
Which of the following spectrochemical series is true?
A.$SC{N^ - } < {F^ - } < N{H_3} < en < CO$
B.$SC{N^ - } < N{H_3} < {F^ - } < en < CO$
C.$SC{N^ - } < {F^ - } < en < N{H_3} < CO$
D.$SC{N^ - } < {F^ - } < en < CO < N{H_3}$
Answer
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Hint:We have studied the spectrochemical series. It is basically a list of ligands arranged on the basis of ligand strength and a list of metal ions based on oxidation number, group and its identity. We must remember that the spectrochemical series ranks the ligands according to the energy difference between their orbitals in their octahedral complexes. The energy difference is generally measured in the spectral transition between the energy levels.
Complete step by step answer:
We must know that the ligands that produce a large splitting are called strong field ligands or strong ligands and those which produce a small splitting are called weak field ligands or weak ligands. In spectrochemical series the ligands are arranged in the increasing order of absorption spectra. The spectrochemical series is given as:
$\begin{array}{*{20}{c}}
{{I^ - } < B{r^ - } < {S^{2 - }} < SC{N^ - } < C{l^ - } < N_3^ - < {F^ - } < NC{O^ - } < O{H^ - } < {C_2}O_4^{2 - } < {O^{2 - }} < {H_2}O < aca{c^ - } < NC{S^ - }}{ < C{H_3}CH < gly < py < N{H_3} < en < bipy < phen < N{O_2} < PP{h_3} < C{N^ - } < CO}
\end{array}$ The weak ligands are generally on the left side and those on the right side in the increasing series are strong ligands. The spectrochemical series is derived from the classification of the ligand’s donor and acceptor abilities.
From the given spectrochemical series we can clearly observe that $SC{N^ - }$ is a weak ligand followed by ${F^ - }$ . Then going higher in the series, we have $N{H_3}$ followed by ethylenediamine $(en)$ followed by $CO$ being the strongest ligand in the series.
Thus, the correct order of ligands according to spectrochemical series is option A.
Note:
We need to remember that the spectrochemical series are defined for metals and ligands both. Ligands that have occupied p orbitals are potentially pi-donors. These types of ligands tend to donate these electrons to the metal along with the sigma bonding electrons, exhibiting stronger metal-ligand interactions and an effective decrease of energy. We must remember that for metal ions the energy difference increases with increasing oxidation number and it decreases down in a group.
Complete step by step answer:
We must know that the ligands that produce a large splitting are called strong field ligands or strong ligands and those which produce a small splitting are called weak field ligands or weak ligands. In spectrochemical series the ligands are arranged in the increasing order of absorption spectra. The spectrochemical series is given as:
$\begin{array}{*{20}{c}}
{{I^ - } < B{r^ - } < {S^{2 - }} < SC{N^ - } < C{l^ - } < N_3^ - < {F^ - } < NC{O^ - } < O{H^ - } < {C_2}O_4^{2 - } < {O^{2 - }} < {H_2}O < aca{c^ - } < NC{S^ - }}{ < C{H_3}CH < gly < py < N{H_3} < en < bipy < phen < N{O_2} < PP{h_3} < C{N^ - } < CO}
\end{array}$ The weak ligands are generally on the left side and those on the right side in the increasing series are strong ligands. The spectrochemical series is derived from the classification of the ligand’s donor and acceptor abilities.
From the given spectrochemical series we can clearly observe that $SC{N^ - }$ is a weak ligand followed by ${F^ - }$ . Then going higher in the series, we have $N{H_3}$ followed by ethylenediamine $(en)$ followed by $CO$ being the strongest ligand in the series.
Thus, the correct order of ligands according to spectrochemical series is option A.
Note:
We need to remember that the spectrochemical series are defined for metals and ligands both. Ligands that have occupied p orbitals are potentially pi-donors. These types of ligands tend to donate these electrons to the metal along with the sigma bonding electrons, exhibiting stronger metal-ligand interactions and an effective decrease of energy. We must remember that for metal ions the energy difference increases with increasing oxidation number and it decreases down in a group.
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