Chemistry ICSE Class 8 Important Questions Chapter 5 Language of Chemistry
Question 1.
Explain the term ‘symbol’. State a reason why – the symbol of calcium is ‘Ca’ & of copper is ‘Cu’
Answer:
Symbol is a short form or abreviated name – of the element.
OR
“Symbol of an element is the ‘first letter’ or the ‘first letter and another ‘letter’ of the English name or Latin name of the element”.
As the name calcium and copper start with same letter of English alphabet ‘C’ so another letter from the name of the element is added to ‘C’. Hence symbol of calcium is ‘Ca’ and symbol of copper or cuprum (Latin name of copper) is ‘Cu’.
Question 2.
Define the term ‘valency’. With reference to water & ammonia as compounds respectively, state the valency of oxygen & nitrogen. Magnesium [2, 8, 2] has valency 2+. Give reasons.
Answer:
Valency is the number of hydrogen atoms which combine with (or displace) one atom of the element forming a compound. Compound water H2O, two atoms of hydrogen combine with one atom of oxygen to form water. Hence valency of oxygen atom is 2.
Compound ammonia NH3
Valency of atom nitrogen is 3, as it combines with 3 atoms of hydrogen.
Valency of magnesium is 2+, Mg [2, 8, 2] Electronic configuration of Mg is 2, 8, 2 i.e. there are 2 valence electrons which Mg can lose and achieve stable configuration of nearest noble gas Neon. Hence is cation Mg2+ has valency 2+.
Question 3.
Explain the term ‘variable valency‘ Copper having electronic configuration 2, 8,18,1 exhibits variable valency. Give a reason for the same & name the compound CuCl & CuCl2.
Answer:
Variable valency : “When an element has more than one valency, its valency is called variable valency.” Copper exhibit valency ‘1’ and ‘2’ i.e. has variable valency reason for variable valency Valency is – the number of electrons lost or gained from the outer shell of an atom of an element – during chemical reaction. Variation in this gain or loss results in ‘variable valency’.
Lower valency ends with – ous
Higher valency ends with – ic
CuCl is named as cuprous chloride (lower valency)
OR
Cu [I] Cl i.e. Copper [I] chloride and
CuCl2 – Cupric chloride (higher valency)
i. e. Copper [II] chloride Cu[II]Cl2
Question 4.
State the valencies of the following metallic elements –
(a) Potassium
(b) Sodium
(c) Calcium
(d) Magnesium
(e) Zinc
(f) Aluminium
(g) Chromium
[write each symbol with the valency
Answer:
Metals have positive valency 1, 2 or 3 Valency of
(a) Potassium is
K+
(b) Sodium 1+
Na+
(c) Calcium 2+
Ca2+
(d) Magnesium 2+
Mg2+
(e) Zinc 2+
Zn2+
(f) Aluminium 3+
Al3+
(g) Chromium 3+
Cr3+
Question 5.
Certain metals exhibit variable valencies which include valencies: 1+, 2+, 3+ & 4+. State the variable valency of the following metals –
(a) Copper
(b) Silver
(c) Mercury
(d) Iron
(e) tin
(f) Lead
[write each symbol with the variable valency]
Answer:
(a) Copper : exhibits valency 1 and 2
(b) Silver : exhibits valency 1 and 2
(c) Mercury exhibits valency 1 and 2
(d) Iron has variable valency 2 and 3
(e) Tin has variable valency 2 and 4
(f) Lead has variable valency 2 and 4
Question 6.
State which of the following ions or radicals given below of non-metallic elements exhibit-valency: 1–, 2– & 3–
(a) Chloride
(b) Bromide
(c) Iodide
(d) Nitrate
(e) Hydroxide
(f) Bicarbonate
(g) Bisulphite
(h) Bisulphate
(i) Aluminate
(j) Permanganate
(k) Oxide
(l) Sulphide
(m) Sulphite
(n) Sulphate
(o) Carbonate
(p) Dichromate
(q) Zincate
(r) Plumbite
(s) Phosphate
(t) Nitride
[write each ion or radical with the correct valency]
Answer:
Question 7.
Differentiate between the terms – ‘Ion’ & ‘radical’ with suitable examples.
Answer:
Ion “is an atom or a group of atoms carrying a positive or negative charge due to loss or gain of electrons.” e.g. cation Na+ and Cl– is anion.
Radical “is group of atoms of elements that behaves like a single unit & show valency.”
Positive radical [NH41+] (Ammonium)
Negative radical [HCO31-] (Bicarbonate), [NO31-](Nitrate)
Question 8.
Write the chemical formula of the following compounds in a step-by-step manner –
(a) Potassium chloride
(b) Sodium bromide
(c) Potassium nitrate
(d) Calcium hydroxide
(e) Calcium bicarbonate
(f) Sodium bisulphate
(g) Potassium sulphate
(h) Zinc hydroxide
(i) Potassium permanganate
(j) Potassium dichromate
(k) Aluminium hydroxide
(l) Magnesium nitride
(m) Sodium zincate
(n) Copper [II] oxide
(o) Copper [I] sulphide
(p) Iron [III] chloride
(q) Iron [II] hydroxide
(r) Iron [III] sulphide
(s) Iron [III] oxide.
Answer:
Write the formula of
Step 1 – Write each symbol with its valency positive ion is written first
K1+ Cl1-
Step II – Inter change the valencies
(+ and -ve) signs ignored
and Ignore 1 in KCl
Therefore Formula is KCl
Step 1 – Write each symbol with valency.
Na1+ Br1-
Positive ion is written first
Step II – Inter change the valencies and ignore the signs
The Formula is NaBr
Step 1 – Write each symbol with valency Positive ion is written first
K1+ NO31-
Step II – Inter change the valencies and ignoring the +ve and -ve signs
The Formula is KNO3
Step 1 – Write each symbol with valency Positive ion first
Ca2+ [OH]1-
Step II – Inter change the valencies and ignoring positive and negative signs.
The Formula is Ca[OH]2
Step 1 – Write each symbol with valency Positive ion is written first
Ca2+ [HCO3]1-
Step II – Inter change the valencies and ignoring (+) and (-) signs
The Formula is Ca[HCO3]2
0
Step 1 – Write each symbol with valency Positive ion is written first
Na1+ [HSO3]1-
Step II – Interchange the valencies ignoring the signs.
The Formula is NaHSO4
Step 1 – Write each symbol with valency Positive ion is written first
K1++ [SO3]2-
Step II – Inter change the valencies ignoring the signs.
The Formula is K2SO4
Step 1 – Write each symbol with valency Positive ion is written first
Zn2+ [OH]1-
Step II – Inter change the valencies ignoring the signs.
The Formula is Zn[OH]2
Step 1 – Write each symbol with valency Positive ion is written first
K1+ [MnO4]1-
Step II – Inter change the valencies ignoring the (+) and (-) signs.
The Formula is KMnO4]<
Step 1 – Write each symbol with valency Positive ion is written first
K1+ + [Cr2O7]2-
Step II – Inter change the valencies ignoring the (+) and (-) signs.
The Formula is K2Cr2O7
Step 1 – Write each symbol with valency Positive ion is written first
Al3+ [OH]1-
Step II – Inter change the valencies positive ion is write first.
The Formula is Al[OH]3
Step I – Write each symbol with valency Positive ion is written first. Mg2+ N3-
Step II – Inter change the valencies positive ion is write first.
The Formula is Mg3N2
Step I – Write each symbol with valency Positive ion is written first
Na1+ [ZnO2]2-
Step II – Inter change the valencies ignoring (+) and (-) signs
The Formula is Na2ZnO2
Step 1 – Write each symbol with valency Positive ion is written first
Cu2+ O2-
Step II – Inter change the valencies ignoring (+) and (-) signs
The Formula is CuO
Step 1 – Write each symbol with valency Positive ion is written first
Cu1+ S2-
Step II – Inter change the valencies ignoring the signs
The Formula is Cu2O
Step 1 – Write each symbol with valency Positive ion is written first
Fe3+ Cl1-
Step II – Inter change the valencies ignoring (+) and (-) signs
The Formula is FeCl3
Step 1 – Write each symbol with valency Positive ion is written first
Fe2+ [OH]1-
Step II – Inter change the valencies ignoring sign
The Formula is Fe[OH]2
Step 1 – Write each symbol with valency Positive ion is written first
Fe3+ S2-
Step II – Inter change the valencies ignoring (+) and (-) signs
The Formula is Fe2 S2
Step 1 – Write each symbol with valency Positive ion is written first
Fe3+ O2-
Step II – Inter change the valencies ignoring (+) and (-) sign
The Formula is Fe2O3
Question 9.
What is a chemical equation. How is it represented. Differentiate between a ‘word equation’ and a ‘molecular equation’ with a suitable example.
Answer:
“Chemical equation is a short form – representing the result of a chemical change.”
OR
“Is the representation of a chemical change.”
Word equation tells which substances react (take part) in chemical reaction and which substances are produced where as molecular equation symbols and molecular formulas are used for both reactants and products.
Example :
When zinc reacts with dilute sulphuric acid, both being reactants, products produced are zinc sulphate and hydrogen gas, which are shown as below :
Word equation :
Zinc + Sulphuric acid → Zinc sulphate + Hydrogen
Molecular equation :
Question 10.
State the information provided by a chemical equation.
Chemical equations suffer from a number of limitations.
State the main limitations of a chemical equation.
Answer:
Information provided by a chemical equation :
(a) It tells us the formulas and symbols of the reactants and products.
(b) It tells us the ratio in which substances reacts or are produced. If limitations are covered.
(c) It tells the physical state of substances i.e. solid, liquid, gas.
(d) Whether the reaction is endothermic or exothermic.
(e) Conditions for starting the reaction i.e. if catalyst is needed or not.
(f) If reaction is reversible or not.
Limitations :
(a) Physical states of the reactants or products. But now we write along with substances (1) for liquids, (s) for solid and (g) for gas.
(b) Conditions that effect the reaction, i.e. temp, pressure or catalyst.
(c) Concentration of reactants and products we use dil. for dilute and cone, for concentrated.
(d) Nature of the chemical reaction.
(e) Speed – reaction is fast or slow.
(f) Exothermic or endothermic we write + heat or – heat towards products for exothermic and endothermic.
(g) The completion of the reaction.
Question 11.
State what is a balanced equation with a relevant example.
Give a reason why an equation is balanced with reference to the law of conservation of matter.
Answer:
A balanced equation : “An equation is said to be balanced if the number of atoms of each element of the reactant is equal to the number of atoms of each element of the product.”
Example :
Reactans have 2 atoms of Mg and 2 atoms of oxygen. Products have 2 atoms of magnesium and 2 atoms of oxygen.
.’. Number of atoms of each element of reactants = Number of atoms of each element of product.
Reason for balancing equation :
Law of conservation of matter says that “Matter is neither created nor destroyed during a chemical reaction.” This is possible only if number of atoms of each element on both sides of → are equal i.e. in reactants and also in products. Hence a reaction is balanced.
Question 12.
Write balanced molecular equations for the following word equations :
(a) Calcium + oxygen → Calcium oxide
(b) Calcium + water → Calcium + hydrogen
(c) Zinc + sulphuric hydroxide → Zinc sulphate + hydrogen
(d) Lead sulphate acid + ammonium hydroxide → Ammonium sulphate + lead hydroxide
(e) Copper hydroxide + nitric acid → Copper nitrate + water
(f) Lead nitrate + sodium Chloride → Sodium nitrate + lead Chloride
Answer:
(a) 2Ca + O2 → 2CaO chloride
(b) Ca + 2H2O → Ca(OH)2 + H2
(c) Zn + H2SO4 → ZnSO4 + H2
(d) PbSO4 + 2NH4OH → Pb[OH]2 + [NH4]2SO4
(e) CU[OH]2 + 2HNO3 → CU[NO3]2+ 2H2O
(f) Pb[NO3]2 + 2NaCl → 2NaNO3 + PbCl2
Question 13.
Balance the following equations :
(a) P + O2 → P2 O5
(b) Na2O + H2O → NaOH
(c) K + H2O → KOH + H2
(d) Fe + H2O ⇌ Fe3O4 + H2
(e) CaO + HCl2 → CaCl2 + H2O
(f) Fe + Cl → FeCl3
(g) Al + H3O → Al3O2 + H2
(h) Al + H2SO4 → Al2(SO4)4+ H2
(i) Fe2O3 + H2 → Fe + H2O
(j) C + H2SO4 → CO2 + H2O + SO2
(k) Pb3O4 → PbO + O2
(l) Al + O2 → Al2O3
(m) NO + O2 → NO2
(n) ZnS + O2 → ZnO + SO2
(o) Pb3O4 + HCl →PbCl2 + H2O + Cl2
(p) ZnO + NaOH → Na2ZnO2 + H2O
(q) H2S + Cl2 → S + HCl
(r) FeCI3+ NaOH →NaCl + Fe(OH)3
(s) Fe2 O2 + CO → Fe + CO2
(t) KHCO3 → K2 CO3 + H2 O + CO2
(u) CuO + NH3 → Cu + H2O + N2
Answer:
(a) P + O2 → P2 O2
To balance O atoms on both sides
P + 5O2 → P2O5
There are 4 P atoms on RHS
Make P atoms 4 on LHS
4P + 5O2 → 2P2O5
Atoms of P and O are equal on both sides.
(b) Na2O + H2O → NaOH
To make Na atoms equal to 2 on RHS
Na2O + H2O → 2NaOH
Atoms of Na, O and H are balance on both sides.
(c) K + H2O KOH + H2
Multiply H2O by 2 on left side and KOH by 2 on right to make H and O equal.
K + 2H2O → 2KOH + H2 To make K equal make 2K on left.
2K + 2H2O → 2KOH + H2
There are 2K atoms, 2(O) atoms and 4H atoms on both side.
(d) Fe + H2O ⇌ Fe3O4 + H2
To make Fe3 mutliply Fe by 3 on left and to make O4 multiply H2O by 4
∴ 3Fe + 4H2O ⇌ Fe3O4 + 4H2
(e) CaO + HCl → CaCl2 + H2O
There 2 Cl atoms on right : Use 2HCl on left side.
CaO + 2HCl → CaCl2 + H2O
(f) Fe + Cl2 → FeCl3
To make Cl atoms equal i.e. 6
Fe + 3Cl2 → 2FeCl3 Make 2 Fe on left side.
2Fe + 3Cl2 → 2FeCl3
(g) Al + H2O → Al2O3+ H2
To balance A1 and O atoms equal 2Al + 3H2O → Al2O3 + H2
But there are 6 atoms of H on left and 2 atoms of H on right.
2Al + 3H2O → Al2O3 + 3H2
(h) Al + H2SO4 → Al2(SO4)3 + H2
To make atoms of Al and (SO4) equal
2Al + 3H2SO4 → Al2(SO4) + H2
To H atoms 6 multiply H2 on right by 3
2Al + 3H2SO4 → Al2(SO4)3 + 3H2
Atoms of each element on left are equal on right side.
(i) Fe2O3 + H2 → Fe + H2O There are 2 atoms of Fe on left.
Fe2O2 + H2 → 2Fe + H2O
Now equate O atoms and use 3H2O on right.
Fe2O2 + H2 → 2Fe + 3H2O
But H atoms on right are 6 make 3H2 on left.
∴ Fe2O3 + 3H2 → 2Fe + 3H2O It is balanced.
(j) C + H2SO4 → CO2 + H2O + SO2
Atoms of oxygen are not equal on both sides.
To make atoms equal, we multiply H2SO2 on left by 2 and equate.
C + 2H2SO4 → CO2 +2H2O + 2SO2 Atoms of each element are equal on both sides.
(k) Pb3O4 → PbO + O2
Multiply PbO by 3 to make Pb equal.
Pb3O4 → 3PbO + O2 Multiply Pb3O4 by 2 and 3PbO by 2
2Pb3O4 → 6PbO + O2lit is balanced equation.
(l) Al + O2 → Al2O3
Multiply Al by 4 and Al2O3by 2 to make atoms of Al equal.
4Al + O2 → 2Al2O3 To make oxygen atoms equal 4Al + 3O2 → 2Al2O3
(m) NO + O2 → NO2
Multiply NO and NO2 by 2
2NO + O2 → 2NO2
(n) ZnS + O2→ ZnO + SO2
Multiply O2 by 3 and to make atoms of oxygen equal on both sides multiply right side by 2.
ZnS + 3O2 → 2ZnO + 2SO2
Multiply ZnS by 2 to make balanced euqation.
2ZnS + 3O2 → 2ZnO + 2SO2
Atoms of every element ae equal on both sides.
(o) Pb3O4 + HCl → PbCl2 + H2O + Cl2
There are 4 atoms of oxygen in Pb3O4 so make H atoms 8 in HCl.
Pb3O4 + 8HCl → PbCl2 + H2O + Cl2
Multiply PbCl2 by 3 to make Pb atoms equal and H2O by 4.
Pb3O4 + 8HCl → 3PbCl2 + 4H2O + Cl2
(p) ZnO + NaOH → Na2ZnO2 + H2O
Multiply NaOH on left side by 2 to make Na and O equal.
ZnO + 2NaOH → Na2ZnO2 + H2O
(q) H2S + Cl2 → S + HCl
Multiply HCl on right side by 2 and atoms of H and Cl are balanced.
H2S + Cl2 → S + 2HCl
(r) FeCl3 + NaOH → NaCl + Fe(OH)3
Multiply NaOH by 3 and NaCl by 3
FeCl3 + 3NaOH → 3NaCl + Fe(OH)3
(s) Fe2O3+ CO → Fe + CO2
Multiply Fe by 2 to make atoms of Fe equal and to make CO2 atoms equal by using 3 atoms of oxygen in Fe2O3
Fe2O3 + 3CO → 2Fe + 3CO2
(t) KHCO → K2CO3 + H2O + CO2
Multiply KHCO2 by 2 on left side
2KHCO3 → K2CO3 + H2O + CO2
(u) CuO + NH3 → Cu + H2O + N2
Multiply CuO by 3, NH3 by 2, Cu by 3 and H2O by 3
3CuO + 2NH3 → 3Cu + 3H2O + N2
Objective Type Questions
Question 1.
Complete the statements given below by filling in the blank with the correct words.
1. The formula of silver [I] chloride is ……….. [AgCl/AgCl2].
Answer:
The formula of silver [I] chloride is AgCl.
2. The basic unit of an element is a/an ……….. [molecule/ atom/ion]
Answer:
The basic unit of an element is a/an atom.
3. Atoms contains ……….. [netron/nucleus/, with positively charged ………. [electrons/protons].
Answer:
Atoms contains nucleus with positively charged protons.
4. Element ……….. [calcium/lead/carbon] has the symbol derived from its Latin name ‘plumbum’.
Answer:
Element lead has the symbol derived from its Laltin name ‘plumbum’.
5. From the elements – He, Br, Pt & O; the element which forms a polyatomic molecule is ……….. & which is liquid
at room temperature is …………
Answer:
From the elements – He, Br, Pt & O; the element which forms a polyatomic molecule is O & which is liquid at room temperature is Br.
6. The valency of iron in FeO is ……….. [2+/l+] chlorine [chloride] in CaCl2 is ……….. [l–/2–] and of dichromate in K2Cr2O7is ………… [2+/2–]
Answer:
The valency of iron in FeO is 2+ of chlorine [chloride] in CaCl2 is 1– and of dichromate in K2Cr2O7 is 2–.
Question 2.
Match the statements – 1 to 10 below with their correct . ‘ answers from – A to J.
1. Elements having valency of two – J: K1+
2. An anion – B: Divalent
3. A gaseous non-metal – C: Reactants
4. A cation – D: Ammonium
5. The term used for the substances which take part in the chemical reaction – E: Nitric oxide
6. The meaning of the symbol ‘A’ over the arrow in a chemical equation – F: Nitrogen
7. The chemical name for nitrogen monoxide – G: Zero
8. A radical containing nitrogen & hydrogen only – H: Nitrous oxide
9. The chemical name for dinitrogen oxide – I: Heat required
10. The valency of noble gases – J: K1+
Answer:
Elements having valency of two – B: Divalent
An anion – A: Br
A gaseous non-metal A cation – F: Nitrogen
The term used for the substances which take part in the chemical reaction – J: K1+
The meaning of the symbol ‘A’ over the arrow in a chemical equation – C: Reactants
The chemical name for nitrogen monoxide – I: Heat required
A radical containing nitrogen & hydrogen only – D: Ammonium
The chemical name for dinitrogen oxide – H: Nitrous oxide
10. The valency of noble gases – G: Zero
Question 3.
Match the compounds in List I – 1 to 20 with their correct formulas in List II – A to T.
1. Copper [I] sulphide – A. KMnO4
2. Potassium permanganate – B. Mg3N2
3. Phosphoric acid – C. Mg(NO3)2
4. Copper [I] oxide – D. Al2(SO4)3
5. Carbonic acid – E. Na2ZnO2
6. Aluminium sulphide – F. N2O
7. Iron [II] oxide – G H2CO3
8. Iron [III] sulphide – H. Al2S3
9. Iron [II] sulphate – I. NO
10. Sodium zincate – J. FeS
11. Nitrous oxide – K. Fe2S3
12. Aluminium sulphate – L. H3PO4
13. Magnesium nitride – M. Cu2S
14. Iron [III] sulphate – N. CuS
15. Copper [II] oxide – O. Fe2O3
16. Iron [III] oxide – P. FeO
17. Nitric oxide – Q. FeSO4
18. Copper [II] sulphide – R. Fe2(SO4)3
19. Iron [II] sulphide – S. CuO
20. Magnesium nitrate – T. Cu2O
Answer:
1. Copper [I] sulphide – M. Cu2S
2. Potassium permanganate – A. KMnO4
3. Phosphoric acid – L. H3PO4
4. Copper [I] oxide – T. Cu2O
5. Carbonic acid – G H2CO3
6. Aluminium sulphide – H. Al2S3
7. Iron [II] oxide – P. FeO
8. . Iron [III] sulphide – K. Fe2S3
9. Iron [II] sulphate – Q. FeSO4
10. Sodium zincate – E. Na2ZnO2
11. Nitrous oxide – F. N2O
12. Aluminium sulphate – D. Al2(SO4)3
13. Magnesium nitride – B. Mg3N2
14. Iron [III] sulphate – R. Fe2(SO4)3
15. Copper [II] oxide – S. CuO
16. Iron [III] oxide – O. Fe2O3
17. Nitric oxide – I. NO
18. Copper [II] sulphide – N. CuS
19. Iron [II] sulphide – J. FeS
20. Magnesium nitrate – C. Mg(NO3)2
Question 4.
Underline the incorrectly balanced compound in each equation & rewrite the correct equation.
1. 2Na + 3H2O → 2NaOH + H2
Answer:
Correct equation is :
2Na + 2H2O → 2NaOH + H2
2. 4P + 4O2 →2P2OS
Answer:
Correct equation is :
4P + 5O2 → 2P2OS
3. Fe2O3 + 2H2 →2Fe + 3H2O
Answer:
Correct equation is :
Fe2O3 + 3H2 →2Fe + 3H2O
4. 2Al + 2H2SO4 →Al2(SO4)3 + 3H2
Answer:
Correct equation is :
2Al + 3H2SO4 →Al2(SO4)2 + 3H2
5. N2 + 3H2 ⇌ NH3
Answer:
Correct equation is :
N2 + 3H2 ⇌ 2NH3
6. ZnO + 3NaOH → Na2ZnO2 + H2O
Answer:
Correct equation is :
ZnO + 2NaOH → Na2ZnO2 + H2O
7. FeCl3 + 3NH4OH → 2NH4Cl + Fe(OH)3
Answer:
Correct equation is :
8. FeS + 2HCl → 2FeCl2 + H2S
Answer:
Correct equation is :
FeS + 2HCl → FeCl2 + H2S
9. 3NH3 + H2SO4 → (NH4)2SO4
Answer:
Correct equation is :
2NH2 + H2SO4 → (NH2)2SO2
10. PbO2 + 4HCl → PbCl2 + H2O + Cl2
Answer:
Correct equation is :
PbO2 + 4HCl → PbCl2+ 2H2O + Cl2
ICSE Class 8 Chemistry Important Questions