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In [Cr(O(2))(NH(3))(4)H(2)O]Cl(2) oxidat...

In `[Cr(O_(2))(NH_(3))_(4)H_(2)O]Cl_(2)` oxidation number of Cr is +3 then oxygen will be the form:

A

dioxo

B

peroxo

C

superoxo

D

oxo

Text Solution

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The correct Answer is:
To determine the form of oxygen in the complex `[Cr(O2)(NH3)4H2O]Cl2` where the oxidation number of chromium (Cr) is +3, we can follow these steps: ### Step 1: Identify the components of the complex The complex consists of: - Cr (Chromium) - O2 (Oxygen) - NH3 (Ammonia) - H2O (Water) - Cl2 (Chloride) ### Step 2: Assign oxidation numbers - The oxidation number of Cr is given as +3. - NH3 and H2O are neutral ligands, so their contribution to the overall charge is 0. - Cl in Cl2 has an oxidation number of -1, and since there are two Cl atoms, the total contribution from Cl is -2. ### Step 3: Set up the equation Let the oxidation number of O in O2 be x. The overall charge of the complex is neutral (0). Therefore, we can set up the equation based on the contributions of each component: \[ +3 + 2x + 0 + 0 - 2 = 0 \] ### Step 4: Simplify the equation Now, simplify the equation: \[ 3 + 2x - 2 = 0 \] \[ 2x + 1 = 0 \] ### Step 5: Solve for x Now, solve for x: \[ 2x = -1 \] \[ x = -\frac{1}{2} \] ### Step 6: Determine the form of oxygen The oxidation number of oxygen in O2 is -1/2. In coordination chemistry, this indicates that the oxygen is in the superoxo form (O2^−), where each oxygen atom has an oxidation state of -1/2. ### Conclusion Thus, the oxygen in the complex `[Cr(O2)(NH3)4H2O]Cl2` is in the superoxo form.
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