For the octahedral complexes of `Fe^(3+)` in `SCN^(-)` (thiocyanato -S) and in `CN^(-)` ligand environments, the difference between the spin only magnetic moments in Bohr magnetons (when approximated to the nearest integer) is [atomic number of `Fe = 26` ]

Answer

To solve the problem, we need to determine the difference in the spin-only magnetic moments of the octahedral complexes of `Fe^(3+)` with `SCN^(-)` and `CN^(-)` ligands.

### Step-by-Step Solution:

1. **Identify the oxidation state of Iron in both complexes**:
- For both complexes, `Fe` is in the +3 oxidation state. This means that the electronic configuration of `Fe^(3+)` is `3d^5`.

2. **Determine the number of unpaired electrons in the `Fe(SCN)6^(3-)` complex**:
- `SCN^(-)` is a weak field ligand, which does not cause pairing of electrons.
- In the `3d` subshell, there are 5 electrons:
- Configuration: ↑ ↑ ↑ ↑ ↑ (5 unpaired electrons)
- Therefore, the number of unpaired electrons (n) = 5.

3. **Calculate the magnetic moment for `Fe(SCN)6^(3-)`**:
- The formula for the spin-only magnetic moment (μ) is given by:
\[ \mu = \sqrt{n(n + 2)} \text{ Bohr Magnetons} \]
- Substituting n = 5:
\[ \mu = \sqrt{5(5 + 2)} = \sqrt{5 \times 7} = \sqrt{35} \approx 5.92 \text{ BM} \approx 5.8 \text{ BM} \text{ (approximated to nearest integer)} \]

4. **Determine the number of unpaired electrons in the `Fe(CN)6^(3-)` complex**:
- `CN^(-)` is a strong field ligand, which causes pairing of electrons.
- In the `3d` subshell, the 5 electrons will pair up:
- Configuration: ↑↓ ↑↓ ↑ (1 unpaired electron)
- Therefore, the number of unpaired electrons (n) = 1.

5. **Calculate the magnetic moment for `Fe(CN)6^(3-)`**:
- Using the same formula:
\[ \mu = \sqrt{n(n + 2)} \text{ Bohr Magnetons} \]
- Substituting n = 1:
\[ \mu = \sqrt{1(1 + 2)} = \sqrt{3} \approx 1.73 \text{ BM} \text{ (approximated to nearest integer)} \]

6. **Calculate the difference in magnetic moments**:
- The difference between the magnetic moments of the two complexes is:
\[ \Delta \mu = \mu_{Fe(SCN)6^{3-}} - \mu_{Fe(CN)6^{3-}} \]
- Substituting the values:
\[ \Delta \mu = 5.8 \text{ BM} - 1.73 \text{ BM} \approx 4.07 \text{ BM} \]
- Approximating to the nearest integer gives us:
\[ \Delta \mu \approx 4 \text{ BM} \]

### Final Answer:
The difference between the spin-only magnetic moments in Bohr magnetons (when approximated to the nearest integer) is **4 BM**.