Let's take a couple that are easier to write to give examples. Na and Zn.
11Na is 1s2 2s2 2p6 3s1. It loses the one valence electron (the 3s) to become 1s2 2s2 2p6 = 10 e and that is Ne. That is the Na +1 ion is isoelectronic with Ne and it has a filled outside shell. Thus Na atom has become a Na +1 ion that looks like Ne and this is a noble gas so we say Na has attained a noble gas configuration.
30Zn is 1s2 2s2 2p6 3s2 3p6 3d10 4s2. Zn loses the two outside valence electrons to become
1s2 2s2 2p6 3s2 3p6 3d10 = 28 electrons. Notice that the Zn +2 ion has all filled shells BUT 28 is not a noble gas. So we call that a psuedo noble gas (it has filled shells as a noble gas does; however, it isn't a noble gas). Now apply that to your problems. You see the element Sn loses the 5s2 and 5p2 (the four outside valence electrons) for a +4 ion but it is a pseudo noble gas (because the 4d shell is filled BUT it isn't a noble gas). Count up the electrons. It would need to have either 54 or 36 to be a noble gas configuration.
Can you explain what the pseudo-gas configuration is? I have the definition but I still don't understand.
I have the example..
Sn([Kr] 5s24d105p2) Sn4+ ([Kr]4d10)+4e-
I understand the first part but I don't understand the second.
1 answer