Problem 37 - Chemistry in 30 days

Question

The successive ionization energies in kJ/mol of an element \(\mathrm{P}\) are \(740,1500,7000,10500\), \(13600,18000,21700 .\) Which ion is the most likely to be formed when \(X\) reacts with chloride ? (a) \(\underline{p^{2}}\) (b) \(\mathrm{P}^{+}\) (c) \(\mathrm{P}^{+2}\) (d) \(\mathrm{P}^{+3}\)

Step-by-Step Solution

Answer

The most likely ion formed is B (c).

Step 1: Understand Ionization Energy

Ionization energy is the energy required to remove an electron from an atom in the gas phase. The successive ionization energies increase because it's consistently harder to remove an electron from a positively charged ion.

Step 2: Analyze Given Ionization Energies

The sequence given is 740, 1500, 7000, 10500, 13600, 18000, and 21700 kJ/mol. Notice a significant jump in ionization energy after the second ionization energy (from 1500 to 7000 kJ/mol).

Step 3: Determine Likely Charge Based on Energy Jump

The large increase in ionization energy indicates that after the second electron is removed, removing the next electron is considerably harder, signifying a stable electron configuration resembling a noble gas structure after the second electron is removed.

Step 4: Predict Likely Ion Formation

Based on the large increase after removing two electrons, the element is likely to form a B charge, as further ionizations are much higher in energy, indicating a transition into an unfavorable electron configuration.

Key Concepts

Successive Ionization Energies

When learning about ionization energies, remember they refer to the energy needed to remove electrons one by one from an atom. Each removal process requires energy, and this energy is different for each electron. As each electron is removed, the remaining atom becomes more positively charged. Hence, the atom holds onto its remaining electrons more tightly, requiring more energy for subsequent removals.

In this concept:

• The first ionization energy is the energy needed to remove the first electron, turning a neutral atom into a positive ion.
• The second ionization energy is the energy needed to remove the second electron, now from a positively charged ion.
• As you continue removing electrons, the energy required increases because the ion is getting more positively charged.

Analyzing the trend can help determine an element's reactivity and the stability of its electron configuration.

Noble Gas Configuration

A noble gas configuration refers to a situation where an atom's electron configuration is so stable, it resembles that of a noble gas. Noble gases, like helium and neon, are known for their low reactivity due to their full outer electron shells.

In chemical reactions, atoms often lose or gain electrons to reach a more stable, noble gas-like state. This is because a full valence shell corresponds to lower energy and higher stability.

• For example, magnesium likes to lose two electrons to resemble the configuration of neon.
• Achieving this configuration makes the atom less reactive and more stable.

Recognizing significant jumps in ionization energy helps identify when an atom reaches a noble gas configuration, indicating higher stability.

Electron Removal

Electron removal can happen in single steps, and each step could need more energy than the last. Each removal results in a positively charged ion, and with each step, further electron removal becomes increasingly challenging due to increased positive charge.

The process includes:

• Analyzing energy values related to electron removal can help predict the natural stopping point for ion formation.
• Large increases in ionization energies often indicate a stable electronic arrangement after certain electrons have been removed.

This concept directly connects to the behavior of atoms during chemical reactions and their tendency to lose electrons in certain predictable patterns.

Chemical Reactions with Chlorides

When we talk about chemical reactions involving chlorides, it's essential to understand that chloride ions are negatively charged (Cl⁻). Metal ions often react with chloride ions to form compounds.

During these reactions:

• Atoms might lose electrons to form positively charged ions, which then combine with chloride ions.
• By assessing ionization energies, you can predict which form of an atom will likely react with chlorine to form stable chlorides.

Realizing the importance of ion formation helps predict how elements will behave in reactions with chlorides. For element \( P \), it means reaching a stable ion charge that balances well with chloride for compound formation.