Question

What could be the fourth quantum number for one of the electrons in the 4p energy sublevel of bromine?

Answer 1

Answer : The fourth quantum number for one of the electrons in the 4p energy sublevel of bromine is, $m_s=+\frac{1}{2}\text{ or }-\frac{1}{2}$

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.

In the bromine atom the one electron in 4p energy sublevel, $4p^1$.

Now we have to calculate the values of all the quantum numbers.

For $4p^1$ :

$n=4$       (Because it is in 4rd shell)

$l=1$         (Because it is in 'p' orbital)

$m_l=-1,0,1$     (Because l = 1)

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

Therefore, the fourth quantum number for one of the electrons in the 4p energy sublevel of bromine is, $m_s=+\frac{1}{2}\text{ or }-\frac{1}{2}$