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For the configuration 1s^(2)2s^(1), the ...

For the configuration `1s^(2)2s^(1)`, the quantum numbers for the outermost electron are

A

2,1,0,-1/2

B

2,0,0,+1/2

C

2,1,0,+1/2

D

2,0,1,+1/2

Text Solution

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The correct Answer is:
To determine the quantum numbers for the outermost electron in the electron configuration \(1s^2 2s^1\), we will follow these steps: ### Step 1: Identify the outermost electron The given electron configuration is \(1s^2 2s^1\). The outermost electron is the one in the highest energy level, which in this case is in the \(2s\) subshell. ### Step 2: Determine the principal quantum number (\(n\)) The principal quantum number (\(n\)) indicates the energy level of the electron. For the \(2s\) subshell, the value of \(n\) is 2. ### Step 3: Determine the azimuthal quantum number (\(l\)) The azimuthal quantum number (\(l\)) defines the shape of the orbital. For an \(s\) orbital, \(l\) is equal to 0. Therefore, for the \(2s\) electron, \(l = 0\). ### Step 4: Determine the magnetic quantum number (\(m_l\)) The magnetic quantum number (\(m_l\)) can take values from \(-l\) to \(+l\). Since \(l = 0\), the only possible value for \(m_l\) is 0. ### Step 5: Determine the spin quantum number (\(s\)) The spin quantum number (\(s\)) can be either \(+\frac{1}{2}\) or \(-\frac{1}{2}\). We can choose either value, but typically, we can denote it as \(+\frac{1}{2}\) for the outermost electron. ### Summary of Quantum Numbers Thus, the quantum numbers for the outermost electron in the configuration \(1s^2 2s^1\) are: - Principal quantum number (\(n\)): 2 - Azimuthal quantum number (\(l\)): 0 - Magnetic quantum number (\(m_l\)): 0 - Spin quantum number (\(s\)): \(+\frac{1}{2}\) (or \(-\frac{1}{2}\)) ### Final Answer The quantum numbers for the outermost electron are: \(n = 2\), \(l = 0\), \(m_l = 0\), \(s = +\frac{1}{2}\). ---
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