The ground state electron configuration of chromium (Cr, $ Z = 24 $) is:

$$ \text{Cr: } [\text{Ar}]\,3d^5\,4s^1 $$

Let's break down the configuration:

The noble gas core $[\text{Ar}]$ represents:

$1s^2$ (with $l=0$)

$2s^2$ (with $l=0$)

$2p^6$ (with $l=1$)

$3s^2$ (with $l=0$)

$3p^6$ (with $l=1$)

Electrons with azimuthal quantum number $l = 1$ are found in $p$ orbitals:

$2p^6$: 6 electrons

$3p^6$: 6 electrons

Total $p$-electrons: $6 + 6 = 12$.

Electrons with azimuthal quantum number $l = 2$ are found in $d$ orbitals:

$3d^5$: 5 electrons

Thus, there are 12 electrons with $l = 1$ and 5 electrons with $l = 2$.

The correct option is Option B.