallic character.

Answer

Rust is hydrated ferric oxide (Fe2O3⋅xH2OFe_2O_3 \cdot xH_2O) formed as 4Fe+3O2+xH2O→2Fe2O3⋅xH2O4Fe + 3O_2 + xH_2O \rightarrow 2Fe_2O_3 \cdot xH_2O.

Q4. State two conditions necessary for rusting of iron.

  1. Presence of air (oxygen): oxygen is needed to oxidise the iron surface.

  2. Presence of moisture (water): rusting is a wet or electrochemical corrosion that occurs only in moist air.

  3. When both are present together, iron forms a brown layer of hydrated ferric oxide (rust) on its surface.

Answer

Iron rusts only in the presence of both air (oxygen) and moisture (water).

VII. Long answer questions

Question 1

(a)

State the reason for addition of caustic alkali to bauxite ore during purification of bauxite.

Show steps (7)
  1. Bauxite ore contains impurities (gangue) such as iron oxide and silica along with alumina.

  2. In Baeyer's process the finely ground ore is heated under pressure with concentrated caustic soda (NaOH) solution at 150 °C.

  3. Alumina is amphoteric, so it dissolves in the hot alkali and forms soluble sodium meta aluminate: Al2O3+2NaOH→2NaAlO2+H2OAl_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O

  4. The impurities do not dissolve in caustic alkali; they remain as solid residue and are removed by filtration.

  5. The filtrate (sodium meta aluminate) on dilution with water gives a precipitate of aluminium hydroxide: NaAlO2+2H2O→Al(OH)3+NaOHNaAlO_2 + 2H_2O \rightarrow Al(OH)_3 + NaOH

  6. This precipitate is filtered, washed, dried and ignited at 1000 °C to give pure alumina: 2Al(OH)3→1000∘CAl2O3+3H2O2Al(OH)_3 \xrightarrow{1000^\circ C} Al_2O_3 + 3H_2O

  7. Thus caustic alkali is added to dissolve alumina selectively and separate it from the insoluble impurities.

Answer

Caustic alkali dissolves the amphoteric alumina as soluble sodium meta aluminate while the impurities stay insoluble and are filtered off, giving pure alumina.

(b)

Along with cryolite and alumina, another substance is added to the electrolyte mixture. Name the substance and give one reason for the addition.

Show steps (6)
  1. The substance added is fluorspar, CaF2CaF_2.

  2. In Hall's process the electrolyte is a mixture of pure alumina + molten cryolite (Na3AlF6Na_3AlF_6) + fluorspar.

  3. Pure alumina melts only at a very high temperature, which makes electrolysis costly and difficult.

  4. Fluorspar lowers the fusion (melting) temperature of the electrolyte, so the cell can be worked at only 900–950 °C.

  5. It also keeps the molten mixture freely flowing and conducting, so aluminium is deposited at the graphite-lined iron cathode and oxygen is liberated at the graphite anodes.

  6. Overall reaction: 2Al2O3→4Al+3O2↑2Al_2O_3 \rightarrow 4Al + 3O_2 \uparrow

Answer

Fluorspar (CaF2CaF_2) is added because it lowers the fusion temperature of the electrolyte mixture.

Q2. The electronic configuration of metal A is 2,8,18,1. …

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\mathrm{H_2SO_4} forms C and D along with water. D is a gaseous compound. Find A,B,C and D.

Show steps (6)
  1. A: The electronic configuration 2, 8, 18, 1 corresponds to atomic number 29, which is copper (Cu).

  2. B: Copper exposed to air and moisture (in presence of CO2CO_2) gets covered with a green layer of basic copper carbonate. So B is CuCO3⋅Cu(OH)2CuCO_3 \cdot Cu(OH)_2.

  3. Reaction: 2Cu+O2+CO2+H2O→CuCO3⋅Cu(OH)22Cu + O_2 + CO_2 + H_2O \rightarrow CuCO_3 \cdot Cu(OH)_2

  4. C and D: Copper reacts with hot concentrated sulphuric acid giving copper sulphate, sulphur dioxide gas and water.

  5. Reaction: Cu+2H2SO4→CuSO4+SO2↑+2H2OCu + 2H_2SO_4 \rightarrow CuSO_4 + SO_2 \uparrow + 2H_2O

  6. So C is copper sulphate (CuSO4CuSO_4) and D is the gaseous compound sulphur dioxide (SO2SO_2).

Answer

A = Copper (Cu), B = basic copper carbonate CuCO3⋅Cu(OH)2CuCO_3\cdot Cu(OH)_2, C = copper sulphate CuSO4CuSO_4, D = sulphur dioxide SO2SO_2.

Q3. Explain smelting process.

Show steps (8)
  1. Definition: Smelting is the process of reducing the roasted metallic oxide to the metal in its molten condition.

  2. It is one of the important steps of metallurgy, carried out in a furnace (blast furnace or reverberatory furnace) at high temperature.

  3. The roasted ore is mixed with a reducing agent such as coke (carbon) or carbon monoxide, which removes oxygen from the metal oxide.

  4. A flux is added along with the charge to remove the earthy (gangue) impurities. A basic flux (CaO) is used for acidic gangue and an acidic flux (silica) for basic gangue.

  5. The flux combines with the gangue to form an easily fusible product called slag: Flux + Gangue → Slag.

  6. The slag is lighter, so it floats on the molten metal and is drawn off separately, leaving the molten metal at the bottom of the furnace.

  7. Example (copper): the roasted ore mixed with powdered coke and sand is heated in a blast furnace to give matte (Cu2S+FeSCu_2S + FeS) and slag; FeO+SiO2→FeSiO3FeO + SiO_2 \rightarrow FeSiO_3 (slag).

  8. Example (iron): in the blast furnace, CaCO3→CaO+CO2CaCO_3 \rightarrow CaO + CO_2, CaO+SiO2→CaSiO3CaO + SiO_2 \rightarrow CaSiO_3 (slag), and Fe2O3+3CO→2Fe+3CO2Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2.

Answer

Smelting is the reduction of a roasted metal oxide to molten metal using a reducing agent like coke, the impurities being removed as slag by adding a suitable flux.

VIII. HOT questions

Q1. Metal A belongs to period 3 and group 13. A in red …

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

  1. Element in period 3, group 13 has atomic number 13 — so A is aluminium (Al).

  2. Red hot aluminium reacts with steam to give aluminium oxide and hydrogen, so B is aluminium oxide (Al2O3Al_2O_3): 2Al+3H2O→Al2O3+3H2↑2Al + 3H_2O \rightarrow Al_2O_3 + 3H_2 \uparrow

  3. Aluminium reacts with a strong alkali (caustic soda) to form sodium meta aluminate, so C is sodium meta aluminate (NaAlO2NaAlO_2): 2Al+2NaOH+2H2O→2NaAlO2+3H2↑2Al + 2NaOH + 2H_2O \rightarrow 2NaAlO_2 + 3H_2 \uparrow

  4. Hydrogen gas is liberated in both reactions.

Answer

A = Aluminium, B = Aluminium oxide (Al2O3Al_2O_3), C = Sodium meta aluminate (NaAlO2NaAlO_2).

Q2. Name the acid that renders aluminium passive. Why?

  1. Nitric acid (HNO3HNO_3), both dilute and concentrated, renders aluminium passive.

  2. Nitric acid does not attack aluminium; instead it oxidises the surface of the metal.

  3. A thin, continuous and non-porous oxide film (Al2O3Al_2O_3) is formed on the surface.

  4. This film prevents further contact of the acid with the metal below, so the metal becomes chemically inactive (passive).

Answer

Nitric acid (HNO3HNO_3) makes aluminium passive because a protective oxide film forms on its surface.

Question 3

(a)

Identify the bond between H and F in HF molecule.

  1. Electronegativity values: F=4.0F = 4.0 and H=2.1H = 2.1.

  2. Difference in electronegativity =4.0−2.1=1.9= 4.0 - 2.1 = 1.9.

  3. Since the difference is greater than 1.7, the bond is considered to be more ionic in character.

  4. Fluorine, being highly electronegative, pulls the shared pair of electrons strongly towards itself.

Answer

The H–F bond is highly polar and, with an electronegativity difference of 1.9 (> 1.7), it is more ionic in character.

(b)

What property forms the basis of identification?

  1. The property used is electronegativity.

  2. Electronegativity is the measure of the tendency of an atom to attract the shared pair of electrons towards itself in a covalent bond.

  3. The Pauling scale of electronegativity is used to predict the nature of the bond: difference = 1.7 means 50% ionic and 50% covalent; less than 1.7 means more covalent; greater than 1.7 means more ionic.

Answer

Electronegativity (measured on the Pauling scale) is the basis of identification.

(c)

How does the property vary in periods and in groups?

  1. Along a period (left to right): electronegativity increases, because the nuclear charge increases and the atom attracts the shared electrons more strongly.

  2. Down a group (top to bottom): electronegativity decreases, because the number of valence shells increases and the shared electrons are farther from the nucleus.

  3. Hence fluorine, at the top right of the table, is the most electronegative element (4.0).

Answer

Electronegativity increases along a period and decreases down a group.

Quick check

  1. Chemical formula of rust is ________.

    1. (1) FeO.xH2O\mathrm{FeO.xH_2O}
    2. (2) FeO4.xH2O\mathrm{FeO_4.xH_2O}
    3. (3) Fe2O3.xH2O\mathrm{Fe_2O_3.xH_2O}
    4. (4) FeO\mathrm{FeO}
    Show answer

    Answer: (3) Fe2O3.xH2O\mathrm{Fe_2O_3.xH_2O}

    1. Rusting is the formation of a brown layer when iron is exposed to moist air.

    2. Rust is chemically hydrated ferric oxide, Fe2O3.xH2O\mathrm{Fe_2O_3.xH_2O}.

    Question 1 of 5
  2. Which of the following have inert gases 2 electrons in the outermost shell.

    1. (1) He
    2. (2) Ne
    3. (3) Ar
    4. (4) Kr
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    Answer: (1) He

    1. Helium has only 2 electrons, both in its single outermost shell.

    2. Ne, Ar and Kr have 8 electrons in their outermost shell.

    Question 2 of 5
  3. _____ group contains the member of halogen family.

    1. (1) 17th17^{th}
    2. (2) 15th15^{th}
    3. (3) 18th18^{th}
    4. (4) 16th16^{th}
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    Answer: (1) 17th17^{th}

    1. In the family-wise grouping, group 17 elements are called halogens.

    2. Group 15 is nitrogen family, 16 is oxygen family and 18 is noble gases.

    Question 3 of 5
  4. The number of periods and groups in the periodic table are______.

    1. (1) 6,16
    2. (2) 7,17
    3. (3) 8,18
    4. (4) 7,18
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    Answer: (4) 7,18

    1. The horizontal rows are periods and the vertical columns are groups.

    2. The modern periodic table has 7 periods and 18 groups.

    Question 4 of 5
  5. In the alumino thermic process the role of Al is _____.

    1. (1) oxidizing agent
    2. (2) reducing agent
    3. (3) hydrogenating agent
    4. (4) sulphurising agent
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    Answer: (2) reducing agent

    1. In the aluminothermic process: Fe2O3+2Al→2Fe+Al2O3+Heat\mathrm{Fe_2O_3 + 2Al \rightarrow 2Fe + Al_2O_3 + Heat}.

    2. Aluminium removes oxygen from ferric oxide, so it acts as a powerful reducing agent.

    Question 5 of 5

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