Which of the following is the most stable monatomic ion formed from aluminum?

• A. Al^{+}
• B. Al^{3+}
• C. Al^{-}
• D. Al^{2+}

Answer: (B)

Verified step by step guidance

1. Step 1: Understand the concept of ion stability in terms of electron configuration. Atoms tend to form ions that achieve a noble gas electron configuration, which is particularly stable due to a full valence shell.
2. Step 2: Write the electron configuration of neutral aluminum (Al). Aluminum has an atomic number of 13, so its electron configuration is \$1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{1}$.
3. Step 3: Consider the possible ions and their electron configurations: - \(Al^{+}\) means losing 1 electron, resulting in \$1s^{2} 2s^{2} 2p^{6} 3s^{2}\(. - \)Al^{2+}\( means losing 2 electrons, resulting in \)1s^{2} 2s^{2} 2p^{6} 3s^{1}\(. - \)Al^{3+}\( means losing 3 electrons, resulting in \)1s^{2} 2s^{2} 2p^{6}$, which is the electron configuration of neon, a noble gas.
4. Step 4: Analyze the stability of each ion based on their electron configurations. The \(Al^{3+}\) ion has a full octet in its outer shell, matching the stable noble gas configuration, making it the most stable ion.
5. Step 5: Conclude that among the given options, \(Al^{3+}\) is the most stable monatomic ion formed from aluminum because it achieves a noble gas electron configuration.