Question

Write the values of all the quantum numbers of the valence electrons of sodium atom?

Answer 1

For Sodium atom, the Valence electron lies in 3s. ( 1s2 2s2 2p6 3s1 )

For 3s,
n=3
l=0
m=2l+1 = 1
s= +1/2, -1/2

Hope you get your answer.

Answer 2

Answer : The values of all the quantum numbers of the valence electrons of sodium atom are, $(n=3,l=0,m_l=0,m_s=+\frac{1}{2})$

Explanation :

There are 4 quantum numbers :

Principle Quantum Numbers : It describes the size of the orbital. It is represented by n. n = 1,2,3,4....

Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...

Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as $m_l$. The value of this quantum number ranges from $(-l\text{ to }+l)$. When l = 2, the value of $m_l$ will be -2, -1, 0, +1, +2.

Spin Quantum number : It describes the direction of electron spin. This is represented as $m_s$ The value of this is $+\frac{1}{2}$ for upward spin and $-\frac{1}{2}$ for downward spin.

The electronic configuration of sodium atom is, $1s^22s^22p^63s^1$

In the sodium atom the last electron present in $3s^1$.

Now we have to calculate the values of all the quantum numbers of the valence electrons of sodium atom.

For $3s^1$ :

$n=3$       (Because it is in 3rd shell)

$l=0$         (Because it is in 's' orbital)

$m_l=0$     (Because l = 0)

$m_s=+\frac{1}{2}$     (Because of the sign convention)

Therefore, the values of all the quantum numbers of the valence electrons of sodium atom are, $(n=3,l=0,m_l=0,m_s=+\frac{1}{2})$