A $d^3$ ion has no decision to make: its three electrons go one to each $t_{2g}$ orbital. The fourth electron has two choices. It can go up into $e_g$, costing $\Delta_o$ in orbital energy, or it can pair with an electron already in $t_{2g}$, costing the **pairing energy** $P$, the extra repulsion of two electrons sharing one orbital. It takes whichever costs less. - If $\Delta_o < P$, climbing is cheaper: the ion is **high spin**, $t_{2g}^3 e_g^1$, with four unpaired electrons. - If $\Delta_o > P$, pairing is cheaper: the ion is **low spin**, $t_{2g}^4 e_g^0$, with two unpaired electrons. The same choice appears at $d^5$, $d^6$ and $d^7$. Below $d^4$ and above $d^7$ there is only one filling: $d^8$, $d^9$ and $d^{10}$ must use $e_g$ whatever the field, because $t_{2g}$ holds only six. ![The crystal field stabilization, in units of the octahedral splitting and before pairing energy is counted, for d4 to d7 in high-spin and low-spin arrangements. Low spin gains far more: 1.6 against 0.6 at d4, 2.0 against 0 at d5, 2.4 against 0.4 at d6, and 1.8 against 0.8 at d7. It pays for that with extra paired electrons.](img_chem_spin_state_cfse) The chart compares the crystal field stabilization of the two fillings before pairing is counted. Low spin always gains more, because it keeps electrons out of the raised $e_g$ set. But count the pairs. At $d^6$, high spin $t_{2g}^4 e_g^2$ has one pair and low spin $t_{2g}^6$ has three: two extra pairs, and a gain of $2.4 - 0.4 = 2.0\,\Delta_o$. Every $d$ count gives the same pattern: **the low-spin filling gains exactly one $\Delta_o$ for each extra pair it makes.** Moving one electron from $e_g$ to $t_{2g}$ gains $0.6 + 0.4 = 1.0\,\Delta_o$ and creates one pair. So the energy advantage of low spin is $$n(\Delta_o - P),$$ where $n$ is the number of extra pairs, one at $d^4$ and $d^7$ and two at $d^5$ and $d^6$. Its sign is the sign of $\Delta_o - P$, which is why the whole decision comes down to comparing two numbers. What makes $\Delta_o$ large? Above all the ligand. The **spectrochemical series**, found by comparing the spectra of complexes of the same metal, ranks ligands by the splitting they cause: $$\mathrm{I^- < Br^- < Cl^- < F^- < OH^- < H_2O < NH_3 < en < NO_2^- < CN^- < CO}.$$ Ligands at the left are **weak-field** and give high-spin complexes of first-row metals; those at the right are **strong-field** and usually give low-spin ones.