Question

Identify all allowable combinations of quantum numbers for an electron.

=5, =4, =−4, s=+1/2

=2, =2, =−1, s=−1/2

=2, =1, =1, s=1

=3, =1, =0, s=+1/2

=3, =−1, =0, s=−1/2

=4, =3, =4, s=−12

Answer 1

The selection rules allow to find the results for the correct answers of the combinations are:

      a) n = 4, l = 4, m = -4, s = + ½

      d) n = 3, l = 1,  m = 0, s = + ½

The solution of the Schrodrigen equations gives the energy levels of the electrons in the atoms, with the solution in three dimensions there are three constant that are related, these constant are called quantum numbers as a whole:

• Principal (n). The first constant is related to the energy of the electron and can have values ​​from 1 to infinity.

• Secondary or orbital (l). It is related to the shape of the orbit of the electron, in general it is represented by letters: s, p, d, f. It can have values ​​from 0 to (n-1).

• Orbital magnetic moment (m). It is related to the magnetic field due to the movement of the electron around the nucleus. It can have values ​​from -l to l.

Independent measurements found a fourth quantum number.

• Spin magnetic moment (s). Due to the spin of the electron, it can be visualized by the movement on itself. It can have two values ​​+ ½ and -½

The above restrictions are collectively called selection rules and determine the allowed combinations for electrons.

Let's examine the answers to see which one is true.

a) n = 5 l = 4 m = -4 s = + ½

True. The principal quantum number can have any value.

           The orbital quantum number can go up to (n-1) that is l = 4

           The magnetic orbital goes from -4

           The magnetic spin can have two values.

           This combination of quantum numbers is allowed by the selection rules.

b) n = 2  l = 2  m = 1 s = -1/2

False The principal quantum number can be 2

             The orbital quantum number must be less than the principal.

c) n = 2 l = 1 m = 1 s = 1

False. The number of spin can only have two values.

d) n = 3 l = 1 m = 0 s = + ½

True     The number n can be 3

           The orbital number goes from 0 to 2, so it is correct

           The magnetic quantum number m can have the value 0

           The spin is worth + ½

This combination of quantum numbers is allowed

e) n = 3 l = -1 m = 0 s = - ½

False  The orbital quantum number cannot be negative.

f) n = 4 l = 3 m = 4 s = - ½

False The orbital moment quantum number cannot be greater than the orbital quantum number.

In conclusion using the selection rules we can find the results for the correct answers of the combinations are:

      a) n = 4, l = 4, m = -4, s = + ½

      d) n = 3, l = 1, m = 0, s = + ½

Learn more here: brainly.com/question/23780112