Question 2A.23 7th Edition

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Layal Suboh 1I
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Joined: Fri Sep 28, 2018 12:23 am

Question 2A.23 7th Edition

Postby Layal Suboh 1I » Wed Oct 31, 2018 6:32 pm

Hi,

I'm having difficulty with solving this question from the homework:

On the basis of the expected charges of the monatomic ions, give the chemical formula of each of the following compounds: a) manganese arsenide; b) indium (III) sulfide; c) aluminum hydride; (d) hydrogen telluride; e) bismuth (III) fluoride.

How do you know what the charge is of each of the atoms in order to write the compound?

Thanks

Andonios Karas 4H
Posts: 30
Joined: Fri Sep 28, 2018 12:27 am

Re: Question 2A.23 7th Edition

Postby Andonios Karas 4H » Thu Nov 01, 2018 12:03 am

You know the charges of monatomic cations from group 1,2, and 13 by looking at the number of valence electrons. They want to lose enough electrons to have the same electron configuration as the previous noble gas. So +1,+2, and +3 respectively.

The charges of monatomic cations from groups 3 through 12 are given by the roman numerals next to the metals. Metals have different oxidation states because they have both a s-Orbital and d-Orbital. They will always give up their s-Orbital electrons which can range from 1-2. However, they may give up between none and all of their d-Orbital electrons
i.e. I = +1 II = +2 and III = +3

You know the charges of monatomic anions from groups 15, 16, and 17 by looking at the number of valence electrons. They want to gain enough electrons to have the same electron configuration as the next noble gas. So -3, -2, and -1.

The question is asking you to create a compound of the two ions with a neutral charge, so you need to figure out how many of each is needed.
e.g. for a) Magnesium Arsenide
Magnesium ion has a charge of +2 because it lost its two valence electrons. Arsenic ion has a charge of -3 because it gained 3 valence electrons for an octet.
The stable compound they would make is Mg3As2


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