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Cr in [Cr(NH3)6] Br3 has number of unpai...

Cr in `[Cr(NH_3)_6] Br_3` has number of unpaired electron :

A

4

B

3

C

1

D

2

Text Solution

AI Generated Solution

The correct Answer is:
To determine the number of unpaired electrons in the complex ion \([Cr(NH_3)_6]Br_3\), we will follow these steps: ### Step 1: Determine the oxidation state of Chromium (Cr) In the complex \([Cr(NH_3)_6]Br_3\): - Ammonia (NH₃) is a neutral ligand, contributing 0 to the oxidation state. - Bromide (Br⁻) has a charge of -1. Since there are three bromide ions, they contribute a total of -3. Let the oxidation state of chromium be \(x\). The equation for the oxidation state is: \[ x + 0 - 3 = 0 \] \[ x = +3 \] ### Step 2: Write the electronic configuration of Cr in its +3 oxidation state The atomic number of chromium (Cr) is 24, and its electronic configuration is: \[ \text{Cr: } [Ar] 4s^1 3d^5 \] When chromium is in the +3 oxidation state, it loses three electrons. The electrons are removed first from the 4s orbital and then from the 3d orbital: - Remove 1 electron from 4s: \(4s^0\) - Remove 2 electrons from 3d: \(3d^3\) Thus, the electronic configuration of \(Cr^{3+}\) is: \[ \text{Cr}^{3+}: [Ar] 3d^3 \] ### Step 3: Determine the number of unpaired electrons In the \(3d\) subshell, the distribution of electrons is as follows: - The \(3d\) subshell can hold a maximum of 10 electrons, but in \(Cr^{3+}\), there are 3 electrons. - According to Hund's rule, the electrons will occupy the orbitals singly before pairing up. The configuration for \(3d^3\) will be: - \(3d_{xy}^1\) - \(3d_{xz}^1\) - \(3d_{yz}^1\) This means all three electrons are unpaired. ### Conclusion The number of unpaired electrons in \([Cr(NH_3)_6]Br_3\) is **3**. ### Final Answer **3 unpaired electrons.** ---
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