7th edition 1F.3

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Sophia Fox 4B
Posts: 29
Joined: Fri Sep 28, 2018 12:27 am

7th edition 1F.3

Postby Sophia Fox 4B » Sat Oct 27, 2018 4:46 pm

How would you figure this out? "Place the following ions in order of increasing ionic radius: S^2-, Cl^-, P^3-." Its 7th edition 1F.3. Thanks!

Adrian C 1D
Posts: 60
Joined: Fri Sep 28, 2018 12:19 am

Re: 7th edition 1F.3

Postby Adrian C 1D » Sat Oct 27, 2018 5:22 pm

Based on ionic radii trends, the ionic radius increase down a group and decrease across a period. Since P^-3 is the leftmost option is has the largest ionic radius. The rightmost option is Cl^-1, which would have the smallest ionic radius. SInce S^-2 is between them on the periodic table, its ionic radius is also between the two.

Leela_Mohan3L
Posts: 44
Joined: Fri Sep 28, 2018 12:26 am

Re: 7th edition 1F.3

Postby Leela_Mohan3L » Sat Oct 27, 2018 5:25 pm

All of these atoms have the same amount of electrons, so you can look at the number of protons as an indicator of atomic radius. When going across a period, more protons --> smaller atomic radius. This is because the amount of protons increases the nuclear charge which increases the attraction of the electrons to the nucleus, thus decreasing the atomic radius. So the order would be Cl-, S^2-, P^3-.


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