00:01
This question asks us to consider the isoelectronic ions, fluorine minus, and sodium plus.
00:07
First, it asks us which ion is smaller? since they're isoelectronic, they have the same number of electrons, therefore the same screening constant.
00:14
So the only thing that's going to change here is the nuclear charge, and that's determined by the number of protons.
00:21
Since sodium is further along on the periodic table than fluorine, it would have a greater number of protons, greater nuclear charge.
00:28
That means the electrons are held more tightly.
00:31
Smaller atom.
00:32
So sodium plus is the smaller of those two ions.
00:36
Then it says using equation 7 .1 and assuming that core electrons contribute 1 .0 and valence electrons contribute 0 .0 to the screening constant s, calculate z effective for the 2 p electrons in both ions.
00:50
So here's equation 7 .1.
00:52
Z effective equals z minus s or the effective nuclear charge equals the nuclear charge which minus the screening constant.
01:00
And so for f, the nuclear charge or the number of protons is nine.
01:05
And for sodium, there are 11 protons.
01:08
And then we're going to subtract the screening constant, which is the number of valence electrons, or the number of core electrons times one in f minus.
01:18
There are only two core electrons in the 1s orbital.
01:22
So 2 times 1 .0.
01:26
This gives us an effective nuclear charge of 7.
01:31
And for sodium, it has the same screening constant because they're isoelectronic...