Calculate the fraction of lattice sites that are Schottky defects for cesium chloride at 623 oC (this temperature is below the melting temperature (645oC)). Assume an energy for defect formation of 1.86 eV.

5.9 × 10^-6.

Explanation:

In the arrangements of crystal solids there is likely going to be an imperfection or defect and one of the defect or imperfections in the arrangements of solids is known as the Schottky defects. The Schottky defects is a kind of lattice arrangements imperfection that occurs when positively charged ions and negatively charged ions leave their position.

So, let us delve right into the solution of the question. We will be making use of the formula below;

Wb/ W = e^ - c/ 2kT.

Where Wb/ W= fraction of lattice sites, c= energy for defect formation = 1.86 eV, and T = temperature= 623° C= 896 k.

So, Wb/ W = e ^ -1.86/ (2 × 896 × 8.62 × 10^ -5).

Wb/ W= 0.000005896557435956372.

Wb/ W=5.9 × 10^-6.

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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.