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 \(\displaystyle -2\).

Generally, oxygen has an oxidation number of \(\displaystyle -2\), and hydrogen has an oxidation number of \(\displaystyle +1\). If we multiply \(\displaystyle -2\) by the number of oxygen atoms in the compound (\(\displaystyle 4\)), we get an overall charge of \(\displaystyle -8\) on the oxygen. If we multiply \(\displaystyle +1\) by the number of hydrogen atoms in the compound (\(\displaystyle 1\)), we get an overall charge of \(\displaystyle +1\) on the hydrogen. The sum of \(\displaystyle +1\) and \(\displaystyle -8\) is \(\displaystyle -7\). We can see that \(\displaystyle HClO_{4}\) is a neutral molecule, so the charge on \(\displaystyle Cl\) would have to balance out the \(\displaystyle -7\) charge on the rest of the molecule. To obtain a charge of \(\displaystyle 0\), the oxidation number on \(\displaystyle Cl\) must then be \(\displaystyle +7\).

The answer is .

Lindlar's catalyst enables catalytic hydrogenation, specifically for an alkyne to reduce to a cis-alkene (though the reaction does not continue to produce an alkane). The rest of the answer options are oxidizing agents.

Water, \(\displaystyle H_2O\), has an oxygen atom with an oxidation number of \(\displaystyle -2\) whereas hydrogen peroxide, \(\displaystyle H_2O_2\), has an oxygen atom with an oxidation number of \(\displaystyle -1\). A less negative oxidation number suggests that the oxygen lost an electron. Recall that oxidation is the loss of electrons whereas reduction is the gain of electrons; therefore, the oxygen atom lost an electron and was oxidized when water was converted to hydrogen peroxide.

Oxidation is the process of removing electrons from an atom. This increases the oxidation number (makes oxidation number more positive). An atom undergoes oxidation by losing electrons and donating them to another atom. Since it is facilitating reduction (gain of electrons) of another atom, an atom that is oxidized is also called a reducing agent. Oxidizing agents, on the other hand, are reduced (gain electrons) and facilitate the oxidation of other atoms (removal of electrons from other atoms). Oxidation of an atom does not depend on the Gibbs free energy; therefore, oxidation can be spontaneous or nonspontaneous.