Given below are two statements:

Statement (I): In octahedral complexes, when $$\Delta_o \lt P$$ high spin complexes are formed. When $$\Delta_o \gt P$$ low spin complexes are formed.

Statement (II): In tetrahedral complexes because of $$\Delta_t \lt P$$, low spin complexes are rarely formed.

In the light of the above statements, choose the most appropriate answer from the options given below:

Crystal‐field splitting energy ($$\Delta$$) and pairing energy ($$P$$) together decide whether a coordination complex will be high spin (maximum unpaired electrons) or low spin (minimum unpaired electrons).

Case 1: Octahedral complexes
In an octahedral field the $$d$$ orbitals split into a lower $$t_{2g}$$ set and an upper $$e_g$$ set with an energy gap of $$\Delta_o$$.

If $$\Delta_o \lt P$$, it costs less energy to promote an electron from $$t_{2g}$$ to $$e_g$$ than to pair it inside $$t_{2g}$$. Hence electrons occupy the $$e_g$$ orbitals before pairing - a high-spin configuration results.
If $$\Delta_o \gt P$$, promotion is costlier than pairing, so electrons pair up within $$t_{2g}$$ first - a low-spin configuration results.
Therefore Statement I is correct.

Case 2: Tetrahedral complexes
For a tetrahedral field the splitting $$\Delta_t$$ is much smaller: $$\Delta_t \approx \tfrac{4}{9}\Delta_o$$. Numerically, $$\Delta_t$$ is almost always less than the usual pairing energy $$P$$.

Because $$\Delta_t \lt P$$ in nearly every tetrahedral complex, promotion of electrons into the higher-energy $$e$$ set is always cheaper than pairing inside the lower $$t_2$$ set. Thus tetrahedral complexes almost invariably adopt high-spin configurations; low-spin tetrahedral complexes are extremely rare.
Hence Statement II is also correct.

Since both statements are correct, the appropriate option is Option D.