Question

Compare the elements \(\mathrm{Na}, \mathrm{B}, \mathrm{Al},\) and \(\mathrm{C}\) with regard to the following properties: (a) Which has the largest atomic radius? (b) Which has the most negative electron attachment enthalpy? (c) Place the elements in order of increasing ionization energy.

Short answer

(a) Na has the largest atomic radius. (b) C has the most negative electron attachment enthalpy. (c) Increasing ionization energy order is Na < Al < B < C.

Step-by-step solution

Determine Atomic Radius Trend

Atomic radius generally increases down a group and decreases across a period in the periodic table. Sodium (Na) and aluminum (Al) are in the same period, with Na being to the left of Al. Boron (B) and carbon (C) are also in that period, with B to the left of C. Therefore, Na has the largest atomic radius among these elements.

Determine Electron Attachment Enthalpy Trend

Electron attachment enthalpy becomes more negative as you move from left to right across a period, meaning elements further right are more likely to gain electrons. Among Na, B, Al, and C, carbon (C) is farthest right in the period, indicating it has the most negative electron attachment enthalpy.

Determine Ionization Energy Trend

Ionization energy generally increases across a period and decreases down a group. Arranging the elements Na, B, Al, and C, we find Na has the lowest ionization energy, and C has the highest. The order from lowest to highest is Na < Al < B < C.

Key concepts

Atomic Radius

The concept of atomic radius refers to the size of an atom. It’s essentially the distance from the nucleus of an atom to its outermost electron shell. When comparing elements based on their position on the periodic table, certain trends become apparent:

• The atomic radius increases as you move down a group. This occurs because each subsequent element has an additional electron shell.
• Conversely, the atomic radius decreases from left to right across a period. This happens because as the number of protons increases with elements to the right, the positive charge of the nucleus increases, pulling the electron shells closer.

Given these trends, for the elements in question—sodium (\(\mathrm{Na}\)) and aluminum (\(\mathrm{Al}\))—we see that sodium, being to the left of aluminum in the same period, possesses a larger atomic radius. Likewise, boron (\(\mathrm{B}\)) and carbon (\(\mathrm{C}\)) also fit this trend, with boron having a larger atomic radius than carbon, but smaller than sodium.

Electron Attachment Enthalpy

Electron attachment enthalpy is a measure of the energy change when an electron is added to a neutral atom in the gas phase to form a negative ion. A more negative electron attachment enthalpy means that the atom releases more energy when gaining an electron, implying it's more favorable for the atom to gain that electron.

• As you move from left to right across a period, the electron attachment enthalpy generally becomes more negative. This is because elements towards the right have nearly filled electron shells, making them more eager to accept additional electrons to achieve a stable electron arrangement.
• Although there are exceptions to this trend, in general, elements towards the right within the same period show a stronger affinity for electrons.

For the given elements, carbon (\(\mathrm{C}\)) is furthest to the right, indicating it has the most negative electron attachment enthalpy. This means carbon is most likely to gain electrons compared to sodium, boron, or aluminum.

Ionization Energy

Ionization energy is the energy required to remove an electron from a neutral atom in the gas phase. It indicates how strongly an atom holds onto its electrons.

• Across a period from left to right, ionization energy tends to increase. This is because atoms on the right side of the periodic table have more protons in the nucleus, creating a stronger attraction for electrons.
• On the other hand, ionization energy decreases as you move down a group. The outermost electrons are further from the nucleus in larger atomic radii, making them easier to remove.

Putting these trends to use with the elements sodium (\(\mathrm{Na}\)), boron (\(\mathrm{B}\)), aluminum (\(\mathrm{Al}\)), and carbon (\(\mathrm{C}\)), we find that sodium, located further left, has the lowest ionization energy. Meanwhile, carbon, situated further right, requires the most energy to remove an electron. The sequence from lowest to highest ionization energy for these elements is: \(\mathrm{Na < Al < B < C}\).