Question

Calculate the difference in net stabilisation energies (in kJ mol$^{-1}$) of $Fe^{2+}$ complexes with $CN^-$ ligands ($\Delta_0$ = 25000 cm$^{-1}$) and with $H_2O$ ligands ($\Delta_0$ = 10000 cm$^{-1}$). Given that the pairing energy, P = 15000 cm$^{-1}$ for $Fe^{2+}$ ion for both the complexes.

(Given that h = 6 x $10^{-34}$ J sec., Avogadro Number = 6 x $10^{23}$, Speed of light in vaccum = 3 x $10^8$ m sec$^{-1}$).

Answer 1

-280.80

Answer 2

The problem asks to calculate the difference in net stabilization energies (in kJ mol$^{-1}$) of two $Fe^{2+}$ complexes: one with $CN^-$ ligands and another with $H_2O$ ligands.

1. Determine the electronic configuration of $Fe^{2+}$: Iron (Fe) has atomic number 26. Its electronic configuration is $[Ar] 3d^6 4s^2$. $Fe^{2+}$ is formed by losing the two 4s electrons, so its electronic configuration is $[Ar] 3d^6$. We need to consider the $d^6$ configuration in an octahedral crystal field.

2. Calculate the Crystal Field Stabilization Energy (CFSE) for each complex: The general formula for CFSE in an octahedral complex, including pairing energy, is: $\text{CFSE}_{\text{net}} = [-0.4 \times n_{t2g} + 0.6 \times n_{eg}] \Delta_0 + mP$ where: $n_{t2g}$ = number of electrons in $t_{2g}$ orbitals $n_{eg}$ = number of electrons in $e_g$ orbitals $\Delta_0$ = crystal field splitting energy $P$ = pairing energy $m$ = number of extra electron pairs formed in the complex compared to the free ion.

For a free $d^6$ ion, according to Hund's rule, there are 4 unpaired electrons and 1 paired electron (1 pair).

a) $Fe^{2+}$ complex with $CN^-$ ligands (e.g., $[Fe(CN)_6]^{4-}$): Given: $\Delta_0 = 25000 \text{ cm}^{-1}$, $P = 15000 \text{ cm}^{-1}$. Since $\Delta_0 > P$ ($25000 \text{ cm}^{-1} > 15000 \text{ cm}^{-1}$), $CN^-$ is a strong field ligand, and the complex will be low spin. For a $d^6$ low spin configuration in an octahedral field, all 6 electrons will occupy the $t_{2g}$ orbitals before any go into $e_g$. Electronic configuration: $t_{2g}^6 e_g^0$. $n_{t2g} = 6$, $n_{eg} = 0$. In $t_{2g}^6 e_g^0$, all 6 electrons are paired, resulting in 3 pairs. Number of extra pairs ($m$) = (Pairs in complex) - (Pairs in free ion) = 3 - 1 = 2.

$\text{CFSE}_{\text{CN}} = [-0.4 \times 6 + 0.6 \times 0] \Delta_0 + 2P$ $\text{CFSE}_{\text{CN}} = -2.4 \Delta_0 + 2P$ $\text{CFSE}_{\text{CN}} = -2.4 \times (25000 \text{ cm}^{-1}) + 2 \times (15000 \text{ cm}^{-1})$ $\text{CFSE}_{\text{CN}} = -60000 \text{ cm}^{-1} + 30000 \text{ cm}^{-1}$ $\text{CFSE}_{\text{CN}} = -30000 \text{ cm}^{-1}$

b) $Fe^{2+}$ complex with $H_2O$ ligands (e.g., $[Fe(H_2O)_6]^{2+}$): Given: $\Delta_0 = 10000 \text{ cm}^{-1}$, $P = 15000 \text{ cm}^{-1}$. Since $\Delta_0 < P$ ($10000 \text{ cm}^{-1} < 15000 \text{ cm}^{-1}$), $H_2O$ is a weak field ligand, and the complex will be high spin. For a $d^6$ high spin configuration in an octahedral field, electrons will singly occupy all $t_{2g}$ and $e_g$ orbitals before pairing up. Electronic configuration: $t_{2g}^4 e_g^2$. $n_{t2g} = 4$, $n_{eg} = 2$. In $t_{2g}^4 e_g^2$, there is 1 pair in $t_{2g}$ and 4 unpaired electrons (2 in $t_{2g}$ and 2 in $e_g$). So, 1 pair in total. Number of extra pairs ($m$) = (Pairs in complex) - (Pairs in free ion) = 1 - 1 = 0.

$\text{CFSE}_{\text{H2O}} = [-0.4 \times 4 + 0.6 \times 2] \Delta_0 + 0P$ $\text{CFSE}_{\text{H2O}} = [-1.6 + 1.2] \Delta_0$ $\text{CFSE}_{\text{H2O}} = -0.4 \Delta_0$ $\text{CFSE}_{\text{H2O}} = -0.4 \times (10000 \text{ cm}^{-1})$ $\text{CFSE}_{\text{H2O}} = -4000 \text{ cm}^{-1}$

3. Calculate the difference in net stabilization energies: Difference = $\text{CFSE}_{\text{CN}} - \text{CFSE}_{\text{H2O}}$ Difference = $(-30000 \text{ cm}^{-1}) - (-4000 \text{ cm}^{-1})$ Difference = $-30000 \text{ cm}^{-1} + 4000 \text{ cm}^{-1}$ Difference = $-26000 \text{ cm}^{-1}$

4. Convert the difference from cm$^{-1}$ to kJ mol$^{-1}$: We use the relationship $E = h c \bar{\nu}$, where $\bar{\nu}$ is the wavenumber in m$^{-1}$. Given $\bar{\nu}$ in cm$^{-1}$, we convert it to m$^{-1}$ by multiplying by 100 ($1 \text{ m} = 100 \text{ cm}$). Energy per molecule in Joules: $E (\text{J/molecule}) = h \times c \times (\text{wavenumber in cm}^{-1} \times 100)$ $E (\text{J/molecule}) = (6 \times 10^{-34} \text{ J s}) \times (3 \times 10^8 \text{ m s}^{-1}) \times (-26000 \text{ cm}^{-1} \times 100 \text{ cm/m})$ $E (\text{J/molecule}) = (18 \times 10^{-26} \text{ J m}) \times (-2.6 \times 10^6 \text{ m}^{-1})$ $E (\text{J/molecule}) = -46.8 \times 10^{-20} \text{ J/molecule}$

Now, convert to energy per mole using Avogadro's number ($N_A = 6 \times 10^{23} \text{ mol}^{-1}$): $E (\text{J/mol}) = E (\text{J/molecule}) \times N_A$ $E (\text{J/mol}) = (-46.8 \times 10^{-20} \text{ J/molecule}) \times (6 \times 10^{23} \text{ molecules/mol})$ $E (\text{J/mol}) = (-46.8 \times 6) \times 10^{(-20+23)} \text{ J/mol}$ $E (\text{J/mol}) = -280.8 \times 10^3 \text{ J/mol}$ $E (\text{J/mol}) = -280800 \text{ J/mol}$

Finally, convert to kJ mol$^{-1}$: $E (\text{kJ/mol}) = \frac{-280800 \text{ J/mol}}{1000 \text{ J/kJ}}$ $E (\text{kJ/mol}) = -280.8 \text{ kJ/mol}$

The difference in net stabilization energies is -280.8 kJ mol$^{-1}$.