Medium MCQ +4 / -1 PYQ · JEE Mains 2025

The correct order of the following complexes in terms of their crystal field stabilization energies is :

  1. A $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\right]^{2+}<\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{2+}<\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}<\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}$ Correct answer
  2. B $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\right]^{2+}<\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{2+}<\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}<\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$
  3. C $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{2+}<\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}<\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\right]^{2+}<\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}$
  4. D $\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}<\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}<\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{2+}<\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\right]^{2+}$

Solution

<p>The crystal field stabilization energy (CFSE) of a complex depends on two main factors:</p> <p><p><strong>Charge/Oxidation State of the Central Metal Atom</strong>: Higher oxidation states generally lead to greater crystal field splitting energy (Δ). </p></p> <p><p><strong>Field Strength of the Ligand</strong>: Stronger field ligands also result in greater splitting. Additionally, chelating ligands typically increase the field strength.</p></p> <p>For octahedral complexes, the CFSE is calculated as:</p> <p>$ \text{CFSE} = [-0.4 \, t_{2g} + 0.6 \, e_g ] \, \Delta_\circ $</p> <p>Considering the complexes:</p> <p><p><strong>$[\text{Co(en)}_3]^{3+}$</strong>:</p></p> <p><p>Co exists as Co$^{3+}$ with electronic configuration $t_{2g}^6 e_g^0$.</p></p> <p><p>CFSE = $-2.4 \, (\Delta_0)_1$.</p></p> <p><p><strong>$[\text{Co(NH}_3)_6]^{3+}$</strong>:</p></p> <p><p>Co also as Co$^{3+}$ with $t_{2g}^6 e_g^0$.</p></p> <p><p>CFSE = $-2.4 \, (\Delta_0)_2$.</p></p> <p><p><strong>$[\text{Co(NH}_3)_6]^{2+}$</strong>:</p></p> <p><p>Co with Co$^{2+}$ configuration $t_{2g}^5 e_g^2$.</p></p> <p><p>CFSE = $-0.8 \, (\Delta_0)_3$.</p></p> <p><p><strong>$[\text{Co(NH}_3)_4]^{2+}$</strong>:</p></p> <p><p>Co as Co$^{2+}$ with different configuration $e^4 t_2^3$.</p></p> <p><p>CFSE = $-1.2 \, \Delta_t$, where $\Delta_t = \frac{4}{9} (\Delta_0)_3$.</p></p> <p>The order of crystal field splitting energies, based on our criteria, is: </p> <p>$ \Delta_t < (\Delta_0)_3 < (\Delta_0)_2 < (\Delta_0)_1 $ </p> <p>Therefore, the CFSEs of these complexes compare as follows: </p> <p><p>$[\text{Co(en)}_3]^{3+}$ has the highest CFSE due to the $\Delta_0$ of the strongest field ligand and oxidation state.</p></p> <p><p>$[\text{Co(NH}_3)_6]^{3+}$ follows with similarly high CFSE due to Co$^{3+}$.</p></p> <p><p>$[\text{Co(NH}_3)_6]^{2+}$ has a moderate CFSE.</p></p> <p><p>$[\text{Co(NH}_3)_4]^{2+}$ has the lowest CFSE with $\Delta_t$.</p></p> <p>This hierarchy of CFSE is determined by both the oxidation state of the cobalt metal and the field strength of the ligands involved.</p>

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Subject: Chemistry · Chapter: Coordination Compounds · Topic: Ligands and Coordination Number

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