Given below are two statements: Statement I: In octahedral complexes, when o < P high spin complexes are formed. When o…

🧠 Spin State Rule

For any complex, comparing $\Delta$ to pairing energy $P$:

• $\Delta > P$ → low-spin (electrons pair in lower set rather than promote).
• $\Delta < P$ → high-spin (electrons go to upper set rather than pair).

🗺️ Statement I (Octahedral)

$\Delta_o$ vs $P$:

• $\Delta_o < P$ → HS ✓
• $\Delta_o > P$ → LS ✓

This is the textbook rule. Statement I is correct.

🗺️ Statement II (Tetrahedral)

For Td:

• $\Delta_t \approx (4/9) \Delta_o$ — much smaller than $\Delta_o$.
• For typical first-row TM ions, $\Delta_t < P$ holds almost universally.

→ Td complexes are essentially always HS. Statement II is correct.

⚡ Why Td Almost Always HS

Td has only 4 ligands (vs 6 in oct), and they don't point directly at d-orbitals (closer to gap directions). Smaller orbital overlap → smaller $\Delta_t$. Combined with $\Delta_t \approx 4/9 \Delta_o$, $\Delta_t$ is typically below the pairing energy.

⚠️ Rare LS Td Exceptions

LS Td complexes do exist (with second/third row metals and very strong field ligands), but they are extremely rare. NCERT/JEE accept Statement II as correct.

$\boxed{\text{Answer: (4) Both Statement I and Statement II are correct}}$