Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
The ion has a radius of . The radii of the 3+ ions of the heavy lanthanoids are: Gd , Tb , Dy , Ho , Er , Tm , Yb and Lu . Find the lanthanoid whose ion is closest in size to , its atomic number, and the difference in radius.
Going down a group adds a shell, and the radius rises. Between La and Hf, 14 electrons enter the 4f subshell, which shields badly. The effective charge on the outer electrons rises by about at each step, and the ions shrink: . The ionic potential is charge divided by radius. A larger value means a stronger pull on and a weaker base.
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. The gaps. Gd , Tb , Dy , Ho , Er , Tm , Yb and Lu .
Step 2. The smallest. Holmium has the smallest gap, . Its atomic number is .
Answer: Holmium (atomic number ), with a difference of .
Check. Dysprosium, away, and erbium, away, are also close. Yttrium is found in the same minerals as these heavy lanthanoids, and it follows them in separations.
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