Problem 63 - Chemistry and Chemical Reactivity

Question

Answer the questions below about the elements \(A\) and \(\mathrm{B},\) which have the ground state electron configurations shown. $$ \mathrm{A}=[\mathrm{Kr}] 5 s^{2} \quad \mathrm{B}=|\mathrm{Kr}| 4 d^{10} 5 s^{2} 5 p^{5} $$ (a) Is element \(A\) a metal, nonmetal, or metalloid? (b) Which element has the greater ionization energy? (c) Which element has a larger atomic radius? (d) Which element has the more negative electron attachment enthalpy? (e) Which is more likely to form a cation? (f) What is a likely formula for a compound formed between A and B?

Step-by-Step Solution

Answer

(a) Metal, (b) B, (c) A, (d) B, (e) A, (f) \(\text{SrI}_2\)

Step 1: Analyze Element A

Element \( A = [\text{Kr}] 5s^2 \). This configuration places element A in the 5th period and in the group of alkaline earth metals, specifically Strontium (Sr). Metals are located on the left side of the periodic table.

Step 2: Analyze Element B

Element \( B = [\text{Kr}] 4d^{10} 5s^2 5p^5 \). This configuration corresponds to a halogen, specifically Iodine (I), which are nonmetals found on the right side of the periodic table.

Step 3: Determine Element with Greater Ionization Energy

Ionization energy increases across a period and decreases down a group. Since iodine is to the right of strontium in the same period, iodine (element B) has a greater ionization energy than strontium (element A).

Step 4: Determine Element with Larger Atomic Radius

Atomic radius increases down a group and decreases across a period. Strontium (element A) is in Group 2 while iodine (element B) is in Group 17; strontium will have a larger atomic radius since it's an alkaline earth metal.

Step 5: Determine Element with More Negative Electron Affinity

More negative electron affinity is a characteristic of nonmetals. Iodine (element B) is a halogen, typically having more negative electron affinities than metals like strontium (element A).

Step 6: Determine Element More Likely to Form a Cation

Metals tend to lose electrons and form cations. Therefore, strontium (element A), being a metal, is more likely to form a cation than iodine (element B).

Step 7: Determine Possible Compound Formula

The alkaline earth metal strontium tends to form a \(2^+\) cation \(\text{Sr}^{2+}\), while halogen iodine forms a \(1^-\) anion \(\text{I}^-\). To achieve charge neutrality, two iodide ions are needed. Hence, the formula for the compound is \(\text{SrI}_2\).

Key Concepts

Electron Configuration

Electron configuration refers to the arrangement of electrons within an atom. This determines many properties of the element. Electrons fill orbitals in order of increasing energy levels: 1s, 2s, 2p, 3s, 3p, and so on.
For element A, the electron configuration is \([\text{Kr}] 5s^2\). This configuration signifies that element A is in the fifth period and belongs to the alkaline earth metals, such as Strontium (Sr).
Element B has the configuration \([\text{Kr}] 4d^{10} 5s^2 5p^5\). This is recognized as a configuration for halogens, specifically Iodine (I). The valence shell has electrons in the p-orbital, indicating that Iodine is very reactive.

• "[Kr]" signifies the electron configuration of Krypton, which is a core configuration used as a reference.
• After Krypton, element A has electrons fill the 5s orbital, while element B fills 4d, 5s, and 5p orbitals.

Ionization Energy

Ionization energy is the amount of energy required to remove an electron from an atom. It provides insight into how strongly an atom holds onto its electrons. Generally, elements on the right side of the periodic table have higher ionization energies.
In the case of elements A and B:

• Iodine (element B), situated further right on the periodic table, has a higher ionization energy than Strontium (element A).
• Higher ionization energy in Iodine means it is less likely to lose an electron compared to Strontium.
• This tendency is because as you move across a period from left to right, additional protons increase the nuclear charge without an increase in shielded electron shells, pulling electrons more tightly.

Atomic Radius

Atomic radius is a measure of the size of an element's atoms, typically the distance from the nucleus to the outer electron shell. Generally, the atomic radius decreases across a period and increases down a group.
In this scenario:

• Strontium (element A) is expected to have a larger atomic radius compared to Iodine (element B). This is because, as an alkaline earth metal, it resides in Group 2.
• The larger atomic radius results from Strontium's fewer protons in its period and being earlier in the periodic layout, leading to less pull on its outer electrons.

The greater size in elements like Strontium is due to additional electron shells being filled as opposed to an increase in nuclear charge, which dramatically impacts the atomic size.

Electron Affinity

Electron affinity refers to the amount of energy released when an electron is added to a neutral atom, illustrating how much an atom wants to gain an electron.
For these elements:

• Iodine (element B), as a halogen, generally has a more negative electron affinity than Strontium (element A), an alkaline earth metal.
• This tendency is due to the halogens’ nearly complete valence shell, making them highly eager to gain an electron to reach a stable configuration.
• The concept implies how desirous an element is to accept an electron: the more negative the electron affinity, the more readily the atom will accept an electron.

Understanding electron affinity helps predict element reactivity and chemical bonding potential.