Natural rubidium has the average mass of 85.4678 and is composed of isotopes 85 rb(mass = 84.9117) and 87 rb. the ratio of atoms 85rb/87rb in natural rubidium is 2.591. calculate the mass of 87 rb.

Mass of Rb-87 is 86.913 amu.

Explanation:

Given data:

Average mass of rubidium = 85.4678 amu

Mass of Rb-85 = 84.9117

Ratio of 85Rb/87Rb in natural rubidium = 2.591

Mass of Rb = ?

Solution:

The ration of both isotope is 2.591 to 1. Which means that for 2.591 atoms of Rb-85 there is one Rb-87.

For 100% naturally occurring Rb = 2.591 + 1 = 3.591

% abundance of Rb-85 = 2.591/ 3.591 = 0.722

% abundance of Rb-87 = 1 - 0.722= 0.278

84.9117 × 0.722 + X × 0.278 = 85.4678

61.306 + X × 0.278 = 85.4678

X × 0.278 = 85.4678 - 61.306

X × 0.278 = 24.1618

X =  24.1618 / 0.278

X = 86.913 amu

the electronic configuration of an atom with eight electrons is 1s²2s²2p⁴. the orbital diagrams are shown in the figure.

as per the hunds rule:

1. electrons tend to avoid being in the same orbital as per as practicable.

2. electrons in different orbitals of the same energy will have their spin parallel.

this can be illustrated from the multiplicity of the orbitals of the atom. as there are four electrons in the p orbital and p orbital has 3 sub-shells. the electrons can be fed in different ways as shown in the figure but the correct orbital diagram will be as per the hunds rule.

it takes on an electronical energy