Question

The crystal field stabilization energy (CFSE) of the complex $\left[Mn(CN)_6\right]^{4-}$ is:

(Given: $\Delta_o$ is the splitting energy of octahedral field, P is pairing energy)

• A. 0
• B. $-2\Delta_o$
• C. $-3\Delta_o + 2P$
• D. $-2\Delta_o + 2P$

Answer 1

Answer: (D)

-2Δo + 2P

Answer 2

To determine the Crystal Field Stabilization Energy (CFSE) of the complex $\left[Mn(CN)_6\right]^{4-}$, we follow these steps:

1. Determine the oxidation state of the central metal ion.
   Let the oxidation state of Mn be x. The cyanide ligand (CN-) has a charge of -1. The overall charge of the complex is -4.
   $x + 6(-1) = -4$
   $x - 6 = -4$
   $x = +2$
   So, the central metal ion is $Mn^{2+}$.

2. Determine the electronic configuration of the central metal ion.
   Manganese (Mn) has atomic number 25, with an electronic configuration of $[Ar] 3d^5 4s^2$.
   For $Mn^{2+}$, two electrons are removed from the 4s orbital.
   Therefore, the electronic configuration of $Mn^{2+}$ is $3d^5$.

3. Identify the nature of the ligand and geometry.
   The ligand is $CN^-$, which is a strong field ligand. Strong field ligands cause a large crystal field splitting ($\Delta_o$ is large, meaning $\Delta_o > P$, where P is the pairing energy).
   The complex has 6 ligands, indicating an octahedral geometry.

4. Determine the crystal field splitting and electron filling.
   In an octahedral crystal field, the five d-orbitals split into two sets:

   • Three lower energy orbitals: $t_{2g}$ (with energy $-0.4\Delta_o$ relative to the barycenter).
   • Two higher energy orbitals: $e_g$ (with energy $+0.6\Delta_o$ relative to the barycenter).

   Since $CN^-$ is a strong field ligand, the 5 electrons in the $d^5$ system will first fill the lower energy $t_{2g}$ orbitals, pairing up before occupying the higher energy $e_g$ orbitals.

   • 1st electron goes to $t_{2g}$.
   • 2nd electron goes to $t_{2g}$.
   • 3rd electron goes to $t_{2g}$.
   • 4th electron pairs with the 1st electron in $t_{2g}$ (costs P energy).
   • 5th electron pairs with the 2nd electron in $t_{2g}$ (costs P energy).

   Thus, the electronic configuration in the crystal field is $(t_{2g})^5 (e_g)^0$.

5. Calculate the CFSE.
   CFSE is calculated as the sum of the energy contributions from electrons in the $t_{2g}$ and $e_g$ orbitals, plus any pairing energy penalty.
   $\text{CFSE} = (n_{t_{2g}} \times \text{Energy of } t_{2g}) + (n_{e_g} \times \text{Energy of } e_g) + (N_p \times P)$
   Where:

   • $n_{t_{2g}}$ = number of electrons in $t_{2g}$ orbitals = 5
   • $n_{e_g}$ = number of electrons in $e_g$ orbitals = 0
   • Energy of $t_{2g} = -0.4\Delta_o$
   • Energy of $e_g = +0.6\Delta_o$
   • $N_p$ = number of additional electron pairs formed in the complex compared to the free gaseous metal ion.

   For a $d^5$ free ion, all 5 electrons are unpaired (0 pairs).
   In the low spin $(t_{2g})^5 (e_g)^0$ configuration, there are two electron pairs ($\uparrow\downarrow \uparrow\downarrow \uparrow$).
   So, the number of additional pairs formed is $2 - 0 = 2$.

   Now, substitute these values into the CFSE formula:
   $\text{CFSE} = (5 \times -0.4\Delta_o) + (0 \times +0.6\Delta_o) + (2 \times P)$
   $\text{CFSE} = -2.0\Delta_o + 0 + 2P$
   $\text{CFSE} = -2\Delta_o + 2P$