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The complex showing a spin -magnetic mom...

The complex showing a spin -magnetic momnet of `2.82 BM` is .

A

`Ni(CO)_(4)`

B

`[NiCI_(4)]^(2-)`

C

`Ni(PPh_(3))_(4)`

D

`[Ni(CN)_(4)]^(2-)`

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To solve the problem of identifying the complex that shows a spin magnetic moment of 2.82 Bohr magnetons (BM), we can follow these steps: ### Step 1: Understand Magnetic Moment The magnetic moment (μ) of a complex can be calculated using the formula: \[ \mu = \sqrt{n(n + 2)} \] where \( n \) is the number of unpaired electrons. ### Step 2: Determine the Value of n Given that the magnetic moment is 2.82 BM, we can set up the equation: \[ 2.82 = \sqrt{n(n + 2)} \] Squaring both sides gives: \[ (2.82)^2 = n(n + 2) \] Calculating \( (2.82)^2 \): \[ 7.9524 = n(n + 2) \] ### Step 3: Solve the Quadratic Equation Rearranging the equation: \[ n^2 + 2n - 7.9524 = 0 \] Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 2, c = -7.9524 \): \[ n = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot (-7.9524)}}{2 \cdot 1} \] Calculating the discriminant: \[ n = \frac{-2 \pm \sqrt{4 + 31.8096}}{2} \] \[ n = \frac{-2 \pm \sqrt{35.8096}}{2} \] Calculating \( \sqrt{35.8096} \): \[ n = \frac{-2 \pm 5.983}{2} \] This gives two possible values for \( n \): 1. \( n = \frac{3.983}{2} \approx 1.9915 \) (approximately 2) 2. \( n = \frac{-7.983}{2} \) (not a valid solution since n cannot be negative) Thus, \( n \) is approximately 2. ### Step 4: Identify the Complex Now that we know there are 2 unpaired electrons, we can analyze the types of ligands: - Weak field ligands (like Cl⁻) do not cause pairing of electrons, allowing for unpaired electrons, while strong field ligands (like CN⁻, CO) tend to cause pairing. Given that we have 2 unpaired electrons, we can conclude that the complex is likely to contain a weak field ligand. ### Step 5: Conclusion The complex that shows a spin magnetic moment of 2.82 BM is likely one that contains nickel in the +2 oxidation state with weak field ligands like chloride (Cl⁻). ### Final Answer The complex showing a spin magnetic moment of 2.82 BM is likely a nickel(II) complex with chloride ligands. ---

To solve the problem of identifying the complex that shows a spin magnetic moment of 2.82 Bohr magnetons (BM), we can follow these steps: ### Step 1: Understand Magnetic Moment The magnetic moment (μ) of a complex can be calculated using the formula: \[ \mu = \sqrt{n(n + 2)} \] where \( n \) is the number of unpaired electrons. ...
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