CBSE 11th Standard Chemistry Subject Structure of Atom Ncert Exemplar 3 Mark Questions 2021
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CBSE 11th Standard Chemistry Subject Structure of Atom Ncert Exemplar 3 Mark Questions 2021
11th Standard CBSE
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Reg.No. :
Chemistry
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According to de-Broglie, matter should exhibit dual behaviour, that is both particle and wave like properties.However, a cricket ball of mass 100g does not move like a wave when it is thrown by a bowler at a speed of 100km/h.Calculate the wavelength of the ball and explain why it does not show wave nature?
(a) -
Why was the change in the Bohr model of atom is required, due to which important development(s), concept of movement of an electron in an orbit was replaced by the concept of probability of finding an electron in an orbital?What is the name given to the changed model of an atom?
(a) -
The arrangement of orbitals on the basis of energy is based upon their (n + 1), lower is the energy. For orbitals having same values of (n + l), the orbitals with lower value of n will have lower energy.
Based upon the above information, arrange the following orbitals in the increasing order of energy.
a) 1s 2s, 3s, 2p
b) 4s, 3s, 3p, 4d
c) 5p, 4d, 5d, 4f, 6s
d) 5f, 6d, 7s, 6p(a) -
The arrangement of orbitals on the basis of energy is based upon their (n + 1), lower is the energy. For orbitals having same values of (n + l), the orbitals with lower value of n will have lower energy.
Based upon the above information, solve the questions given below.
a) Which of the following orbitals has the lowest energy?
4d, 4f, 5s, 5p
b) Which of the following orbitals has the highest energy?
5p, 5d, 5f, 6s, 6p(a) -
An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.
(a)
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CBSE 11th Standard Chemistry Subject Structure of Atom Ncert Exemplar 3 Mark Questions 2021 Answer Keys
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Given, m = 100g = 0.1kg
v = 100km/h = \(\frac { 100\times 1000 }{ 60\times 60 } =\frac { 1000 }{ 36 } { ms }^{ -1 }\)
From de-Broglie equation, wavelength,
\(\lambda =\frac { h }{ mv } =\frac { 6.626\times { 10 }^{ -34 }{ kgm }^{ 2 }{ s }^{ -1 } }{ 0.1kg\times \frac { 100 }{ 36 } { ms }^{ -1 } } =238.5\times { 10 }^{ -36 }m\)
As the wavelength is very small so wave nature cannot be detected. -
In Bohr model, an electron is regarded as a charged particle moving in well defined circular orbits about the nucleus. An orbit can completely be defined only if both the position and the velocity of the electron are known exactly at the same time. This is not possible according to the Heisenberg uncertainty principle. Further more, the wave character of the electron is not considered in Bohr model.
Therefore, concept of movement of an electron in an orbit was replaced by the concept of probability of finding electron in an orbital due to de-Broglie concept of dual nature of electron and Heisenberg's uncertainty principle. the changed model is called quantum mechanical model of the atom. -
a) (n + 1) values are
1s = 1 + 0 = 1, 2s = 2+ 0 = 2, 3s = 3 + 0 = 3, 2p = 2 + 1 = 3
Hence, increasing order of their energy is 1s < 2s < 2p < 3s.
b) 4s = 4 + 0 =4, 3s = 3 + 0 = 3, 3p = 3 + 1 = 4, 4d = 4 + 2 = 6.
Hence 3s < 3p < 4s < 4d.
c) 5p = 5 + 1 = 6, 4d = 4 + 2 = 6, 5d = 5 + 2 = 7, 4f = 4 + 3 = 7, 6s = 6 + 0 = 6.
Hence, 4d < 5p < 6s < 4f < 5d.
d) 5f = 5 + 3 = 8, 6d = 6 + 2 = 8, 7s = 7 + 0 = 7, 7p = 7 + 1 = 8.
Hence, 7s < 5f < 6d < 7p. -
a) 4d = 4 + 2 = 6, 4f = 4 + 3 = 7, 5s = 5 + 0 = 5, 7p = 7 + 1 = 8
Hence, 5s has the lowest energy.
b) 5p = 5 + 1 = 6, 5d = 5 + 2 = 7, 5f = 5 + 3, 6s = 6 + 0 = 6, 6p = 6 + 1 = 7
Hence, 5f has highest energy. -
We know that mass number of the element,
A = p + n = 81 .....(i)
Let the number of protons, p = x
Then, number of neutrons,
\(n=x+\frac { 31.7 }{ 100 } x=1.317x\)
(As number of neutrons are 31.7% more than the protons.)
Hence, from Eq.(i)
\(x+1.317x=81\)
or \(2.317x=81=\frac { 81 }{ 2.317 } =34.958\approx 35\)
Therefore, number of protons = 35 and the symbol is. \(_{ 35 }^{ 81 }{ Br }\) (Number of protons = atomic number)
