The electronic configuration of lawrencium ($\ce{Lr}$) is $\mathrm{[Rn] 7s^2 5f^{14} 6d^1}$. As its last electron enters the $\mathrm{6d}$ sub shell, it should be a part of $\mathrm{d}$-block elements, but in the periodic table it is given as a part of $\mathrm{f}$-block elements. Why?

I noticed that if we place lawrencium in $\mathrm{d}$-block elements the symmetry and similarity in properties of many elements will be broken. Is that the reason to place it like this? Or something else? Is lawrencium an exception? How? Please explain.

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    $\begingroup$ As noted in en.wikipedia.org/wiki/Block_(periodic_table), "Not everyone agrees on the exact membership of each set of elements." and "Sometimes the f-block is considered to be between groups 2 and 3 instead of 3 and 4, which means that lanthanum and actinium are f-block elements, and lutetium and lawrencium are d-block elements instead located under yttrium." Ultimately, what you call it doesn't really matter much at all. $\endgroup$ – Jon Custer Jun 25 '20 at 15:45

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