The ground state of sulfur-33 (33S16) has an atomic number of 16, which means it has 16 protons. The ground state electron configuration of sulfur is 1s22s22p63s23p4.

To determine the ground state magnetic moment, we need to consider the spin and orbital angular momentum of the electrons.

In the ground state electron configuration, the 1s and 2s orbitals are fully filled with two electrons each, and the 2p orbital is half-filled with only one electron. Therefore, the total spin angular momentum is given by the spin of the 2p electron:

s = 1/2

The total orbital angular momentum is given by the sum of the individual orbital angular momenta of the electrons in the 3s and 3p orbitals:

l = 0 (for the 3s orbital) + 1 (for each of the three 3p orbitals) = 3

The total angular momentum (J) is given by the vector sum of the spin and orbital angular momenta:

J = |l - s| to |l + s|

J = |3 - 1/2| to |3 + 1/2|

J = 5/2 to 7/2

The ground state magnetic moment (μ) is given by the formula:

μ = gJμB

where g is the Landé g-factor, J is the total angular momentum, and μB is the Bohr magneton.

The Landé g-factor for sulfur is approximately 2.

Using the minimum value of J (5/2), we can calculate the minimum ground state magnetic moment:

μmin = (2)(5/2)(μB)

μmin ≈ 5(μB)

Therefore, the ground state magnetic moment of 33S16 is approximately 5 Bohr magnetons (μB).