Question on 2A.23 part a

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Alvaro Chumpitaz 4D
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Joined: Wed Sep 11, 2019 12:17 am

Question on 2A.23 part a

Postby Alvaro Chumpitaz 4D » Sat Oct 26, 2019 9:44 pm

On the basis of the expected charges on the monoatomic ions, give the chemical formula for each of the following compounds
a) magnesium arsenide
How can you figure the amount of each element in the formula (aka the subscripts)?

JChen_2I
Posts: 107
Joined: Fri Aug 09, 2019 12:17 am

Re: Question on 2A.23 part a

Postby JChen_2I » Sat Oct 26, 2019 10:18 pm

Magnesium has 2 valence electrons so it will generally give them up to atoms with almost full electrons. Arsenic has 5 valence electrons so it needs 3 more to complete its shell. Therefore if you have 3 magnesium atoms, there will be 6 electrons to give. And with 2 arsenic atoms, it needs 6 electrons to fill its shells. So the chemical formula is Mg3As2

Helen Struble 2F
Posts: 97
Joined: Sat Aug 24, 2019 12:17 am

Re: Question on 2A.23 part a

Postby Helen Struble 2F » Sun Oct 27, 2019 3:13 pm

Because magnesium is a group 2 element, we can expect it to take on a 2+ charge, making it isoelectronic to neon and thus more stable. Though the p block is a little harder to predict, in an ionic compound, we can make an educated guess that arsenic will take on a 3- electronic charge, making it isoelectronic to krypton. Now we just need to find the least common multiple of 2 and 3, which is 6. This lets us know there is a total exchange of 6 electrons in our ionic compound—3 Mg giving up 2 e- to each take on a 2+ charge, and 2 As gaining 3 e- to each take on a 3+ charge. Therefore, our ionic compound is: Mg3As2.


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