Q48 / 180 · KCET 2016 · Chemistry
Consider the following sets of quantum numbers. Which of the below setting is not permissible arrangement of electrons in an atom?
- A\(\mathrm{n}=4,1=0, \mathrm{~m}=0, \mathrm{~s}=-2\)
- B\(\mathrm{n}=5,1=3, \mathrm{~m}=0, \mathrm{~s}=+2\)
- C\(\mathrm{n}=3,1=2, \mathrm{~m}=-2, \mathrm{~s}=-2\)
- Correct answerD\(\mathrm{n}=3,1=2, \mathrm{~m}=-3, \mathrm{~s}=+2\)
Solution
The value of azimuthal quantum number \( l=n-1 \).
The value of magnetic quantum number \( m_{1}=-l \) to \( +l \)
\begin{array}{|l|l|l|l|l|} \hline Quantum Numbers & \( \mathrm{n} \) & \( 1 \) & \( m_{1} \) & \( m_{s} \) \\ \hline PossibleValues & \( 3 \) & \( 0,1,2 \) & \( -2,-1,0,+1,+2 \) & \( +\frac{1}{2} \) and \( -\frac{1}{2} \) \\ \hline \end{array}
Therefore, \( m_{1}=-3 \) is not possible for \( n=3 \).
The value of magnetic quantum number \( m_{1}=-l \) to \( +l \)
\begin{array}{|l|l|l|l|l|} \hline Quantum Numbers & \( \mathrm{n} \) & \( 1 \) & \( m_{1} \) & \( m_{s} \) \\ \hline PossibleValues & \( 3 \) & \( 0,1,2 \) & \( -2,-1,0,+1,+2 \) & \( +\frac{1}{2} \) and \( -\frac{1}{2} \) \\ \hline \end{array}
Therefore, \( m_{1}=-3 \) is not possible for \( n=3 \).
Hence, the correct answer is option: 'D'.