Which of the following is correct order of ligand field strength?

We need to find the correct order of ligand field strength (spectrochemical series) for the given ligands.

First, recall the spectrochemical series.

The spectrochemical series arranges ligands in order of increasing field strength (ability to cause crystal field splitting). The standard order is:

$$I^- < Br^- < S^{2-} < Cl^- < N_3^- < F^- < OH^- < C_2O_4^{2-} < H_2O < NCS^- < CH_3CN < NH_3 < en < NO_2^- < CN^- < CO$$

Next, extract the relevant ligands and their order.

From the options, the ligands to be ordered are: $$S^{2-}$$, $$C_2O_4^{2-}$$, $$NH_3$$, $$en$$ (ethylenediamine), and $$CO$$.

From the spectrochemical series:

$$S^{2-}$$ — A weak field ligand. It has large, diffuse orbitals and acts as a pi-donor, which reduces the crystal field splitting energy.

$$C_2O_4^{2-}$$ (oxalate) — A moderate field ligand. It is a bidentate ligand but still a pi-donor.

$$NH_3$$ — A moderately strong field ligand. It is a strong sigma-donor with no pi-bonding ability.

$$en$$ (ethylenediamine) — A strong field ligand. It is a bidentate sigma-donor, slightly stronger than $$NH_3$$ due to the chelate effect and stronger sigma donation.

$$CO$$ — The strongest field ligand. It is both a strong sigma-donor and an excellent pi-acceptor (pi-backbonding), which greatly increases crystal field splitting.

Now, write the correct increasing order of ligand field strength.

$$S^{2-} < C_2O_4^{2-} < NH_3 < en < CO$$

From this, match with the options.

Option A: $$CO < en < NH_3 < C_2O_4^{2-} < S^{2-}$$ — This is the reverse (decreasing) order. Incorrect.

Option B: $$S^{2-} < C_2O_4^{2-} < NH_3 < en < CO$$ — This matches the spectrochemical series. Correct.

Option C: $$NH_3 < en < CO < S^{2-} < C_2O_4^{2-}$$ — Incorrect ordering.

Option D: $$S^{2-} < NH_3 < en < CO < C_2O_4^{2-}$$ — Places oxalate above CO, which is incorrect.

The correct answer is Option B: $$S^{2-} < C_2O_4^{2-} < NH_3 < en < CO$$.