Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
The effective charge on the 5s and 5p electrons of is , and the radius of is . Each lanthanoid adds one proton and one 4f electron, and the 4f electron cancels of the proton's charge. Use the model to find the effective charge and the radius of (atomic number 64). By how many picometres does your radius differ from the measured ?
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 steps. .
Step 2. The effective charge. It rises by at each step. .
Step 3. The radius. .
Step 4. The comparison. .
Answer: , , and a difference of .
Check. The model radius is a little larger than the measured one. With the effective charge would be and the radius , which is too small. The value is a good choice.
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