Electronic Geometry of Lead(II) Chromate

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The discussion centers on the electronic geometry of Lead(II) Chromate, specifically the valence electrons of chromium, lead, and oxygen. There is confusion about lead's valence electrons, with a suggestion that Lead(II) should have 6 due to its +2 charge. Participants clarify that Lead(II) actually has 4 valence electrons, leading to the conclusion that the compound is ionic, involving the transfer of electrons from the chromate ion to the lead ion. The need for a correct Lewis dot structure is emphasized, particularly regarding the placement of chromium and the representation of lead. Overall, the conversation highlights the complexities of drawing accurate Lewis structures for ionic compounds.
The_ArtofScience
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I was wondering if anyone can solve this odd problem. Chromium is known to vary in the number of valence electrons and in this case, I would hope it has 6. Lead has 4 and oxygen 6. If you were to draw its Lewis dot structure with Chromium in the middle you would leave out Pb without any filled electrons which is to say, it's left with only a stick figure representing 2 electrons.


How does one go about in solving this problem?
 
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Shouldn't Lead(II) have 6 valence electrons since that is what gives it the +2 charge?
I'm not quite sure though, maybe someone who knows more can confirm that.
 
This compound is ionic in nature.
 
Then that would just mean a transfer of 2 electrons from the chromate ion to the lead ion.

Wow. My Chemistry is pretty rusty after a year of not doing it.
 

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