First, let"s look at those because that the neutral atom, and then work-related our means to the cation.

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VANADIUM (V)

#"V"# is atom number #23#, for this reason its configuration is #1s^2 2s^2 2p^6 3s^2 3p^6 3d^3 4s^2#. In shorthand the is # 3d^3 4s^2#. This is an expected configuration; not an oddball element.

Since the #4s# orbit is greater in energy, that electrons will be gotten rid of first. No that it problem here, though, due to the fact that exactly #5# electrons room removed:

#color(blue)("V"^(5+) -> 1s^2 2s^2 2p^6 3s^2 3p^6 -= )#

GOLD (III)

#"Au"# is atom number #79#, for this reason its construction is #1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6 4f^14 5d^10 6s^1#. In shorthand the is # 4f^14 5d^10 6s^1#.

This is strange, not only since gold desires to acquire a filled #5d# subshell, but an ext likely additionally something to carry out with the high amount of relativistic contraction effects; the #s# electrons relocate close to the speed of light and are regularly farther native the nucleus 보다 closer, shrinking the radius of the #1s# orbital by #~22%#, and other orbitals by a bit as well.

Let"s research the radial density distribution of the #6s#, #5d#, and #4f# valence orbitals come see exactly how that transforms out:

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By a lengthy shot, the #4f# is the many penetrating, meaning that most of that electron thickness is focused near the yellow nucleus. Further out are the #5d# electrons, and also then the #6s# electrons.

This argues that the #6s# orbital, which has most electron thickness farther out from the nucleus, has actually electron(s) that are less attracted to the cell nucleus (smaller #Z_(eff)#) 보다 those in either the #5d# or #4f# orbitals, and thus includes the first electron that is simplest to remove during ionization.

The #5d# includes the next two electrons that will be removed throughout the second and 3rd ionizations.

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So, we would get:

#color(blue)("Au"^(3+) -> 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6 4f^14 5d^8 -= 4f^14 5d^8)#

IRON (II)

#"Fe"# is atomic number #26#, therefore its construction is #1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2#. In shorthand that is # 3d^6 4s^2#. This is an supposed configuration; no an oddball element.

The very first and second ionizations would eliminated the #4s# electrons, i m sorry are greater in power than the #3d# electrons, so us get: