# Which set of the three quantum numbers (n, l, ml) corresponds to a 3d orbital?

Dec 20, 2016

n=3, l=2, ${m}_{l}$ = 2,1,0,-1,-2.

#### Explanation:

n is the principle quantum number. It is the number that governs all the other quantum numbers. It limits l: l < n. and ${m}_{l}$ goes from l to -l in unit increments.

Fortunately, it is given to you: 3d tells you that n = 3 for this set of orbitals.

l is the angular momentum quantum number. One way you can look at it is that l is the number of nodes (points of zero density) in the electron wave function. For l =0, there are no nodes os you get one big (for certain values of big) spherical orbital: an s orbital! for l = 1, you get orbitals with two lobes and a node centered at the nucleus. that is the description of a p orbital! for l =2, you get two nodes. For example, the ${d}_{x y}$ orbital has four lobes with nodal planes between the x and y axis and between the -x and y axis. That is the d orbitals.

${m}_{l}$ is the magnetic quantum number and can have values from +l to -l. so for l=0, there is only one possible orbital, ${m}_{l}$ = 0. But for l=1 there are three possible orbitals with ${m}_{l}$ = 1, 0, and -1: that gives us the three p orbitals. For l = 2, ${m}_{l}$ can be 2, 1, 0, -1, or -2, and that gives us the five d orbitals.