Consider four atoms from the second period: lithium, beryllium boron carbon and nitrogen which of these element has the lowest e
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5 bromo 1.1.1 -trichloro -5 ethyl 2.3 dimethyl heptane
Answer:
.
Explanation:
Consider the oxidation state on each of the element:
Left-hand side:
- O: -2 (as in most compounds);
- Cr:
;
- Fe: +2 (from the charge of the ion);
Right-hand side:
- Cr: +3;
- Fe: +3.
Change in oxidation state:
- Each Cr atom: decreases by 3 (reduction).
- Each Fe atom: increases by 1 (oxidation).
Changes in oxidation states shall balance each other in redox reactions. Thus, for each Cr atom on the left-hand side, there need to be three Fe atoms.
Assume that the coefficient of the most complex species is 1. There will be two Cr atoms and hence six Fe atoms on the left-hand side. Additionally, there are going to be seven O atoms.
Atoms are conserved in chemical reactions. As a result, the right-hand side of this equation will contain
- two Cr atoms,
- six Fe atoms, and
- seven O atoms.
O atoms seldom appear among the products in acidic environments; they rapidly combine with ions to produce water
. Seven O atoms will make seven water molecules. That's fourteen H atoms and hence fourteen
ions on the product side of this equation. Hence the balanced equation. Double check to ensure that the charges on the ions also balance.
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