Problem 93 - Chemistry: The Central Science

Question

(a) Triazine, \(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{~N}_{3}\), is like benzene except that in triazine every other \(\mathrm{C}-\mathrm{H}\) group is replaced by a nitrogen atom. Draw the Lewis structure(s) for the triazine molecule. (b) Estimate the carbon-nitrogen bond distances in the ring.

Step-by-Step Solution

Answer

The Lewis structure of the Triazine molecule (\(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{~N}_{3}\)) is: H | C--N :/ \: N C : \ / : C--N : : H There are no additional resonance structures. The carbon-nitrogen bond distance in the Triazine molecule is estimated to be approximately 1.36 Å.

Step 1: Determine the total number of valence electrons

First, we need to find the total number of valence electrons in a Triazine molecule. Triazine has the chemical formula \(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{~N}_{3}\). The number of valence electrons for each atom is as follows: - Carbon (C) has 4 valence electrons - Hydrogen (H) has 1 valence electron - Nitrogen (N) has 5 valence electrons Now, we can calculate the total number of valence electrons in the Triazine molecule: \[ 3(4) + 3(1) + 3(5) = 12+3+15 = 30 \]

Step 2: Draw the skeleton structure

Next, we need to draw the skeleton structure of the Triazine molecule by connecting the atoms with single bonds. Remember that in Triazine, every other Carbon-Hydrogen group is replaced by a nitrogen atom. The skeleton structure will look like this: H | C--N / \ N C \ / C--N | H

Step 3: Distribute the remaining valence electrons as lone pairs

Now, we need to distribute the remaining valence electrons as lone pairs. We used 12 valence electrons to create the single bonds in the skeleton structure. So, we have 30-12=18 valence electrons left. Distribute these electrons as lone pairs around the atoms so that each atom has a full outer shell. Each nitrogen atom needs 2 more electrons while each carbon atom needs 4 more electrons: H | C--N :/ \: N C : \ / : C--N : : H

Step 4: Check for resonance structures

A resonance structure occurs when there are multiple valid Lewis structures for the same molecule. In the case of Triazine, there are no additional resonance structures possible, as all atoms have a full outer shell.

Step 5: Estimate the carbon-nitrogen bond distances

The carbon-nitrogen bond distances in Triazine can be estimated by comparing them to the known bond lengths of a single bond (C-N) and a double bond (C=N). - A carbon-nitrogen single bond distance is approximately 1.47 Å. - A carbon-nitrogen double bond distance is approximately 1.25 Å. In Triazine, the bond between carbon and nitrogen appears to be between a single and double bond due to resonance. This bond is called a "one-and-a-half bond" or 1.5 bond. We can estimate the carbon-nitrogen bond distance in Triazine as the average of the single and double bond distances: \[ \frac{1.47Å + 1.25Å}{2} ≈ 1.36 Å \] So, the carbon-nitrogen bond distance in the Triazine molecule is estimated to be approximately 1.36 Å.