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To solve the problem, we need to analyze the given information about the d6 metal ion (M2+) and its complex with aqua ligands. The spin-only magnetic moment is given as 4.90 BM, which we will use to determine the number of unpaired electrons and the geometry of the complex.
Step 1: Calculate the number of unpaired electrons
The formula for calculating the spin-only magnetic moment ($\mu_{s}$) is given by:
$$\mu_{s} = \sqrt{n(n + 2)}$$
where $n$ is the number of unpaired electrons.
Given that the magnetic moment is 4.90 BM, we can set up the equation:
$$4.90 = \sqrt{n(n + 2)}$$
Squaring both sides, we get:
$$24.01 = n(n + 2)$$
This simplifies to the quadratic equation:
$$n^2 + 2n - 24.01 = 0$$
Using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = 2$, and $c = -24.01$, we find:
$$n = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-24.01)}}{2 \cdot 1}$$
$$n = \frac{-2 \pm \sqrt{4 + 96.04}}{2}$$
$$n = \frac{-2 \pm \sqrt{100.04}}{2}$$
$$n = \frac{-2 \pm 10}{2}$$
Calculating the two possible values:
1. $n = \frac{8}{2} = 4$ (valid, as number of unpaired electrons cannot be negative)
2. $n = \frac{-12}{2} = -6$ (not valid)
Thus, we have $n = 4$ unpaired electrons.
Step 2: Determine the geometry of the complex
For a d6 metal ion, the arrangement of electrons can lead to different geometries depending on the ligand field strength. In this case, since we have 4 unpaired electrons, it suggests a tetrahedral geometry. In tetrahedral complexes, the splitting of d-orbitals is such that the energy difference ($\Delta_t$) is less than that in octahedral complexes ($\Delta_0$), which leads to a higher number of unpaired electrons.
Step 3: Calculate the Crystal Field Stabilization Energy (CFSE)
For tetrahedral complexes, the CFSE can be calculated as follows:
$$\text{CFSE} = (-0.6 \Delta_t) \cdot 3 + (0.4 \Delta_t) \cdot 3 + 0 \cdot 0$$
This simplifies to:
$$\text{CFSE} = -0.6 \Delta_t$$
Thus, the final answer is that the geometry of the complex is tetrahedral and the CFSE is $-0.6 \Delta_t$.
