Question

What is the maximum number of electrons in an atom that can have the following quantum numbers?

(a) n = 4, ms = -1/2

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(b) n = 5, l = 3

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(c) n = 6, l = 3, ml = -2

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(d) n = 2, l = 1, ml = -1, ms = +1/2

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Answer 1

(a) For n =4, we can have possible values of l as 0 to (n – 1),i.e, 0,1,2 and 3. Thus, we have the 4s, 4p, 4d and 4f subshells.

Now, we have n = 4, l = 0, m_l = 0; therefore only one orbital which can hold a maximum of 2 electrons, one of them having m_s = -(1/2). This is due to the Pauli exclusion Principle.

For n = 4, l = 1, m_l = -1, 0, +1; three orbitals, each of which can hold a maximum of 2 electrons; So, the 4p subshell has 6 electrons in total out of which three will have m_s = -(1/2)

For n = 4, l = 2, m_l = -2, -1, 0, +1, +2; five orbitals which can hold 10 electrons in total, out of which 5 will have m_s = -(1/2)

For n = 4, l = 3, m_l = -3,-2, -1,0,+1,+2,+3; seven orbitals in total that can accommodate a maximum of 14 electrons out of which 7 will have m_s = -(1/2).

So, the total number of electrons that will have n = 4, m_s = -(1/2) will be (1+3+5+7) = 16.

(b) Here we have n = 5, l = 3. This indicates the 5f subshell. For l = 3, we can have m_l = -3,-2,-1,0,+1,+2,+3. Hence, there are total 7 f orbitals; each orbital can hold 2 electrons, hence the total number of electrons in the 5f subshell is 14.

(c) We have n = 6, l = 3 and m_l = -2. This set of quantum numbers depicts only one of the 6f orbitals. Each f orbital can hold 2 electrons at a maximum, hence the maximum electrons in the orbital denoted as n = 6, l = 3 and m_l = -2 is 2.

(d) n =2 , l = 1, m_l = -1; m_s = +(1/2) : this set of quantum numbers depicts the spin-up electron in one of the three 2p orbitals. The answer here is 1 and that will be the spin up electron in one of the 2p orbitals.