On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
Principal quantum number is number which determines the main energy level or shell in which the electron is present. It gives the average distance of the electron from the nucleus and specifies the value of the energy of the electron.
Azimuthal quantum number is the number which determines the sub shell in a principal energy shell to which an electron belongs.
In the periodic table of the elements, a period indicates the value of the principal quantum number (n) for the outermost shells. Each period begins with the filling of principal quantum number (n). The value of n for the sixth period is 6. For n = 6, azimuthal quantum number (l) can have values of 0, 1, 2, 3, 4.
According to Aufbau’s principle, electrons are added to different orbitals in order of their increasing energies. The energy of the 6d subshell is even higher than that of the 7s subshell.
In the 6th period, electrons can be filled in only 6s, 4f, 5d, and 6p subshells. Now, 6s has one orbital, 4f has seven orbitals, 5d has five orbitals, and 6p has three orbitals. Therefore, there are a total of sixteen (1 + 7 + 5 + 3 = 16) orbitals available. According to Pauli’s exclusion principle, each orbital can accommodate a maximum of 2 electrons. Thus, 16 orbitals can accommodate a maximum of 32 electrons.Hence, the sixth period of the periodic table should have 32 elements.
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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 4: On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 ....
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for people who are confused that why are there 0,1,2,3,4 orbitals instead of 0,1,2,3,4,5 it is because the formula for l=n-1 doesn't tell the sub-shells of the shell, but it tells the NUMBER OF ORBITALS in each shell. example- taking n=5, l should have n-1=4 NUMBER OF SUB-SHELLS i.e 0,1,2,3 and not 0,1,2,3,4
The value of l should be 0,1,2,3 There's no point of putting 4 as we have s,p,d and f orbitals only
L =0 to (n-1) then its value should be 0 1 2 3 4 5
If n=6 then l should be 0,1,2,3,4,5