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Consider the stable elements through lead \((Z=82)\). In how many instances are the atomic weights of the elements in the reverse order relative to the atomic numbers of the elements? What is the explanation for these cases?

Short Answer

Expert verified
In short, there are two instances where the atomic weights of stable elements up to lead (Z=82) are in reverse order relative to their atomic numbers: 1. Argon (Ar) \(Z=18\) and Potassium (K) \(Z=19\) 2. Cobalt (Co) \(Z=27\) and Nickel (Ni) \(Z=28\) The explanation for these instances is due to isotopic abundance, as the atomic weight of an element is a weighted average of the naturally occurring isotopes' masses of the element. In both cases, the relative abundances of the heavier isotope for the lighter element are higher than for the heavier element, resulting in the overall atomic weight being higher for the element with a lower atomic number and lower for the element with a higher atomic number.

Step by step solution

01

Understand Atomic Weight and Atomic Number

Atomic number (Z) is the number of protons in the nucleus of an atom, and it determines the element's place in the periodic table. Atomic weight is the weighted average mass of the naturally occurring isotopes of an element. It should be noted that the atomic weight of an element will always be greater than or equal to its atomic number, as it is the mass of protons and neutrons combined.
02

Identify the Stable Elements up to Lead (Z=82)

A stable element is one that does not undergo radioactive decay. In this exercise, we shall consider elements up to lead (Z=82), which is the heaviest stable element. We will have to analyze the atomic weights and atomic numbers of these elements.
03

Compare Atomic Weights and Atomic Numbers for the Elements

In order to find the instances where the atomic weights of the stable elements are in reverse order relative to the atomic numbers, we have to compare and analyze both quantities for these elements. We need to find instances where the element with a higher atomic number has a lower atomic weight than the element with a lower atomic number.
04

Identify the Instances Where Atomic Weights Are in Reverse Order

There are two instances where the atomic weights of consecutive elements are in reverse order relative to their atomic numbers: 1. Argon (Ar) \(Z=18\) and Potassium (K) \(Z=19\). 2. Cobalt (Co) \(Z=27\) and Nickel (Ni) \(Z=28\). In both these cases, the atomic weight of the element with a higher atomic number is less than that of the element with a lower atomic number.
05

Explain the Reason for These Instances

The explanation for the reverse order in atomic weights relative to the atomic numbers is due to isotopic abundance. The atomic weight of an element is not an integer number, as it is a weighted average of the naturally occurring isotopes' masses of an element. In both instances mentioned above, the relative abundances of the heavier isotope for the lighter element are higher than for the heavier element. This results in the overall atomic weight being higher for the element with a lower atomic number and lower for the element with a higher atomic number. In conclusion, there are two instances where the atomic weights of stable elements up to lead are in reverse order relative to their atomic numbers, and the reason for this is due to the isotopic abundance of the elements involved.

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Most popular questions from this chapter

Consider the change in effective nuclear charge experienced by a \(2 p\) electron as we proceed from \(C\) to \(N\). (a) Based on a simple model in which core electrons screen the valence electrons completely and valence electrons do not screen other valence electrons, what do you predict for the change in \(Z_{\text {eff }}\) from \(C\) to \(N\) ? (b) What change do you predict using Slater's rules? (c) The actual change in \(Z_{\text {eff }}\) from \(C\) to \(N\) is \(0.70+.\) Which approach to estimating \(Z_{\text {eff }}\) is more accurate? (d) The change in \(Z_{\text {dif }}\) from \(N\) to \(O\) is smaller than that from \(C\) to \(N\). Can you provide an explanation for this observation?

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