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The two electrons have the following set of quantum numbers :
\[
\begin{array}{l}
P=3,2,-2,+\frac{1}{2} \\
Q=3,0,0,+\frac{1}{2}
\end{array}
\]
Which of the following statement is true?
Options:
\[
\begin{array}{l}
P=3,2,-2,+\frac{1}{2} \\
Q=3,0,0,+\frac{1}{2}
\end{array}
\]
Which of the following statement is true?
Solution:
1074 Upvotes
Verified Answer
The correct answer is:
\( P \) has greater energy than \( Q \)
For electron $P$, the values of quantum numbers are $n=3, I=2, m l=-2$ and $s=+1 / 2$. For electron $Q$ quantum numbers are $n=3, I=0, m l=0$ and $s=+1 / 2$. Therefore, from the quantum number values, electron $P$ belongs to $3 d-$ orbital and electron $Q$ to $3 s$ - orbital. Energy of the orbital is $s < p < d < f$.
Hence, energy of electron $P$ is higher than electron $Q$.
Hence, energy of electron $P$ is higher than electron $Q$.
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