Question

What is the geometry of ${\left[Cu{\left(H_{2}O\right)}_4\right]}^{+2}$?

Answer 1

The geometry of$(Cu(H_{2}O{)}_4{)}^{2+}$ tetrahedral shape.

we can justify this answer by valence bond theory. The 3d orbitals of $C{u}^{2+}$ will have 9 electrons.So the four molecules of $H_{2}O$ occupies the empty 4s and the three 4p orbitals. Which means the molecule $C{u}^{2+}$ will have hybridization state of $s{p}^3$ This corresponds to the tetrahedral shape whereas square planar will have $ds{p}^2$ hybridization state.