To find the number of electrons with a magnetic quantum number equal to zero present in the outermost orbit of \( \text{Co}^{2+} \), we can follow these steps:
### Step 1: Determine the Electron Configuration of Cobalt
Cobalt (Co) has an atomic number of 27. The electron configuration for neutral cobalt is:
\[
\text{Co}: 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^7 \, 4s^2
\]
### Step 2: Adjust for the \( \text{Co}^{2+} \) Ion
When cobalt loses two electrons to form \( \text{Co}^{2+} \), the electrons are removed from the outermost shell first. The 4s electrons are removed before the 3d electrons. Therefore, the electron configuration for \( \text{Co}^{2+} \) becomes:
\[
\text{Co}^{2+}: 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^7
\]
### Step 3: Identify the Outermost Orbitals
The outermost orbitals for \( \text{Co}^{2+} \) are:
- 3s
- 3p
- 3d
### Step 4: Determine the Magnetic Quantum Number
The magnetic quantum number (\( m_l \)) can take specific values based on the type of orbital:
- For the **s orbital** (3s), the magnetic quantum number can only be \( m_l = 0 \).
- For the **p orbital** (3p), there is one orbital with \( m_l = 0 \).
- For the **d orbital** (3d), there is one orbital with \( m_l = 0 \).
### Step 5: Count the Electrons with \( m_l = 0 \)
- In the **3s orbital**, there are 2 electrons (both have \( m_l = 0 \)).
- In the **3p orbital**, there are 2 electrons (1 orbital with \( m_l = 0 \) containing 2 electrons).
- In the **3d orbital**, there are 2 electrons in the orbital with \( m_l = 0 \) (1 orbital with \( m_l = 0 \) containing 2 electrons).
### Step 6: Total Count
Adding these up:
- From 3s: 2 electrons
- From 3p: 2 electrons
- From 3d: 2 electrons
Total number of electrons with \( m_l = 0 \):
\[
2 + 2 + 2 = 6
\]
### Final Answer
The number of electrons with a magnetic quantum number equal to zero present in the outermost orbit of \( \text{Co}^{2+} \) is **6**.
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