Multiple Choice
How many total electrons are present in a Fe^{2+} ion?
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2. Atoms & Elements
Ions
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Which of the following is the most stable monatomic ion formed from aluminum?
A
Al^{+}
B
Al^{3+}
C
Al^{-}
D
Al^{2+}
Verified step by step guidance1
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.
Step 2: Write the electron configuration of neutral aluminum (Al). Aluminum has an atomic number of 13, so its electron configuration is $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{1}$.
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.
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.
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.
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