2.60  [ENDORSED]

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Alex Nechaev 1I
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Joined: Fri Sep 29, 2017 7:04 am


Postby Alex Nechaev 1I » Sun Oct 29, 2017 12:27 pm

We went over this question in one of the Peer Learning Sessions, and I just need some clarification. The question reads:

Which ion of the following pairs has the larger radius: a) Ca^2+ , Ba^2+ b) As^3- , Se^2- c) Sn^2+ , Sn^4+ ?

What effect do the charges of these ions have on their atomic radius? Should we be able to tell which is larger just by following the trends on the periodic table?

Srbui Azarapetian 2C
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Re: 2.60  [ENDORSED]

Postby Srbui Azarapetian 2C » Sun Oct 29, 2017 1:38 pm

Atomic radii generally decrease from left to right across a period as the effective atomic number increases, and they increase down a group as successive shells are occupied. If the two ions are in the same group, the smaller ion will be the one that lies higher in the group, because its outermost electrons are closer to the nucleus. The smaller ion will also be that of an element that lies farther to the right in a period, because that ion has the greater effective nuclear charge.

The charges don’t have that much of an effect as the number of electrons do, except in part c. Each element is an ion of its most stable noble gas configuration.The periodic table trends should be a guide for ordering ions by their atomic radius. In this case, the largest radius is
A) Ba 2+
B)As 3-
C) Sn 2+

In the case of c, Sn 2+ has two more electrons than that of Sn4+, and while they both have the same number of protons, Sn4+ has a greater effective nuclear charge because there are less electrons governed by the same number of protons, making the ion smaller.

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