I am trying to solve a pset from MIT $5.111$ class. The question is which of vanadium and molybdenium has the largest radius. The right answer (given in the pset answers) is: molybdenum.

In class, we have only seen that the radius across a period gets smaller, and across a column gets bigger.

However, this knowledge doesn't allow us to conclude (because $r_V > r_{Cr}$ but $r_{Cr} < r_{Mo}$).

What kind of reasoning allows us to conclude ?

  • 1
    $\begingroup$ Why not obscure your question even more? I don't care whatever "pset from MIT 5.111 class" is. You need to actually write a question. Not necessarily in original form, but asking why radius of Mo is bigger then Cr in title, would tell whatever the thing is about. $\endgroup$
    – Mithoron
    Commented Jun 21 at 11:27

1 Answer 1


It is hard to derive the answer from basic principles, as it kind of relies on knowledge of empirical data patterns.

  • The relative radius contraction across the period is relatively small between the period neighbors (except for the first two or three elements).
  • The relative radius expansion across the group is relatively big between the periods 4 and 5.
  • Therefore, it can be assumed from these patterns that $(r_\ce{Mo} - r_\ce{Cr}) \gt (r_\ce{V} - r_\ce{Cr})$ and therefore $r_\ce{Mo} \gt r_\ce{V}$.
  • OTOH, due the lanthanide contraction, the differences of the radius (and of chemical properties) are much smaller between the periods 5 and 6, therefore there can be implied that $r_\ce{W} \lt r_\ce{Nb}$ as $r_\ce{W} - r_\ce{Mo} \lt r_\ce{Nb} - r_\ce{Mo} $.

Note that there are pattern glitches and shift in trends, so the above empirical rules are just the general guidance and are not true for all cases.

For more, see Wikipedia links


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