Write the chemical symbols for three different atoms or atomic anions with 10 electrons.

            i)  Neon (Ne)

            ii)  Fluoride Ion (F⁻¹)

             iii)  Oxide Ion  (O⁻²)

Explanation:

                   Ions are those charged species which are either positively charged (by loosing electrons) called as cations or negatively charged (by gaining electrons) called as anions.

                  In given examples, Neon is a neutral atom which has an atomic number 10. It contains 10 electrons in its neutral state with the electronic configuration 1s², 2s², 2p⁶.

                  Fluorine atom has an atomic number of 9. Therefore, it contains 9 electrons in its neutral state with an electronic configuration of 1s², 2s², 2p⁵. When Fluorine atom gains one electron it gets 10 electrons with electronic configuration of 1s², 2s², 2p⁶.

                 Oxygen atom has an atomic number of 8. Therefore, it contains 8 electrons in its neutral state with an electronic configuration of 1s², 2s², 2p⁴. When Oxygen atom gains two electron it gets 10 electrons with electronic configuration of 1s², 2s², 2p⁶ forming an Oxide Ion.

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years