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NCERT Class 11 Chemistry Textbook Chapter 2 With Answer PDF Free Download
Chapter 2: Structure of Atom
2.1 Sub-Atomic Particles
Dalton’s atomic theory was able to explain the law of conservation of mass, the law of constant composition, and the law of multiple proportions very successfully.
However, it failed to explain the results of many experiments, for example, it was known that substances like glass or ebonite when rubbed with silk or fur generate electricity.
Many different kinds of sub-atomic particles were discovered in the twentieth century. However, in this section, we will talk about only two particles, namely electron, and proton.
2.1.1 Discovery of Electron
In 1830, Michael Faraday showed that if electricity is passed through a solution of an electrolyte, chemical reactions occurred at the electrodes, which resulted in the liberation and deposition of matter at the electrodes. He formulated certain laws which you will study in class XII.
These results suggested the particulate nature of electricity. An insight into the structure of the atom was obtained from the experiments on electrical discharge through gases.
Before we discuss these results we need to keep in mind a basic rule regarding the behavior of charged particles: “Like charges repel each other and unlike charges attract each other”.
In the mid-1850s many scientists mainly Faraday began to study electrical discharge in partially evacuated tubes, known as cathode ray discharge tubes.
It is depicted in Fig. 2.1. A cathode-ray tube is made of glass containing two thin pieces of metal, called electrodes, sealed in it.
The electrical discharge through the gases could be observed only at very low pressures and at very high voltages. The pressure of different gases could be adjusted by evacuation.
When sufficiently high voltage is applied across the electrodes, current starts flowing through a stream of particles moving in the tube from the negative electrode (cathode) to the positive electrode (anode).
These were called cathode rays or cathode ray particles. The flow of current from cathode to anode was further checked by making a hole in the anode and coating the tube behind the anode with phosphorescent material zinc sulfide.
When these rays, after passing through the anode, strike the zinc sulfide coating, a bright spot on the coating is developed(the same thing happens in a television set)
| Author | NCERT |
| Language | English |
| No. of Pages | 44 |
| PDF Size | 2.8 MB |
| Category | Chemistry |
| Source/Credits | ncert.nic.in |
NCERT Solutions Class 11 Chemistry Chapter 2 Structure of Atom
Q.1. What is the lowest value of n that allows g orbitals to exist?
Ans. For g-orbitals, l = 4.
For any given value of ‘n’, the possible values of ‘l’ range from 0 to (n-1).
∴ For l = 4 (g orbital), least value of n = 5.
Q.2. An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron.
Ans: For the 3d orbital:
Possible values of the Principal quantum number (n) = 3
Possible values of the Azimuthal quantum number (l) = 2
Possible values of the Magnetic quantum number (ml) = – 2, – 1, 0, 1, 2
Q.3. An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.
Ans:
(i)
In a neutral atom, a number of protons = the number of electrons.
∴ Number of protons present in the atoms of the element = 29
(ii)
The electronic configuration of this element (atomic number 29) is 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 10 . The element is copper (Cu).
Q.4.Give the number of electrons in the species , H2+, H2 and O2+
Ans: No. electrons present in H2 = 1 + 1 = 2.
∴ Number of electrons in H2+ = 2 – 1 = 1
Number of electrons in H2 = 1 + 1 = 2
Number of electrons in O2 = 8 + 8 = 16.
∴ Number of electrons in O2+= 16 – 1 = 15
Q.5. (I)An atomic orbital has n = 3. What are the possible values of l and ml ?
(II)List the quantum numbers (ml and l) of electrons for 3d orbital.
(III) Which of the following orbitals are possible? 1p, 2s, 2p and 3f
Ans.
(I)
The possible values of ‘l’ range from 0 to (n – 1). Thus, for n = 3, the possible values of l are 0, 1, and 2.
The total number of possible values for ml = (2l + 1). Its values range from -l to l.
For n = 3 and l = 0, 1, 2:
m0 = 0
m1 = – 1, 0, 1
m2 = – 2, – 1, 0, 1, 2
(II)
For 3d orbitals, n = 3 and l = 2. For l = 2 , possible values of m2 = –2, –1, 0, 1, 2
(III)
It is possible for the 2s and 2p orbitals to exist. The 1p and 3f cannot exist.
For the 1p orbital, n=1 and l=1, which is not possible since the value of l must always be lower than that of n.
Similarly, for the 3f orbital, n =3 and l = 3, which is not possible.
Q.6. Using s, p and d notations, describe the orbital with the following quantum numbers.
(a)n = 1, l = 0;
(b)n = 3; l =1
(c) n = 4; l = 2;
(d) n = 4; l =3.
Ans:
(a)n = 1, l = 0 implies a 1s orbital.
(b)n = 3 and l = 1 implies a 3p orbital.
(c)n = 4 and l = 2 implies a 4d orbital.
(d)n = 4 and l = 3 implies a 4f orbital.
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