Which of the following best describes the Lewis dot structure for the neutral compound SiBr4?

• A. Silicon atom in the center with four single bonds to four bromine atoms; each bromine has three lone pairs.
• B. Silicon atom in the center with four single bonds to four bromine atoms; each bromine has one lone pair.
• C. Silicon atom in the center with two double bonds to two bromine atoms; each bromine has two lone pairs.
• D. Silicon atom in the center with one double bond and three single bonds to bromine atoms; each bromine has two lone pairs.

Answer: (A)

Verified step by step guidance

1. Identify the central atom and the surrounding atoms in the compound SiBr4. Silicon (Si) is less electronegative than bromine (Br), so Si will be the central atom with four Br atoms bonded to it.
2. Determine the total number of valence electrons available. Silicon has 4 valence electrons, and each bromine atom has 7 valence electrons. Calculate the total: \$4 + 4 \times 7 = 32$ valence electrons.
3. Draw single bonds between the silicon atom and each bromine atom. Each single bond represents 2 electrons, so 4 bonds use \$4 \times 2 = 8$ electrons.
4. Distribute the remaining electrons as lone pairs on the bromine atoms to complete their octets. Since each Br needs 8 electrons total and already shares 2 in the bond, each Br will have 6 electrons as lone pairs, which corresponds to 3 lone pairs per Br atom.
5. Check the silicon atom's octet. Silicon shares 8 electrons through the four single bonds, satisfying the octet rule. Therefore, the Lewis structure is silicon in the center with four single bonds to bromine atoms, each bromine having three lone pairs.