Question

Which of the following is the correct order of increasing field strength of ligands to form coordination compounds ?

• A. $SCN^- < F^- < C_2O_4^{2-} < CN^-$
• B. $SCN^- < F^- < CN^- < C_2O_4^{2-}$
• C. $F^- < SCN^- < C_2O_4^{2-} < CN^-$
• D. $CN^- < C_2O_4^{2-} < SCN^- < F^-$

Answer 1

The Correct Option is A

In coordination chemistry, the strength of ligands is typically discussed in terms of the spectrochemical series. This series ranks ligands based on their ability to split the d-orbitals of a central metal ion in a coordination compound. The greater the splitting, the stronger the ligand.

The typical order from weaker field ligands to stronger field ligands is as follows:

\(I^- < Br^- < S^{2-} < SCN^- < Cl^- < F^- < OH^- < C_2O_4^{2-} < H_2O < NH_{3} < en < CN^- < CO\)

From the options provided, we need to identify the correct order of increasing field strength of ligands:

1. \(SCN^- < F^- < C_2O_4^{2-} < CN^−\)
2. \(SCN^- < F^- < CN^- < C_2O_4^{2-}\)
3. \(F^- < SCN^- < C_2O_4^{2-} < CN^-\)
4. \(CN^- < C_2O_4^{2-} < SCN^- < F^−\)

Based on the spectrochemical series, the correct increasing order of field strength for the ligands provided in the options is:

\(SCN^- < F^- < C_2O_4^{2-} < CN^−\)

Explanation:

• \(SCN^−\) is a relatively weak ligand compared to \(F^-\) and \(C_2O_4^{2-}\).
• \(F^−\) is stronger than \(SCN^−\) but weaker than \(C_2O_4^{2-}\).
• \(C_2O_4^{2-}\) is stronger than both \(SCN^−\) and \(F^−\), but weaker than \(CN^−\).
• \(CN^−\) is a strong field ligand and the strongest among the choices given.

 

Therefore, the correct answer is option 1: \(SCN^- < F^- < C_2O_4^{2-} < CN^−\).