Among the following series of transition metal ions the one where all metal ions have `3d^2` electronic configuration is
A
`Ti^(2+), V^(3+), Cr^(4+), Mn^(5+)`
B
`Ti^(3+), V^(2+), Cr^(3+), Mn^(4+)`
C
`Ti^(+), V^(4+), Cr^(6+), Mn^(7+)`
D
`Ti^(4+), V^(3+), Cr^(2+), Mn^(3+)`
Text Solution
AI Generated Solution
The correct Answer is:
To solve the question of which series of transition metal ions have the electronic configuration of `3d^2`, we will follow these steps:
### Step 1: Identify the Ground State Electronic Configurations
First, we need to determine the ground state electronic configurations of the transition metals involved. The relevant transition metals and their atomic numbers are:
1. **Titanium (Ti)** - Atomic number 22
- Electronic configuration: `Ar 3d^2 4s^2`
2. **Vanadium (V)** - Atomic number 23
- Electronic configuration: `Ar 3d^3 4s^2`
3. **Chromium (Cr)** - Atomic number 24
- Electronic configuration: `Ar 3d^5 4s^1`
4. **Manganese (Mn)** - Atomic number 25
- Electronic configuration: `Ar 3d^5 4s^2`
### Step 2: Determine the Required Electron Removal for `3d^2`
Next, we will analyze how many electrons need to be removed from each metal to achieve the `3d^2` configuration.
1. **Titanium (Ti)**:
- Starting configuration: `Ar 3d^2 4s^2`
- To achieve `3d^2`, remove 2 electrons from the `4s` orbital.
- Resulting ion: `Ti^2+` with configuration `Ar 3d^2`.
2. **Vanadium (V)**:
- Starting configuration: `Ar 3d^3 4s^2`
- To achieve `3d^2`, remove 1 electron from `3d` and 2 from `4s`.
- Total electrons removed: 3.
- Resulting ion: `V^3+` with configuration `Ar 3d^2`.
3. **Chromium (Cr)**:
- Starting configuration: `Ar 3d^5 4s^1`
- To achieve `3d^2`, remove 1 electron from `4s` and 3 from `3d`.
- Total electrons removed: 4.
- Resulting ion: `Cr^4+` with configuration `Ar 3d^2`.
4. **Manganese (Mn)**:
- Starting configuration: `Ar 3d^5 4s^2`
- To achieve `3d^2`, remove 2 electrons from `4s` and 3 from `3d`.
- Total electrons removed: 5.
- Resulting ion: `Mn^5+` with configuration `Ar 3d^2`.
### Step 3: Conclusion
All the ions that we have analyzed (Ti^2+, V^3+, Cr^4+, Mn^5+) have the `3d^2` electronic configuration. Therefore, the series of transition metal ions where all metal ions have `3d^2` electronic configuration is:
**Answer: A (Ti^2+, V^3+, Cr^4+, Mn^5+)**
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NARENDRA AWASTHI ENGLISH|Exercise Match the column|1 Videos
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