Selina ICSE Solutions for Class 9 Chemistry – Atomic Structure

Selina ICSE Solutions for Class 9 Chemistry – Atomic Structure

ICSE SolutionsSelina ICSE Solutions

Download Formulae Handbook For ICSE Class 9 and 10

Selina ICSE Solutions for Class 9 Chemistry Chapter 7 Atomic Structure

Exercise 7

Solution 1.

The latest research on atom has proved that most of the postulates of Dalton’s atomic theory contradict. But Dalton was right that atoms take part in chemical reactions. Comparison of Dalton’s atomic theory with Modem atomic theory.

Dalton’s atomic theoryModern atomic theory
1. Atoms are indivisible particles.1. Atoms are divisible into sub-atomic particles like protons, neutrons and electrons.
2. Atoms can neither be created nor destroyed.2. Atoms can be created and destroyed by nuclear fusion and fission.
3. The atoms of an element are alike in all respect and differ from atoms of other elements.3. The atoms of an element may not be alike in all respects, as it is seen in the case of isotopes. Isotopes which are atoms of the same element having the same atomic number but different mass numbers.

Solution 2.

(a) Inert elements: The elements which have complete outer most shell i.e. 2 or electrons. They ordinarily do not enter into any reaction.

(b) These exist as monoatoms because molecules of these elements contain only one atom.

(c) Valence electrons: The number of electrons present in the outermost shell or valence shell is known as valence electrons.

Solution 3.

The three isotopes of hydrogen differ only due to their mass number which is respectively 1,2 and 3 and named protium, deuterium and tritium.
Selina ICSE Solutions for Class 9 Chemistry - Atomic Structure image - 1

Solution 4.

Atomic numberName with valency
4
15A solid non-metal of valency 3.
8A gas of valency 2.
19A metal of valency 1.
14A non-metal of valency 4.

Solution 5.

AtomAtomic numberAtomic massNo. of ProtonsNo. of ElectronsNo. of NeutronsElectronic configuration
(a) Sodium11231111122, 8, 1
(b) Chlorine17351717182, 8, 7
(c) Oxygen8168882, 6
(d) Carbon6126662, 4

Selina ICSE Solutions for Class 9 Chemistry - Atomic Structure image - 2

Solution 6.

The significance of the number of protons found in the atoms in each of different element is fixed its place in periodic table.

Solution 7.

Atomic numbers of –

XYZ
6912
(2,4)(2,7)(2,8,2)

(a) Y (2,7) forms Anion.
(b) Z (2, 8, 2) forms Cation.
(c) X (2,4) has four electrons in the valence shell.

Solution 8.

(a) X+1
(b) Oxidising agent, because it has ability to donate electron.

Solution 9.

(a) Mass number:The atomic mass number is defined as the sum of the number of protons and neutrons contained in the nucleus of an atom of that element. It is denoted by the symbol A.

(b) Ion: An atom or molecule that carries a positive or negative charge because of loss or gain of electrons.

(c) Cation: It is positively charged ion that is formed when an atom loses one or more electrons e.g. Na+, Hg2+, Ca2+ etc.

(d) Atom: It is defined as the smallest unit of matter which takes part in a chemical reaction.

(e) Element: It is a substance which cannot be split up into two or simpler substances by usual chemical methods of applying heat, light or electric energy. e.g. Hydrogen, Oxygen, Chlorine etc.

(f) Orbit: It is defined as a circular path around the nucleus in which electrons of the atom revolve.

Solution 10.

Atomic number = 2 Mass number = 4

Solution 11.

(a) (i) Atom E contains 7 protons.
(ii) Atom E has an electronic configuration 2, 7.

(b) Atom C Stands for 73Li
Atom D stands for 816O
Compound formula = Li2O

(c) Metals are: A and C Non-metals are: E, D, E

Solution 12.

No of electrons in M Shell = 2
Number of electrons in K and L shell will be 2, 8 and respectively.
Therefore, Electronic configuration will be: 2, 8, 2.
Atomic number = 2 + 8 + 2 = 12, Since, atomic number = Number of Protons
No of Protons = 12

Solution 13.

Selina ICSE Solutions for Class 9 Chemistry - Atomic Structure image - 3
(a) (ii) Electronic configuration = 2, 8, 2
(b) Mass numbers are different of two isotopes of magnesium because of different number of neutron, that is, 12 and 14 respectively.

Solution 14.

Nucleons: Particles which constitute nucleus are called nucleons. Proton and neutrons are the nucleons.
At. weight of phosphorus = 31 Atomic number = 15
Total number of nucleons = 31 (No. of P + No. of N) No. of neutrons = 31 – 15 = 16
Electronic configuration
No. of electrons = 15 = 2,8,5
Selina ICSE Solutions for Class 9 Chemistry - Atomic Structure image - 4

Solution 15.

(a) Atoms of the same elements differing in the number of neutrons in their nuclei are known as isotopes. Thus, isotopes of an element have the same atomic number but different atomic mass number.
The fundamental particles is Neutrons which differs.

Uses of isotopes:

  1. Some isotopes are Redioactive; isotopes of cobalt are used for treating cancer and other diseases.
  2. An isotopes of uranium 235U is used as a fuel in nuclear reactor.

(b)

Selina ICSE Solutions for Class 9 Chemistry - Atomic Structure image - 5

Solution 16.

In chemical reactions only electrons take part. The chemical properties depend upon the electronic configuration.

The isotopes of element 1735Cl and 1737Cl have same atomic number and hence, the same configuration. So they have same chemical properties. These differ only in physical contents and weights because neutrons contribute to the mass of an atom.

1735Cl and 1737Cl have different number of neutrons 18 and 20 respectively.

Solution 17.

The atomic masses of the isotopes of chlorine are 35 and 37. However in any given sample of chlorine gas, the isotopes occur in approximate 3 : 1, 75% of Cl35, and 25% of Cl37. Therefore, the relative atomic mass or atomic weight of chlorine is 35.5.

Fractional atomic weight of chlorine

Selina ICSE Solutions for Class 9 Chemistry - Atomic Structure image - 6

Solution 18.

(a) Atomic number: The number of protons present in the nucleus of an atom is the atomic number of that atom. It is represented by the symbol Z.
Atomic number (Z) = Number of protons (p)

(b)

No. of ProtonsNo. of ElectronsNo. of NeutronsAtomic numberMass number
35 Cl
17
1717181735
37 Cl
17
1717201737

(c) Electronic configuration of chlorine is 2, 8, 7.

Solution 19.

(a) Hydrogen
(b) Element of zero group i.e. He (Helium)
(c) Calcium (2, 8, 8, 2). Therefore, 2 electrons in valance shell. Hence valency is 2.
(d) Chlorine atoms: 1735Cl and 1737Cl
(e) K shell.

Solution 20.

(a) Physical properties depend on the Atomic mass and isotopes have different mass number. (A) i.e. they have different number of neutrons. So, isotopes have different physical properties.

(b) Argon does not react as, Argon has completely filled outer-most orbit. The atomic number of argon is 18. Therefore, electronic configuration is 2, 8, 8. There are 8 electrons in the outermost or valance shell. Therefore, argon does not react.

(c) Actual Atomic Mass is greater than mass number (P + N) since mass number is a whole number approximation of atomic mass unit. In fact Neutrons are slightly heavier than protons and atom includes the existence of over 200 sub-atomic particles.

(d) 1735Cl and 1737Cl are isotopes of chlorine element which differ in number of neutrons. Whereas chemical properties are determined by electronic configuration of an atom. Isotopes of an element are chemically alike.

Solution 21.

Element A

Atomic number = 7
Electronic configuration: = 2, 5
Valency of element A = 8-5 = (3)

Element B

Electronic configuration 2,8,8 Valency of element B = Zero Element C
Number of Electrons 13 Electronic configuration: 2,8,3
Valancy of element C = 3+

Element D

Protons = 18 = electrons Electronic configuration: 2,8, 8
Valency of element D = Zero

Element E

Electronic configuration = 2, 8, 8, 1
Valency of element E = 1+
[ii] C and E are metals [iii] A is a Non-metal
[iv] B and D are inert gases. (A, C and E are not inert gases.)

Solution 22.

(a) C. Atomic nucleus
(b) A. 6
(c) C. 2,8,8,1

Solution 23.

Elements tend to combine with one another to attain the stable electronic configuration of the nearest noble gas/inert gas (Duplet or Octet)

(a) Sodium Chloride

The electronic configuration of Sodium is 2, 8, Thus, Sodium atom tends to lose one electron to attain the stable electronic configuration of neon gas 2, 8.

The electronic configuration of Chlorine is 2, 8, 7. Thus, Chlorine atom needs one electron to complete its octet and achieve the stable electronic configuration of Argon 2, 8, 8.

When sodium and chlorine atoms approach each other, the sodium atom, Na, loses an electron to form sodium ion, The cation, Na+, carries a single positive charge.
Na → Na+ + e

The electron lost by sodium atom is transferred to the chlorine atom forming chloride ion, Cl.
Cl + e → Cl

The sodium and chloride ions thus, achieve stable electronic configurations.
Selina ICSE Solutions for Class 9 Chemistry - Atomic Structure image - 7

(b) Hydrogen H2

Hydrogen atom has one electron in its valence shell. Hydrogen atom needs one more electron to complete its duplet and attain the stable electronic configuration of Helium.

In case of hydrogen H2 molecule, each of the two hydrogen H atoms contributes one electron so as to have one shared pair of electrons. Both the hydrogen atoms attain stable duplet structure resulting in the formation of a single covalent bond [H – H] between them.

Both the hydrogen atoms have equal attraction for the electrons. Thus, the shared pair of electrons remains equidistant from both the atoms.

Selina ICSE Solutions for Class 9 Chemistry - Atomic Structure image - 8

Solution 24.

Element SymbolAtomic NumberMass NumberNumber of NeutronsNumber of ElectronsNumber of Protons
Li36433
Cl1735181717
Na1123121211
Al1327141313
S1532161515

ChemistryPhysicsBiologyMaths

Selina Concise Chemistry Class 9 ICSE Solutions The Periodic Table

Selina Concise Chemistry Class 9 ICSE Solutions The Periodic Table

ICSE SolutionsSelina ICSE Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 9 Chemistry Chapter 5 The Periodic Table. You can download the Selina Concise Chemistry ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Chemistry for Class 9 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Download Formulae Handbook For ICSE Class 9 and 10

Selina ICSE Solutions for Class 9 Chemistry Chapter 8 The Periodic Table

Page No. 79

Question 1.
What is the need for classification of elements?
Solution:
It is impossible for a chemist to study all the elements and their compounds. Hence, classification is a must.
Following are the reasons for the classification of elements:

  1. To study elements better
  2. To correlate the properties of the elements with some fundamental properties which are characteristic of all the elements
  3. To reveal relationships between elements

Question 2.
What was the basis of the earliest attempts made for classification and grouping of elements?
Solution:
The first classification of elements was into 2 groups-metals and non-metals.

Question 3(a).
A, B and C are the elements of a Dobereiner’s triad. If the atomic mass of A is 7 and that of C is 39, what should be the atomic
mass of B?
Solution:

(a)
Selina Concise Chemistry Class 9 ICSE Solutions The Periodic Table image - 1

Question 3(a).
Why was Dobereiner’s triad discarded?
Solution:
(b) Döbereiner failed to arrange all the known elements in the form of triads.
In the triad of fluorine (19), chlorine (35.5) and bromine (80), it is observed that the mean of the atomic masses of fluorine and bromine is ½(19 + 80) = 49.5, not 35.5.

Question 4.
Explain ‘Newland’s Law of Octaves.’ Why was the law discarded?
Solution:
Elements when arranged in the increasing order of their atomic weights are similar to the eighth and the first note of the musical scale. For example, the eighth element from lithium is sodium. Similarly, the eighth element from sodium is potassium. Thus, lithium and sodium provide any specific place for hydrogen.

  1. This classification did not work with heavier elements.
  2. Newland adjusted two elements Cobalt (Co) and Nickel (Ni) in the same slot.
  3. Fe, which resembles Co and Ni in properties, has been placed far away.

Question 5.
Did Dobereiners triads also exist in the columns of Newland’s Octaves? Compare and find out.
Solution:
Yes, Döbereiner’s triads also exist in the columns of Newland’s octaves. For example, the second column of Newlands classification has the elements Lithium (Li), Sodium (Na) and Potassium (K), which constitute a Döbereiner’s triad.

Question 6(a).
Lithium, sodium and potassium elements were put in one group on the basis of their similar properties. What are those similar properties?
Solution:
(a) Elements of lithium, sodium and potassium have the following similar properties:

  1. All these have one electron in the outermost shell.
  2. They form unipositive ions.
  3. They are good reducing agents.
  4. They are soft metals.
  5. They impart colour to the flame.
  6. Common name of the group is alkali metals [Group 1A].

Question 6(b).
The elements calcium, strontium and barium were put in one group or family on the basis of their similar properties.
What were those similar properties?
Solution:
(b)

  1. All of them are metals.
  2. Oxide of each of them is alkaline in nature.
  3. Each has valency 2.

Question 7(a).
What was Mendeleev’s basis for classification of elements?
Solution:
(a) Mendeleev’s basis for periodic classification:

  1. Similarities in the chemical properties of elements.
  2. Increasing order of atomic weights of elements.

Question 7(b).
Mendeleev’s contributions to the concept of periodic table laid the foundation for the Modern Periodic Table. Give reasons.
Solution:
(b) Mendeleev laid the foundation for the modern periodic table by showing periodicity of the properties of the elements by arranging the elements (63) then known into 8 groups, by leaving gaps for undiscovered elements and predicting their properties. He made separate groups for metals and non-metals. He also created periods in which the element gradually changes from metallic to non-metallic character. He was also able to show that the element in the same sub-group had the same valency.

Question 8.
State Mendeleev’s periodic law.
Solution:
Mendeleev’s periodic law: The physical and chemical properties of all the elements are a periodic function of their atomic masses.

Question 9(a).
Use Mendeleev’s Periodic Table to predict the formula of hydrides of carbon and silicon.
Solution:
(a) C is in Group 4. So, the hydride will be CH4 (Methane).
Si is in Group 4. So, the hydride will be SiH4 (Silane).

Question 9(b).
Use Mendeleev’s Periodic Table to predict the formula of oxides of potassium, aluminium and barium.
Solution:
(b)
K is in Group 1. So, the oxide will be K2O (Potassium oxide).
Al is in Group 3. So, the oxide will be Al2O3 (Aluminium oxide).
Ba is in Group 2. So, the oxide will be BaO (Barium oxide).

Question 10.
Which group of elements was missing from Mendeleev’s original periodic table?
Solution:
Anomalous pairs of elements were missing from Mendeleev’s periodic table.

Question 11.
State the merits of Mendeleev’s classification of elements.
Solution:
Merits of Mendeleev’s classification of elements:

  1. Grouping of elements
  2. Gaps for undiscovered elements: Mendeleev left some gaps in his periodic table for subsequent inclusion of elements not known at that time.
  3. He predicted the properties of the then unknown elements on the basis of properties of elements lying adjacent to the vacant slots (eka-aluminium and eka-silicon).

Question 12.
Why did Mendeleev’s leave some gaps in his periodic table os elements? Explain your answer with an example.
Solution:
He left gaps in the table for the undiscovered elements. He discovered the properties of such elements with the help of neighboring elements.
He discovered eka-silicon with atomic mass of 72 which was later named Germanium with atomic mass 72.6.

Question 13.
The atomic number of an element is more important to the chemist than its relative atomic mass. Why?
Solution:
Henry Moseley found that when cathode rays struck anodes of different metals, the wavelength of these metals was found to decrease in a regular manner of changing the metal of anode in the order of its position in the periodic table. By this, he concluded that the number of positive charges present in the nucleus due to protons (atomic number) is the most fundamental property of the element.

So, Henry Moseley found that the atomic number is a better fundamental property of an element compared to its atomic mass. This lead to the modern periodic law.

This law gave explanations for anomalies in Mendeleev’s classification of elements such as

  1. Position of isotopes with the same atomic number can be put in one place in the same group.
  2. Position of argon and potassium: Potassium with higher atomic number should come later, and argon with lower atomic number should come first.

Question 14.
Consider the following elements: Be, Li, Na, Ca, K. Name the elements of (a) same group (b) same period.
Solution:

ElementAt. No.Electronic distribution
Be42, 2
Li32, 1
Na112, 8, 1
Ca202, 8, 8
K192, 8, 8, 2
  1. Same IA group (Li, Na, K) and IIA group (Be, Ca)
  2. In the second period (Be, Li) and in the fourth period (K, Ca)

Question 15(a).
Name an element whose properties were predicted on the basis of its position in Mendeleev’s periodic table.
Solution:
a. Eka-silicon

Question 15(b).
Name two elements whose atomic weights were corrected on the basis of their positions in Mendeleev’s periodic table.}
Solution:
b. Gold and Platinum

Question 15(c).
How many elements were known at the time of Mendeleev’s classification of elements?
Solution:
c. Only 63 elements were discovered at the time of Mendeleev’s classification of elements.

Page No. 86

Question 1(a).
State the modern periodic law.
Solution:
a. Modern periodic law: The physical and chemical properties of all elements are a periodic function of their atomic numbers.

Question 1(b).
How many periods and groups are there in the modern periodic table?
Solution:
b. Eighteen groups and seven periods

Question 2.
What is the main characteristic of the last elements in the periods of a periodic table? What is the general name of such elements?
Solution:
Last elements of each period have their outermost shell complete, i.e. 2 or 8 electrons.
The general name is inert gases or noble gases.

Question 3(a).
What is meant in the periodic table by a group?
Solution:
a. Vertical columns in a periodic table which have the same number of valence electrons and similar chemical properties are called a group.

Question 3(b).
What is meant in the periodic table by a period?
Solution:
b. In a periodic table, elements are arranged in the order of increasing atomic numbers in horizontal rows called periods.

Question 4.
From the standpoint of atomic structure, what determines which elements will be the first and which the last in a period of the periodic table?
Solution:
Atomic number determines which element will be the first and which will be the last in a period of the periodic table.

Question 5(a).
What are the following groups known as?

  1. Group 1
  2. Group 17
  3. Group 18

Solution:

(a)

  1. Group 1 is known as the alkali metals.
  2. Group 17 is known as the halogens.
  3. Group 18 is known as the transition elements.

Question 5(b).
Name two elements of each group.
Solution:
(b)

  1. Group 1: Lithium (Li), Sodium (Na)
  2. Group 17: Chlorine (Cl), Iodine (I)
  3. Group 18: Helium (He), Neon (Ne)

Question 6(a).
What is the number of elements in the 1st period?
Solution:
(a) There are two elements in the first period.

Question 6(b).
What is the number of elements in the 3rd period, of the modern periodic table?
Solution:
(b) There are eight elements in the third period.

Question 7(a).
How does number of (i) valence electrons (ii) valency; vary on moving from left to right in the second period of a periodic table?
Solution:

(a)

  1. The valence electrons in the same shell (outermost shell) increase progressively by one across the period. The first element hydrogen has one valence electron and helium has two valence electrons.
  2. On moving from left to right in a period, valency increases from 1 to 4, then falls to one and ultimately to zero in the last group.

Question 7(b).
How does number of (i) valence electrons (ii) valency; vary on moving from left to right in the third period of a periodic table?
Solution:

(b)

  1. Valence electrons in the same shell (outermost shell) increase progressively by one across the period. The first element sodium has one valence electron and magnesium has two valence electrons.
  2. On moving from left to right in the third period, valency increases from 1 to 7 and ultimately to zero in the last group.

Question 8.
How do atomic structures (electron arrangements) change in a period with increase in atomic numbers moving left to right?
Solution:
The size of atoms decreases when moving from left to right in a period. Thus, in a particular period, the alkali metal atoms are the largest and the halogen atoms are the smallest.

Question 9(a).
This question refers to elements of the periodic table with atomic numbers from 3 to 18. In the table below, some elements are shown by letters, even though the letters are not the usual symbols of the elements.
Selina Concise Chemistry Class 9 ICSE Solutions The Periodic Table image - 2
Which of these is:

    1. a noble gas?
    2. a halogen?
    3. an alkali metal?
    4. an element with valency 4?

Solution:

(a)

  1. H and P are noble gases.
  2. G and O are halogens.
  3. A and I are alkali metals.
  4. D and L have valency 4.

Question 9(b).
This question refers to elements of the periodic table with atomic numbers from 3 to 18. In the table below, some elements are shown by letters, even though the letters are not the usual symbols of the elements.
Selina Concise Chemistry Class 9 ICSE Solutions The Periodic Table image - 2
If A combines with F, what would be the formula of the resulting compound?
Solution:
(b) Li2O. A stands for lithium and F stands for oxygen. The valence of lithium is +1 and the valence of O is -2, i.e. A2F.

Question 9(c).
This question refers to elements of the periodic table with atomic numbers from 3 to 18. In the table below, some elements are shown by letters, even though the letters are not the usual symbols of the elements.
Selina Concise Chemistry Class 9 ICSE Solutions The Periodic Table image - 2
Solution:
(c) G has atomic number 9; therefore, its electronic arrangement is 2, 7.

Question 10.
Sodium and aluminium have atomic numbers 11 and 13, respectively. They are separated by one element in the periodic table, and have valencies 1 and 3 respectively. Chlorine and potassium are also separated by one element in the periodic table (their atomic numbers being 17 and 19, respectively) and yet both have valency 1. Explain.
Solution:
Na and Al have the capacity to donate an electron due to which the valency is positive, whereas Cl and K can only gain or lose one electron due to which their valency is -1 and +1, respectively. This is the only difference between these two.

Question 11.
Helium is an unreactive gas and neon is a gas of extremely low reactivity. What, if anything, do their atoms have in common.
Solution:
These elements have a full outermost subshell, which results in high stability. They only react with other elements in extreme circumstances.

Question 12(a).
In which part of a group would you separately expect the elements to have the greatest metallic character?
Solution:
a. The greatest metallic character can be expected at the bottom of the group.

Question 12(b).
In which part of a group would you separately expect the elements to have the largest atomic size?
Solution:
b. The largest atomic size can be expected at the lower part of the group.

Question 13.
What happens to the number of valence electrons in atoms of elements as we go down a group of the periodic table?
Solution:
The number of valence electrons remains the same as we go down a group.

Question 14(a).
The position of elements A, B, C, D and E in the periodic table are shown below:
Selina Concise Chemistry Class 9 ICSE Solutions The Periodic Table image - 4
State which are metals, non-metals and noble gas in this table.
Solution:
(a) Metals: A and B; Non-metals: C; Noble gases: D and E

Question 14(b).
The position of elements A, B, C, D and E in the periodic table are shown below:
Selina Concise Chemistry Class 9 ICSE Solutions The Periodic Table image - 5
State which is most reactive (i) metal (ii) non-metal
Solution:
(b) Most reactive

  1. Metals: Alkali metals (Group I); Caesium
  2. Non-metals: Halogens (Group 17); Fluorine

Question 14(c).
The position of elements A, B, C, D and E in the periodic table are shown below:
Selina Concise Chemistry Class 9 ICSE Solutions The Periodic Table image - 6
Which type of ion will be formed by element A, B and C.
Solution:

(c)

Element A will form a positive ion 1+ (cation).
Element B will form a positive ion 2+ (cation).
Element C will form a negative ion 1 (anion).

Question 14(d).
The position of elements A, B, C, D and E in the periodic table are shown below:
Selina Concise Chemistry Class 9 ICSE Solutions The Periodic Table image - 7
Which is larger in size (i) D or E (ii) B or C
Solution:
(d)

  1. E
  2. B

Question 15.
Write the electronic configuration of element 17T35.

  1. What is the group number of T?
  2. What is the period number of T?
  3. How many valence electrons are there in an atom of T?
  4. What is the valency of T?
  5. Is it a metal or non-metal?
  6. State number of protons and neutrons in T.

Solution:
K L M
Electronic configuration = 2, 8, 7

a. VIIA
b. Third period
c. Seven
d. Valency of T = -1
e. Non-metal
f. Protons = 17, Neutrons = 18

Page No. 91

Question 1.
Element P has atomic number 19. To which group and period, does P belong? Is it a metal or a non-metal? Why?
Solution:
Atomic number of P = 19
Electronic configuration = 2, 8, 8, 1
Group number of the element = 1
A Period number of the element = 4
P is a metal.

Question 2.
An element belongs to the third period and Group IIIA (13) of the periodic table. State:
a. the number of valence electrons,
b. the valency,
c. if it is a metal or non-metal?
d. the name of the element.
Solution:
a. 3
b. +3
c. Metal
d. Aluminium

Question 3.
Name and state the following with reference to the elements of the first three periods of the periodic table.

  1. Noble gas with duplet arrangement of electrons.
  2. Metalloid in Period 3.
  3. Valency of elements in Group 14 and 15.
  4. Noble gas having electronic configuration: 2, 8, 8.
  5. Group whose elements have zero valency.
  6. A covalent compound formed by an element in period 2 and ahalogen.
  7. Non-metallic element present in Period 3 of Groups 15 and 16
  8. An electrovalent compound formed by an alkaline earth metal and a halogen.
  9. Bridge elements of Period 3 of Group 1,2 and 3.
  10. Alkali metal in period 3 that dissolves in water giving a strong alkali.
  11. Typical elements of Groups 14 and 15.
  12. Alkaline earth metal in period 3.

Solution:

(a) Helium
(b) Silicon
(c) 4, 3
(d) Argon
(e) Noble gases
(f) Carbon tetrachloride (CCl4)
(g) Silicon, Phosphorus
(h) Sodium chloride (Na+Cl)
(i) Li and Mg; Be and Al; B and Si
(j) Sodium
(k) Typical elements of Period 2 belonging to group 14 and 15 are carbon and nitrogen.
Typical elements of’ Period 3 belonging to group 14 to 15 are silicon and phosphorus.
(l) Beryllium

Question 4.
Match column A with column B
Selina Concise Chemistry Class 9 ICSE Solutions The Periodic Table image - 8
Solution:

Column AColumn B
(a) Elements short of 1 electron in octet(v) Halogens
(b) Highly reactive metals(iii) Alkali metals
(c) Unreactive elements(ii) Noble gases
(d) Elements of groups 3 to 12(i) Transition elements
(e) Radioactive elements(vi) Actinides
(f) Elements with 2 electrons in outermost orbit(iv) Alkaline earth metals

Question 5.
Complete the table:
Selina Concise Chemistry Class 9 ICSE Solutions The Periodic Table image - 9
Solution:

Atomic numberElementElectronic configurationSelect element of the same group
11Sodium2, 8, 1K
15Phosphorus2, 8, 5N
16Sulphur2, 8, 6O
9Fluorine2, 7Cl

Question 6.

  1. Write down the word that will correctly complete the following sentences:
  2. Relative atomic mass of a light element up to calcium is approximately_____ its atomic number.
  3. The horizontal rows in a periodic table are called _____ .
  4. Going across a period left to right, atomic size _____ .
  5. Moving left to right in the second period, number of valence electrons _____ .
  6. Moving down in the second group, number of valence electrons___________.

Solution:

a. Relative atomic mass of a light element up to calcium is approximately 20 its atomic number.
b. The horizontal rows in a periodic table are called periods.
c. Going across a period left to right, atomic size increases.
d. Moving left to right in the second period, number of valence electrons increases from 1 to 8.
e. Moving down in the second group, number of valence electrons remain same.

Question 7(a).
Name the alkali metals, How many electron(s) they have in their outermost orbit.
Solution:
Name of the alkali metals: Lithium, sodium, potassium, rubidium, cesium and francium
Electrons in the outermost orbit: 1

Question 7(b).
Take any one alkali metal and write its reaction with (i)oxygen (ii)water (iii)acid.
Solution:

  1. Reaction of alkali metal with oxygen – React rapidly with oxygen
    4Na + O2 → 2Na2O
  2. Reaction of alkali metal with water – React with water violently and produce hydrogen
    2M + 2H2O → 2MOH + H2
  3. Reaction of alkali metal with acid – React violently with dil. HCl and dil. H2SO4 to produce hydrogen
    2M + 2HCl → 2MCl + H2

Page No. 92

Question 8(a).
Name the method by which alkali metals can be extracted.
Solution:
a. Alkali metals can be extracted by the electrolysis of their molten salts.

Question 8(b).
What is the colour of the flame of sodium and potassium?
Solution:
b. The colour of the flame of sodium is golden yellow, and the colour of the flame of potassium is pale violet.

Question 9(a)
Name the first three alkaline earth metals.
Solution:
a. The first three alkaline earth metals are Beryllium, Magnesium and Calcium.

Question 9(b)
Write the reactions of first three alkaline earth metals with dilute hydrochloric acid.
Solution:
b. Reactions of the first three alkaline earth metals with dilute hydrochloric acid:
Be + 2HCl → BeCl2 + H2
Mg + 2HCl → MgCl2 + H2
Ca + 2HCl → CaCl2 + H2

Question 10(a)
How do alkaline earth metals occur in nature?
Solution:
a. Alkaline earth metals occur in nature in the combined state and not in the free state as they are very reactive.

Question 10(b)
Write the electronic configuration of the first two alkaline earth metals.
Solution:
b. Electronic configuration of the first two alkaline earth metals:
4Be: 1s22s2
12Mg: 1s22s22p63s2

Solution 11.
a. Group 17 elements are called halogens. The name halogens is from Greek halo (sea salt) and gens (producing, forming) and thus means ‘sea salt former’.

b. Group 17 elements or halogens:

  1. Reactivity: Halogens are the most reactive non-metals, their reactivity decreases down the group. Fluorine is the most reactive halogen and iodine is the least reactive halogen.
  2. Colour: Fluorine is a pale yellow gas, chlorine is a greenish yellow gas, bromine is a reddish brown liquid and iodine is a violet solid.
  3. Physical state: Gaseous

Solution 12.
(a)

  1. Group 17 elements react with metals to form metal halides which are neutral in nature.
  2. Group 17 elements react with non-metals to form acidic compounds such as hydrogen halides.

(b) Group 17 elements are highly reactive because of their closeness to the noble or stable gas configuration. They can easily achieve a noble gas electron structure.

Solution 13.
a. All the noble or inert gases have 8 electrons in their valence shell except helium which has two electrons in its valence shell.
b. Xenon or krypton from Group 18 can form compounds.

Solution 14.

  1. Helium gas is used in airships and balloons.
  2. Neon gas is used in neon lights. The brightly coloured advertising light works when an electric discharge is passed in a tube containing a little of a noble gas such as neon.

Question 9.
An element A has 2 electrons in its fourth shell. State:

  1. its atomic number
  2. its electronic configuration
  3. its valency
  4. position in the periodic table
  5. is it a metal or non metal
  6. is it an oxidising or reducing agent

Solution 15.

a. Ca
b. 1s22s22p63s23p64s2
c. 2
d. Group 2 Period 4
e. Metal
f. Reducing agent

More Resources for Selina Concise Class 9 ICSE Solutions

Selina Concise Chemistry Class 9 ICSE Solutions The Language of Chemistry

Selina Concise Chemistry Class 9 ICSE Solutions The Language of Chemistry

ICSE SolutionsSelina ICSE Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 9 Chemistry Chapter 1 Language of Chemistry. You can download the Selina Concise Chemistry ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Chemistry for Class 9 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Download Formulae Handbook For ICSE Class 9 and 10

Selina ICSE Solutions for Class 9 Chemistry Chapter 4 The Language of Chemistry

Page No: 8

Question 1.
What is a symbol? What information does it convey?
Solution:
A symbol is the short form which stands for the atom of a specific element or the abbreviations used for the names of elements.

  1. It represents a specific element.
  2. It represents one atom of an element.
  3. A symbol represents how many atoms are present in its one gram (gm) atom.
  4. It represents the number of times an atom is heavier than one atomic mass unit (amu) taken as a standard.
Question 2
Why is the symbol S for sulphur, but Na for sodium and Si for silicon?
Solution:
In most cases, the first letter of the name of the element is taken as the symbol for that element and written in capitals (e.g. for sulphur, we use the symbol S). In cases where the first letter has already been adopted, we use a symbol derived from the Latin name (e.g. for sodium/Natrium, we use the symbol Na). In some cases, we use the initial letter in capital together with a small letter from its name (e.g. for silicon, we use the symbol Si).
Question 3.
Write the full form of IUPAC. Name the elements represented by the following symbols:
Au, Pb, Sn, Hg
Solution:
The full form of IUPAC is International Union of Pure and Applied Chemistry.
Names of the elements:
Au – Gold
Pb – Lead
Sn – Tin
Hg – Mercury
Question 4.
If the symbol for Cobalt, Co, were written as CO, what would be wrong with it?
Solution:
Co stands for Cobalt. If we write CO, then it would mean that it is a compound containing two non-metal ions, i.e. carbon and oxygen, which forms carbon monoxide gas.

Question 5(d).
2H2
Solution:
(a) H stands for one atom of hydrogen.
(b) H2 stands for one molecule of hydrogen.
(c) 2H stands for two atoms of hydrogen.

Question 6.
What is meant by atomicity? Name the diatomic element.
Solution:
The number of atoms of an element that join together to form a molecule of that element is known as its atomicity. Diatomic molecules: H2, O2, N2, Cl2
Question 7(a).
Explain the terms ‘valency‘ and ‘variable valency‘.
Solution:
  1. Valency of Na is +1 because it can lose one electron.
  2. Valency of O is -2 because it can accept two electrons.

Variable valency: It is the combining capacity of an element in which the metal loses more electrons from a shell next to a valence shell in addition to electrons of the valence shell.

Question 7(b).
How are the elements with variable valency named? Explain with an example.
Solution:
If an element exhibits two different positive valencies, then
  1. for the lower valency, use the suffix -OUS at the end of the name of the metal
  2. for the higher valency, use the suffix -IC at the end of the name of the metal.

Example:

ElementLower valencyHigher valency
Ferrum (Iron)Ferrous (Fe2+)Ferric (Fe3+)

Question 8.
Give the formula and valency of:

  1. aluminate ………………… .
  2. chromate ………….…….. .
  3. aluminium ………………. .
  4. cupric  ………………… .

Solution:

NameFormulaValency
a.AluminateAlO2-2
b.ChromateCrO4-2
c.AluminiumAl+3
d.CupricCu+2

Question 9.b
What is the significance of formula?
Solution:
Chemical formula: The chemical formula of a substance (element or compound) is a symbolic representation of the actual number of atoms present in one molecule of that substance.
It also indicates the fixed proportion by weight in which atoms combine.
Rules:
(i) The positive and negative radicals are written side by side (+ve first) with their charge as a superscript on the right side.
(ii) Charges are then interchanged and written as a subscript.
(iii) The final formula is written without the sign of charge, e.g. Hg2O

  1. Hg1+O2-
  2. Selina Concise Chemistry Class 9 ICSE Solutions The Language of Chemistry image - 1
  3. Hg2O

Question 10(a).
What do you understand by the following terms?
Acid radical
Solution:
Acid radical: The electronegative or negatively charged radical is called an acid radical.
Examples: Cl, O2-

Question 10(b).
What do you understand by the following terms? Basic radical
Solution:
Basic radical: The electropositive or positively charged radical is called a basic radical.
Examples: K+, Na+

Question 11.

Match the following:

CompoundFormula
(a) Boric acid i. NaOH
(b) Phosphoric acid ii. SiO2
(c) Nitrous acid iii. Na2CO3
(d) Nitric acid iv. KOH
(e) Sulphurous acid v. CaCO3
(f) Sulphuric acid vi. NaHCO3
(g) Hydrochloric acid vii. H2S
(h) Silica (sand) viii. H2O
(i) Caustic soda (sodium hydroxide) ix. PH3
(j) Caustic potash (potassium hydroxide) x. CH4
(k) Washing soda(sodium carbonate) xi. NH3
(l) Baking soda(sodium bicarbonate) xii. HCl
(m) Lime stone.(calcium carbonate) xiii. H2SO3
(n) Water xiv. HNO3
(o) Hydrogen sulphide xv. HNO2
(p) Ammonia xvi. H3BO3
(q) Phosphine xvii. H3PO4
(r) Methane xviii. H2SO4

Solution:

CompoundFormula (Ans)
(a) Boric acidxvi. H3BO3
(b) Phosphoric acidxvii. H3PO4
(c) Nitrous acidxv. HNO2
(d) Nitric acidxiv. HNO3
(e) Sulphurous acidxiii. H2SO3
(f) Sulphuric acidxviii. H2SO4
(g) (a) Hydrochloric acidxii. HCl
(h) Silica (sand)ii. SiO2
(i) Caustic soda
(sodium hydroxide)
i. NaOH
(j) Caustic potash (potassium hydroxide)iv. KOH
 (k) Washing soda
(sodium carbonate)
iii. Na2CO3

 (l) Baking soda
(sodium bicarbonate)

vi. NaHCO3
 (m) Lime stone
(calcium carbonate)
v. CaCO3
(n) Waterviii. H2O
(o) Hydrogen sulphidevii. H2S
(p) Ammoniaxi. NH3
(q) Phosphineix. PH3
(r) Methanex. CH4

Question 12.
Select the basic and acidic radicals in the following compounds.

  1. MgSO4
  2. (NH4)2SO4
  3. Al2(SO4)3
  4. ZnCO3
  5. Mg(OH)2

Solution:

Acidic radicalBasic radical
MgSO4SO4Mg+
(NH4)2SO4SO4NH4+
Al2(SO4)3SO4Al3+
ZnCO3CO3Zn2+
Mg(OH)2OHMg2+
Question 13.
Write chemical formula of the sulphate of Aluminium, Ammonium and Zinc.
Solution:
Valencies of aluminium, ammonium and zinc are 3, 1 and 2, respectively.
The valency of sulphate is 2.
Hence, chemical formulae of the sulphates of aluminium, ammonium and zinc are Al2(SO4)3, (NH4)2SOand ZnSO4.
Question 14.
The valency of an element A is 3 and that of element B is 2. Write the formula of the compound formed by the combination of A and B
Solution:
Formula of the compound = A2B3

Question 15.
Write the chemical names of the following compounds:

  1. Ca3(PO4)2
  2. K2CO3
  3. K2MnO4
  4. Mn3(BO3)2
  5. Mg(HCO3)2
  6. Na4Fe(CN)6
  7. Ba(ClO3)2
  8. Ag2SO3
  9. (CH3COO)2Pb
  10. Na2SiO3

Solution:
Chemical names of compounds:

  1. Ca3(PO4)2 – Calcium phosphate
  2. K2CO3 – Potassium carbonate
  3. K2MnO– Potassium manganate
  4. Mn3(BO3)2 – Manganese (II) borate
  5. Mg(HCO3)2 – Magnesium hydrogen carbonate
  6. Na4Fe(CN)6 – Sodium ferrocyanide
  7. Ba(ClO3)2 – Barium chlorate
  8. Ag2SO3 – Silver sulphite
  9. (CH3COO)2Pb – Lead acetate
  10. Na2SiO3 – Sodium silicate

Question 16
Write the basic radicals and acidic radicals of the following and then write the chemical formulae of these compounds.

  1. Barium sulphate
  2. Bismuth nitrate
  3. Calcium bromide
  4. Ferrous sulphide
  5. Chromium sulphate
  6. Calcium silicate
  7. Potassium ferrocyanide
  8. Stannic oxide
  9. Magnesium phosphate
  10. Sodium zincate
  11. Stannic phosphate
  12. Sodium thiosulphate
  13. Potassium manganate
  14. Nickel bisulphate

Solution:

CompoundsAcidic
radical
Basic
radical
Chemical
formulae
Barium sulphateSO42-Ba2+BaSO4
Bismuth nitrateNO3Bi3+Bi(NO3)3
Calcium bromideBrCa2+CaBr2
Ferrous sulphideS2-Fe2+FeS
Chromium sulphateSO42-Cr3+Cr2(SO4)3
Calcium silicateSiO42-Ca2+Ca2SiO4
Potassium ferrocyanide[Fe(CN)6]4-K1+K4[Fe(CN)6]
Stannic oxideO2-Sn2+SnO2
Magnesium phosphate(PO4)3-Mg2+Mg3(PO4)2
Sodium zincateZnO2-Na1+Na2ZnO2
 Stannic phosphate(PO4)3-Sn4+Sn3(PO4)4
 Sodium thiosulphate(S2O3)2-Na1+Na2S2O3
Potassium manganateMnO42-K1+K2MnO4
Nickel bisulphateHSO41-Ni3+Ni(HSO4)3

Question 16.
Give the names of the following compounds.

  1. NaClO
  2. NaClO2
  3. NaClO3
  4. NaClO4

Solution:

  1. NaClO – Sodium hypochlorite
  2. NaClO2 – Sodium chlorite
  3. NaClO3 – Sodium chlorate
  4. NaClO4 – Sodium perchlorate

Question 18(a).
Complete the following statements by selecting the correct option :
The formula of a compound represents
i. an atom
ii. a particle
iii. a molecule
iv. a combination
Solution:
iii. The formula of a compound represents a molecule.

Question 18(b).
Complete the following statements by selecting the correct option :
The correct formula of aluminium oxide is
i. AlO3
ii. AlO2
 iii. Al2O3
Solution:
iii. The correct formula of aluminium oxide is Al2O3.

Question 18(c).
Complete the following statements by selecting the correct option :
The valency of nitrogen in nitrogen dioxide (NO2) is
i. one
ii. two
iii. three
iv. four
Solution:
iv. The valency of nitrogen in nitrogen dioxide (NO2) is four.

Page No: 13

Question 1.
Balance the following equations:
  1. Fe + H2O → Fe3O4 + H2
  2. Ca + N2 → Ca3N2
  3. Zn + KOH → K2ZnO2 + H2
  4. Fe2O3 + CO → Fe + CO2
  5. PbO + NH3 → Pb + H2O + N2
  6. Pb3O4 → PbO + O2
  7. PbS + O2 → PbO + SO2
  8. S + H2SO→ SO2 + H2O
  9. S + HNO3 → H2SO4 + NO2 + H2O
  10. MnO2 + HCl → MnCl2 + H2O + Cl2
  11. C + H2SO4 → CO2 + H2O + SO2
  12. KOH + Cl2 → KCl + KClO + H2O
  13. NO2 +H2O → HNO2 + HNO3
  14.  Pb3O4 + HCl → PbCl2 + H2O + Cl2
  15.  H2O + Cl2 → HCl + O2
  16.  NaHCO3 → Na2CO3 + H2O + CO2
  17. HNO3 + H2S → NO2 + H2O + S
  18. P + HNO3 → NO2 + H2O + H3PO4
  19. Zn + HNO3  → Zn(NO3)2 + H2O + NO2

Solution:
Balanced chemical equations:

  1. 3Fe + 4H2O → Fe3O4 + 4H2
  2. 3Ca + N2 → Ca3N2
  3. Zn + 2KOH → K2ZnO2 + H2
  4. Fe2O3 + 3CO → 2Fe + 3CO2
  5. 3PbO + 2NH3 → 3Pb + 3H2O + N2
  6. 2Pb3O4 → 6PbO + O2
  7. 2PbS + 3O2 → 2PbO + 2SO2
  8. S + 2H2SO→ 3SO2 + 2H2O
  9. S + 6HNO3 → H2SO4 + 6NO2 + 2H2O
  10. MnO2 + 4HCl → MnCl2 + 2H2O + Cl2
  11. C + 2H2SO4 → CO2 + H2O + SO2
  12. 2KOH + Cl2 → KCl + KClO + H2O
  13. 2NO2 + H2O → HNO2 + HNO3
  14. Pb3O4 + 8HCl → 3PbCl2 + 4H2O + Cl2
  15. 2H2O + 2Cl2 → 4HCl + O2
  16. 2NaHCO3 → Na2CO3 + H2O + CO2
  17. 2HNO3 + H2S → 2NO2 + 2H2O + S
  18. P + 5HNO3 → 5NO2 + H2O + H3PO4
  19. Zn + 4HNO3  → Zn(NO3)2 + 2H2O + 2NO2

Page No: 17

Question 1.
Fill in the blanks:

  1. Dalton used symbol _____ for oxygen _____ for hydrogen.
  2. Symbol represents _____ atom(s) of an element.
  3. Symbolic expression for a molecule is called _____.  .
  4. Sodium chloride has two radicals. Sodium is a _____ radical while chloride is _____ radical.
  5. Valency of carbon in CH4 is _____ , in C2H_____, in C2H4 ___ and in C2H2 is ____.
  6. Valency of Iron in FeCl2 is _____  and in FeCl3 it is ____ .
  7. Formula of iron (ill) carbonate is _____ .

Solution:

  1. Dalton used symbol [O] for oxygen,[H] for hydrogen.
  2. Symbol represents gram atom(s) of an element.
  3. Symbolic expression for a molecule is called molecular formula.
  4. Sodium chloride has two radicals. Sodium is a basic radical, while chloride is an acid radical.
  5. Valency of carbon in CH4 is 4, in C2H64, in C2H44 and in C2H2 is 4.
  6. Valency of iron in FeCl2 is 2 and in FeCl3 it is 3.
  7. Formula of iron (III) carbonate is Fe2[CO3]3.

Question 2.
Complete the following table.

 

Acid Radicals

 

Basic Radicals

ChlorideNitrateSulphateCarbonateHydroxidePhosphate
MagnesiumMgCl2Mg(NO3)2MgSO4MgCO3Mg(OH)2 Mg3(PO4)2
Sodium
Zinc
Silver
Ammonium
Calcium
Iron (II)
Potassium

Solution:

Selina Concise Chemistry Class 9 ICSE Solutions The Language of Chemistry image - 3

Question 3.
Sodium chloride reacts with silver nitrate to produce silver chloride and sodium nitrate

  1. Write the equation.
  2. Check whether it is balanced, if not balance it.
  3. Find the weights of reactants and products.
  4. State the law which this equation satisfies.

Solution:
(a) NaCl+ AgNO3 → NaNO3 + AgCl↓
(b) It is a balanced equation.
(c) Weights of reactants:NaCl – 58.44, AgNO3 – 169.87
Weights of products: NaNO3 – 84.99, AgCl – 143.32
NaCl + AgNO3 → NaNO + AgCl
(23+35.5) + (108+14+48) → (23+14+48) + (108+35.5)
58.5 + 170 → 85 + 143.5
228.5 g → 228.5 g
(d) Law of conservation of mass: Matter is neither created nor destroyed in the course of a chemical reaction.

Question 4(a).
What information does the following chemical equation convey? Zn + H2SO4 → ZnSO4+ H2
Solution:
(a) This equation conveys the following information:
  1. The actual result of a chemical change.
  2. Substances take part in a reaction, and substances are formed as a result of the reaction.
  3. Here, one molecule of zinc and one molecule of sulphuric acid react to give one molecule of zinc sulphate and one molecule of hydrogen.
  4. Composition of respective molecules, i.e. one molecule of sulphuric acid contains two atoms of hydrogen, one atom of sulphur and four atoms of oxygen.
  5. Relative molecular masses of different substances, i.e. molecular mass of
    Zn = 65
    H2SO4 = (2+32+64) = 98
    ZnSO= (65+32+64) = 161
    H2 = 2
  6. 22.4 litres of hydrogen are formed at STP.
Question 4(b).
What information do the following chemical equations convey? Mg + 2HCl → MgCl2+ H2
Solution:
(b) This equation conveys the following information:
  1. Magnesium reacts with hydrochloric acid to form magnesium chloride and hydrogen gas.
  2. 24 g of magnesium reacts with 2(1 + 35.5) = 73 g of hydrochloric acid to produce (24 + 71), i.e. 95 g of magnesium chloride.
  3. Hydrogen produced at STP is 22.4 litres.
Question 5(a).
What are polyatomic ions? Give two examples.
Solution:
(a) A poly-atomic ion is a charged ion composed of two or more atoms covalently bounded that can be carbonate (CO32-) and sulphate (SO42-)
Question 5(b).
Name the fundamental law that is involved in every equation.
Solution:
(b) The fundamental laws which are involved in every equation are:
  1. A chemical equation consists of formulae of reactants connected by plus sign (+) and arrow (→) followed by the formulae of products connected by plus sign (+).
  2. The sign of an arrow (→) is to read ‘to form’. It also shows the direction in which reaction is predominant.
Question 6(a).
What is the valency of : fluorine in CaF2
Solution:
(a) Valency of fluorine in CaFis -1.
Question 6(b).
What is the valency of :
sulphur in SF6
Solution:
(b) Valency of sulphur in SF6 is -6.

Question 6(c).

What is the valency of :

phosphorus in PH3
Solution:
(c) Valency of phosphorus in PH3 is +3.

Question 6(d).
What is the valency of : carbon in CH4
Solution:
(d) Valency of carbon in CH4 is +4.
Question 6(e).
What is the valency of :
nitrogen in the following compounds:
(i) N2O3 (ii) N2O5 (iii) NO2 (iv) NO
Solution:
(e) Valency of nitrogen in the given compounds:
  1. N2O = N is +3
  2. N2O= N is +5
  3. NO2 = N is +4
  4. NO = N is +2
Question 7.
Why should an equation be balanced? Explain with the help of a simple equation.
Solution:
According to law of conservation of mass, “matter can neither be created nor be destroyed in a chemical reaction”. This is possible only, if total number of atoms on the reactants side is equals to total number of atoms on products side. Thus, a chemical reaction should be always balanced.
Let us consider an example,
Fe + H2O → Fe3O4 + H2
In this equation number of atoms on both sides is not the same, the equation is not balanced.
The balanced form of this equation is given by,
3Fe + 4H2O  → Fe3O4 + 4H2
Question 8(a).
Write the balanced chemical equations of the following reactions. sodium hydroxide + sulphuric acid → sodium sulphate + water
Solution:
2NaOH + H2SO4 → Na2SO4 + 2H2O

Question 8(b).
Write the balanced chemical equations of the following reactions. potassium bicarbonate + sulphuric acid → potassium sulphate + carbon dioxide + water
Solution:
2KHCO3 + H2SO4 → K2SO4 + 2CO2 + 2H2O

Question 8(c).
Write the balanced chemical equations of the following reactions. iron + sulphuric acid → ferrous sulphate + hydrogen.
Solution:
Fe + H2SO4 → FeSO4 + H2

Question 8(d).
Write the balanced chemical equations of the following reactions. chlorine + sulphur dioxide + water →  sulphuric acid + hydrogen chloride
Solution:
Cl2 + SO2 + 2H2O → H2SO4 + 2HCl

Question 8(e).
Write the balanced chemical equations of the following reactions. silver nitrate → silver + nitrogen dioxide + oxygen”
Solution:
2AgNO3 → 2Ag + 2NO2 + O2

Question 8(f).
Write the balanced chemical equations of the following reactions.
copper + nitric acid → copper nitrate + nitric oxide + water
Solution:
3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O

Question 8(g).
Write the balanced chemical equations of the following reactions.
ammonia + oxygen → nitric oxide + water
Solution:
  Selina Concise Chemistry Class 9 ICSE Solutions The Language of Chemistry image - 9

Question 8(h).
Write the balanced chemical equations of the following reactions.
barium chloride + sulphuric acid → barium sulphate + hydrochloric acid
Solution:
BaCl2 + H2SO4 → BaSO4 + 2HCl

Question 8(i).
Write the balanced chemical equations of the following reactions.
zinc sulphide + oxygen → zinc oxide + sulphur dioxide
Solution:
2ZnS + 3O2 → 2ZnO + 2SO2

Question 8(j).
Write the balanced chemical equations of the following reactions.
aluminium carbide + water → aluminium hydroxide + methane
Solution:
Al4C3 + 12H2O → 4Al(OH)3 + 3CH4

Question 8(k).
Write the balanced chemical equations of the following reactions.
iron pyrites(FeS2) + oxygen → ferric oxide + sulphur dioxide
Solution:
 4FeS2 + 11O2 → 2Fe2O3 + 8SO2

Question 8(l).
Write the balanced chemical equations of the following reactions.
potassium permanganate + hydrochloric acid → potassium chloride + manganese chloride + chlorine + water
Solution:
2KMnO4 + HCl → 2KCl + 2MnCl2 + 5Cl2 + 8H2O

Question 8(m).
Write the balanced chemical equations of the following reactions.
aluminium sulphate + sodium hydroxide → sodium sulphate + sodium meta aluminate + water.
Solution:
Al2(SO4)+ 8NaOH → 3Na2SO4 + 2NaAlO2 + 4H2O

Question 8(n).
Write the balanced chemical equations of the following reactions.
aluminium + sodium hydroxide + water → sodium meta aluminate + hydrogen
Solution:
2Al + 2NaOH + 2H2O → 2NaAlO2 + 3H2

Question 8(o).
Write the balanced chemical equations of the following reactions.
potassium dichromate + sulphuric acid → potassium sulphate + chromium sulphate + water + oxygen.
Solution:
2K2Cr2O7 + 8H2SO4 → 2K2SO4 + 2Cr2(SO4)3 + 8H2O + 3O2

Question 8(p).
Write the balanced chemical equations of the following reactions.
potassium dichromate + hydrochloric acid → Potassium chloride + chromium chloride + water + chlorine
Solution:
K2Cr2O7 + 14HCl → 2KCl + 2CrCl3 + 7H2O + 3Cl2

Question 8(q).
Write the balanced chemical equations of the following reactions.
sulphur + nitric acid → sulphuric acid + nitrogen dioxide + water.
Solution:
S + HNO3 → H2SO4 + NO2 + H2O

Question 8(r).
Write the balanced chemical equations of the following reactions.
sodium chloride + manganese dioxide + sulphuric acid → sodium hydrogen sulphate  + manganese sulphate + water + chlorine.
Solution:
2NaCl + MnO2 + 3H2SO4 → 2NaHSO4 + MnSO4 + 2H2O + Cl2

Question 9(a).
Define atomic mass unit.
Solution:
Atomic mass unit (amu) is equal to one-twelfth the mass of an atom of carbon-12 (atomic mass of carbon taken as 12).

Question 9(b)(ii)
Calculate the molecular mass of the following:
(NH4)2CO3
Given atomic mass of Cu = 63·5, H = 1, O= 16, C = 12, N = 14, Mg = 24, S = 32
Solution:
Molecular mass of (NH4)2CO3
= (2 × 14) + (8 × 1) + 12 + (3 × 16)
= 28 + 8 + 12 + 48
= 96

Question 9(b)(iii)
Calculate the molecular mass of the following:
(NH2)2CO
Given atomic mass of Cu = 63·5, H = 1, O= 16, C = 12, N = 14, Mg = 24, S = 32
Solution:
Molecular mass of (NH2)2CO
= (14 × 2) + (4 × 1) + 12 + 16
= 28 + 4 + 12 + 16
= 60

Question 9(b)(iv)
Calculate the molecular mass of the following:
Mg3N2
Given atomic mass of Cu = 63·5, H = 1, O = 16, C = 12, N = 14, Mg = 24, S = 32
Solution:
Molecular mass of Mg3N2
= (3 × 24) + (2 × 14)
= 72 + 28
= 100

Question 10(a).
Choose the correct answer from the options given below.
Modern atomic symbols are based on the method proposed by
i. Bohr
ii. Dalton
iii. Berzelius
iv. Alchemist
Solution:
iii. Berzelius

Question 10(b).
Choose the correct answer from the options given below.
The number of carbon atoms in a hydrogen carbonate radical is
i. One
ii. Two
iii. Three
iv. Four
Solution:
One

Question 10(c).
Choose the correct answer from the options given below.
The formula of iron (III) sulphate is
i. Fe3SO4
ii. Fe(SO4)3
iii. Fe2(SO4)3
iv. FeSO4
Solution:
iii. Fe2(SO4)3

Question 10(d).
Choose the correct answer from the options given below.
In water, the hydrogen-to-oxygen mass ratio is
i. 1: 8
ii. 1: 16
iii. 1: 32
iv. 1: 64
Solution:
i. 1:8

Question 10(e).
Choose the correct answer from the options given below.
The formula of sodium carbonate is Na2CO3 and that of calcium hydrogen carbonate is
i. CaHCO3
ii. Ca(HCO3)2
iii. Ca2HCO3
iv. Ca(HCO3)3
Solution:
i. Ca(HCO3)2

Solution 11.
(a) A molecular formula represent The Molecule of an element or of a Compound.
(b) The molecular formula of water (H2O) represents 18 parts by mass of water.
(c) A balanced equation obeys the law of conservation of mass wherever unbalanced equation does not obey this law.
(d) CO and Co represent carbon-monoxide and cobalt respectively.

Solution 12.

  1. Relative molecular mass of CHCl3
    = 12 + 1 + (3 × 35.5)
    = 12 + 1 + 106.5
    = 119.5
  2. Relative molecular mass of (NH4)2 Cr2O7
    = (14 × 2) + (1× 8) + (52 × 2) + (16 × 7)
    = 28 + 8 + 104 + 112
    = 252
  3. Relative molecular mass of CuSO4· 5H2O
    = 63.5 + 32 + (16 × 4) + 5(2 + 16)
    = 159.5 + 90
    = 249.5
  4. Relative molecular mass of (NH4)2SO4
    = (2 × 14) + (8 × 1) + 32 + (4 × 16)
    = 28 + 8 + 32 + 64
    = 132
  5. Relative molecular mass of CH3COONa
    = (12 × 2) + (1× 3) + (16 × 2) + 23
    = 24 + 3 + 32 + 23
    = 82
  6. Potassium chlorate (KClO3)
    = 39.1+ 35.5 + (16 × 3)
    = 39.1+ 35.5 + 48
    = 122.6
  7. Ammonium chloroplatinate (NH4)2PtCl6
    = (14 × 2) + (1 × 8) + 195.08 + (35.5 × 6)
    = 28 + 8 + 195.08 + 213
    = 444.08

Solution 13.

CompoundEmpirical formula

(a) Benzene (C6H6)

CH

CompoundEmpirical formula
(b) Glucose (C6H12O6)

CH2O

CompoundEmpirical formula
(c) Acetylene (C2H2)

CH

CompoundEmpirical formula
(d) Acetic acid (CH3COOH)

CH2O

Solution 14.
Relative molecular mass of MgSO4·7H2O
=24 + 32 + (16 × 4) + 7(2 + 16)
=24 + 32 + 64 + 126
=246
26 g of Epsom salt contains 126 g of water of crystallisation.
Hence, 100 g of Epsom salt contains
Selina Concise Chemistry Class 9 ICSE Solutions The Language of Chemistry image - 5
The % of H2O in MgSO4·7H2O = 51.2

Solution 15.
(a)  Relative molecular mass of Ca(H2PO4)2
= 40.07 + (1 × 4) + (30.9 × 2) + (16 × 8)
= 40.07 + 4 + 61.8 + 128
= 233.87
233.87 g Ca(H2PO4)contains 61.8 g P
So, 100 g Ca(H2PO4)contains
Selina Concise Chemistry Class 9 ICSE Solutions The Language of Chemistry image - 6
The % of P in Ca(H2PO4)2 is 26.42%.
(b) Relative molecular mass of Ca3(PO4)2
= (40.07 × 3) + (30.9 × 2) + (16 × 8)
= 120.21 + 61.8 + 128
= 310.01
310.01 g Ca3(PO4)contains 61.8 g P
So, 100 g Ca(H2PO4)contains
(IMAGE)
The % of P in Ca(H2PO4)2 is 19.93%.

Solution 16.
Relative molecular mass of KClO3
= 39.09 + 35.5 + (3 × 16)
= 122.59 g
Selina Concise Chemistry Class 9 ICSE Solutions The Language of Chemistry image - 7
The percentages of K, Cl and O in KClO3 are 31.9%, 28.9% and 39.1%, respectively.

Solution 17.
Relative molecular mass of urea is

ElementNo. of atomsAtomic massTotal
N21428
C11212
H414
O11616

[12 + 16 + 28 + 4] = 60
Hence, relative molecular mass of urea = 60
Selina Concise Chemistry Class 9 ICSE Solutions The Language of Chemistry image - 8

More Resources for Selina Concise Class 9 ICSE Solutions