Why is #"Sn"^"2+"# diamagnetic?

1 Answer
Mar 30, 2016

http://www.cmu.edu/

If you look at the periodic table, you should see that the electron configuration for #"Sn"^0# is

#color(green)([Kr] 4d^10 5s^2 5p^2)#,

so its valence shell has ten #4d#, two #5s#, and two #5p# electrons.

Upon ionization, #"Sn"# will lose two of its highest-energy electrons to form #"Sn"^(2+)#.

The highest-energy electrons here are the #5p# electrons (by about #"8.55 eV"# above the energy of the #5s# electrons), so we now achieve an ionic electron configuration of

#color(blue)([Kr] 4d^10 5s^2)#.

Since the #4d# and #5s# subshells are entirely filled, there are no unpaired electrons and #"Sn"^(2+)# by definition is diamagnetic.