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