Question #3b3d4
1 Answer
The spin-only magnetic moment is most applicable to light-enough transition metals with minimal orbital contribution to
bb(mu_S = 2.00023sqrt(S(S+1)))
where
[Ar]3d^9 ,"Cu"^(2+)
[Ar]3d^8 ,"Ni"^(2+)
[Ar]3d^6 ,"Co"^(3+)
[Ar]3d^6 ,"Fe"^(2+)
Write out the orbital diagrams.
ul(uarr darr)" "ul(uarr darr)" "ul(uarr darr)" "ul(uarr darr)" "ul(uarr color(white)(darr))
ul(uarr darr)" "ul(uarr darr)" "ul(uarr darr)" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))
ul(uarr darr)" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))
One can clearly see the increasing number of unpaired electrons going from
color(green)(mu_(S,3d^9)) = mu_(S,"Cu"^(2+)) = 2.00023sqrt(1/2(1/2+1)) = color(green)(1.732)
color(green)(mu_(S,3d^8)) = mu_(S,"Ni"^(2+)) = 2.00023sqrt(2/2(2/2+1)) = color(green)(2.829)
color(green)(mu_(S,3d^6)) = mu_(S,"Co"^(3+),"Fe"^(2+)) = 2.00023sqrt(4/2(4/2+1)) = color(green)(4.900)
Alternatively,
bb(mu_(S+L) = 2.00023sqrt(S(S+1) + 1/4L(L+1)))
where
color(green)(mu_(S+l,"Cu"^(2+))) = 2.00023sqrt(1/2(1/2+1) + 1/4*2(2+1)) = color(green)(3.000)
color(green)(mu_(S+l,"Ni"^(2+))) = 2.00023sqrt(2/2(2/2+1) + 1/4*3(3+1)) = color(green)(4.473)
color(green)(mu_(S+l,"Co"^(3+),"Fe"^(2+))) = 2.00023sqrt(4/2(4/2+1) + 1/4*2(2+1)) = color(green)(5.478)
For instance,
Do note that the observed magnetic moments for each of these cations are
So,