Animated Solution for Chemistry - Coordination Compounds: The species that has a spin-only magnetic moment of $5.9\text{ BM}$, is ($T_d=\text{tetrahedral}$)

The journey to solving this problem begins with a fundamental concept in coordination chemistry: the spin-only magnetic moment. This property gives us a direct window into the electronic structure of a complex, specifically telling us how many unpaired electrons are dancing around the central metal ion.

The relationship between the magnetic moment ($\mu$) and the number of unpaired electrons ($n$) is elegantly captured by the formula:

We are given a magnetic moment of $5.9\text{ BM}$. If we plug this into our equation, we get:

Squaring both sides gives us approximately $35=n(n+2)$. It doesn't take long to see that $n=5$ is the perfect fit ($5\times 7=35$). So, our mission is clear: we need to find the complex that houses exactly five unpaired electrons.

Let's put each option under the microscope, examining the oxidation state of the metal and the nature of the ligands.

1. The Tetracyanonickelate Ion: $[Ni(CN{)}_4{]}^{2-}$ Here, Nickel is in a $+2$ oxidation state, leaving it with a $3d^8$ configuration. The cyanide ion ($C{N}^{-}$) is a notoriously strong field ligand. It forces the electrons to pair up against their will, leaving us with $0$ unpaired electrons ($n=0$). This complex is diamagnetic and adopts a square planar geometry. Not our guy.

2. The Tetrachloridonickelate Ion: $[NiC{l}_4{]}^{2-}$ Again, we have $N{i}^{2+}$ with a $3d^8$ configuration. However, chloride ($C{l}^{-}$) is a weak field ligand. It doesn't have the muscle to force pairing, so the electrons follow Hund's rule, leaving $2$ unpaired electrons ($n=2$). Close, but no cigar.

3. Tetracarbonylnickel: $[Ni(CO{)}_4]$ This one is a classic trap! Nickel is in a $0$ oxidation state, meaning its valence shell is $4s^{2}3d^8$. The carbonyl ($CO$) ligand is incredibly strong. It forces the $4s$ electrons to migrate and pair up in the $3d$ orbitals, resulting in a completely filled $3d^{10}$ configuration. With all orbitals full, there are $0$ unpaired electrons ($n=0$).

4. The Tetrabromidomanganate Ion: $[MnB{r}_4{]}^{2-}$ Finally, we arrive at Manganese. In this complex, Manganese is in a $+2$ oxidation state. A quick look at the periodic table tells us that $M{n}^{2+}$ has a $3d^5$ configuration. The bromide ion ($B{r}^{-}$) is a weak field ligand, meaning it won't cause any forced pairing.

Because there is no forced pairing, all five electrons will singly occupy the five available $3d$ orbitals. This gives us exactly five unpaired electrons ($n=5$).

Let's verify this with our master equation:

The math aligns perfectly with the physical reality. The $[MnB{r}_4{]}^{2-}$ complex is our culprit, proudly displaying a magnetic moment of $5.9\text{ BM}$.