What best describes bromide ion that forms?

A  potassium atom (atomic number 19) and a bromine atom (atomic number 35) can form a chemical bond through a transfer of one electron. The potassium ion that forms has 18 electrons. What best describes the bromide ion that forms?

Explanation:

Original Question:

A  potassium atom (atomic number 19) and a bromine atom (atomic number 35) can form a chemical bond through a transfer of one electron. The potassium ion that forms has 18 electrons. What best describes the bromide ion that forms?

Ionic Bond (Bromide ion with 36 electrons and unit negative charge)

Explanation:

Whenever an atom gains or loses one or more electrons, it converts into either a positive or negative charge which are termed as ions.

Potassium atom has atomic number 19, which means it has 19 electrons and 19 protons. Potassium is a very unstable metal and it reacts with exothermic reaction when comes in contact with air or water. Potassium atom loses an electron making a positive ion because the number of protons were balancing number of electrons before, and after losing an electron, number of proton exceeded number of electrons i.e. 19 - 18 = +1 which is the unit charge of each potassium atom.

Similarly, Bromine atom has atomic number 35. Before chemical reaction, it had 35 protons and 35 electrons. Bromine comes in halogen group (Group VII-A) and is a salt former. Bromine atom gains one electron from potassium and achieve noble gas configuration of Krypton (Kr), thereby having unit negative charge because of an electron gain which overwhelmed the 35 protons. i.e. 35 - 36 = -1

2K + Br2 > 2KBr

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