The problem asks for the sum of the spin-only magnetic moments of two manganese coordination complexes: $\left[\mathrm{Mn}(\mathrm{Br})_6\right]^{3-}$ and $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}$. We will calculate the magnetic moment for each complex separately and then add them.

The spin-only magnetic moment ($\mu_s$) is calculated using the formula: $\mu_s = \sqrt{n(n+2)} \text{ B.M.}$ where 'n' is the number of unpaired electrons.

Step 1: Analyze the complex $\left[\mathrm{Mn}(\mathrm{Br})_6\right]^{3-}$

1. Determine the oxidation state of Mn: Let the oxidation state of Mn be 'x'. The bromide ligand (Br⁻) has a charge of -1. The overall charge of the complex is -3. $x + 6(-1) = -3$ $x - 6 = -3$ $x = +3$ So, the manganese is in the +3 oxidation state (Mn³⁺).

2. Determine the electronic configuration of Mn³⁺: The atomic number of Mn is 25. Its electronic configuration is $\left[\mathrm{Ar}\right] 3d^5 4s^2$. To form Mn³⁺, we remove three electrons (two from the 4s orbital and one from the 3d orbital). The electronic configuration of Mn³⁺ is $\left[\mathrm{Ar}\right] 3d^4$.

3. Determine the number of unpaired electrons (n₁): The bromide ligand (Br⁻) is a weak-field ligand. In the presence of a weak-field ligand, the crystal field splitting energy ($\Delta_o$) is small, and the complex will be high-spin (pairing energy P > $\Delta_o$). For a $d^4$ configuration in a high-spin octahedral complex, the electrons will occupy the orbitals to maximize the spin multiplicity. The configuration is $t_{2g}^3 e_g^1$. The electrons in the d-orbitals are arranged as: $t_{2g}$: ↑ ↑ ↑ $e_g$: ↑ The number of unpaired electrons is $n_1 = 4$.

4. Calculate the magnetic moment (μ₁): $\mu_1 = \sqrt{4(4+2)} = \sqrt{4 \times 6} = \sqrt{24} \approx 4.90 \text{ B.M.}$

Step 2: Analyze the complex $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}$

1. Determine the oxidation state of Mn: Let the oxidation state of Mn be 'y'. The cyanide ligand (CN⁻) has a charge of -1. The overall charge of the complex is -3. $y + 6(-1) = -3$ $y - 6 = -3$ $y = +3$ Again, the manganese is in the +3 oxidation state (Mn³⁺), so its electronic configuration is $\left[\mathrm{Ar}\right] 3d^4$.

2. Determine the number of unpaired electrons (n₂): The cyanide ligand (CN⁻) is a strong-field ligand. In the presence of a strong-field ligand, the crystal field splitting energy ($\Delta_o$) is large, and the complex will be low-spin (pairing energy P < $\Delta_o$). For a $d^4$ configuration in a low-spin octahedral complex, the electrons will pair up in the lower energy $t_{2g}$ orbitals first. The configuration is $t_{2g}^4 e_g^0$. The electrons in the d-orbitals are arranged as: $t_{2g}$: ↑↓ ↑ ↑ $e_g$: The number of unpaired electrons is $n_2 = 2$.

3. Calculate the magnetic moment (μ₂): $\mu_2 = \sqrt{2(2+2)} = \sqrt{2 \times 4} = \sqrt{8} \approx 2.83 \text{ B.M.}$

Step 3: Calculate the sum of the magnetic moments

The total sum of the spin-only magnetic moments is $\mu_1 + \mu_2$. $\text{Sum} = \sqrt{24} + \sqrt{8} \approx 4.90 + 2.83 = 7.73 \text{ B.M.}$ Using more precise values: $\sqrt{24} = 4.8989...$ $\sqrt{8} = 2.8284...$ $\text{Sum} = 4.8989... + 2.8284... = 7.7273... \text{ B.M.}$

Rounding to two decimal places, the sum is 7.73. This value lies within the given range of 7.5 to 7.8.