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What is the set of four quantum numbers for last electron of aluminum and beryllium?

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In atoms and electrons, quantum numbers are used to describe the orbitals in atoms, mathematical representations of the likelihood of finding electrons in a given space. There are four quantum numbers:

Principal quantum number (n) - is generally related to the size of the orbital, or the distance of the electron from the nucleus. This also correspond to the energy level. Allowed values from n are 1, 2, 3, ... and where an increasing number of n refer to an increasing size or distance.

Angular quantum number (l) - describes the shape of the orbital. Orbitals can be spherical (s), can contain two lobes with a node in the middle (p), or four equally-spaced lobes with a node in the center (d), among others. These shapes correspond to a number: s is 0, p is 1, d is 2, f is 3, g is 4, ... Allowed numbers for this quantum number are 0 up to n-1. Hence, if the principal quantum number is 1, the only allowed shape is an s-orbital.

Magnetic quantum number (ml) - describes the orientation of an orbital in space, and may take on values from -l to +l. The d orbital, with l=2, can therefore have 5 different orientations each represented by a magnetic quantum number of -2, -1, 0, 1, 2.

Spin quantum number (ms) - describes the spin state of an electron in this case, and can only take on values of +1/2 and -1/2.

These quantum numbers may be derived from the electron configuration corresponding to an electron - for instance, the last electron. The electron configuration of aluminum is:

[Ne] 3s2 3p1,

Hence, the last electron is in 3p1. The n quantum number then is 3, as this is the 3rd energy level. The l quantum number is 1, which corresponds to the p orbital. The p orbital has three orientations - -1, 0, +1, and there is only one electron - the -1 orientation is conventionally filled out first (though these have equal likelihood to be filled out and these numbers are simplified representations). Moreover, if it is the first electron in a sub-shell, +1/2 is typically assigned. Hence, the last electron of aluminum will have the following QN:

n = 3, l = 1, ml = -1, ms = +1/2

Meanwhile, the last electron of beryllium has the following electron configuration:

[He] 2s2,

Hence, the n quantum number is 2, and the l quantum number is 0. There is only one possible ml for the s orbital and that is 0 - there are no other possible unique orientations for a sphere in 3D space. There are two electrons in this orbital - the first gets +1/2 spin, and the second -1/2. Hence, the last electron will have the following QN: