# 6.4 Electronic structure of atoms (electron configurations)  (Page 5/15)

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Beginning with the transition metal scandium (atomic number 21), additional electrons are added successively to the 3 d subshell. This subshell is filled to its capacity with 10 electrons (remember that for l = 2 [ d orbitals], there are 2 l + 1 = 5 values of m l , meaning that there are five d orbitals that have a combined capacity of 10 electrons). The 4 p subshell fills next. Note that for three series of elements, scandium (Sc) through copper (Cu), yttrium (Y) through silver (Ag), and lutetium (Lu) through gold (Au), a total of 10 d electrons are successively added to the ( n – 1) shell next to the n shell to bring that ( n – 1) shell from 8 to 18 electrons. For two series, lanthanum (La) through lutetium (Lu) and actinium (Ac) through lawrencium (Lr), 14 f electrons ( l = 3, 2 l + 1 = 7 m l values; thus, seven orbitals with a combined capacity of 14 electrons) are successively added to the ( n – 2) shell to bring that shell from 18 electrons to a total of 32 electrons.

## Quantum numbers and electron configurations

What is the electron configuration and orbital diagram for a phosphorus atom? What are the four quantum numbers for the last electron added?

## Solution

The atomic number of phosphorus is 15. Thus, a phosphorus atom contains 15 electrons. The order of filling of the energy levels is 1 s , 2 s , 2 p , 3 s , 3 p , 4 s , . . . The 15 electrons of the phosphorus atom will fill up to the 3 p orbital, which will contain three electrons:

The last electron added is a 3 p electron. Therefore, n = 3 and, for a p -type orbital, l = 1. The m l value could be –1, 0, or +1. The three p orbitals are degenerate, so any of these m l values is correct. For unpaired electrons, convention assigns the value of $+\phantom{\rule{0.2em}{0ex}}\frac{1}{2}$ for the spin quantum number; thus, ${m}_{s}=+\frac{1}{2}.$

Identify the atoms from the electron configurations given:

(a) [Ar]4 s 2 3 d 5

(b) [Kr]5 s 2 4 d 10 5 p 6

(a) Mn (b) Xe

The periodic table can be a powerful tool in predicting the electron configuration of an element. However, we do find exceptions to the order of filling of orbitals that are shown in [link] or [link] . For instance, the electron configurations (shown in [link] ) of the transition metals chromium (Cr; atomic number 24) and copper (Cu; atomic number 29), among others, are not those we would expect. In general, such exceptions involve subshells with very similar energy, and small effects can lead to changes in the order of filling.

In the case of Cr and Cu, we find that half-filled and completely filled subshells apparently represent conditions of preferred stability. This stability is such that an electron shifts from the 4 s into the 3 d orbital to gain the extra stability of a half-filled 3 d subshell (in Cr) or a filled 3 d subshell (in Cu). Other exceptions also occur. For example, niobium (Nb, atomic number 41) is predicted to have the electron configuration [Kr]5 s 2 4 d 3 . Experimentally, we observe that its ground-state electron configuration is actually [Kr]5 s 1 4 d 4 . We can rationalize this observation by saying that the electron–electron repulsions experienced by pairing the electrons in the 5 s orbital are larger than the gap in energy between the 5 s and 4 d orbitals. There is no simple method to predict the exceptions for atoms where the magnitude of the repulsions between electrons is greater than the small differences in energy between subshells.

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