To draw the Lewis structure for C2H2Br2, we start by determining the total number of valence electrons. Carbon (C) has 4 valence electrons, hydrogen (H) has 1, and bromine (Br) has 7. Therefore, the total is:
- 2 Carbon: 2 x 4 = 8
- 2 Hydrogen: 2 x 1 = 2
- 2 Bromine: 2 x 7 = 14
Total = 8 + 2 + 14 = 24 valence electrons.
Next, we arrange the atoms in the structure. Since C is central, we can line up the two C atoms in the center, with H and Br atoms around them. A typical arrangement is:
H – C – C – H
Now we will add the Br atoms. Placing one Br atom on each of the carbon atoms, we have:
H – C – Br
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H – C – Br
After placing the atoms, we proceed to complete the octets for each of the carbons and bromines. Each carbon will share three bonding pairs with the hydrogens and bromines. This gives us:
H – C(Br) – C(Br) – H
And the structure can be illustrated as:
H – C – Br
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H – C – Br
This structure gives us a total of 12 bonding electrons used (6 pairs) which is consistent with our initial count.
Now let’s assess the molecular geometry. With 2 carbons each connected to 2 different elements (H and Br), we have a linear shape around the carbon-carbon bond due to sp2 hybridization. The bond angles will be approximately 120 degrees.
As for polarity, the C-H and C-Br bonds differ in electronegativity, leading to a polar bond. However, due to symmetry in the linear arrangement, the dipole moments from the hydrogen and bromine atoms will cancel each other out. Therefore, C2H2Br2 is considered a nonpolar molecule.
In summary, the Lewis structure for C2H2Br2 shows a linear geometry, and the molecule is nonpolar.