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

Solution

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

$\mathrm{AsCl}_{4}^{-}$:
- 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).
$\mathrm{ICl}_{4}^{-}$:
- 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).
$\mathrm{ICl}_{4}^{+}$:
- 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).
$\mathrm{AlCl}_{4}^{-}$:
- 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).