# Why are the atomic radii of iron, cobalt, nickel, and copper almost the same?

We have to select the set of elements which will have almost the same atomic radii:

1. $\ce{O, S, Se, Te}$
2. $\ce{Li, Be, B, C}$
3. $\ce{Fe, Co, Ni, Cu}$

I googled the atomic radii of the third set and found them to be extremely close to one another: all of them are nearly $\pu{125 pm}$ (according to empirical data from Wikipedia). Is there a specific reason for this, or is it just a coincidence?

Atomic radius is inversely proportional to the effective nuclear charge. As we move from left to right in a period the effective nuclear charge increases. This will decrease the radius of an atom. At the same time, in transition elements the number of electrons in the 3d sub-shell will increase. This will repel the already present 4s electrons. This repulsion will increase the atomic radius. In Fe, Co, Ni & Cu both the effects nearly balance each other keeping the atomic radius same.

I think the point of this question is for you to realise that options 1 and 2 can't be the correct answer.

As you go down a group, new electron shells are occupied which extend further from the nucleus, increasing the atomic radius. Therefore option 1 must be wrong.

Effective nuclear charge increases across a period because the nuclear charge increases but the shielding stays roughly the same (at least until you get to transition metals). Therefore atomic radius will decrease across the group and so option 2 cannot be correct and the answer must be option 3.

## protected by Community♦Nov 1 '17 at 17:03

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