The metals one column to the right of the first have a charge of 2-. This is due to the element having an extra electron in comparison to the number of neutrons.
In this case, according to the given information about the oxidation numbers anf the compounds given, it turns out possible to figure out the oxidation number of manganese in both MnI2, manganese (II) iodide and MnO2, manganese (IV) oxide, by using the concept of charge balance.
Thus, we can define the oxidation state of iodine and oxygen as -1 and -2, respectively, since the former needs one electron to complete the octet and the latter, two of them.
Next, we can write the following
, since manganese has five oxidation states, and it is necessary to calculate the appropriate ones:

Next, we multiply each anion's oxidation number by the subscript, to obtain the following:

Thus, the correct choice is Manganese has an oxidation number of +2 in Mnl2 and +4 in MnO2.
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Six step guide to help you solve problems
Step 1: Identify and define the problem. State the problem as clearly as possible. ...
Step 2: Generate possible solutions. ...
Step 3: Evaluate alternatives. ...
Step 4: Decide on a solution. ...
Step 5: Implement the solution. ...
Step 6: Evaluate the outcome
<u>Answer:</u> The solubility of B is high than the solubility of A.
<u>Explanation:</u>
The solubility is defined as the amount of substance dissolved in a given amount of solvent. More the solute gets dissolved, high will be the solubility and less the solute dissolved, low will be the solubility.
For the given observations:
Mass of undissolved substance of substance A is more than Substance B at every temperature. This implies that less amount of solute gets dissolved in the given amount of solvent.
Hence, substance B has high solubility than substance A.
I can't answer this question if the structural formula is not given. However, I found a similar problem in terms of wording. Taking this problem to be solved, let's take a look at the structural formula as shown in the second picture. First, you must know the parent chain, which is the longest chain. This is a trial-and-error process. The longest chain which has a branching group that is nearest to the head is the correct numbering. In this case, the longest chain has 8 carbon atoms. Thus, the base of the name if octane. Because a 3-carbon chain is branching from the 4th carbon, the IUPAC name of the compound shown is:
<em>4-propyloctane.</em>