Describe all 4 quantum numbers for last electron of 3d^5?

2 Answers
Jun 6, 2018

see below

Explanation:

it is Mn (manganese)
N= 3 third energetic level
l= 3 shape of orbitals of type d : (quatrefoil)
5 = the 5 orbitals d are all full means with only one electron in every orbitals (M,magnetic orbitatals = -2,-1, 0 +1 +2 what is the last, you don't know because are isoenergetic)
s =intrinsic magnetic moment = +- h/(2 xx pi)=±h2×π(what is the last, you don't know, for convention the negative one)

Jun 6, 2018

The four quantum numbers are 3, 2, +2, +½.

Explanation:

The four quantum numbers are

  • n color(white)(ll)=nll= principal quantum number; determines energy and size of orbital
  • l color(white)(m)= lm= secondary quantum number; determines shape of orbital
  • m_text(l) =ml=magnetic quantum number; determines orientation of orbital in magnetic field
  • m_text(s) =ms= spin quantum number; determines direction of electron spin

The rules for quantum numbers are:

  • n color(white)(ll)=nll= integers from 1 to ∞.
  • lcolor(white)(m)= lm= integers from 0 to n - 1n1
  • m_text(l) =ml= integers from -ll to +l+l
  • m_text(s) =ms= +½ or -½

Here is a table of quantum numbers for the first five "3d"3d electrons. The quantum numbers for the fifth electron are red.

ulbb(ncolor(white)(m)lcolor(white)(ml)m_text(l)color(white)(ml)m_text(s))
3color(white)(m)2color(white)(mll)"-2"color(white)(m)"+½
3color(white)(m)2color(white)(mll)"-1"color(white)(m)"+½
3color(white)(m)2color(white)(mm)"0"color(white)(m)"+½
3color(white)(m)2color(white)(m)"+1"color(white)(m)"+½
color(red)(3color(white)(m)2color(white)(m)"+2"color(white)(m)"+½)

Here's how the quantum numbers fit in the Periodic Table

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