11th Standard Syllabus & Materials
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Published on: 04/10/2019
Quantum Mechanical Model of Atom
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1.
Write a note on the shape of f orbitals
2.
Determine the values of all the four quantum numbers of the 8th electron in O- atom and 15th electron in Cl atom.
3.
Calculate the uncertainty in position of an electron, if Δv = 0.1% and \(\upsilon \) = 2.2 x 106 ms-1.
4.
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wave length associated with the electron revolving around the nucleus.
5.
Calculate the total number of angular nodes and radial nodes present in 3p-orbital.
6.
Explain briefly the time independent schrodinger wave equation?
7.
Which one among the following salts is more stable? Ferrous and ferric salts
8.
Describe the Aufbau principle
9.
State and explain pauli exclusion principle.
10.
The quantum mechanical treatment of the hydrogen atom gives the energy value:
\({ E }_{ n }=\frac { -13.6 }{ { n }_{ 2 } } ev{ \ atom }^{ -1 }\)
(i) use this expression to find ΔE between n = 3 and n = 4
(ii) Calculate the wavelength corresponding to the above transition.
1.
f-orbitals :
For 'f' orbital, l = 3 and the m values are -3, -2, -1, 0, + 1, +2, +3 corresponding to seven f orbitals \({ f }_{ { z }^{ 3 }, }{ f }_{ { xz }^{ 2 }, }{ f }_{ { yz }^{ 3 }, }{ f }_{ { xz }y },{ f }_{ { z({ x }^{ 2 }-y^{ 2 }) } },{ f }_{ { x({ x }^{ 2 }-3y^{ 2 }) } },{ f }_{ { y({ 3x }^{ 2 }-y^{ 2 }) } }\)
There are 3 nodal planes in the f-orbitals.
2.
Electronic configuration of oxygen

ஃ 8th electron present in 2px orbital and the quantum numbers are
n = 2,l = 1,m1 = either + 1 or -1 and s = -1/2
Electronic configuration of chlorine
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15th electron present in 3Pz orbital and the quantum numbers are n = 3, l = 1, m1 = either +1 or -1 and ms = +1/2
3.
\(\triangle x.\triangle p\ge \frac { h }{ 4\pi } \)
\(\triangle x.\triangle p\ge 5.28 \times{ 10 }^{ -35 }Kg{ m }^{ 2 }{ s }^{ -1 }\)
\(\triangle x.(m\triangle v)\ge 5.28 \times{ 10 }^{ -35 }Kg{ m }^{ 2 }{ s }^{ -1 }\)
Given \(\triangle\)v = 0.1%
v = 2.2 x 106 ms-1
m = 9.1 x 10-31Kg
\(\triangle\)v = \(\frac{0.1}{100}\times2.2\times{10}^{6}ms^{-1}\)
= \(2.2\times{10}^{6}ms^{-1}\)
\(\therefore \triangle x\ge \frac { { 5.28\times 10 }^{ -35 }{ Kgm }^{ 2 }{ s }^{ -1 } }{ 9.1\times { 10 }^{ -31 }Kg\times 2.2\times { 10 }^{ 3 }m{ s }^{ -1 } } \)
\(\\ \triangle x\ge 2.64\times { 10 }^{ -8 }m\)
4.
Circumference of the orbit - 2\(\pi\)r .... ...(1)
Circumference of the orbit of H - atom (n = 1) - n\(\lambda\), .........(2)
(1) = (2) \(2 \pi r=\lambda \text { (or) } 2 \pi r=\frac{\mathrm{h}}{\mathrm{mv}}\)
5.
For 3p-orbital n = 3, l = 1
Number of angular nodes = l = 1
Number of radial nodes = n - l -1 = 3 - 1 - 1 =1
6.
Erwin Schrodinger expressed the wave nature of electron in terms of a differential equation. This equation determines the change of wave function in space depending on the field of force in which the electron moves. The time independent Schrodinger equation can be expressed as,
\(\overset { \wedge }{ H } \psi =E\psi \) .........(1)
Where \(\overset { \wedge }{ H } \) is called Hamiltonian operator, \(\psi \) is the wave function and is a function of position coordinates of the particle and is denoted as \(\psi \) (x, y, z) E is the energy of the system
\(\overset { \wedge }{ H } =\left[ \frac { { -h }^{ 2 } }{ 8{ \pi }^{ 2 } } \left( \frac { { \partial }^{ 2 }\psi }{ { \partial x }^{ 2 } } +\frac { { \partial }^{ 2 }\psi }{ { \partial y }^{ 2 } } +\frac { { \partial }^{ 2 }\psi }{ { \partial z }^{ 2 } } \right) +V \right] \)
can be written as
\(\left[ \frac { { -h }^{ 2 } }{ 8{ \pi }^{ 2 }m } \left( \frac { { \partial }^{ 2 }\psi }{ { \partial x }^{ 2 } } +\frac { { \partial }^{ 2 }\psi }{ { \partial y }^{ 2 } } +\frac { { \partial }^{ 2 }\psi }{ { \partial z }^{ 2 } } \right) +V\Psi \right] =E\Psi \)
Multiply by \(\frac { 8{ \pi }^{ 2 }m}{ { -h }^{ 2 } } \)and rearranging
\(\frac { { \partial }^{ 2 }\psi }{ { \partial x }^{ 2 } } +\frac { { \partial }^{ 2 }\psi }{ { \partial y }^{ 2 } } +\frac { { \partial }^{ 2 }\psi }{ { \partial z }^{ 2 } } +\frac { 8{ \pi }^{ 2 }m }{ { -h }^{ 2 } } (E-V)\Psi =0\) ........(2)
The above Schrodinger wave equation does not contain time as a variable and is referred to as time independent Schrodinger wave equation. This equation can be solved only for certain values of E, the total energy. i.e. the energy of the system is quantised. The permitted total energy values are called eigen values and corresponding wave functions represent the atomic orbitals.
7.
Ferrous and ferric salts: In ferrous salts Fe2+, the configuration is 1s22s2,2p6,3s2, 3p6, 3d6.In ferric salts Fe3+ ,the configuration is 1s22s2,2p6,3s2, 3p6, 3d5.
As half-filled 3d5 configuration is more stable therefore ferric salts are more stable than ferrous salts.
8.
The word Aufbau in German means 'building up'. In the ground state of the atoms, the orbitals are filled in the order of their increasing energies. That is the electrons first occupy the lowest energy orbital available to them.
Once the lower energy orbitals are completely filled, then the electrons enter the next higher energy orbitals. The order of filling of various orbitals as per the Aufbau principle which is in accordance with (n + l) rule.

9.
Statement : "No two electrons in an atom can have the same set of values of all four quantum numbers"
Explanation : It means that, each electron must have unique values for the four quantum numbers (n, l, m and s).
For the lone electron present in hydrogen atom, the four quantum numbers are: n = 1; l = 0; m = 0 and s = +1/2. For the two electrons present in helium, one electron has the quantum numbers same as the electron of hydrogen atom, n = 1.
l = 0, m = 0 and s = +1/2. For other electron, the fourth quantum number is different i.e., n = 1, l = 0, m = 0 and s = -1/2.
As we know that the spin quantum number can have only two values +1/2 and - 1/2, only two electrons can be accommodated in a given orbital in accordance with pauli exclusion principle.
| Atom | e- | n | l | m | s |
| Helium | First | 1 | 0 | 0 | +1/2 |
| Second | 1 | 0 | 0 | +1/2 |
10.
\({ E }_{ n }=\frac { -13.6 }{ { n }_{ 2 } } ev{ \quad atom }^{ -1 }\)
n = 3 E3 = \(\frac { -13.6 }{ { 3 }^{ 2 } } =\frac { -13.6 }{ 9 } \)
= -1.51 ev atom-1
n = 4 E4 =\(\frac { -13.6 }{ { 4 }^{ 2 } } =\frac { -13.6 }{ 16 } \)
= -0.85 ev atom-1
\(\triangle \)E = (E4-E3) = (-0.85) - (-1.51) ev atom-1
= (-0.85 + 1.51)
= 0.66eV atom-1
(1eV = 1.6 x 10-19J)
\(\triangle \)E = 0.66 x 1.6 x 10-19J
\(\triangle \)E = 1.06 x 10-19J
hv = 1.06 x 10-19J
\(\frac { hv }{ \leftthreetimes } \) = 1.06 x 10-19J
\(\therefore\)\( \leftthreetimes\) = \(\frac { hc }{ 1.06\times { 10 }^{ -19 }J } \)
= \(\frac { 6.626\times { 10 }^{ -34 }JS\times 3\times { 10 }^{ 8 }{ ms }^{ -1 } }{ 1.06\times { 10 }^{ -19 }J } \)
\(\lambda=1.875\times10^{-6}m\)
11th Standard Syllabus & Materials
11th Standard
TN 11th Tamil பீடு பெற நில் - செய்யுள் - காவடிச்சிந்து Important Questions And Answers Study Material - QB365 Set A
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