Easy MCQ +4 / -1 PYQ · JEE Mains 2024

The four quantum numbers for the electron in the outer most orbital of potassium (atomic no. 19) are

  1. A $\mathrm{n}=3, l=0, \mathrm{~m}=1, \mathrm{~s}=+\frac{1}{2}$
  2. B $\mathrm{n}=4, l=0, \mathrm{~m}=0, s=+\frac{1}{2}$ Correct answer
  3. C $\mathrm{n}=2, l=0, \mathrm{~m}=0, s=+\frac{1}{2}$
  4. D $\mathrm{n}=4, l=2, \mathrm{~m}=-1, s=+\frac{1}{2}$

Solution

<p>The four quantum numbers for the outermost electron of potassium can be determined by first understanding the electronic configuration of potassium (atomic number 19). The electron configuration is as follows:</p> <p>$1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^1$</p> <p>The outermost electron resides in the 4s orbital. </p> <p>The four quantum numbers are as follows:</p> <p><p>Principal quantum number ($n$): This indicates the energy level or shell number of the electron and for the 4s orbital, $n = 4$.</p></p> <p><p>Azimuthal quantum number ($l$): This indicates the subshell or shape of the orbital. For an $s$ orbital, $l = 0$.</p></p> <p><p>Magnetic quantum number ($m$): This indicates the orientation of the orbital in space. Since there is only one orientation for $s$ orbitals, $m = 0$.</p></p> <p><p>Spin quantum number ($s$): This can be either $+\frac{1}{2}$ or $-\frac{1}{2}$. Typically, the first electron in an orbital has a spin of $+\frac{1}{2}$.</p></p> <p>Thus, the four quantum numbers for the outermost electron of potassium are:</p> <p>$n = 4, \, l = 0, \, m = 0, \, s = +\frac{1}{2}$</p> <p>This matches with Option B:</p> <p>$\mathrm{n}=4, l=0, \mathrm{~m}=0, s=+\frac{1}{2}$</p>

About this question

Subject: Chemistry · Chapter: Atomic Structure · Topic: Bohr's Model

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