2A23 Part E

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Rachel Yu 1G
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Joined: Fri Aug 09, 2019 12:15 am

2A23 Part E

Postby Rachel Yu 1G » Mon Oct 28, 2019 9:27 pm

2A23 asks for the chemical formula of bismuth (III) fluoride which ends up being BiF3, an ionic bond between Bi 3+ and 3F-. Why does Bismuth (III) have a charge of 3+ when it is a Group 5 element? I though Group 5 ions would have a charge of 3-.

DarrenKim_1H
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Re: 2A23 Part E

Postby DarrenKim_1H » Mon Oct 28, 2019 9:40 pm

Bismuth is undergoing the inert-pair effect

Brian Tangsombatvisit 1C
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Re: 2A23 Part E

Postby Brian Tangsombatvisit 1C » Mon Oct 28, 2019 9:54 pm

Bismuth is one of the weird elements that doesn't always follow the typical rules or have the usual characteristics of one of the first 18 atoms. When you see the roman numeral "III" just know that it is a cation with a charge of +3. I don't think we would have to memorize this cation of Bismuth since we mostly focus on the top half of the periodic table when it comes to elements.

Lisa Wang 2J
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Joined: Wed Sep 30, 2020 9:51 pm

Re: 2A23 Part E

Postby Lisa Wang 2J » Tue Oct 20, 2020 2:19 pm

Yes, in this case although you would count normally forward to Radon and have Bismuth have the charge 3-, since it is denoted (III), it counts as 3+ and thus 3 F charges would cancel out. The answer would be BiF3

isha dis1k
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Re: 2A23 Part E

Postby isha dis1k » Wed Oct 21, 2020 3:23 pm

The group 5 elements have five valence electrons in their highest-energy orbitals so they can form ionic compounds by gaining three electrons, forming anions, but I believe they more frequently form compounds through covalent bonding. Bismuth can lose either their outermost p electrons to form 3+ charges, or their outermost s and p electrons to form 5+ charges. In this case, it is losing its p electrons in a covalent bond.


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