select the lewis structure that most accurately represents carbon tetrabromide (cbr₄). refer to a periodic…

Explanation:

Step1: Determine valence electrons

Carbon (C) has 4 valence electrons, Bromine (Br) has 7 valence electrons. In ( \text{CBr}_4 ), total valence electrons: ( 4 + 4\times7 = 32 ).

Step2: Analyze bonding and lone pairs

Carbon forms 4 single bonds with Br (each bond uses 2 electrons). Each Br has 3 lone pairs (6 electrons) + 1 bonding pair. The correct Lewis structure should have C in the center, 4 single bonds to Br, each Br with 3 lone pairs (6 electrons) and a negative formal charge? Wait, no—formal charge: For Br, ( \text{Formal Charge} = 7 - (6 + 1) = 0 )? Wait, no, each Br in CBr₄: bonded to C (1 bond), so 6 lone electrons (3 pairs) + 1 bond (2 electrons, but Br contributes 1 to the bond). Wait, maybe better: C has 4 bonds (8 electrons), each Br has 3 lone pairs (6 electrons) and 1 bond (so total 8 electrons around Br, satisfying octet). The structure with C in center, 4 single bonds to Br (each Br with 3 lone pairs) is the correct one. Looking at the options, the first structure (top - left) has C with 4 single bonds (each Br connected by single bond, each Br with 3 lone pairs) – wait, no, the second structure (top - right) is a tetrahedral arrangement? Wait, no, the first structure (top - left) is a cross - like (square planar? No, CBr₄ is tetrahedral). Wait, no, the correct Lewis structure for ( \text{CBr}_4 ) is carbon in the center, four single bonds to bromine atoms, each bromine atom has three lone pairs. The second structure (top - right) shows a tetrahedral arrangement (four bonds from C to Br, each Br with 3 lone pairs). Wait, the first structure (top - left) has C with four single bonds (one up, one down, one left, one right), each Br with 3 lone pairs. Wait, actually, both the first and second? No, let's check valence electrons. Each single bond is 2 electrons. For ( \text{CBr}_4 ), 4 bonds (8 electrons) + 4 Br atoms with 3 lone pairs each (4×6 = 24 electrons) → total 32 electrons, which matches. Now, formal charge: C: ( 4 - 4 = 0 ), Br: ( 7 - (6 + 1) = 0 ). So the structure where C is in the center, four single bonds to Br (each Br with 3 lone pairs) – the second structure (top - right) is the tetrahedral arrangement (correct geometry for ( \text{CBr}_4 ), which is tetrahedral, so the Lewis structure should show the tetrahedral bonding, with C in center and four Br atoms bonded to it, each with 3 lone pairs. Wait, the first structure (top - left) is a planar (square) arrangement, but ( \text{CBr}_4 ) is tetrahedral. So the second structure (top - right) is the correct one? Wait, no, maybe I messed up. Wait, the first structure: C is in center, with four single bonds (up, down, left, right), each Br with 3 lone pairs. The second structure: C in center, four bonds (tetrahedral), each Br with 3 lone pairs. Wait, the problem is about Lewis structure, not molecular geometry. Lewis structure can be drawn with the bonds as lines, and lone pairs as dots. The correct Lewis structure for ( \text{CBr}_4 ) has carbon in the center, four single bonds to bromine, each bromine has three lone pairs. So the second structure (top - right) is the correct one? Wait, no, let's count electrons. Each Br has 6 lone electrons (3 pairs) and 1 bond (2 electrons, but Br contributes 1, so 7 electrons? No, octet: 8 electrons. So each Br needs 8 electrons: 6 lone (3 pairs) + 2 in the bond (so 1 from Br, 1 from C). So each Br has 6 + 2 = 8. C has 4 bonds (8 electrons). So total electrons: 4×8 (Br) + 8 (C) - 4×2 (bonds, counted twice) = 32 + 8 - 8 = 32, which matches. So the structure with C in center, four single bonds to Br (each Br with 3 lone pairs) – the second structure (top - right) is the correct Lewis structure? Wait, no, the first structure (top - left) also has four single bonds. Wait, maybe the difference is the arrangement. ( \text{CBr}_4 ) is tetrahedral, so the Lewis structure should represent the bonding as four single bonds from C to Br, with each Br having 3 lone pairs. The second structure (top - right) shows a tetrahedral arrangement (four bonds from C, each Br with 3 lone pairs), so that's the correct one. Wait, but the user's image: let's re - examine. The first structure (top - left): C is in center, with four single bonds (up, down, left, right), each Br with 3 lone pairs. The second structure (top - right): C in center, four bonds (tetrahedral), each Br with 3 lone pairs. The third structure (bottom - left): C with some bonds, but maybe wrong. The fourth (bottom - right): double bonds, which is wrong (C can't form double bonds with Br here, since C has 4 valence electrons, Br has 7, so single bonds are better). So the correct structure is the second one (top - right) or the first? Wait, no, in Lewis structure, the arrangement can be drawn with the bonds as lines, and the tetrahedral geometry is represented by the four bonds from C. Wait, actually, the correct Lewis structure for ( \text{CBr}_4 ) is carbon in the center, four single bonds to bromine, each bromine has three lone pairs. So the second structure (top - right) is the correct one. Wait, but maybe I made a mistake. Let's check formal charge. For C: ( 4 - 4 = 0 ). For each Br: ( 7 - (6 + 1) = 0 ). So that's correct. So the second structure (top - right) is the correct Lewis structure.

Answer:

The Middle - Top Graph (the second structure from the left in the top row)