Questions: Consider the following XCl4 ions: AsCl4^-, ICl4^-, ICl4^+, and AlCl4^-. Which of the ions will have an octahedral electron-domain geometry? Check all that apply. AsCl4^- ICl4^- ICl4^+ AlCl4^-

Consider the following XCl4 ions: AsCl4^-, ICl4^-, ICl4^+, and AlCl4^-. Which of the ions will have an octahedral electron-domain geometry? Check all that apply. AsCl4^- ICl4^- ICl4^+ AlCl4^-
Transcript text: Review Constan Part C Consider the following $\mathrm{XCl}_{4}$ ions: $\mathrm{AsCl}_{4}{ }^{-}$, $\mathrm{ICl}_{4}^{-}, \mathrm{ICl}_{4}^{+}$, and $\mathrm{AlCl}_{4}^{-}$. Which of the ions will have an octahedral electron-domain geometry? Check all that apply. $\mathrm{AsCl}_{4}{ }^{-}$ $\mathrm{ICl}_{4}{ }^{-}$ $\mathrm{ICl}_{4}{ }^{+}$ $\mathrm{AlCl}_{4}{ }^{-}$ Submit Request Answer
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Solution

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Solution Steps

Step 1: Determine the Electron-Domain Geometry

To determine the electron-domain geometry, we need to consider the number of electron domains around the central atom. This includes both bonding pairs and lone pairs of electrons.

Step 2: Analyze Each Ion

  1. $\mathrm{AsCl}_{4}^{-}$:

    • Arsenic (As) has 5 valence electrons.
    • Each chlorine (Cl) contributes 1 electron, and there are 4 chlorines.
    • The negative charge adds 1 more electron.
    • Total electrons: \(5 + 4 \times 1 + 1 = 10\) electrons.
    • This corresponds to 5 electron pairs (4 bonding pairs and 1 lone pair).

  2. $\mathrm{ICl}_{4}^{-}$:

    • Iodine (I) has 7 valence electrons.
    • Each chlorine (Cl) contributes 1 electron, and there are 4 chlorines.
    • The negative charge adds 1 more electron.
    • Total electrons: \(7 + 4 \times 1 + 1 = 12\) electrons.
    • This corresponds to 6 electron pairs (4 bonding pairs and 2 lone pairs).

  3. $\mathrm{ICl}_{4}^{+}$:

    • Iodine (I) has 7 valence electrons.
    • Each chlorine (Cl) contributes 1 electron, and there are 4 chlorines.
    • The positive charge removes 1 electron.
    • Total electrons: \(7 + 4 \times 1 - 1 = 10\) electrons.
    • This corresponds to 5 electron pairs (4 bonding pairs and 1 lone pair).

  4. $\mathrm{AlCl}_{4}^{-}$:

    • Aluminum (Al) has 3 valence electrons.
    • Each chlorine (Cl) contributes 1 electron, and there are 4 chlorines.
    • The negative charge adds 1 more electron.
    • Total electrons: \(3 + 4 \times 1 + 1 = 8\) electrons.
    • This corresponds to 4 electron pairs (4 bonding pairs and 0 lone pairs).
Step 3: Identify Octahedral Geometry

An octahedral electron-domain geometry requires 6 electron pairs around the central atom.

  • $\mathrm{AsCl}_{4}^{-}$: 5 electron pairs (not octahedral).
  • $\mathrm{ICl}_{4}^{-}$: 6 electron pairs (octahedral).
  • $\mathrm{ICl}_{4}^{+}$: 5 electron pairs (not octahedral).
  • $\mathrm{AlCl}_{4}^{-}$: 4 electron pairs (not octahedral).

Final Answer

\[ \boxed{\mathrm{ICl}_{4}^{-}} \]

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