A student is using the periodic table to find information for a report on elements. One of the statements about fluorine in her report contains an error. Find the error, and write your replacement for the error beneath the statement.

Flourine is a halogen in group 17. Its chemical symbol is F. It is atomic number 9 and has 7 valence electrons. Flourine has a low reactivity compared to many other elements.

Fluorine has a low reactivity compared to many other elements. False

Fluorine has a high reactivity compared to many other elements. True

Explanation:

The elements of group 17 are called halogens. These are six elements Fluorine, Chlorine, Bromine, Iodine, Astatine. Halogens are very reactive these elements can not be found free in nature. Their chemical properties are resemble greatly with each other. As we move down the group in periodic table size of halogens increases that's way fluorine is smaller in size as compared to other halogens elements and highly reactive because of high electronegativity. Their boiling points also increases down the group which changes their physical states. i.e fluorine is gas while iodine is solid.

Fluorine:

1. it is yellow in color.

2. it is flammable gas.

3. it is highly corrosive.

4. fluorine has pungent smell.

5. Its atomic number of 9.

6. Its atomic weight is 18.998 amu.

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.