The atomic weight of iodine is less than the atomic weight of tellurium. However, Mendeleev listed iodine after tellurium in his original periodic table because he suspected the atomic weights for these elements were inaccurate. What was Mendeleev's reason for this suspicion?

Explanation :

As we know that Mendeleev arranged the elements in horizontal rows and vertical columns of a table in order of their increasing relative atomic weights.

He placed the elements with similar nature in the same group.

According to the question, the atomic weight of iodine is less than the atomic weight of tellurium. So according to this, iodine should be placed before tellurium in Mendeleev's tables. But Mendeleev placed iodine after tellurium in his original periodic table.

However, iodine has similar chemical properties to chlorine and bromine. So, in order to make iodine queue up with chlorine and bromine in his periodic table, Mendeleev exchanged the positions of iodine and tellurium.

As we know that the positions of iodine and tellurium were reversed in Mendeleev's table because iodine has one naturally occurring isotope that is iodine-127  and tellurium isotopes are tellurium-128 and tellurium-130.

Due to high relative abundance of tellurium isotopes gives tellurium the greater relative atomic mass.

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.