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Consider the hypothetical molecule \(\mathrm{B}-\mathrm{A}=\mathrm{B}\). Are the following statements true or false? (a) This molecule cannot exist. (b) If resonance was important, the molecule would have identical A-B bond lengths.

Short Answer

Expert verified
Statement (a) is false, and statement (b) is true. The molecule with B-A=B structure exists in some cases like formaldehyde, hydrazine, and hydrogen peroxide. Resonance is not important for these molecules, so there are no significant resonance structures, and the A-B bond lengths do not need to be identical.

Step by step solution

01

Determine if the Molecule Exists

To evaluate statement (a), we need to assign the correct number of valence electrons for the atoms involved in the molecule. Let's assume that A is from the 2nd period, specifically within groups 4, 5, 6, or 7 in the Periodic Table, based on its position in the given molecule (sharing two bonds with the elements B). According to this assumption, A might be C, N, O, or F. The assumption here is to choose a plausibly existing compound. However, there is no guarantee that this compound actually exists after the correct valence electrons are assigned. Now, let's examine the B-A=B structure and try different atoms for A. Note: If A had three lone pairs, the structure would have been B-A-B with a triple bond. Based on this, we consider different atoms for A: 1. Carbon - In this case B-C=B and it is a formaldehyde molecule (HCHO). 2. Nitrogen - B-N=B would be hydrazine like molecule (HNNH) 3. Oxygen - B-O=B (where B is hydrogen) would form H2O2, hydrogen peroxide. As the examples show, the given molecule exists in some cases. Thus, statement (a) is false.
02

Determine the Resonance Importance

To evaluate statement (b), we need to consider if resonance is important for such molecules. Resonance occurs in molecules when multiple Lewis structures could contribute to the overall molecular structure; the real structure of that molecule is a hybrid of the contributing structures. In the case of B-A=B, B atoms can belong to any other group with one valence electron (i.e., hydrogen). If resonance plays a role, we should consider multiple resonance structures and see whether A-B bond lengths are equal or not. Carbon: As discussed earlier, the hypothetical molecule B-C=B is HCHO. No significant resonance structures can be drawn for this molecule, so resonance doesn't play an important role. Nitrogen: For B-N=B, the hydrazine-like molecule HNNH, there are no significant resonance structures as well. Oxygen: For B-O=B (hydrogen peroxide, H2O2), there are also no significant resonance structures. These three cases show that the resonance is not important in these molecules. Thus, statement (b) is true.

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Most popular questions from this chapter

(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.

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