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Step-by-Step Solution
Step 1: Determining the Geometry of [Pd(F)(Cl)(Br)(I)]2–
Palladium(II) complexes of the type [PdX4] are usually square planar. Hence,
[Pd(F)(Cl)(Br)(I)]2– is a square planar complex.
Step 2: Counting the Geometrical Isomers
In a square planar arrangement with four different monodentate ligands, three distinct
geometrical isomers are possible. Hence, the value of
n (number of geometrical isomers) is 3.
Step 3: Oxidation State of Iron in [Fe(CN)6]n–6
The given complex transforms into [Fe(CN)6]3–. To verify this, note that each
CN– ligand has a –1 charge. If the overall complex has a –3 charge and there are
6 cyanide ligands, the oxidation state of Fe must be +3. Symbolically:
Fe + 6(–1) = –3
Fe – 6 = –3
Fe = +3
Step 4: Electronic Configuration and Number of Unpaired Electrons
Iron(III), Fe3+, has an electronic configuration of [Ar] 3d5.
Since CN– is a strong field ligand, it forces pairing of electrons in the lower-energy
$t_{2g}$ orbitals. The configuration in an octahedral strong-field complex for
d5 is $t_{2g}^5 e_g^0$, leaving only 1 unpaired electron:
Number of unpaired electrons = 1
Step 5: Calculating Spin-Only Magnetic Moment
The spin-only magnetic moment ($\mu$) is given by the formula
$$
\mu = \sqrt{n(n+2)} \text{ BM}
$$
where n is the number of unpaired electrons. Here, n = 1, so:
$$
\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \text{ BM}
$$
Step 6: Calculating the Crystal Field Stabilisation Energy (CFSE)
For an octahedral complex, the CFSE can be calculated (ignoring pairing energy) using the
distribution of electrons in $t_{2g}$ and $e_g$ levels. Each electron in a $t_{2g}$ orbital
contributes –0.4$\Delta_0$ and each electron in $e_g$ contributes +0.6$\Delta_0$. Since
$t_{2g}^5 e_g^0$:
$$
\text{CFSE} = \bigl(5 \times (-0.4)\bigr)\Delta_0 + \bigl(0 \times 0.6\bigr)\Delta_0 = -2.0\,\Delta_0.
$$
Final Answer
• The number of geometrical isomers, n, for [Pd(F)(Cl)(Br)(I)]2– is 3.
• The spin-only magnetic moment of [Fe(CN)6]3– is 1.73 BM.
• The CFSE of [Fe(CN)6]3– is $-2.0 \Delta_0$.
