Sapling #28

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luludaly2B
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Joined: Wed Sep 30, 2020 9:42 pm
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Sapling #28

Postby luludaly2B » Sun Nov 01, 2020 10:12 pm

Arrange these ions according to ionic radius.
P^(3-)
Cl^(-)
Ca^(2+)
S^(2-)
K^(+)


Could someone explain this concept to me? Thanks!

Kyla Villarama 1E
Posts: 85
Joined: Wed Nov 20, 2019 12:18 am

Re: Sapling #28

Postby Kyla Villarama 1E » Sun Nov 01, 2020 10:17 pm

The solution provided in Sapling was really helpful to me in understanding how to compare the radius sizes! In short, it basically says that the most negative ion has the largest radius while the most positive ion has the smallest radius (less protons=large radius; more protons=small radius).

Lillian
Posts: 69
Joined: Wed Sep 30, 2020 9:48 pm

Re: Sapling #28

Postby Lillian » Sun Nov 01, 2020 10:24 pm

With this specific set of ions, we can see that all of them have the same number of total electrons. The only difference is their nuclear mass and charge. The more protons the atom has, the more attraction it holds on the electrons, which brings them closer to the nucleus (like a magnet). :D Hope this helps!

Sam Wentzel 1F 14B
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Re: Sapling #28

Postby Sam Wentzel 1F 14B » Sun Nov 01, 2020 10:25 pm

The periodic trends for atomic radius are, going left to right, atomic radius decreases. Going from top to bottom, atomic radius increases. This is because in a horizontal ROW on the periodic table, the number of protons in the atom goes up by one with each successive element, but the shell (n = 2, 3 etc) stays the same within the whole row, so there is not as much electron-electron repulsion.

Hope this helps!


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