The four quantum numbers for the valence shel | vedclass

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Question

(B) The atomic number of sodium $(Na)$ is $11$. The electronic configuration is $1s^2, 2s^2, 2p^6, 3s^1$. The valence electron is in the $3s$ orbital.

Vedclass · Accepted Answer

$n = 3, l = 0, m = 0, s = +\frac{1}{2}$

Vedclass · Answer

(B) The atomic number of sodium $(Na)$ is $11$. The electronic configuration is $1s^2, 2s^2, 2p^6, 3s^1$. The valence electron is in the $3s$ orbital. For the $3s$ orbital,the principal quantum number $(n) = 3$. Since it is an $s$-orbital,the azimuthal quantum number $(l) = 0$. Consequently,the magnetic quantum number $(m) = 0$. For the single electron in the $s$-orbital,the spin quantum number $(s) = +\frac{1}{2}$. Thus,the set of quantum numbers is $(n = 3, l = 0, m = 0, s = +\frac{1}{2})$.

The four quantum numbers for the valence shell electron or last electron of sodium $(Z = 11)$ are:

Similar Questions

Given below are the quantum numbers for $4$ electrons.
$A. \; n = 3, l = 2, m_l = 1, m_s = +1/2$
$B. \; n = 4, l = 1, m_l = 0, m_s = +1/2$
$C. \; n = 4, l = 2, m_l = -2, m_s = -1/2$
$D. \; n = 3, l = 1, m_l = -1, m_s = +1/2$
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