What are the set of four quantum numbers that represent the electron gained to form the #Br# ion from #Br# atom?

1 Answer
Jun 13, 2016

#n = 4, l = 1, m_l = 0, m_s = -1/2#

Explanation:

As you know, there are four quantum numbers used to describe the position and spin of an electron in an atom.

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Your goal here will be to use the information provided by the electron configuration of a neutral bromine atom, #"Br"#, to determine the quantum numbers associated with the electron needed to form the bromide anion, #"Br"^(-)#.

So, bromine is located in period 4, group 17 of the periodic table and has an atomic number equal to #35#. This tells you that a neutral bromine atom has a total of #35# electrons surrounding its nucleus.

The electron configuration of a neutral bromine atom looks like this

#"Br: " 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 color(red)(4) s^2 color(red)(4)p^5#

Notice that bromine's outermost electrons, i.e. its valence electrons, are located on the fourth energy level, #n = color(red)(4)#.

The incoming electron will be added to this energy level, so right from the start you know that it must have #n=4#.

The angular momentum quantum number, #l#, tells you the subshell in which the electron is located. In this case, the incoming electron will be added to the 4p-subshell, which is characterized, much like any p-subshell, by #l=1#.

The magnetic quantum number, #m_l#, tells you the exact orbital in which the electron is located. The 4p-subshell contains a total of #3# p-orbitals

  • #4p_x -> m_l = -1#
  • #4p_y -> m_l = +1#
  • #4p_z -> m_l = color(white)(-)0#

This is used by convention because the wave function associated with #m_l = 0# is said to be symmetric about its axis, I.e. it has no component of angular momentum about its axis, which is usually chosen as the #z# axis.

Since the neutral bromine atom already has 5 electrons in its 4p-subshell, you can say that its #4p_x# and #4p_y# orbitals are completely filled and the #4p_z# contains one electron.

The incoming electron will thus be added to the half-empty #4p_z# orbital, and so it will have #m_l = 0#.

Finally, the spin quantum number, #m_s#, tells you the spin of the electron. Because the #4p_z# already contained one electron that has spin-up, or #m_s = +1/2#, the incoming electron must have an opposite spin, and so #m_s = -1/2#.

The quantum number set that describes the incoming electron will thus be

#color(green)(|bar(ul(color(white)(a/a)color(black)(n = color(red)(4), l = 1, m_l = 0, m_s = -1/2)color(white)(a/a)|)))#

SIDE NOTE You'll sometimes see this notation used for the magnetic quantum number

  • #4p_x -> m_l = -1#
  • #4p_y -> m_l = color(white)(-)0#
  • #4p_z -> m_l = +1#

You can use this if you want, but make sure that you are consistent.

In this notation, the wave function associated with #m_l = 0# has symmetry about the #y# axis, so make sure that you specify and keep track of this for other orbitals such as the d-orbitals of f-orbitals.