Asked by Sami
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.
I have the example..
Sn([Kr] 5s24d105p2) Sn4+ ([Kr]4d10)+4e-
I understand the first part but I don't understand the second.
Answers
Answered by
DrBob222
Let's take a couple that are easier to write to give examples. Na and Zn.
<sub>11</sub>Na 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.
<sub>30</sub>Zn 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.
<sub>11</sub>Na 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.
<sub>30</sub>Zn 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.
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