Question

The following sets of quantum numbers, listed in the order n, l, ml, and ms were written for the last electrons added to an atom. Identify which sets are valid and classify the others by the rule or principle that is violated.
choices for these answers are: "other violation," "pauli violation," and "valid."
a. 5 0 0 +1/2
5 0 0 -1/2
b. 4 1 -1 +1/2
4 1 0 +1/2
4 1 +1 +1/2
c. 3 2 -1 +1/2
3 2 0 +1/2
3 2 +1 +1/2
3 2 0 +1/2
3 2 +2 +1/2
d. 3 1 -1 +1/2
3 1 0 +1/2
3 3 +1 +1/2

Answer 1

Answer:

a. 5 0 0 +1/2

5 0 0 -1/2

Valid

b. 4 1 -1 +1/2

4 1 0 +1/2

4 1 +1 +1/2

Valid

c. 3 2 -1 +1/2

3 2 0 +1/2

3 2 +1 +1/2

3 2 0 +1/2

3 2 +2 +1/2

Pauli violation

d. 3 1 -1 +1/2

3 1 0 +1/2

3 3 +1 +1/2

other violation

Explanation:

The four quantum number and possible values are:

n = 1,2,3.....

l = 0 , (n-1), (n-2).....

m = +l , 0 , -l

s = $+\frac{1}{2}$ or $-\frac{1}{2}$

Pauli's exclusion principle: No two electrons in an atom can have all the four quantum numbers same.

Let us check each case:

a. 5 0 0 +1/2

5 0 0 -1/2

Valid

b. 4 1 -1 +1/2

4 1 0 +1/2

4 1 +1 +1/2

Valid

c. 3 2 -1 +1/2

3 2 0 +1/2

3 2 +1 +1/2

3 2 0 +1/2

3 2 +2 +1/2

Pauli violation

The two electrons have same four quantum numbers

3 2 0 +1/2

3 2 0 +1/2

d. 3 1 -1 +1/2

3 1 0 +1/2

3 3 +1 +1/2

other violation

As mentioned above in the condition the value of "l" can be only less than "n"

So for 3 3 +1 +1/2 : n = 3 and l= 3, which is not valid.

Answer 2

The arrangement of electrons in orbitals are showed by four sets of quantum numbers.

According to the Pauli exclusion theory, no two electrons in an atom should have all the four quantum numbers as the same. According to this principle, the spin quantum number of electrons in an atom must differ even if they are in the same orbital.

For the first set;

5 0 0 +1/2

5 0 0 -1/2

This set, correctly corresponds to the 5s  orbital so it is valid.

For the second set;

4  1  -1  +1/2

4  1  0  +1/2

4  1  +1  +1/2

This should have corresponding to the 4p orbital so it is valid.

For the third set;

3  2 -1  +1/2

3  2  0  +1/2

3  2  +1  +1/2

3  2  0  +1/2

3  2 +2  +1/2

This set should correspond to a 3d orbital but we can see that are two electrons in the set that has exactly the same quantum numbers of 3 2 0 +1/2. This violates the Pauli exclusion theory so we should mark "Pauli violation".

For the fourth set:

3  1  -1  +1/2

3   1   0  +1/2

3  3  +1  +1/2

This set should have corresponded to a 3p orbital but remember that the values of l only range from 0 to (n - 1). This means that we can not have n =3, l=3 so the arrangement 3  3  +1  +1/2 is not possible. We should mark  "other violation".

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