# How is the tetrachloridocobaltate(II) anion stable if its crystal field stabilization energy comes to be positive?

I tried to calculate the ligand field stabilisation enthalpy of $$\ce{[CoCl4]^2-}$$ (as shown in the image) but found it to be positive. What could be the reason behind this? How is it stable with positive crystal field stabilization energy? You didnt put the right values. Higher orbitals have energy value of $+0.4 \Delta_t$ and the lower orbitals have the energy value of $-0.6\Delta_t$.

This complex has a coordination number of 4, so it can have tetrahedral or square planar geometry. Tetrahedral complexes are favored where attainment of regular complexes is important. For the tetrahedral complexes $\mathrm{d^0, d^2, d^5, d^7}$, and $\mathrm{d^{10}}$ the configurations are regular.

In tetrahedral complexes, mostly $\Delta t < P\ (\Delta_t \approx 4/9 \Delta_o)$ and hence they generally form high spin complexes. The splitting will be into $3\ \mathrm{t_2}$ (not $\mathrm{t_{2g}}$) higher energy orbitals and $2\ \mathrm e$ (not $\mathrm{e_g}$) lower energy orbitals. The effect of the ligand wont matter here because the value of $\Delta_t$ is small. So the filling will be same for strong field and weak field.

\begin{align}\text{CFSE} &= +0.4(3) -0.6(4)\\ &= -1.2\end{align}

• Thanks for the answer... It was really a silly mistake.. – Abhi Feb 12 '17 at 13:56
• ‘Tetrahedral complexes are favored where attainment of regular complexes is important.’ — This is incorrect. Tetrahedral complexes are formed if the overall complex is better stabilised in a tetrahedral manner than in a square-planar, octahedral (having added ligands), linear (having removed ligands) or any other manner – Jan Feb 12 '17 at 19:43