I was wondering why the electronegativity among transition metals are so similar. I can't find anything on the internet, just vague answers about the d orbitals that don't seem to provide a sufficient justification.

  • $\begingroup$ Working on it, but a very general answer is that since there is an xs sublevel and a (x-1)d sublevel that are very close to each other in energy, there are only small differences in the relative stability of different electron configurations of transition metals. Rare earth metals gain much more stability by losing s electrons, but transition metals don't because they still have a partially filled d, but they are not as unstable in the first place because the xs and (x-1d) electrons have a variety of arrangements that are more stable than for a base metal that only has 1 or 2 s electrons. $\endgroup$ – Joseph Hirsch Dec 4 '16 at 4:19
  • $\begingroup$ For example, Chromium: $4s^1$ $3d^5$ is more stable than would be $4s^2$ $3d^4$. By being able to use both sublevels there are more stable configurations "in between". Interesting that the electronegativity tends to rise and then fall across the period, because you start to run out of different ways to configure the electrons. By the way, this gave me a question. Why are do the rare earth metals have lower electonegativity than the d-block. It seems to be due to a lack of overlap in energy level between xs and (x-2)f sublevel, so fewer "in between" configurations. $\endgroup$ – Joseph Hirsch Dec 4 '16 at 4:31

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