The correct answer to this question is the unique atomic number which would be B
You have to calculate the oxidation estates of the atoms in each compound.
I will start with K2Cr2O7 because I believe that Cr is the best candidate to reduce its oxidation number in 3 units.
In K2Cr2O7:
- K has oxidation state of 1+, then K2 has a charge of 2* (1+) = 2+.
- O has oxidation state of 2*, then O7 has a charge of 7* (2-) = 14-.
That makes that Cr2 has charge of 14 - 2 = +12, so each Cr has +12/2 = +6 oxidation state.
In Cr2O3:
- O has oxidation state of 2-, then O3 has charge 3 * (2-) = - 6
- Then, Cr2 has charge 6+, and each Cr has charge 6+ / 2 = 3+.
So, we have seen that Cr reduced its oxidation state in 3 units, from 6+ to 3+.
Answer: Cr has a change in oxidation number of - 3.
Since there is one carbon with 4 Fluorines attached to it, and both compounds are no metals, we use the covalent method for naming,
Here we ignore the prefix for the first element if it is 1. Mono. Then pay attention to the second one, it would be tetra, because tetra means 4. Here there are 4 fluorines.
Drop ine and place ide
CF4 = carbon tetrafluoride.
Answer:
P-positive
N-negative
E-no charge
