Question

draw the lewis structure for bei₂. then compare your drawing to the answer choices, and select the correct answer choice. select all correct options

Explanation:

To determine the correct Lewis structure for \( \text{BeI}_2 \):

1. Valence Electrons:

• Beryllium (Be) has 2 valence electrons.
• Iodine (I) has 7 valence electrons, and there are 2 I atoms: \( 2 \times 7 = 14 \) valence electrons.
• Total valence electrons: \( 2 + 14 = 16 \).

1. Bonding and Lone Pairs:

• Be is the central atom. It forms single bonds with each I atom (2 electrons per bond, using \( 2 \times 2 = 4 \) electrons).
• Remaining electrons: \( 16 - 4 = 12 \), which are distributed as lone pairs on I atoms (each I needs 6 lone electrons to satisfy the octet, \( 2 \times 6 = 12 \) electrons).

1. Analyzing the Options:

• The correct structure should have Be in the center, single bonds to two I atoms, and 3 lone pairs (6 electrons) on each I. The orange - colored option (I - Be - I) is incorrect because it lacks lone pairs on I. The yellow - colored option has incorrect atom placement. The teal - colored option has incorrect lone pair distribution. The purple - colored option (with Be in the center, single bonds to two I atoms, and 3 lone pairs on each I) matches the electron distribution for \( \text{BeI}_2 \).

Answer:

The correct option is the purple - colored card (with the structure showing Be in the middle, single bonds to two I atoms, and 3 lone pairs on each I atom). If we assume the purple - colored option is, for example, the second card (from top) with the structure \( \ddot{\text{I}} - \text{Be} - \ddot{\text{I}} \) (with 3 lone pairs on each I, though the drawing shows the correct lone pair count), the correct choice is the purple - colored card (the one with Be in the center, single bonds to two I atoms, and 3 lone pairs on each I).