Question

select the lewis structure that most accurately represents xenon tetrafluoride (xef₄). refer to a periodic table.

Explanation:

Step1: Determine valence electrons

Xenon (Xe) is in group 18, so it has 8 valence electrons. Fluorine (F) is in group 17, so each F has 7 valence electrons. For \( \text{XeF}_4 \), total valence electrons = \( 8 + 4\times7 = 8 + 28 = 36 \).

Step2: Determine bonding and lone pairs

Xe is the central atom. Each F forms a single bond with Xe (2 electrons per bond), so 4 bonds use \( 4\times2 = 8 \) electrons. Remaining electrons: \( 36 - 8 = 28 \). Each F needs 6 more electrons (to complete octet), so 4 F atoms use \( 4\times6 = 24 \) electrons. Remaining for Xe: \( 28 - 24 = 4 \) electrons (2 lone pairs on Xe).

The molecular geometry of \( \text{XeF}_4 \) is square planar (due to 4 bonding pairs and 2 lone pairs on Xe, arranged in octahedral electron geometry with lone pairs opposite each other). The correct Lewis structure should show Xe with 4 single bonds to F (each F with 3 lone pairs) and Xe with 2 lone pairs (though sometimes lone pairs on Xe are not shown in the structure but the bonding arrangement is square planar). Looking at the options, the top - left structure (with Xe in the center, four F atoms bonded in a square - like arrangement, each F with 3 lone pairs) is incorrect in geometry? Wait, no—wait, the correct structure for \( \text{XeF}_4 \) has Xe with 4 single bonds to F (each F has 3 lone pairs) and Xe has 2 lone pairs (which are in the axial positions of the octahedral electron geometry, but the molecular geometry is square planar, so the F atoms are in the equatorial plane). Wait, among the given options, the second structure (top - right) shows Xe with four bonds (two axial and two equatorial? No, the correct structure's Lewis representation: Let's re - evaluate.

Wait, the correct Lewis structure for \( \text{XeF}_4 \): Xe is central, 4 F atoms bonded (single bonds), each F has 3 lone pairs, and Xe has 2 lone pairs (which are above and below the plane of the F atoms). But in the given options, the second structure (top - right) has Xe with four bonds (two vertical, one top - right, one bottom - right F) and the lone pairs on Xe? No, maybe I made a mistake. Wait, no—let's count the electrons in each structure.

First structure (top - left): Xe bonded to four F (each with 3 lone pairs). Let's count electrons: Each F - Xe bond is 2 electrons, 4 bonds: 8. Each F has 6 lone pair electrons: 4 F × 6 = 24. Xe has 0 lone pairs? But Xe should have 2 lone pairs. Wait, no—Xe has 8 valence electrons. In the first structure, Xe is using 4×2 = 8 electrons for bonding, so it has 0 lone pairs? That's wrong.

Second structure (top - right): Xe bonded to four F (two vertical, one top - right, one bottom - right). Let's count electrons: 4 bonds (8 electrons). Each F has 3 lone pairs (6 electrons each): 4×6 = 24. Total used: 8 + 24 = 32. Xe has 4 electrons (2 lone pairs) left. So total electrons: 32+4 = 36, which matches. And the geometry: the two lone pairs on Xe are not shown, but the bonding arrangement with four F atoms (two in the axial and two in the equatorial? No, the correct molecular geometry is square planar, so the four F atoms are in a square, and the two lone pairs are above and below the square. But in the Lewis structure, the representation of the square planar arrangement (four F atoms in a plane around Xe) is better represented by the second structure? Wait, no—wait, the correct Lewis structure for \( \text{XeF}_4 \) has Xe with four single bonds to F (each F has 3 lone pairs) and Xe has two lone pairs (which are usually not shown in the structure but the bonding is such that the F atoms are in a square. Wait, maybe the second structure (top - right) is correct. Wait, no, let's check the number of bonds and lone pairs again.

Wait, the correct Lewis structure for \( \text{XeF}_4 \):

Valence electrons: Xe (8) + 4F (4×7 = 28) = 36.

Number of bonding pairs: 4 (since 4 F atoms, each single - bonded to Xe).

Number of lone pairs on Xe: \( \frac{8 - 4\times1}{2}=\frac{8 - 4}{2}=2 \) (using the formula: lone pairs on central atom=\( \frac{\text{valence electrons of central atom}-\text{number of bonding atoms}\times1}{2} \), since each bond is a single bond, so each bonding atom contributes 1 electron to the bond from the central atom's perspective).

So Xe has 2 lone pairs and 4 bonding pairs. The Lewis structure should show Xe with 4 single bonds to F (each F has 3 lone pairs) and Xe with 2 lone pairs. Among the given options, the second structure (top - right) has Xe with four bonds (two vertical, one top - right, one bottom - right F) and the lone pairs on Xe? No, the lone pairs on Xe are not shown, but the electron count: 4 bonds (8 electrons), 4 F atoms with 3 lone pairs each (24 electrons), and Xe with 4 electrons (2 lone pairs) – total 36. The first structure (top - left) has Xe with four bonds and no lone pairs on Xe (electron count: 4 bonds (8) + 4F×6 lone pair electrons (24) = 32, which is less than 36 – wrong). The third structure (bottom - left) has Xe with four bonds but the arrangement is tetrahedral? No, \( \text{XeF}_4 \) is square planar, not tetrahedral. The fourth structure has double bonds, which is wrong because F does not form double bonds (it has 7 valence electrons, needs 1 more, so single bond is sufficient).

So the correct structure is the second one (top - right) with Xe bonded to four F (two vertical, one top - right, one bottom - right F), each F with 3 lone pairs, and Xe with 2 lone pairs (not shown in the structure but the electron count is correct). Wait, maybe I was wrong earlier. Let's re - check:

In the second structure (top - right):

- Xe is bonded to four F: two F are bonded vertically (top and bottom), and two F are bonded at 90 degrees to the vertical (right - top and right - bottom).

- Each F has 3 lone pairs (6 electrons), so 4 F × 6 = 24.

- Each Xe - F bond is 2 electrons, 4 bonds: 8.

- Total electrons from bonds and F lone pairs: 24 + 8 = 32.

- Xe has 8 - 4 (bonds) = 4 electrons (2 lone pairs), so total electrons: 32+4 = 36, which matches the total valence electrons.

The first structure (top - left) has Xe bonded to four F in a square (all in the same plane), each F with 3 lone pairs. But then Xe has 8 - 4 = 4 electrons (2 lone pairs), but in this structure, Xe has no lone pairs shown. Wait, maybe the lone pairs on Xe are not shown in the structure, but the bonding arrangement. However, the correct molecular geometry is square planar, and the Lewis structure should represent the square planar arrangement of the F atoms (with Xe in the center, four F in a square, and two lone pairs on Xe above and below the square). The second structure (top - right) shows a more accurate representation of the electron geometry (octahedral) with two axial F (top and bottom) and two equatorial F (right - top and right - bottom), which is consistent with the octahedral electron geometry (4 bonding pairs and 2 lone pairs).

So the correct Lewis structure is the second one (top - right) among the given options.

Answer:

The second Lewis structure (top - right) with Xe in the center, bonded to four F atoms (two vertically, one top - right, one bottom - right), each F with 3 lone pairs.