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The number of d – electrons in [Cr(H2O)6...

The number of d – electrons in `[Cr(H_2O)_6]^(3+)` [ATOMIC NO. OF Cr = 24] is

A

2

B

3

C

4.159 g

D

5

Text Solution

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The correct Answer is:
To find the number of d-electrons in the complex ion \([Cr(H_2O)_6]^{3+}\), we will follow these steps: ### Step 1: Determine the oxidation state of chromium in the complex In the complex \([Cr(H_2O)_6]^{3+}\), water (H₂O) is a neutral ligand, which means it does not contribute any charge. Since the overall charge of the complex is +3, the oxidation state of chromium must also be +3. **Hint:** Remember that the charge of the complex is the sum of the oxidation states of the metal and the ligands. ### Step 2: Write the electron configuration of chromium The atomic number of chromium (Cr) is 24. The electron configuration of neutral chromium is: \[ Cr: [Ar] 4s^1 3d^5 \] This means chromium has 1 electron in the 4s subshell and 5 electrons in the 3d subshell. **Hint:** Use the periodic table to find the atomic number and write the electron configuration based on the order of filling. ### Step 3: Adjust the electron configuration for the oxidation state Since we determined that chromium is in the +3 oxidation state, we need to remove electrons from its electron configuration. Electrons are removed first from the 4s subshell before the 3d subshell. From the configuration: - Remove 1 electron from the 4s subshell: \(4s^1\) becomes \(4s^0\). - Remove 2 electrons from the 3d subshell: \(3d^5\) becomes \(3d^3\). Thus, the electron configuration for \([Cr]^{3+}\) becomes: \[ [Ar] 3d^3 \] **Hint:** Remember that when determining the oxidation state, the order of electron removal is important: 4s electrons are removed before 3d electrons. ### Step 4: Count the d-electrons From the adjusted electron configuration \([Ar] 3d^3\), we can see that there are 3 electrons in the 3d subshell. **Hint:** Simply count the number of electrons in the d subshell after adjusting for the oxidation state. ### Conclusion The number of d-electrons in \([Cr(H_2O)_6]^{3+}\) is **3**. **Final Answer:** 3 d-electrons
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