Give the number of d electrons (n of d^n) for the central metal ion in (a) [Cr(H2O)6](ClO3)2; (b)[Mn(CN)6]2; (c) [Ru(NO)(en)2Cl]Br

e/ only thing i can't really get

This post was edited by Barcelona1011 on Apr 1 2014 06:31pm

So for each of the coordination complexes, figure out what oxidation state the central metal ion is in. from there, determine number of electrons in outer orbital.

[Cr(H2O)6](ClO3)2 -> Chlorate ion is -1, so Cr(H2O)6 is +2, or Cr+2

Chromium has an electron config of [Ar] 3d5 4s1, and with 2 less electrons, the electron configuration should be [Ar] 3d4 . for transition metals, iirc, the highest s orbital is highest energy/outermost and will be taken from first.



I believe LFT deals with the order of filling d orbitals only (iirc) and therefore you shouldnt have to worry about it.

for ruthenium, make sure when counting d electrons that you count both 3d and 4d if required to

This post was edited by cialda on Apr 1 2014 09:27pm