11th Standard Syllabus & Materials
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TN 11th Tamil இயற்கை வேளாண்மை,சுற்றுச்சூழல் -செய்யுள் - மனோன்மணீயம் Important Questions And Answers Study Material - QB365 Set A
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Published on: 08/06/2021
QB365 provides detailed and simple solution for every Book back Questions in class 11 Chemistry Subject. It will helps to get more idea about question pattern in every book back questions with solution.
Download Tamil Nadu 11th Standard Chemistry question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.
Questions + Answers key
Take MCQ Chemistry Test1.
State and explain pauli exclusion principle.
2.
Determine the values of all the four quantum numbers of the 8th electron in O- atom and 15th electron in Cl atom.
3.
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.
4.
Explain briefly the time independent schrodinger wave equation?
5.
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.
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 |
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.
Circumference of the orbit - 2\(\pi\)r .... ...(1)
Circumference of the orbit of H - atom (n = 1) - n\(\lambda\), .........(2)
(1) = (2) \(2 \pi r=\lambda \text { (or) } 2 \pi r=\frac{\mathrm{h}}{\mathrm{mv}}\)
4.
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.
5.
\({ E }_{ n }=\frac { -13.6 }{ { n }_{ 2 } } ev{ \quad atom }^{ -1 }\)
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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NEW11th Standard
TN 11th Tamil மாமழை போற்றுதும் - செய்யுள் - ஐங்குறுநூறு Important Questions And Answers Study Material - QB365 Set A
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