GThe atomic radii of a divalent cation and a monovalent anion are 0.085 nm and 0.125 nm, respectively. (a) Calculate the force of attraction between these two ions at their equilibrium interionic separation (i. e., when the ions just touch one another)./homework-help/electron-su bshell-filled-rare-earth-series-ele ments-periodic-chapter-2-problem-14 qp-solution-9781118324578-exc

1.05 × 10⁻⁸ N

Explanation:

Using Coulomb's law

F force of attraction between them = k = k

K coulomb constant = 8.99 × 10⁹ N m²C⁻²

Z₁ = + 2 ( valency electron)

Z₂ = -1 ( valency electron)

r is the distance between then since they just touch one another = 0.085 nm + 0.125 nm = 0.21 nm = 0.21 × 10⁻⁹ m

q charge = 1.602 × 10⁻¹⁹ C of an electron

F = ( 2 × 1 × 8.99 × 10⁹ N m²C⁻² × (1.602 × 10⁻¹⁹ C)² / ( 0.21 × 10⁻⁹ m )² = 1046.34 × 10 ⁻¹¹ = 1.05 × 10⁻⁸ N

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.

a:

by the combination of an unlimited number of elements