Question

Identify the period 2 element that is described by the ionization data below. \(\mathrm{M}(\mathrm{g}) \rightarrow \mathrm{M}^{+} 1 \mathrm{e}^{-} \quad \mathrm{IE}(1)=1.40 \times 10^{3} \mathrm{~J} / \mathrm{mol}\) \(\mathrm{M}^{+}(\mathrm{g}) \rightarrow \mathrm{M}^{2+} 1 \mathrm{e}^{-} \mathrm{IE}(2)=2.86 \times 10^{3} \mathrm{~J} / \mathrm{mol}\) \(\mathrm{M}^{2+}(\mathrm{g}) \rightarrow \mathrm{M}^{3+} 1 \mathrm{e}^{-} \mathrm{IE}(3)=4.58 \times 10^{3} \mathrm{~J} / \mathrm{mol}\) \(\mathrm{M}^{3+}(\mathrm{g}) \rightarrow \mathrm{M}^{4+} 1 \mathrm{e}^{-} \mathrm{IE}(4)=7.48 \times 10^{3} \mathrm{~J} / \mathrm{mol}\) \(\mathrm{M}^{4+}(\mathrm{g}) \rightarrow \mathrm{M}^{5+} 1 \mathrm{e}^{-} \mathrm{IE}(5)=9.44 \times 10^{3} \mathrm{~J} / \mathrm{mol}\) \(\mathrm{M}^{5+}(\mathrm{g}) \rightarrow \mathrm{M}^{6+} 1 \mathrm{e}^{-} \mathrm{IE}(6)=5.33 \times 10^{4} \mathrm{~J} / \mathrm{mol}\) \(\mathrm{M}^{6+}(\mathrm{g}) \rightarrow \mathrm{M}^{7+} 1 \mathrm{e}^{-} \mathrm{IE}(7)=6.44 \times 10^{4} \mathrm{~J} / \mathrm{mol}\)

Short answer

The period 2 element described by the ionization data is nitrogen, as there is a significant jump in ionization energy between IE(5) and IE(6), indicating that it has 5 valence electrons in its outer electron shell. Nitrogen's electron configuration is \(1s^2 2s^2 2p^3\), showing the presence of 5 valence electrons in the second energy level (n=2).

Step-by-step solution

Analyze the given ionization energies

Observe the ionization energy values and identify where there is a significant change. The jump in ionization energy indicates that the next electron removed is from a more stable electron shell. IE(1) = 1.40 x 10^3 J/mol IE(2) = 2.86 x 10^3 J/mol IE(3) = 4.58 x 10^3 J/mol IE(4) = 7.48 x 10^3 J/mol IE(5) = 9.44 x 10^3 J/mol IE(6) = 5.33 x 10^4 J/mol (significant jump) IE(7) = 6.44 x 10^4 J/mol

Relate the ionization energies to electron configuration

According to the data, the first five ionization energies show a gradual increase, but there is a significant jump in energy between IE(5) and IE(6). This indicates that M has 5 valence electrons in its outer electron shell.

Find the element with 5 valence electrons in period 2

In period 2, the element with 5 valence electrons is nitrogen (atomic number 7). Its electron configuration is 1s² 2s² 2p³, showing that the outer shell (n=2) has five electrons. Therefore, the period 2 element described by the ionization data is nitrogen.

Key concepts

Period 2 Elements

Period 2 elements are the second row of the periodic table, stretching from lithium ( Li ) to neon ( Ne ). These elements are characterized by having two electron shells, referred to as energy levels. The first shell can hold up to 2 electrons and the second shell can hold up to 8 electrons.
Understanding these elements is crucial because their chemical properties change as you move from left to right. These changes are due to the increasing number of protons in the nucleus, which attracts electrons more strongly. This results in elements that exhibit diverse chemical behaviors. For example, lithium is a highly reactive metal, while neon is an inert gas.
In this exercise, the focus is on identifying a specific element in period 2 using ionization energy data. The analysis centers on recognizing changes in ionization energy to infer the number of valence electrons, helping to pinpoint the element in question.

Electron Configuration

Electron configuration refers to the arrangement of electrons in an atom's shells and subshells. It provides valuable insight into an element's reactivity, stability, and overall properties. Each period on the periodic table corresponds to a new, higher principal energy level.
For period 2 elements, electrons fill the 1s and then the 2s orbitals, followed by the 2p orbitals. For example, nitrogen, the element identified in this exercise, has the electron configuration of 1s^2 2s^2 2p^3 . It indicates that there are five electrons in the outer shell (the second shell).
The electron configuration helps us understand why nitrogen has five valence electrons, as well as the significant jump in ionization energy after the removal of the fifth electron. This pattern suggests that this electron removal leads to electron configurations in an inner, more stable shell.

Valence Electrons

Valence electrons are of paramount importance as they determine an element's chemical properties and reactivity. These electrons reside in the outermost shell of an atom. For period 2 elements, this shell is the second energy level.
In general, the number of valence electrons ranges from 1 to 8 as you move across a period. Elements strive for a full valence shell, typically 8 electrons (known as the octet rule), to achieve maximum stability. This is why elements form bonds—to gain, lose, or share electrons to complete their valence shell.
In the exercise, the significant ionization energy jump indicates the removal of the sixth electron involves breaking into a more stable, completed shell. Knowing this, we identify that nitrogen must have initially had 5 valence electrons. Thus, understanding the concept of valence electrons is crucial for analyzing ionization energy trends and predicting elemental behavior.