High Spin vs Low Spin Complexes: Strong Field and Weak Field Ligands

Look ONLY at the ligand. If the ligand is strong-field (CN-, CO, NH3, NO2-, en), the electrons pair up in the lower t2g orbitals, so the complex is LOW SPIN. If the ligand is weak-field (I-, Br-, Cl-, F-, H2O, C2O4 2-), electrons stay spread out and unpaired, so the complex is HIGH SPIN. You do not need to do the full orbital diagram to answer most NEET questions. Just check: strong = low spin, weak = high spin.

A strong-field ligand pushes hard on the d-orbitals. This makes the energy gap (called delta, the splitting energy) very large. When the gap is bigger than the energy needed to pair two electrons in one orbital (the pairing energy P), the electrons prefer to pair up in the low t2g orbitals rather than jump up to the high eg orbitals. When delta is large (delta > P), you get low spin. When delta is small (delta < P), it is cheaper to go up, so electrons stay unpaired = high spin.

No. The high spin / low spin choice only appears for d4, d5, d6 and d7 metal ions in an octahedral field. For d1, d2, d3 there is only one way to fill (the lower t2g fills first, no choice). For d8, d9, d10 the lower set is already full so there is again only one arrangement. So a NEET question about high vs low spin will almost always involve Fe2+/Fe3+/Co3+/Mn3+/Co2+ type ions, which are d4 to d7.

Not exactly. High spin always has MORE unpaired electrons, so it is more strongly paramagnetic. Low spin has FEWER unpaired electrons. Low spin becomes fully diamagnetic (zero unpaired) only when all electrons pair, like d6 low spin (t2g^6 eg^0), which is why [Co(NH3)6]3+ is diamagnetic. But a low spin d5 like [Fe(CN)6]3- still has 1 unpaired electron, so it is still paramagnetic. So say it carefully: low spin = fewer unpaired = weaker paramagnetism (sometimes zero).

Both are Co3+ (d6) and both are octahedral, so the ONLY difference is the ligand. NH3 is a strong-field ligand, so [Co(NH3)6]3+ is low spin: t2g^6 eg^0, zero unpaired, diamagnetic. F- is a weak-field ligand, so [CoF6]3- is high spin: t2g^4 eg^2, four unpaired, paramagnetic. This exact pair was asked in NEET 2024. Same metal ion, different ligand strength = different spin state = different magnetic behaviour.

Yes, for NEET treat all tetrahedral complexes as high spin. In a tetrahedral field the splitting delta_t is small (delta_t = 4/9 of delta_o for the same metal and ligand). This small gap is almost never bigger than the pairing energy, so electrons never pair up early. So tetrahedral complexes are high spin by default. The high spin / low spin decision really only matters for octahedral complexes.

High spin has the maximum number of unpaired electrons (electrons stay spread out). Low spin has the minimum number of unpaired electrons (electrons pair up in the lower orbitals). Weak-field ligands give high spin; strong-field ligands give low spin.

Strong-field ligands give low spin. The common NEET ones are CO, CN-, NO2- (N-bonded), NH3, and en (ethylenediamine). Remember them as the strong end of the spectrochemical series.

Weak-field ligands give high spin. The common NEET ones are I-, Br-, Cl-, SCN- (S-bonded), F-, and H2O (H2O is weak to moderate). These are the weak end of the spectrochemical series.

Only for d4, d5, d6 and d7 in octahedral complexes. For d1, d2, d3 and d8, d9, d10 there is only one possible filling, so the terms high spin and low spin do not apply.

Low spin. Fe3+ is d5, and CN- is a strong-field ligand, so the electrons pair in t2g giving t2g^5 eg^0 with just 1 unpaired electron. It is low spin but still weakly paramagnetic.