Question

Among the following complexes, the one which shows zero crystal field stabilization energy (CFSE)?
(A) $[Mn(H_{2}O{)}_6{]}^{3+}$
(B) $[Fe(H_{2}O{)}_6{]}^{3+}$
(C) $[Co(H_{2}O{)}_6{]}^{2+}$
(D) $[Co(H_{2}O{)}_6{]}^{3+}$

Answer 1

Hint: The complex in which the metal ion has five or ten electrons in their valence shell has zero CFSE value. So, first find the oxidation state of the metal and then find the number of electrons present in the valence shell.

Complete answer:
We know that the CFSE is the acronym for crystal field stabilization energy. Here, we need to find the complex in which the CFSE value is zero.
- We can see that all the given complexes are octahedral complexes. $H_{2}O$ (water) molecules is a weak ligand. So, it produces weak ligand fields and thus the complex formed will be high spin.
- Now, it is proven from theoretical calculations that the complexes in which the metals have 5 or 10 valence electrons show zero CFSE value.
- Mn in $[Mn(H_{2}O{)}_6{]}^{3+}$ has an oxidation number of +3. We know that the electronic configuration of $M{n}^{3+}$ is $[Ar]3d^4$. So, there are 4 electrons in its valence shell. Thus, its CFSE will not be zero.
- Fe in $[Fe(H_{2}O{)}_6{]}^{3+}$ has an oxidation number of +3. We know that the electronic configuration of $F{e}^{3+}$ is $[Ar]3d^5$. So, there are 5 electrons in its valence shell. Thus, its CFSE value will be zero.
This can be explained by the following equation.
     $$CFSE=(0.6{\mathrm{\Delta }}_o\times e_g)-(0.4{\mathrm{\Delta }}_o\times t_{2g})$$
Here $e_g$ is the number of electrons in doubly degenerate orbitals. $t_{2g}$ is the number of electrons in triply degenerate orbitals.
- We know that for $F{e}^{3+}$, the configuration of 5 electrons is $t_{2g}^3{e}_g^2$ . Thus, the equation will become
     $$CFSE=(0.6{\mathrm{\Delta }}_o\times 2)-(0.4{\mathrm{\Delta }}_o\times 3)=0$$
- Co metal has oxidation state of +2 and +3 in $[Co(H_{2}O{)}_6{]}^{2+}$ and $[Co(H_{2}O{)}_6{]}^{3+}$ respectively. The electronic configuration of $C{o}^{2+}$ and $C{o}^{3+}$ is $[Ar]3d^7\text{ and [Ar]3}{\text{d}}^6$ respectively. Thus, we can say that $[Fe(H_{2}O{)}_6{]}^{3+}$ has zero CFSE value.

So the option B is the correct answer.

Note:
Remember that when a transition metal atom loses an electron to become a positive ion, it loses the electron from 4s orbital first and then the 3d electrons are lost. However, according to Aufbau principle 3d orbital has higher energy than 4s orbital.