An alternative definition of electronegativity is electronegativity= constant (i. e — e. a.) where i. e. is the ionization energy and e. a. is the electron affinity using the sign conventions of this book. use data in chapter 12 to calculate the (i. e. - e. a.) term for f, cl, br, and i. do these values show the same trend as the electronegativity values given in this chapter? the first ionization energies of the halogens are 1678, 1255, 1138, and 1007 kj/mol, respectively.

The electronengativity values of given elements is as follows.

Fluorine - 4

Chlorine -3

Bromine - 2.9

Iodine- 2.5

Explanation:

Electronegativity  =consant (I.E-E.A)

The electron affinity and ionization energy values of the given elements is as follows.

(In attachment)

First we have to find the value of constant by using the fluorine atom to whom the electronengativity taken as "4".

Fluorine:

By using this constant values we can find electronegatvity values of remaining elements.

Chlorine:

Therefore, electronegativity of chlorine is 3.

Bromine:

Therefore, electronegativity of bromine is 2.9.

Iodine:

Therefore, electronegativity of iodine is 2.5.

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