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In an atom, total number of electrons having quantum numbers n=4,| ml|=1n=4,| m𝑙|=1 and ms=−12ms=−12 is______

Question

In an atom, total number of electrons having quantum numbers n=4,| ml|=1n=4,| m𝑙|=1 and ms=−12ms=−12 is______

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Solution

The quantum numbers describe the properties of the atomic orbitals and the properties of the electrons in the atom.

The principal quantum number (n) = 4, indicates the energy level or shell in which the electron is located.

The magnetic quantum number (ml) = ±1, indicates the orientation of the orbital in space.

The spin quantum number (ms) = -1/2, indicates the direction of the electron spin (up or down).

For n=4 and |ml|=1, the possible orbitals are 4p and 4d.

In the 4p orbital, ml can be -1, 0, or 1. But we are only interested in |ml|=1, so only two possibilities exist (ml=1 and ml=-1). Each of these has two possible spin states (ms=1/2 and ms=-1/2), but we are only interested in ms=-1/2. So, there are 2 electrons in the 4p orbital with these quantum numbers.

In the 4d orbital, ml can be -2, -1, 0, 1, or 2. Again, we are only interested in |ml|=1, so only two possibilities exist (ml=1 and ml=-1). Each of these has two possible spin states (ms=1/2 and ms=-1/2), but we are only interested in ms=-1/2. So, there are 2 electrons in the 4d orbital with these quantum numbers.

Therefore, the total number of electrons having quantum numbers n=4, |ml|=1 and ms=-1/2 is 2 (from 4p) + 2 (from 4d) = 4.

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The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. List if any of these combination(s) has/have the same energy(i) n = 4, l = 2, ml = -2, ms = -1/2(ii) n = 3, l = 2, ml = 1, ms = +1/2(iii) n = 4, l = 1, ml = 0, ms = +1/2(iv) n = 3, l = 2, ml = -2, ms = -111(v) n = 3, l = l, ml = -1, ms = +1/2(vi) n = 4, l = 1, ml = 0, ms = +1/2

The maximum number of orbitals which can be identified with n=4n=4 and ml=0𝑚𝑙=0 is ______

How many electrons in an atom can have n + l = 6 ?

Maximum number of electrons that can be accommodated in shell with n = 4 are :

1/3

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