Compile by :Abdul hye
st Helen's school and college
The main postulates of Dalton atomic theory are as follows.
In 1911, Lord Rutherford was studying the radiations, given out by polonium. During his study, he bombarded alpha particles, which are doubly positively charged on a gold foil.
He placed a zinc sulphide plate detector behind the gold foil, to detect the particles emitting from the radioactive material. He observed that most of the particles passed through the foil undeflected. Some of these particles were deflected at a larger angle while few particles bounced back.
From the deflection of the alpha particles, Rutherford concluded there is a positively nucleus present in the centre of the atom. Since most of the particles passed through the foil undeflected, so he concluded that most of the space in the atom is empty.
Based on his experiment, Rutherford presented the following model for the structure of the atom.
The discovery of Rutherford’s atomic model of the atom was a key step forward to understand the atomic structure. However, his model could not explain the arrangement of electrons around the nucleus.
The major objections raised against his model were the following.
Main postulates of Bohr’s atomic theory are:
Let an electron jumps from a higher energy level E2 to a lower energy level E1.The energy is emitted in the form of light .Amount of energy released is given by:
∆E = E2-E1
E2 - E1= hυ
Where
h =Plank'sconstant (h=6.6256x10-34 js)
υ = Frequency of radiant light
m v r =
Where n =1, 2, 3, ………..
m = mass of electron
v = velocity of electron
r = radius of orbit
Electron configuration is the distribution of electrons in shells or sub shells of anatom
These shells are designated as K, L, M, N, O, P, Q etc. While subshells are designated as s, p, d and f. electron are arranged in shells according to 2n2 rule while in subshells they are arranged according to Aufbua principle.
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p……………
s-orbital can accommodate maximum two electrons
p-orbital can accommodate maximum six electrons
d-orbital can accommodate maximum ten electrons
f-orbital can accommodate maximum fourteen electrons
Energy Level
The possible locations around an atom where electrons having specific energy values is called energy level. Bohr observed the transitions of electron in hydrogen atom and name them as energy level. He states that electron emit energy when electron jumps from higher to lower energy level. This radiated energy gives a line spectrum. Each energy level is associated with specific spectrum showing that each shell has a specific value of energy.
Sub Energy Level
Space around the nucleus where probability of finding of an electron is maximum is called sub energy level or orbital. A subshell is a subdivision of electron shells separated byelectron orbitals.Subshells are labelled s, p, d, and f, which stand for sharp, principal, diffused and fundamental respectively. The s orbital is spherical; p orbital is dumbbell, d orbital is double dumbbell while f is complex.
Name |
Symbol |
Electron Configuration |
Hydrogen |
H |
1s1 |
Helium |
He |
1s2 |
Lithium |
Li |
1s2 2s1 |
Beryllium |
Be |
1s2 2s2 |
Boron |
B |
1s2 2s2 2p1 |
Carbon |
C |
1s2 2s2 2p2 |
Nitrogen |
N |
1s2 2s2 2p3 |
Oxygen |
O |
1s2 2s2 2p4 |
Fluorine |
F |
1s2 2s2 2p5 |
Neon |
Ne |
1s2 2s2 2p6 |
Sodium |
Na |
1s2 2s2 2p63s1 |
Magnesium |
Mg |
1s2 2s2 2p63s2 |
Aluminum |
Al |
1s2 2s2 2p63s2 3p1 |
Silicon |
Si |
1s2 2s2 2p63s2 3p2 |
Phosphorus |
P |
1s2 2s2 2p63s2 3p3 |
Sulphur |
S |
1s2 2s2 2p63s2 3p4 |
Chlorine |
Cl |
1s2 2s2 2p63s2 3p5 |
Argon |
Ar |
1s2 2s2 2p63s2 3p6 |
Potassium |
K |
1s2 2s2 2p63s2 3p64s1 |
Calcium |
Ca |
1s2 2s2 2p63s2 3p64s2 |
Scandium |
Sc |
1s2 2s2 2p63s2 3p63d1 4s2 |
Titanium |
Ti |
1s2 2s2 2p63s2 3p63d2 4s2 |
Vanadium |
V |
1s2 2s2 2p63s2 3p63d3 4s2 |
Atoms of the same elements having the same atomic number but different atomic masses are called isotopes. The difference in mass of the atoms in an element is due to the difference in the number of neutrons. Isotopes have the same chemical properties but different physical properties.
Examples of isotopes
Isotopes of hydrogen
Hydrogen have three isotopes. These are protium1H1, Deutrium1D2 and tritium 1T3 with atomic masses 1, 2 and three respectively. All the three isotopes have one proton but difference is in number of neutrons.
Name |
Symbol |
Atomic Number |
Mass Number |
Relative Abundance |
Nature |
Protium or Hydrogen |
1H1 or H |
1 |
1 |
99.99% |
Non-Radioactive |
Deutrium |
1H2 or D |
1 |
2 |
0.017% |
Non-Radioactive |
Tritium |
1H3 or T |
1 |
3 |
Rare |
Radioactive |
Isotopes of carbon
Atomic number of carbon is six. Carbon has three isotopes and represented by
Name |
Symbol |
Atomic Number |
Mass Number |
Relative Abundance |
Nature |
C-12 |
6C12 |
6 |
12 |
89.89% |
Non-Radioactive |
C-13 |
6C13 |
6 |
13 |
11.11% |
Non-Radioactive |
C-14 |
6C14 |
6 |
14 |
Rare |
Radioactive |
It is clear that all three forms (C-12, C-13 and C-14) have same number of protons and electrons but different number of neutrons. The C-12 contains six neutrons, the C-13 contains 7 neutrons and C-14 contains 8 neutrons.
Isotopes of chlorine
Chlorine is a member of halogen family and its atomic number is 17. Chlorine has two isotopes and represented by Cl-35 and Cl-37. It is clear that both forms (Cl-35 and Cl-37) have same number of protons and electrons but different number of neutrons. The Cl-35 contains 18 neutrons, the Cl-37 contains 20 neutrons.
Name |
Symbol |
Atomic Number |
Mass Number |
Relative Abundance |
Nature |
Cl-35 |
17Cl35 |
17 |
35 |
75.53% |
Non-Radioactive |
Cl-37 |
17Cl37 |
17 |
37 |
24.47% |
Non-Radioactive |
Isotopes of Uranium
Uranium atomic number 92 has three isotopes uranium 234, uranium 235 and uranium238.
Name |
Symbol |
Atomic Number |
Mass Number |
Relative Abundance |
Nature |
U-234 |
92U234 |
92 |
234 |
0.005% |
Non-Radioactive |
U-235 |
92U235 |
92 |
235 |
0.75% |
Radioactive |
U-238 |
92U238 |
92 |
238 |
99.245% |
Radioactive |
Uses of Isotopes
Isotopes arc mainly used in chemical, agricultural and medical research and for diagnosing and treatment of diseases. For instance:
Exercise
(i) H has a positive charge.
(a) Neutron (b) Proton
(c) Electron (d) Atom
(ii) The K-shell can accommodate maximum number of electrons.
(a) 2 (b) 6
(c) 8 (d) 18
(iii) The maximum number of electrons in third energy level is:
(a) 10 (b) 18
(c) 32 (d) 64
(iv) Which one is the lightest?
(a) An alpha particle (b) a hydrogen atom
(c) An electron (d) a proton
(v) Electron in its ground state does not:
(a) Spin (b) revolve
(c) Radiate energy (d) reside in orbit
(vi) Alpha particles are
(a) neutral (b) negatively charged
(c) protons (d) doubly positively charged
(vii) Electron does not radiate energy when it:
(a) reside in an orbit (b) jumps from lower to higher orbit
(c) falls in the nucleus (d) jumps from higher to lower orbit
(viii) Isotopes of an element have
(a) different number of electrons (b) different number of protons
(c) different number of neutrons (d) none of these
(ix) 6C12, 6C13, 6C14 are
(a) isomers (b) isobars
(c) allotropes (d) isotopes
|
Short Questions
(i) How many energy levels are there in L-energy level? Give their names and number of electrons, they can accommodate.
Answer: As we know that, L-energy level is the second orbit of atom therefore n=2. There are two sub energy levels in L-shell. These energy levels are‘s’ and ‘p’ which stand for sharp and principal respectively. The s-orbital can only accommodate 2 electrons and p-orbital 6 electrons. Therefore, the total number of electrons L-energy level can accommodate will be 8.
(ii) State the observations of Rutherford’s after he bombarded gold foil with alpha particles.
Answer: In 1911, Lord Rutherford was studying the radiations, given out by polonium. During his study, he bombarded alpha particles, which are doubly positively charged on a gold foil.
He placed a zinc sulphide plate detector behind the gold foil, to detect the particles emitting from the radioactive material. He observed that most of the particles passed through the foil undeflected. Some theof the particles were deflected at a larger angle while few particles bounced back.
From the deflection of the alpha particles, Rutherford concluded there is a positively nucleus present in the centre of the atom. Since most of the particles passed through the foil undeflected, so he concluded that most of the space in the atom is empty.
(iii) What is Rutherford’s model of atom?
Answer: Based on his experiment, Rutherford observed the following model for the structure of the atom.
(iv). State two shortcomings of Rutherford’s atomic model.
Answer: The major short comings raised against his model were the following.
(v). State whether energy of an electron in an orbit/energy level is same or different.
Answer: According to Bohr’s atomic model, Energy of an electron in one of its allowed orbits is fixed. As long as an electron remains in one of its allowed orbit, it cannot absorb or radiate energy.
(vi). What do you mean by the term electronic configuration?
Answer Electron configuration
Electron configuration is the distribution of electrons in shells or sub shells of an atom.These shells are designated as K, L, M, N, O, P, Q etc. While subshells are designated as s, p, d and f. electron are arranged in shells according to 2n2 rule while in subshells they are arranged according to Aufbua principle.
(vii) Give electronic configuration of Ne (Z= 10)
Answer: Electronic configuration of Neis1s22s22p6
(viii). Draw the structure of hydrogen isotopes
Answer
(ix). Explain the uses of isotopes.
Answer Uses of Isotope
Isotopes arc mainly used in chemical, agricultural and medical research and for diagnosing and treatment of diseases. For instance:
(x). There are three isotopes of uranium having atomic number 92 and mass number 234, 235 and 238. Calculate number of neutrons in their nuclei.
Answer: Number of neutrons can be calculated by subtracting atomic number from mass number.
Number of neutrons in U-234= mass number – atomic number
234 - 92 = 142
Number of neutrons in U-235= mass number – atomic number
235 - 92 = 143
Number of neutrons in U-238= mass number – atomic number
238 - 92 = 146
Long Question
Answer Rutherford’s Atomic Model
In 1911, Lord Rutherford was studying the radiations, given out by polonium. During his study, he bombarded alpha particles, which are doubly positively charged on a gold foil.
He placed a zinc sulphide plate detector behind the gold foil, to detect the particles emitting from the radioactive material. He observed that most of the particles passed through the foil undeflected. Some of these particles were deflected at a larger angle while few particles bounced back.
From the deflection of the alpha particles, Rutherford concluded there is a positively nucleus present in the centre of the atom. Since most of the particles passed through the foil undeflected, so he concluded that most of the space in the atom is empty.
Based on his experiment, Rutherford presented the following model for the structure of the atom.
The discovery of Rutherford’s atomic model of the atom was a key step forward to understand the atomic structure. However, his model could not explain the arrangement of electrons around the nucleus.
Answer Bohr’s atomic theory
Main postulates of Bohr’s atomic theory are:
Let an electron jumps from a higher energy level E2 to a lower energy level E1.The energy is emitted in the form of light .Amount of energy released is given by:
∆E = E2-E1
E2 - E1= hυ
Where
h =Plank's constant (6.6256x10-34 js)
υ = Frequency of radiant light
m v r =
Where n =1, 2, 3, ………..
m = mass of electron
v = velocity of electron
r = radius of orbit
Answer Energy Level
The possible locations around an atom where electrons having specific energy values. Bohr observed the transitions of electron in hydrogen atom. He states that electron emit energy when electron jumps from higher to lower energy level. This radiated energy gives a line spectrum. Each energy level is associated with specific spectrum showing that each shell has a specific value of energy.
Sub Energy Level
A subshell is a subdivision of electron shells separated by electron orbitals.
Subshells are labelled s, p, d, and f, which stand for sharp, principal, diffused and fundamental respectively. The s orbital is spherical; p orbital is dumbbell, d orbital is double dumbbell while f is complex.
Answer
Element |
Symbol |
Electronic Configuration |
Boron (Z=5) |
B |
1s2 2s2 2p1 |
Carbon (Z=6) |
C |
1s2 2s2 2p2 |
Nitrogen (Z=7) |
N |
1s2 2s2 2p3 |
Oxygen (Z=8) |
O |
1s2 2s2 2p4 |
Fluorine (Z=9) |
F |
1s2 2s2 2p5 |
Neon (Z=10) |
Ne |
1s2 2s2 2p6 |
(i) Only a few high energy α-particles rebound after striking the metal foil in Rutherford’s experiment.
Answer: As major portion of the atom is empty and nuclei only occupies small portion of atom due to which only a few high energy α-particles rebound. The size of positive centre (nucleus) is very small as compared to the whole atom so deflection of only a few alpha particles (double positive) is possible.
Answer: As the number of electrons (negative) in atom is always equal to the number of protons (positive), therefore atom is a neutral particle because cancel the effects of each other.
iii. Electrons continuously revolve around the nucleus but do not fall into the nucleus.
Answer: Electrons continuously revolve around the nucleus but do not fall into the nucleus because electron does not absorb or emit energy when revolves around the nucleus in a fixed orbit. When electrons revolve in a fixed orbit the centrifugal force of an electron is balanced by the force of attraction between the nucleus and the electron
Answer: The maximum number of electrons can be accommodated by the help of 2n2 formula. For K shell n= 1, therefore maximum number of electrons K-shell can accommodate are 2 i.e. 2n2 = 2(1)2 = 2. However, for other shell the n value is greater than 1 due to which accommodate more than 2 electrons. For example when n=2 then it is called L-shell and maximum number of electrons L-shell can accommodate is 8 i.e. 2n2 = 2(2)2 = 8
Answer: The mass of an atom is present in its nucleus because both proton and neutron are heavy particles of atom which are present inside the nucleus. This was the reason that alpha particles bounced back when collided with nucleus because mass of an atom is present in its nucleus.
Answer: Chemical properties depend upon number of electrons and its configuration around the nucleus. Therefore, chemical properties of isotopes are same because isotopes have same number of electron and same configuration around the nucleus. However, physical properties such as mass, density etc. will be different from each other because isotopes have different number of neutrons. For example, all isotopes of carbon when react with oxygen it form carbon dioxide (same chemical properties) but masses are different such as C-12, C-13 and C-14.
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