Integrated Syllabus

FOUNDATION & OLYMPIAD

CHEMISTRY

CLASS – 10

FOUNDATION & OLYMPIAD CHEMISTRY CLASS - X

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ISBN: 978-93-80299-02-0

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Preface Speed and accuracy play an important role in climbing the competitive ladder. Students have to integrate the habit of being able to calculate and function quickly as well as efficiently in order to excel in the learning culture. They need to think on their feet, understand basic requirements, identify appropriate information sources and use that to their best advantage. The preparation required for the tough competitive examinations is fundamentally different from that of qualifying ones like the board examinations. A student can emerge successful in a qualifying examination by merely scoring the minimum percentage of marks, whereas in a competitive examination, he has to score high and perform better than the others taking the examination. This book provides all types of questions that a student would be required to tackle at the foundation level. The questions in the exercises are sequenced as Basic Practice, Further Practice, Multiple Answer Questions, Paragraph Questions, Numerical Problems, Conceptual Questions and Brain Nurtures. Simple questions involving a direct application of the concepts are given in Basic Practice. More challenging questions on direct application are given in Further Practice. Questions involving higher order thinking or an open-ended approach to problems are given in Brain Nurtures. These questions encourage students to think analytically , to be creative and to come up with solutions of their own. Constant practice and familiarity with these questions will not only make him/her conceptually sound, but will also give the student the confidence to face any entrance examination with ease. Valuable suggestions as well as criticism from the teacher and student community are most welcome and will be incorporated in the ensuing edition.

Publisher

1.

Atomic Structure .................................. 01

2.

Periodic Table ....................................... 25

3.

Chemical Bonding ................................ 48

4.

Mole Concept & Stoichiometry ........... 71

5.

Electrochemistry .................................. 92

6.

Chemical Kinetics & Chemical Equilibrium .......................................... 111

7.

Acids, Bases & Salts .............................. 139

8.

Compounds of Nitrogen ..................... 165

9.

Compounds of Sulphur ....................... 184

10. Compounds of Chlorine ...................... 204 11. Organic Chemistry–I ........................... 224 12. Organic Chemistry–II .......................... 250 13. Metallurgy ........................................... 275 Answers ................................................ 297 Solution to Crossword Puzzles ........... 316

IIT Foundation & Olympiad Explorer

Chemistry/ Class - X

Chapter



Atomic Structure

Common Misconception

Fact

1. Electrons are visualized as particles only.

1. Electrons behave as particles as well as waves.

2. The electrons which are farther from the nucleus travel with greater velocities.

2. The velocity of electron decreases with increase in distance from the nucleus.

3. Orbit and orbital are synonymous terms.

3. Orbit is an imaginary path of electron whereas Orbital is space signifying maximum probabilityof finding the electron.

SYNOPSIS INTRODUCTION Discovery of fundamental particles lead to the study of pattern of arrangement of these particles inside the atom. Various atomic models visualizing various patterns of arrangement have been proposed by different scientists such as Thomson, Rutherford, Niels Bohr etc.,

ATOMIC MODELS Many attempts were made to propose various atomic models to describe the position of the fundamental particles in the atom.

J.J. Thomson’s atomic model J.J. Thomson proposed the atomic model (the discoverer of electrons) in 1898. According to him an atom is considered as a sphere of positive charge in which negatively charged particles (electrons) are embedded uniformly to neutralise that positive charge.

Rutherford’s atomic model Rutherford performed the α - ray scattering to know about the position of fundamental particles (to test J.J. Thomson’s atomic model). In his experiment he observed that most of the α - particles passed through the gold foil without any deflection from the path, few α – particles were deflected by small angles. One in 10,000 α – particles bounced back in the same direction. On the basis of the above observations, Rutherford drew the following conclusions. 1. Atomic Structure

1

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a) Atom consists of predominately empty space. b) Nucleus is located inside the atom, containing positively charged particles. c) Electrons revolve at high speed in the extra nuclear space around the nucleus in circular orbits like the planets revolve around the sun. The model of an atom is popularly known as planetary model. d) As per the law of electrodynamics of classical mechanics, the charged particles like electron moving around the opposite charge in circular motion should continuously lose energy by emission and spiral down into the nucleus. If this happens atom should collapse, which is not happening. If the electron in an atom continuously radiates energy, the spectrum of an element should be a continuous spectrum, But the atoms give rise to discontinuous line spectra.

Note : The arrangement of different electromagnetic radiation in the order of increasing wavelength or frequency is known as electromagnetic spectrum.

Bohr’s atomic theory Among the various models proposed, Bohr’s model of atom has been proved with some experimental evidences. Bohr’s theory was mainly based on Planck’s quantum theory. According to this, emission or absorption of energy is always in the form of packets called quanta. Each quantum carries an energy equal to E = h ν where h is called Planck’s constant (h = 6.625 ×10–34 joule – s and ν is frequency of radiation). Niels Bohr’s atomic model was based on Planck’s quantum theory retaining Rutherford’s nuclear model. The main postulates of this model are: a) Electrons revolve around the nucleus in specified circular paths with definite amount of energy called stationary orbits or stationary shells. b) These are also called energy levels and are designated as K, L, M, N corresponding to 1,2,3,4 respectively. c) The energy associated with a certain energy level increases with the increase of its distance from the nucleus. Hence, if the energy associated with the K, L, M, N shells are E1 , E2 , E3 ............. respectively, then E1 < E2 < E3 , etc., d) An electron jumps from a lower energy level to a higher energy level by absorbing energy. But when it jumps from a higher to lower energy level, energy is emitted in the form of electromagnetic radiations. The energy emitted or absorbed ( Δ E ) is an integral multiple of ‘h υ ’. e) The electron can revolve only in the orbit in which the angular momentum of the electron (mvr) is quantised, i.e, mvr is a whole number multiple of h / 2π . This is known as principle of quantization of angular momentum. ( mvr =

1. Atomic Structure

2

nh ) 2π

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IIT Foundation & Olympiad Explorer

Chemistry/ Class - X n=6 n=5 n=4 n=3 n=2 n=1 +

K

L M N OP

By applying the concept of quantization of energy Bohr calculated the radii and energy of th the n orbit of single electron species.

rn =

n 2h 2 0.529 × n 2 = °A , 2 2 4 π me Z Z

En =

−2π2 me4 2 .Z n 2 h2

En = −

13.6 2 .Z eV /atom n2

For hydrogen atom Z = 1

For hydrogen atom (Z = 1)

rn = 0.529 × n 2 °A

En =−

Where h = Planck’s constant

m = mass of the electron

e = charge of the electron

Z = atomic number or nuclear charge

13.6 21.72×10−19 J eV/atom =− 2 n n2

The energy of electron in the ground state of hydrogen atom is equal to –13.6 eV or –19 – 21.72 × 10 J. Ze 2 Velocity of electron in an orbit v n = 2 nh with the help of these expressions, Bohr gave a

satisfactory explanation for the spectra of hydrogen and hydrogen like species (ions having one electron e.g. He + , Li+2 , Be +3 ) etc., Merits a) Explanation of spectra of hydrogen and hydrogen like species b) Derivation of equations for radius and energy of orbits Demerits This model could not explain : a) Spectra of multi electron atoms. b) Proper explanation for quantization of angular momentum of orbits. c) Fine structure of hydrogen spectrum. d) Zeeman effect and Stark effect. 1. Atomic Structure

3

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Chemistry/ Class - X

Sommerfeld’s extension of Bohr’s theory n=4, k=4

n=4, k=3

n=4,k=2 n=4,k=1

Sommerfeld’s elliptical electron orbits

Sommerfeld’s elliptical model could explain : a) Concept of elliptical orbits with a major axis and a minor axis (n and k). b)

n length of major axis = . k length of minor axis

c) Presence of subshells. d) Variation of angular momentum of electron in elliptical orbit. e) Azimuthal quantum number (l = k – 1).

QUANTUM MECHANICAL MODEL

de Broglie’s theory Louis de Broglie pointed out that the wave behaviour of an electron is similar to that of a standing wave. The circumference of each Bohr’s orbit must be equal to an integral multiple of the wave length of the wave associated with the moving electron. Orbits whose circumferences are not equal to the integral multiple of the wave length are impossible for an electron in Bohr’s atom.

h The expression is λ = momentum of electron or λ =

h mv

where m = mass of electron v = velocity of electrons h = Planck’s constant This equation is called de Broglie’s equation, this hypothesis was experimentally confirmed by Davison and Germer who proved wave like properties of electron in 1927. During this period on the basis of the wave properties of electrons, Austrian Physicist Erwing Schrondinger derived an equation called Schrondinger wave equation to describe the behaviour and energies of electrons in an atom. 1. Atomic Structure

4

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Based on this wave equation, the three dimensional region in space around the nucleus, where the probability of finding an electron with a particular energy is determined to be maximum, is called the atomic orbital. On the basis of Schrodinger wave equation a quantum mechanical model of atom has been proposed. In this model, the state of an electron in an atom can be explained with respect to four quantum numbers.

Heisenberg’s uncertainity principle Accroding to this principle, both position and velocity of microparticle such as moving electron cannot be determined simultaneously and accurately.

Δx.Δp ≥ h/4 π Demerits: Failed when applied to larger bodies since the wave nature is insignificant. The principle lead to the concept of probability of finding an electron and eventually to the concept of orbital.

Quantum numbers Principal quantum number a) Denoted by n and proposed by Bohr. b) Values are designated as 1, 2, 3, 4 or K, L, M, N. c) It determines the size and energy of orbit. Azimuthal quantum number a) Denoted by l and proposed by Sommerfeld. b) Values are 0 to n – 1. c) Values represent subshells s, p, d, f and their shapes are spherical, dumbbell, double dumbbell for s, p, d subshells respectively. d) It determines the shape of the subshell and angular momentum of subshell. Magnetic quantum number a) Denoted by m and proposed by Lande. b) Values are – l to + l including zero. c) Total number of values are (2l + 1) which give number of orientations in space. d) Each value of m represents an orbital. Spin quantum number Apart from the above three quantum numbers there is another quantum number which is not derived from quantum mechanical treatment of atom. This is based on the direction of spin of electron. a) Clockwise spin and anticlockwise spin are represented by + 1/2 and – 1/2 b) Proposed by Uhlenbeck and Goud Smit.

Concept of orbital A three dimensional space around the nucleus where the probability of finding an electron is maximum. 1. Atomic Structure

5

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b) Schrodinger wave equation lead to quantum mechanical model. c) Position of electron in an atom is given by quantum numbers. z

y

x

1s orbital Z

Y

X

Z

Y

X

  PX

x

PY The boundary surface diagrams of three p orbitals y

y

dxy

y

z

z

dyz

z

x

d zx

PZ

x

dx

-y

dz

The boundary surface diagrams of five d orbitals

ELECTRONIC CONFIGURATION OF ATOMS

Pauli’s exclusion principle No two electrons in an atom can have all the four quantum numbers identical. An orbital contains a maximum of two electrons with opposite spins. The maximum number of electrons in a shell are given by 2n2 Electronic Configurations of elements The distribution of electrons in orbitals and shells. The distribution follows certain rules.

1. Atomic Structure

6

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Aufbau Principle Aufbau means building up in German. This rule explains about the order of filling up of orbitals. The orbitals are filled by electrons in an increasing order of energies. The order of energies is determined based on n + l values. The orbital with greater n + l value has greater energy and lower value has lower energy. Among the orbitals with equal n + l values, the orbital with lower n value possesses lower energy. The various orbitals depicting the order of energies are arranged in the form of a pictorial representation which is called Moeller diagram.

1s 2s

2p

3s

3p

3d

4s

4p

4d

4f

5s

5p

5d

5f

6s

6p

6d

7s

7p

8s Moeller diagram

Hund’s rule of maximum multiplicity Accroding to Hund’s rule of maximum multiplicity, the orbitals in the same subshell possess equal energies and called degenerate orbitals. These degenerate orbitals are singly occupied by electrons till all the orbitals are half filled. Pairing starts only when the orbitals in a subshell are exactly half filled. The orbitals lose their degeneracy in the presence of external electric or magnetic fields. The electronic configurations of all the elements are written based on the above three principles. The method of writing the configuration can be shell method or nlx method. In shell method, the total number of electrons in K, L, M, N shells are written in the same order. In nlx method, n represents the main energy level or principal quantum number, l represents the subshell for the corresponding azimuthal quantum number and x represents number of electrons in the subshell. Anamolous electronic configurations In case of few elements, the actual rules are slightly violated for writing the electronic configurations. this is because of the greater stabilities of half filled and fully filled subshells in comparison to the neighboring configurations. Examples:

Chromium-half filled configuration Copper-fully filled configuration

1. Atomic Structure

7

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Electronic configuration of elements upto Z = 36. Element

At. No.

H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

1. Atomic Structure

K n=1 1s 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

L n=2 2s 2p

M n=3 3s 3p 3d

4s

1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

1 2 2 2 2 1 2 2 2 2 1 2 2 2 2 2 2 2

1 2 3 4 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6

1 2 3 4 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6

1 2 3 5 5 6 7 8 10 10 10 10 10 10 10 10 8

N n=4 4p 4d

4f

5s

O n=5 5p 5s

5f

1 2 3 4 5 6

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SOLVED EXAMPLES Example 1: An electron in ‘H’ atom and the one in He+ ion are present in the 2nd orbits of the respective species. Both the electrons jump to the ground states. Calculate the ratio of wave lengths of the emitted radiation. Solution:

Energy of electron is 2nd orbit of ‘H’ atom E2 =

−21.72 × 10−19 × 12 = –5.43 × 10–19 joules 22

Energy of electron in 1st orbit of ‘H’ atom E1 =

−21.72 × 10−19 × 12 = –21.72 × 10–19 joules 12

( ΔE)H

= –16.29 × 10–19 J

Energy of electron in 2nd orbit of He+ ion E2 =

−21.72 × 10−19 × 22 22

= –21.72 × 10–19 joules Energy of electron in 1st orbit of He+ ion

21.72 × 10−19 E1 = – × 22 12 = – 86.88 × 10–19 joules.

( ΔE)He

+

= – 65.16 × 10–19 J

ΔEH : ΔEHe+ = –16.29 × 10–19 : – 65.16 × 10–19

=1:4 λ H : λ He+ = 4 : 1

Example 2 : The ratio of radii of certain orbits of Hydrogen He+ and Li+2 is 12 : 6 : 4. Calculate the energies of the electrons in those orbits. Solution:

Radius of orbit = rH =

0.529×n2 0A z

0.529 × n 2H 0.529 × n 2 ; rHe+ = 2 1

1. Atomic Structure

rLi+2 =

9

0.529 × n2 3

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UNIQUE ATTRACTIONS l

Concept Maps

l

Cross word Puzzles

l

Graded Exercise n

Basic Practice

n

Further Practice

n

Brain Nurtures

l

Numerical Problems Conceptual Questions

l

Multiple Answer Questions

l

Paragraph Questions

l

Assertion & Reason Type Questions

CLASS

-X

Solutions

`

75

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