The spin only magnetic moment of $[MBr_4]^{2-}$ is 5.9 BM. M is:

Mn

The correct answer is Option (3) → Mn.

To determine which metal \( M \) corresponds to the given spin-only magnetic moment of \([MBr_4]^{2-}\) being 5.9 BM (Bohr Magnetons), we can use the formula for the spin-only magnetic moment:
\(\mu_{s} = \sqrt{n(n + 2)} \, \text{BM}\)

where \( n \) is the number of unpaired electrons.

Given:

\( \mu_{s} = 5.9 \, \text{BM} \)

Calculating \( n \):

We can set up the equation:

\(5.9 = \sqrt{n(n + 2)}\)

Squaring both sides:

\((5.9)^2 = n(n + 2)\)

\(34.81 = n^2 + 2n\)

\(n^2 + 2n - 34.81 = 0\)

Solving the Quadratic Equation: Using the quadratic formula:

\(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Here, \( a = 1, b = 2, c = -34.81 \):

\(n = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-34.81)}}{2 \times 1}\)

\(n = \frac{-2 \pm \sqrt{4 + 139.24}}{2}\)

\(n = \frac{-2 \pm \sqrt{143.24}}{2}\)

\(n = \frac{-2 \pm 11.98}{2}\)

Calculating the two potential solutions:

\( n = \frac{9.98}{2} \approx 4.99 \) (approximately 5, which is physically acceptable).

\( n = \frac{-13.98}{2} \) (not valid, as \( n \) cannot be negative).

With \( n = 5 \), we need to identify the metal \( M \) that can exhibit 5 unpaired electrons:

Iron (Fe): \( [Ar] 3d^6 \, 4s^2 \) (can have 4 unpaired in \( Fe^{2+} \) or 5 in \( Fe^{3+} \))

Chromium (Cr): \( [Ar] 3d^5 \, 4s^1 \) (can have 6 in \( Cr^{2+} \), but this is not typical)

Manganese (Mn): \( [Ar] 3d^5 \, 4s^2 \) (5 unpaired in \( Mn^{2+} \))

Cobalt (Co): \( [Ar] 3d^7 \, 4s^2 \) (can have 3 or 4 unpaired).

Conclusion

The metal \( M \) that corresponds to 5 unpaired electrons in the \( [MBr_4]^{2-} \) complex is: Mn (Manganese).