Question:
Find the value of \( n, \ell, m, \) and \( s \) for the 19th electron of a Cr atom.
Find the value of \( n, \ell, m, \) and \( s \) for the 19th electron of a Cr atom.
Updated On: Apr 9, 2026
- \( n = 3; \ell = 2; m = 1; s = +\frac{1}{2} \)
- \( n = 4; \ell = 0; m = 0; s = +\frac{1}{2} \)
- \( n = 2; \ell = 1; m = 1; s = -\frac{1}{2} \)
- \( n = 3; \ell = 2; m = 0; s = 0 \)
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The Correct Option is B
Solution and Explanation
textbf{Step 1: Understanding the electron configuration.}
The 19th electron of Chromium (Cr) will occupy the 3d orbital. The electron configuration for Chromium is: \[ 1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1 \] So, the 19th electron will be in the 3d orbital with quantum numbers \( n = 3 \), \( \ell = 2 \), and the spin quantum number is \( s = +\frac{1}{2} \). \[ \boxed{n = 3; \ell = 2; m = 1; s = +\frac{1}{2}} \]
The 19th electron of Chromium (Cr) will occupy the 3d orbital. The electron configuration for Chromium is: \[ 1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1 \] So, the 19th electron will be in the 3d orbital with quantum numbers \( n = 3 \), \( \ell = 2 \), and the spin quantum number is \( s = +\frac{1}{2} \). \[ \boxed{n = 3; \ell = 2; m = 1; s = +\frac{1}{2}} \]
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