BR GC Section 8 passage 3 question

[SaCkO]

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In this passage they tell us that
E= k (Charge cation*charge anion)/ distance between ions in lattice(they name it R)

So if we look at question one they ask what happens when we replace Lithium metal for Sodium Metal. In their answer explanation they state that sodium has a bigger Radius than Lithium then R would increase and thus E would decrease.

However In the equation it states that R is the radius between Ions and not the atomic radius, thus when you increase the atomic radius of an atom you are putting these atoms closer together which would make R smaller.

In the last question of the passage they contradict their answer in the first question and they say that Substituting a Larger cation for a smaller one would increase the radius and thus decrease E.

 

[SaCkO]

If Anyone who is working with Berkeley review books and done this passage before , please explain to me if iam wrong or the answer is wrong.
thanks

 

[milski]

As far as I understand it, atoms in a lattice are squished as close to each other as possible and there's no free space between them. That would mean that putting a smaller atom will decrease the distance between atoms (R) and E will increase.

I actually don't see contradiction between the two statements in black in your post:
In the first one, sodium has bigger radius and that leads to smaller E. In the second one they say that putting larger cation in the place of a smaller one will decrease E, which is the same.

Is the part in red your take on the problem or something from the text?

 

[SaCkO]

As far as I understand it, atoms in a lattice are squished as close to each other as possible and there's no free space between them. That would mean that putting a smaller atom will decrease the distance between atoms (R) and E will increase.

I actually don't see contradiction between the two statements in black in your post:
In the first one, sodium has bigger radius and that leads to smaller E. In the second one they say that putting larger cation in the place of a smaller one will decrease E, which is the same.

Is the part in red your take on the problem or something from the text?

The part in Red is my understanding of it.
What iam not understanding is how putting a smaller atom will make distance smaller, if you can explain it to me i really appreciate it , because the way iam visualizing the atom in my head, it seems to me that as the radius of an atom gets bigger, atoms are occupying more space then they are closer to each other, its like having small balls at each corner in the room or big balls in the corner, which balls are closer to each other?

 

[SaCkO]

And btw they state that R is the distance between ions and they didn't say that it is the same as the ion radius in the passage

 

[milski]

The balls are not in the corners of the room - they are pressed against each other. The distance R between two balls is really the distance between their centers, the distance between their surfaces should be virtually zero.

In other words, R in the formula should be best written as R1+R2 where R1 and R2 are the radii of the two atoms.

 

[SaCkO]

The balls are not in the corners of the room - they are pressed against each other. The distance R between two balls is really the distance between their centers, the distance between their surfaces should be virtually zero.

In other words, R in the formula should be best written as R1+R2 where R1 and R2 are the radii of the two atoms.

Oh, I got it now, i was thinking of it from a different perspective, i didnt know that the ions would be attached,
Thank you so much , i appreciate your help