Electronic Configurations of Elements and the Periodic Table

Electronic Configurations of Elements and the Periodic Table

The distribution of electrons into the orbitals of an atom is called its electronic configuration. The position of an element in the Periodic Table is directly linked to the quantum numbers of the last orbital that gets filled. In other words, the long form of the Periodic Table is a direct reflection of the electronic configurations of the elements.

(a) Electronic Configurations in Periods

The period number tells you the value of $n$ for the outermost or valence shell. Each successive period corresponds to the filling of the next higher principal energy level ($n = 1$, then $n = 2$, then $n = 3$, and so on).

The First Period ($n = 1$)

The lowest energy level is the 1s orbital. It can hold a maximum of 2 electrons. So the first period has exactly 2 elements:

• Hydrogen: $1s^1$
• Helium: $1s^2$

When helium is reached, the first shell (K shell) is completely filled.

The Second Period ($n = 2$)

The second period begins with lithium. The third electron enters the 2s orbital. Beryllium has four electrons: $1s^2 2s^2$. Starting from boron, the 2p orbitals begin to fill. The period ends at neon, where the L shell is complete: $2s^2 2p^6$.

The available orbitals in the $n = 2$ level are one 2s orbital and three 2p orbitals — a total of 4 orbitals. Each orbital holds 2 electrons, so the maximum capacity is $4 \times 2 = 8$ electrons. Hence the second period has 8 elements.

The Third Period ($n = 3$)

Sodium starts the third period, with the added electron entering the 3s orbital. Successive filling of 3s and then 3p orbitals gives 8 elements from sodium to argon. The 3d orbitals are not filled in this period because the 4s orbital is actually lower in energy than 3d.

The Fourth Period ($n = 4$)

Potassium begins the fourth period. The added electron fills the 4s orbital. But here something important happens: before the 4p orbital is filled, the 3d orbitals become energetically favourable. This gives rise to the 3d transition series.

The filling order is: $4s \rightarrow 3d \rightarrow 4p$.

• Scandium ($Z = 21$) has the configuration $3d^1 4s^2$
• The 3d orbitals are completely filled at zinc ($Z = 30$): $3d^{10} 4s^2$
• The period ends at krypton with the filling of the 4p orbitals

The $n = 4$ level has 4s, 4p, 4d, and 4f orbitals — but only 4s, 3d, and 4p are filled in this period. That's 1 (4s) + 5 (3d) + 3 (4p) = 9 orbitals, giving $9 \times 2 = 18$ elements.

The Fifth Period ($n = 5$)

This period is similar to the fourth. It begins with rubidium and contains the 4d transition series starting at yttrium ($Z = 39$). The filling order is $5s \rightarrow 4d \rightarrow 5p$. The period ends at xenon with the filling of the 5p orbitals.

Problem 3.2 in this section's Problems group below asks you to justify the presence of 18 elements in the 5th period from exactly these orbital counts.

The Sixth Period ($n = 6$)

This period contains 32 elements. The filling order is:

The 4f-inner transition series (the lanthanoid series) begins at cerium ($Z = 58$) and ends at lutetium ($Z = 71$). The orbitals filled are: one 6s, seven 4f, five 5d, and three 6p — a total of 16 orbitals, giving $16 \times 2 = 32$ elements.

The Seventh Period ($n = 7$)

This period is similar to the sixth. The filling order is:

The 5f-inner transition series (the actinoid series) begins after actinium ($Z = 89$). This period includes most of the man-made radioactive elements. It will end at element with atomic number 118, which will belong to the noble gas family.

(b) Groupwise Electronic Configurations …