I'd like to know why does cobalt(II), $\ce{Co^2+}$, form $\ce{[Co(H2O)6]^2+}$ complex with water while it forms $\ce{[Co(SCN)4]^2-}$ with the thiocyanate? Why is there $\ce{6H2O}$ but only $\ce{4 SCN-}$? And is the cobalt's coordination number 6 in these cases? Thanks.
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$\begingroup$ Welcome to chemistry.SE! Please take some of your time for a visit at the help center or take a look at this page to enhance the formatting in your future posts. $\endgroup$– M.A.R.Jan 10, 2015 at 21:16
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$\begingroup$ the coordination number is the number of coordinate bonds to the metal ion: six for the hexaaqua complex and four for the tetrathiocyanate complex. as for why I too am dying to know as I haven't been able to find any good explanations on the internet $\endgroup$– bonJan 10, 2015 at 21:50
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$\begingroup$ Water, being neutral, has a stabilizing effect on positive ions, such as $H^+$. Thiocyanate has a negative charge. I think that after 4 ions have clustered around the cobalt, that the charge repel more thiocyanates from bonding to the cobalt. $\endgroup$– LDC3Jan 10, 2015 at 22:06
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$\begingroup$ That's a good thought @LDC3, but if that were all there was to it, wouldn't you expect the thiocyanate complex to have two water ligands as well? However, when you add $SCN^-$ to a solution of hexaaquo cobalt (II), you get $[Co(SCN)_4]^2-$, not $[Co(SCN)_4(H_2O)_2]^2-$. I am still scratching my head on this one .. water and thiocyanate are both weak field ligands, so I would expect the same coordination geometry for both of them. I must be missing something. $\endgroup$– dtmoore1971Jan 11, 2015 at 0:23
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$\begingroup$ Maybe you need to take into account the geometry in addition of charge stabilisation. $\ce {H2O}$ is bended while $\ce {SCN-}$ is linear. I'm not sayin' that's the answer but maybe it should be taken into account :x $\endgroup$– BabounetJan 11, 2015 at 10:24
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2 Answers
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Firstly, in water Co(II) is not exclusively $\ce{[Co(H2O)6]^2+}$, there is also a small amount of tetrahedral $\ce{[Co(H2O)4]^2+}$, 0.08% at room temperature increasing substantially at high temperature. Canadian Journal of Chemistry, 1980, 58(14): 1418-1426.
According to Greenwood's "Chemistry of the elements" 2nd edition page 1131, the more polarizable the ligand, the more the tetrahedral form of Cobalt (II) is preferred since fewer ligands are required to neutralize the metal ion's positive charge.
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I don't know whether it's correct or not but what I think is due to steric factors .
Given the SCN- is pretty large ligand as compare to H2O .
Also there might be some additional stabilisation in water as ligand due to intra atomic hydrogen bonding.
