Question

question 21
choose the best lewis structure for xei₂.
○ :i=xe=i:
○ :i—xe—i:
○ :i—xe—i:
○ :i→xe—i:

Explanation:

Step1: Determine valence electrons

Xe has 8 valence electrons, each I has 7, so total valence electrons: \(8 + 2\times7 = 22\).

Step2: Analyze each option

- First option: Double bonds. Each double bond uses 4 electrons, plus lone pairs. Let's count: Each I has 2 lone pairs (4 electrons), Xe has 0? Wait, no, let's calculate formal charges or electron count. Wait, first option: \( \dot{\cdot}\!:\text{I}=\text{Xe}=\text{I}\dot{\cdot}: \) – each double bond is 4 electrons, so between Xe and I: 4 each, so Xe has 8 from bonds? But I's lone pairs: left I has 2 (4 electrons), right I has 2 (4 electrons). Total electrons: \(4 + 4 + 4 + 4 + 4?\) No, better to count total electrons. Valence electrons needed: 22.

- Second option: \( :\!\!\ddot{\text{I}} - \ddot{\text{Xe}} - \ddot{\text{I}}: \) – Let's count electrons. Each I has 6 lone pairs? Wait, no, the dots: left I has \( :\!\!\ddot{} \) (4 electrons), Xe has \( \ddot{} \) (2 electrons), right I has \( : \) (2 electrons)? No, maybe I misread. Wait, third option: \( :\!\!\ddot{\text{I}} - \text{Xe} - \ddot{\text{I}}: \) – Wait, let's do formal charge and electron count properly.

Xe in \( \text{XeI}_2 \) has 8 valence electrons. Each I needs 8 (octet). So Xe is central, bonded to two I. Xe has 8, so after bonding (single bonds: 2 bonds, 4 electrons), Xe has 4 lone pairs? Wait, no: \( \text{XeI}_2 \) has a linear structure with 3 lone pairs on Xe? Wait, no, VSEPR: Xe has 8 valence, 2 bonds (2×2=4 electrons), so lone pairs: \( (8 - 4)/2 = 2 \) lone pairs? Wait, no, formula: \( \text{AX}_2\text{E}_3 \) (2 bonds, 3 lone pairs)? Wait, no, Xe is in period 5, can have expanded octet. Wait, let's calculate total valence electrons: Xe (8) + 2×I (7) = 22.

Now, let's count electrons in each option:

1. \( \dot{\cdot}\!:\text{I}=\text{Xe}=\text{I}\dot{\cdot}: \) – Each double bond is 4 electrons, so two double bonds: 8 electrons. Left I: 2 lone pairs (4 electrons), right I: 2 lone pairs (4 electrons). Xe: 0 lone pairs. Total electrons: 8 + 4 + 4 = 16. Not 22. Wrong.

2. \( :\!\!\ddot{\text{I}} - \ddot{\text{Xe}} - \ddot{\text{I}}: \) – Left I: \( :\!\!\ddot{} \) (4 electrons), Xe: \( \ddot{} \) (2 electrons), right I: \( \ddot{} \) (2 electrons). Bonds: 2 single bonds (2 electrons each? No, single bond is 2 electrons). Wait, total electrons: left I lone pairs (4) + bond (2) + Xe lone pairs (2) + bond (2) + right I lone pairs (2) = 4+2+2+2+2=12. No, wrong.

3. \( :\!\!\ddot{\text{I}} - \text{Xe} - \ddot{\text{I}}: \) – Left I: \( :\!\!\ddot{} \) (4 electrons), bond (2) to Xe, Xe (no lone pairs), bond (2) to right I, right I: \( \ddot{}: \) (4 electrons). Total electrons: 4+2+2+4=12. No.

Wait, maybe I messed up. Let's recall: \( \text{XeI}_2 \) has Xe with 8 valence, 2 I (7 each). Total valence: 8 + 14 = 22.

Now, the correct Lewis structure should have: Xe bonded to two I (single bonds: 2×2=4 electrons), Xe has 3 lone pairs (6 electrons), each I has 3 lone pairs (6 electrons each). Let's check:

Each I: 6 lone pairs (12 electrons) + 2 (bond) = 14? No, I needs 8. Wait, I has 7 valence, so needs 1 more to complete octet. Wait, no: I is in group 17, needs 1 electron to complete octet (8). So each I should have 3 lone pairs (6 electrons) and 1 bond (2 electrons), total 8. Xe: 8 valence, uses 2×2=4 electrons in bonds, so has 4 lone pairs? Wait, 8 - 4 = 4, so 2 lone pairs (4 electrons). Wait, total electrons:

Bonds: 2×2=4.

Lone pairs on I: 2×6=12 (3 pairs each).

Lone pairs on Xe: 4 (2 pairs).

Total: 4 + 12 + 4 = 20. Wait, missing 2. Oh, Xe can have expanded octet. So maybe Xe has 3 lone pairs (6 electrons), so total electrons: 4 (bonds) + 12 (I lone pairs) + 6 (Xe lone pairs) = 22. Perfect.

Now, let's look at the options:

Third option: \( :\!\!\ddot{\text{I}} - \text{Xe} - \ddot{\text{I}}: \) – Wait, left I: \( :\!\!\ddot{} \) (4 electrons? No, \( :\!\!\ddot{} \) is 4? Wait, \( :\!\!\ddot{\text{I}} \) has 6 electrons? Wait, \( :\!\!\ddot{} \) is two dots above, two below: that's 4? No, \( :\!\!\ddot{\text{I}} \) – the colon is two, the two dots above and two below? Wait, maybe the correct structure is the one where Xe has 3 lone pairs? Wait, no, let's check the options again.

Wait, the third option (assuming the options are:

1. \( \dot{\cdot}\!:\text{I}=\text{Xe}=\text{I}\dot{\cdot}: \)

2. \( :\!\!\ddot{\text{I}} - \ddot{\text{Xe}} - \ddot{\text{I}}: \)

3. \( :\!\!\ddot{\text{I}} - \text{Xe} - \ddot{\text{I}}: \)

4. \( :\!\!\ddot{\text{I}} - \dot{\text{Xe}} - \ddot{\text{I}}: \)

Wait, maybe the correct one is the third option: \( :\!\!\ddot{\text{I}} - \text{Xe} - \ddot{\text{I}}: \) – Let's count electrons:

Left I: \( :\!\!\ddot{} \) (6 electrons? Wait, \( :\!\!\ddot{\text{I}} \) – colon (2) + two dots (2) = 4? No, \( :\!\!\ddot{\text{I}} \) is 6 electrons: colon (2) + two pairs (4) = 6? Wait, no, a colon is two electrons, two dots (a pair) is two. So \( :\!\!\ddot{\text{I}} \) has 2 (colon) + 2 (dots above) + 2 (dots below) = 6? No, that's 6, which is 3 lone pairs. Then bond to Xe (single bond, 2 electrons), Xe (no lone pairs? No, Xe needs 3 lone pairs). Wait, maybe the correct option is the third one: \( :\!\!\ddot{\text{I}} - \text{Xe} - \ddot{\text{I}}: \) – Wait, let's check formal charges.

Formal charge on Xe: valence electrons (8) - non-bonding (6) - bonding/2 (2) = 8 - 6 - 1 = +1? No, that's not right. Wait, maybe I made a mistake. Let's do it properly.

Valence electrons: Xe (8) + 2×I (7) = 22.

In the correct Lewis structure, Xe is bonded to two I (single bonds: 2×2=4 electrons). Each I has 6 lone pairs (6×2=12 electrons). Xe has 6 lone pairs? No, 8 - 4 = 4, so 2 lone pairs (4 electrons). Wait, 4 (bonds) + 12 (I lone pairs) + 4 (Xe lone pairs) = 20. Missing 2. So Xe must have an expanded octet: 8 + 2 = 10? No, Xe is in period 5, can have expanded octet. So Xe has 10 valence electrons? Wait, no, \( \text{XeI}_2 \) has a linear structure with 3 lone pairs on Xe (so Xe has 8 + 2 (from bonds) + 6 (lone pairs) = 16? No, I'm confused.

Wait, let's look up \( \text{XeI}_2 \) Lewis structure. Xe is central, bonded to two I (single bonds), Xe has 3 lone pairs (6 electrons). So total electrons: 2×2 (bonds) + 2×6 (I lone pairs) + 6 (Xe lone pairs) = 4 + 12 + 6 = 22. Perfect.

Now, let's check the options:

Option 3: \( :\!\!\ddot{\text{I}} - \text{Xe} - \ddot{\text{I}}: \) – Left I: \( :\!\!\ddot{} \) (6 electrons? Wait, \( :\!\!\ddot{\text{I}} \) has 6 electrons (3 lone pairs), right I: \( \ddot{\text{I}}: \) (6 electrons? No, \( \ddot{\text{I}}: \) is 4? Wait, maybe the correct option is the third one, where Xe has no lone pairs? No, that can't be. Wait, maybe the second option is wrong, third is correct. Wait, the third option is \( :\!\!\ddot{\text{I}} - \text{Xe} - \ddot{\text{I}}: \) – Let's count the dots:

Left I: \( :\!\!\ddot{} \) – that's 6 electrons (3 pairs), bond to Xe (2 electrons), Xe (no dots), bond to right I (2 electrons), right I: \( \ddot{}: \) – 4 electrons? No, that's not 8. Wait, I think the correct option is the third one: \( :\!\!\ddot{\text{I}} - \text{Xe} - \ddot{\text{I}}: \) – Wait, maybe the original options have:

Option 3: \( :\!\!\ddot{\text{I}} - \text{Xe} - \ddot{\text{I}}: \) – Left I has 6 lone pairs (3 pairs), right I has 4? No, I'm stuck. Wait, the key is that in the correct structure, Xe has 3 lone pairs, and each I has 3 lone pairs. So the structure with single bonds, each I has 6 lone electrons (3 pairs), and Xe has 6 lone electrons (3 pairs). So the correct option is the third one: \( :\!\!\ddot{\text{I}} - \text{Xe} - \ddot{\text{I}}: \) (assuming the third option is the one with Xe having no lone pairs? No, maybe I misread the options. Wait, the user's options:

1. \( \dot{\cdot}\!:\text{I}=\text{Xe}=\text{I}\dot{\cdot}: \)

2. \( :\!\!\ddot{\text{I}} - \ddot{\text{Xe}} - \ddot{\text{I}}: \)

3. \( :\!\!\ddot{\text{I}} - \text{Xe} - \ddot{\text{I}}: \)

4. \( :\!\!\ddot{\text{I}} - \dot{\text{Xe}} - \ddot{\text{I}}: \)

Wait, the third option: \( :\!\!\ddot{\text{I}} - \text{Xe} - \ddot{\text{I}}: \) – Let's count electrons:

Left I: \( :\!\!\ddot{} \) (6 electrons: 3 pairs), bond (2 electrons) to Xe, Xe (no lone pairs), bond (2 electrons) to right I, right I: \( \ddot{}: \) (4 electrons: 2 pairs). Total electrons: 6 + 2 + 2 + 4 = 14. No, that's not 22. I must be misinterpreting the dots.

Wait, maybe the correct option is the second one? No, let's try formal charge.

In option 3: \( :\!\!\ddot{\text{I}} - \text{Xe} - \ddot{\text{I}}: \)

Formal charge on I: 7 (valence) - 6 (lone pairs) - 1 (bond) = 0.

Formal charge on Xe: 8 (valence) - 0 (lone pairs) - 2 (bonds) = 6. No, that's wrong.

Wait, I think I made a mistake. Let's start over.

Valence electrons: Xe (8) + 2×I (7) = 22.

We need to arrange 22 electrons.

Xe is central.

Single bonds between Xe and each I: 2 bonds, 4 electrons.

Remaining electrons: 22 - 4 = 18.

Each I needs 6 more electrons (to complete octet: 8 - 2 = 6). So 2×6 = 12 electrons for I lone pairs.

Remaining electrons for Xe: 18 - 12 = 6 (3 lone pairs).

So Xe has 3 lone pairs (6 electrons), each I has 3 lone pairs (6 electrons), and two single bonds.

Now, let's represent this:

\( :\!\!\ddot{\text{I}} - \text{Xe} - \ddot{\text{I}}: \) – Wait, no, Xe has 3 lone pairs, so the structure should be:

\( :\!\!\ddot{\text{I}} - \ddot{\text{Xe}} - \ddot{\text{I}}: \) – No, Xe's lone pairs: \( \ddot{\text{Xe}} \) is 2 lone pairs. Wait, 3 lone pairs would be \( \dddot{\text{Xe}} \), but Xe can have expanded octet.

Wait, maybe the correct option is the third one: \( :\!\!\ddot{\text{I}} - \text{Xe} - \ddot{\text{I}}: \) – But according to the electron count, Xe should have 3 lone pairs. Maybe the diagram has Xe with 3 lone pairs, but in the option, it's drawn as \( \text{Xe} \) with no dots, but that's a mistake. Alternatively, the correct option is the third one, where each I has 6 lone electrons (3 pairs) and Xe has 6 lone electrons (3 pairs), making the total 22.

So the best Lewis structure is the third option: \( :\!\!\ddot{\text{I}} - \text{Xe} - \ddot{\text{I}}: \) (assuming the third option is the one with single bonds and each I having 6 lone electrons and Xe having 6 lone electrons).

Answer:

The third option: \( :\!\!\ddot{\text{I}} - \text{Xe} - \ddot{\text{I}}: \) (or the option labeled as the third one in the list, e.g., if the options are numbered 1-4, the third option: \( :\!\!\ddot{\text{I}} - \text{Xe} - \ddot{\text{I}}: \))