Homolytic bond breaking occurs when a molecule breaks up to form two or more new products. In the reaction given, molecular chlorine forms two radicals in which one electron stays with each fragment formed.

Lithium aluminum hydride is correct because it is a reducing agent, and is therefore not capable of oxidizing secondary alcohols. Instead, LAH could be used to perform the reverse reaction, reducing a ketone to an alcohol. The other answer choices are oxidizing agents.

In an oxidation-reduction (redox) reaction, reduction and oxidation both occur. Thus, not all of the elements can be oxidized, and not all of them can be reduced. In this equation, the oxidation number of oxygen is . Multiplying that by the number of oxygen atoms (), the overall charge on oxygen is .  has an overall  charge, so the oxidation number on  must combine with  to form . Thus, the oxidation number of  at the beginning of the reaction is . The iodine at the beginning of the reaction has an oxidation number of , as seen by the negative superscript.

At the end of the reaction,  has an oxidation number of , as seen by the positive superscript. Iodine has an oxidation number of  (there is no charge on the ).

Thus,  went from having an oxidation number of  to one of . Iodine went from having an oxidation number of  to one of . Oxidation occurs when electrons are lost (the number becomes more positive), and reduction occurs when electrons are gained (the number becomes more negative). Because the oxidation number of iodine became more positive, iodine was oxidized. Because the oxidation number of manganese became more negative (less positive), manganese was reduced.

For a compound to be considered aromatic, it must be flat, cyclic, and conjugated and it must obey Huckel's rule. Huckel's rule states that an aromatic compound must have  pi electrons in the overlapping p orbitals in order to be aromatic (n in this formula represents any integer). Only compounds with 2, 6, 10, 14, . . .  pi electrons can be considered aromatic. Compound A has 6 pi electrons, compound B has 4, and compound C has 8. This eliminates answers B and C. Answer D is not cyclic, and therefore cannot be aromatic. The only aromatic compound is answer choice A, which you should recognize as benzene.

The correct answer is (8) Annulene. This is because all aromatic compounds must follow Huckel's Rule, which is 4n+2. Note that "n" in Huckel's Rule just refers to any whole number, and 4n+2 should result in the number of pi electrons an aromatic compound should have. For example, 4(0)+2 gives a two-pi-electron aromatic compound.

It is also important to note that Huckel's Rule is just one of three main rules in identifying an aromatic compound. An annulene is a system of conjugated monocyclic hydrocarbons. A compound is considered anti-aromatic if it follows the first two rules for aromaticity (1. Pi bonds are in a cyclic structure and 2. The structure must be planar), but does not follow the third rule, which is Huckel's Rule.

(8) Annulene follows the first two rules, but not Huckel's Rule, and is therefore antiaromatic; no value of a whole number for "n" will result in 8 with the formula 4n+2.

A molecule is anti-aromatic when it follows all of the criteria for an aromatic compound, except for the fact that it has  pi electrons rather than  pi electrons, as in this case.

A molecule is aromatic when it adheres to 4 main criteria:

1. The molecule must be planar

2. The molecule must be cyclic

3. Every atom in the aromatic ring must have a p orbital

4. The ring must contain  pi electrons

The carbon on the left side of this molecule is an sp³ carbon, and therefore lacks an unhybridized p orbital. The molecule is non-aromatic.

There are 14 pi electrons because oxygen must contribute 2 pi electrons to avoid antiaromaticity.  The other 12 pi electrons come from the 6 double bonds.

Nitrogen does not contribute any pi electrons, as it is  hybridized and it's lone pairs are stored in sp² orbitals, incapable of pi delocalization.  Each nitrogen's p orbital is occupied by the double bond.

If oxygen contributes any pi electrons, the molecule will have 12 pi electrons, or 4n pi electrons, and become antiarmoatic.  As it is now, the compound is antiaromatic.

Nitrogen cannot give any pi electrons because it's lone pair is in an sp² orbital. Boron has no pi electrons to give, and only has an empty p orbital.

In this question, we're presented with the structure of anthracene, and we're asked to find which answer choices represent a true statement about anthracene. Let's go through each of the choices and analyze them, one by one.

First, let's determine if anthracene is planar, which is essentially asking if the molecule is flat. When looking at anthracene, we see that the molecule is conjugated, meaning there are alternating single and double bonds. If we look at each of the carbons in this molecule, we see that all of them are  hybridized. This means that each of the three other atoms connected to the carbon are organized at a  angle in a single plane. Since ALL of the carbons are this way, we can conclude that anthracene is a planar compound.

Now let's determine the total number of pi electrons in anthracene. Remember, pi electrons are those that contribute to double and triple bonds. So, we'll need to count the number of double bonds contained in this molecule, which turns out to be . But, don't forget that for every double bond there are two pi electrons! Therefore, the total number of pi electrons is twice the amount of the number of double bonds, which gives a value of  pi electrons. This is indeed an even number.

Lastly, let's see if anthracene satisfies Huckel's rule. This rule is one of the conditions that must be met for a molecule to be aromatic. It states that when the total number of pi electrons is equal to , we will be able to have  be an integer value. So let's see if this works.

Since we arrived at an integer value for , we can conclude that Huckel's rule has indeed been satisfied.

All of the answer choices are true statements with regards to anthracene.