= Solution
For the usual \b[octahedral Ni(II) hydration-isomer series], distinguish coordinated <water> from <water> of crystallization:
$$
\boxed{\begin{aligned}
&[\mathrm{Ni(H_2O)_6}]\mathrm{Cl_2},\\
&[\mathrm{NiCl(H_2O)_5}]\mathrm{Cl}\cdot\mathrm{H_2O},\\
&\mathrm{cis}\! -[\mathrm{NiCl_2(H_2O)_4}]\cdot2\mathrm{H_2O},\\
&\mathrm{trans}\! -[\mathrm{NiCl_2(H_2O)_4}]\cdot2\mathrm{H_2O}.
\end{aligned}}
$$
The first two <coordination spheres> have no cis/trans choice because they contain zero or one <chloride>. The last hydration class has two <geometrical isomers>: the <chlorides> occupy adjacent or opposite octahedral positions. Thus there are \b[three hydration classes and four structures when cis/trans isomerism is included]. Every formula retains two <chlorides> and six <waters> overall; the <coordination number> remains six. The schematic in B6(c) displays the distinct octahedral connectivities. This enumeration states the usual six-coordinate interpretation, rather than treating unspecified changes in <coordination number> as further members of the same elementary series.
Back to article page
