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AP EAMCET · Chemistry · Structure of Atom

Which of the following sets of quantum numbers is correct for an electron in \(4 \mathrm{f}\) orbital?

  1. A \(\mathrm{n}=3, \ell=2, \mathrm{~m}_1=-2, \mathrm{~m}_{\mathrm{s}}=+1 / 2\)
  2. B \(\mathrm{n}=4, \ell=3, \mathrm{~m}_1=+1, \mathrm{~m}_{\mathrm{s}}=+1 / 2\)
  3. C \(\mathrm{n}=4, \ell=3, \mathrm{~m}_{\mathrm{l}}=+4, \mathrm{~m}_{\mathrm{s}}=+1 / 2\)
  4. D \(\mathrm{n}=4, \ell=4, \mathrm{~m}_1=+4, \mathrm{~m}_{\mathrm{s}}=-1 / 2\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\mathrm{n}=4, \ell=3, \mathrm{~m}_1=+1, \mathrm{~m}_{\mathrm{s}}=+1 / 2\)

Step-by-step Solution

Detailed explanation

For a 4f- orbital, \(n=4\) and \(1=3\) For a given value of \(l\), there can be \((2 l+1)\) values of \(\mathrm{m}_l\) ranging from \(-l\) to \(+l\). Thus, \(\mathrm{m}_l=+1\) is possible. \[ \mathrm{m}_{\mathrm{s}} \text { can be }+\frac{1}{2} \text { or }-\frac{1}{2} \]
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