\(\displaystyle Na/NH_{3}\) produces a trans-alkene from an alkyne whereas \(\displaystyle H_{2}/Lindlar's\) \(\displaystyle Catalyst\) produces a cis-alkene. \(\displaystyle H_{2}/Pd/C\) reduces an alkyne all the way down to an alkane. \(\displaystyle MCPBA\) is a strong oxidizing agent.

This is a standard organolithium reaction.

The organolithium product can be thought of as a strong nucleophile. The carbon steals an electron from the lithium to create \(\displaystyle H_{3}C^{-}Li^{+}\). From there, the highly reactive carbo-anion is free to attack the ketone at the site of its carbon to form a tertiary alcohol on the cyclohexane.

\(\displaystyle NaBH_{4}\) is the only answer choice that is not an oxidizing agent. In fact, it is a reducing agent because of the lack of oxygen atoms present. This compound adds hydrogen atoms to a compound, thereby reducing it.

Alkenes can be reduced in the presence of \(\displaystyle H_2\) and a metal catalyst like platinum to hydrogenate the alkene to give a saturated alkane. The reaction occurs in a heterogeneous solution rather than a homogenous solution. It occurs on the presence of a solid surface of the metal catalyst.

Note that the three carbon-carbon double bonds in the aromatic ring in the presence of the reducing agent does not get reduced because they are extremely stable due to resonance.

Calcium is in the second group of the periodic table, so its oxidation number will always be \(\displaystyle +2\). This is true for all alkaline earth metals.

As for chromium, the dichromate anion (\(\displaystyle Cr_2O_7^{-2}\)) has an overall charge of \(\displaystyle -2\). Each oxygen atom will always have an oxidation number of \(\displaystyle -2\) as well, so the oxygens would have a total negative charge of \(\displaystyle -14\). The sum of the oxidation numbers have to equal the overall charge, so chromium must be used to balance the negative charges from the oxygen. To bring the overall charge to \(\displaystyle -2\), the two chromiums have to have total charge of \(\displaystyle +12\), giving each an oxidation number of \(\displaystyle +6\).

Manganese is a transition metal, so its oxidation number is more variable than the other elements of the compound. Oxygen is almost always \(\displaystyle -2\). Potassium is almost always \(\displaystyle +1\). The sum of the oxygens and the potassium is \(\displaystyle -7\).

\(\displaystyle KMnO_4\)

\(\displaystyle K+Mn+4(O)=0\)

\(\displaystyle 1+Mn+4(-2)=-7+Mn=0\)

The compound is neutral, so the manganese is \(\displaystyle +7\) to balance the net molecular charge.

To solve this question we need to calculate the oxidation number of oxygen in both molecules. The formula for water is \(\displaystyle H_2O\). The oxidation number of hydrogen is +1. Since there are two of them, the hydrogen atoms contribute to a charge of +2. The water molecule is neutral; therefore, the oxygen must have an oxidation number of \(\displaystyle -2\) to balance the charge. The formula for hydrogen peroxide is \(\displaystyle H_2O_2\). Using the same logic as water, we can determine that hydrogen contributes +2. We have two oxygen atoms in this case; therefore, each oxygen atom will have an oxidation number of \(\displaystyle -1\) to give a charge of \(\displaystyle -2\). This will balance the charges and provide a neutral hydrogen peroxide molecule.

Recall that have a negative charge suggests that an atom has extra valence electrons. A charge of \(\displaystyle -1\) suggests one extra valence electron and a charge of \(\displaystyle -2\) suggests two extra valence electrons. An oxygen typically has six valence electrons. The oxygen in water has \(\displaystyle -2\) oxidation number; therefore, it will have two extra valence electrons (eight total). On the other hand, oxygen in hydrogen peroxide will have one extra valence electron (seven total); therefore, oxygen in hydrogen peroxide has one less valence electron than oxygen in water.

Potassium bromide has a formula of \(\displaystyle KBr\). This molecule is made up of an alkali metal (potassium) and a halogen (bromine). Alkali metals have one valence electron that they readily lose to obtain octet whereas halogens have seven valence electrons and they readily gain an electron to obtain octet. Recall that losing an electron will give you a \(\displaystyle +1\) oxidation number whereas gaining an electron will give you a \(\displaystyle -1\) oxidation number. This means that alkali metals always have an oxidation number of \(\displaystyle +1\) whereas halogens always have an oxidation number of \(\displaystyle -1\); therefore, potassium has an oxidation number of \(\displaystyle +1\).

Manganese(III) oxide has a formula of \(\displaystyle Mn_2O_3\). The oxidation number of oxygen is \(\displaystyle -2\). Since there are three of them, the oxygen atoms contribute to a charge of \(\displaystyle -6\). To maintain the neutrality of the molecule, manganese must balance the \(\displaystyle -6\) charge (meaning contribute \(\displaystyle +6\) charge). There are two manganese atoms; therefore, each manganese atom will have an oxidation number of \(\displaystyle +3\) and contribute \(\displaystyle +6\) charge.

The reaction stated in this question is as follows.

\(\displaystyle NaCl\: + \: LiBr \rightarrow NaBr\: + LiCl\)

This is a double replacement reaction. Both products contain a metal (an alkali metal) and a nonmetal (a halogen). The oxidation number of all alkali metals (first column of periodic table) is \(\displaystyle +1\) and of all halogens (seventh column of periodic table) is \(\displaystyle -1\). The results stated in the question seem invalid because a halogen can never have an oxidation number of \(\displaystyle -2\).