Question

Find the number of strong field ligands from the following-$N{{H}_{3}}, en, C{{l}^{-}}, C{{H}_{3}}CO{{O}^{-}}, C{{N}^{-}}$

Answer 1

Hint: We can estimate the strength of a ligand by their ability to donate electrons and also their metal-ligand orbital overlap. Generally, ligands using their oxygen centres and halogens for overlapping are weaker field ligands and ligands which use carbon or nitrogen are generally stronger field ligands.

Complete answer: Ligands are the species which are attached to the central metal ion and together they form a complex.
We generally divide ligands into two parts depending upon their strength due to electronegativity namely strong field and weak field ligands.
Generally, higher the electronegativity, weaker is the ligand i.e. if a ligand uses its oxygen atom to bind with the central metal atom, it is termed as a weak field ligand and if the ligand uses its carbon, or nitrogen atom for binding with the central metal atom, it is termed as a strong field ligand.
We can arrange the ligands according to their strengths in the form of a series. This series is known as the spectrochemical series-
${{I}^{-}} < B{{r}^{-}} < SC{{N}^{-}} < C{{l}^{-}} < {{F}^{-}} < O{{H}^{-}} < {{H}_{2}}O < NC{{S}^{-}} < Py < N{{H}_{3}} < en < N{{O}_{2}}^{-} < C{{N}^{-}} < CO$
In this series, ligands towards the left (iodide) are weak field ligands and ligands towards right (CO) are stronger field ligands.
In the question, firstly we have ammonia. It is named ‘notorious ligand’ because it forms high spin as well as low-spin complexes depending on the metal and the complex environment. But, as it is a Lewis base, which means it is less electronegative therefore we consider it to be in the stronger side generally.
Next we have ethylenediamine. It lies towards the right in the series and also uses its nitrogen atom to bind with the metal atom therefore ‘en’ is a strong field ligand.
Halogens lie towards the extreme left in the series and chloride is electronegative, therefore it is a weak field ligand.
$C{{H}_{3}}CO{{O}^{-}}$ is a strong field ligand as it can donate its electron.
Cyanide lies towards the extreme right in the spectrochemical series therefore we can say it is a strong field ligand.
Therefore, among $N{{H}_{3}} , en , C{{l}^{-}} , C{{H}_{3}}CO{{O}^{-}} and \text{ }C{{N}^{-}}$, only chloride ion is a weak field ligand and the rest are strong field ligands.

Note: Strong field ligands have a higher splitting and results in higher crystal field stabilisation energy whereas weak field ligands have a lower crystal field splitting. Generally, weak field ligands are known to form high spin complexes and strong field ligands form low spin complexes.