Question

For an octahedral complex of Fe$^{3+}$ the value of C.F.S.E. (Crystal field stabilization energy) is "x$\Delta$o" if magnetic moment of complex is $\sqrt{35}$ B.M. Find out the value of |x|. ($\Delta$o is crystal field splitting energy in octahedral field)

Answer 1

0

Answer 2

To find the value of |x| for the given complex, we follow these steps:

1. Determine the oxidation state and electronic configuration of the central metal ion:
   The complex is of Fe$^{3+}$.
   The electronic configuration of Fe (Z=26) is [Ar] 3d$^6$ 4s$^2$.
   When Fe loses 3 electrons to form Fe$^{3+}$, the two 4s electrons and one 3d electron are removed.
   So, the electronic configuration of Fe$^{3+}$ is [Ar] 3d$^5$.

2. Determine the number of unpaired electrons using the magnetic moment:
   The spin-only magnetic moment ($\mu$) is given by the formula:
   $\mu = \sqrt{n(n+2)}$ B.M.
   where 'n' is the number of unpaired electrons.
   Given $\mu = \sqrt{35}$ B.M.
   $\sqrt{n(n+2)} = \sqrt{35}$
   $n(n+2) = 35$
   $n^2 + 2n - 35 = 0$
   Solving the quadratic equation:
   $(n+7)(n-5) = 0$
   Since the number of unpaired electrons cannot be negative, $n=5$.
   Thus, the Fe$^{3+}$ complex has 5 unpaired electrons.

3. Determine the spin state and electron distribution in the octahedral field:
   For a d$^5$ system, if there are 5 unpaired electrons, it must be a high-spin complex.
   In an octahedral field, the d-orbitals split into a lower energy $t_{2g}$ set and a higher energy $e_g$ set.
   For a high-spin d$^5$ configuration, the 5 electrons are distributed as:
   $t_{2g}^3 e_g^2$ (three electrons in the $t_{2g}$ orbitals and two electrons in the $e_g$ orbitals, all unpaired).

4. Calculate the Crystal Field Stabilization Energy (CFSE):
   The energy of $t_{2g}$ orbitals is lowered by $0.4\Delta_o$ each, and the energy of $e_g$ orbitals is raised by $0.6\Delta_o$ each.
   CFSE = (number of electrons in $t_{2g} \times -0.4\Delta_o$) + (number of electrons in $e_g \times +0.6\Delta_o$)
   CFSE = $(3 \times -0.4\Delta_o) + (2 \times +0.6\Delta_o)$
   CFSE = $-1.2\Delta_o + 1.2\Delta_o$
   CFSE = $0\Delta_o$

5. Find the value of |x|:
   Given that the CFSE is "x$\Delta$o".
   From our calculation, CFSE = $0\Delta_o$.
   Therefore, x = 0.
   The value of |x| = |0| = 0.