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Consider the stable elements through lead \((Z=82)\). In how many instances are the atomic weights of the elements out of order relative to the atomic numbers of the elements?

Short Answer

Expert verified
There are 7 instances where the atomic weights of the elements are out of order relative to their atomic numbers for stable elements through lead (Z=82). These instances are: 1. Argon \((Z=18)\) and potassium \((Z=19)\) 2. Cobalt \((Z=27)\) and nickel \((Z=28)\) 3. Tellurium \((Z=52)\) and iodine \((Z=53)\) 4. Cerium \((Z=58)\) and praseodymium \((Z=59)\) 5. Thulium \((Z=69)\) and ytterbium \((Z=70)\) 6. Gold \((Z=79)\) and mercury \((Z=80)\) 7. Thallium \((Z=81)\) and lead \((Z=82)\)

Step by step solution

01

List the stable elements

List the stable elements with their atomic numbers and atomic weights. To save time, you can use a periodic table or search online for a list of the stable elements and their atomic weights. For this problem, work with elements up to lead (Pb) with atomic number 82.
02

Step 2:Compare the order of atomic numbers and atomic weights

Compare the order of atomic numbers and atomic weights for each stable element. Start by checking if the element with the smallest atomic weight has the smallest atomic number. Then, continue this process for the rest of the elements. If you encounter an element with atomic weights in a different order than their atomic numbers, count that as an instance. For example: Compare the atomic weight of hydrogen (1.008) with the atomic weight of helium (4.0026). Hydrogen has a lower atomic weight and a lower atomic number, so these two elements are not an instance of out-of-order elements.
03

Count the instances of out-of-order elements

Count the total number of instances where atomic weights are not in the same order as their atomic numbers. These are instances where the order of atomic weights does not match the order of atomic numbers for a pair of elements. After going through all stable elements up to lead (82), you will find that there are only 7 instances in which the atomic weights of the elements are not in the same order as their atomic numbers: 1. Argon \((Z=18)\) and potassium \((Z=19)\) 2. Cobalt \((Z=27)\) and nickel \((Z=28)\) 3. Tellurium \((Z=52)\) and iodine \((Z=53)\) 4. Cerium \((Z=58)\) and praseodymium \((Z=59)\) 5. Thulium \((Z=69)\) and ytterbium \((Z=70)\) 6. Gold \((Z=79)\) and mercury \((Z=80)\) 7. Thallium \((Z=81)\) and lead \((Z=82)\) From our analysis, there are 7 instances (out-of-order pairs) where the atomic weight order does not match with the atomic number order.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Periodic Table
The periodic table is a fundamental tool in chemistry that organizes all known chemical elements in an informative array. It is designed to highlight the recurring properties of elements, facilitating the understanding of chemical behavior and relationships among elements. The typical layout consists of rows called periods and columns known as groups or families.

Elements are arranged in order of increasing atomic number, which is the number of protons in an atom's nucleus. This number is unique to each element and is central to the element's identity. As you move from left to right in a period, the atomic weight generally increases. Atomic weight, measured in atomic mass units (amu), is the average mass of atoms of an element, accounting for the presence of isotopes and their relative abundance.

However, it's important to note that the atomic weight doesn't increase uniformly, as it's influenced by the number of neutrons and the natural abundance of an element's isotopes. This can sometimes lead to unexpected orderings, which might seem like a discrepancy if one expects strict atomic weight increase with atomic number. Students typically familiarize themselves with the table's arrangement throughout their study of chemistry, using it as a key reference for understanding elemental properties and predicting how different elements will react with one other.
Stable Elements on the Periodic Table
In the context of the periodic table, stable elements are those that do not undergo radioactive decay over observable timescales. In essence, they have a balance of protons and neutrons in their nucleus that confers stability. In our exercise, we consider elements up to and including lead (Pb), with atomic number 82, because all elements beyond lead are radioactive to some degree.

Stability in elements is also related to their atomic structure and the completion of electron shells, with noble gases being prime examples of naturally stable elements thanks to their full valence shells. Understanding stability is crucial not just for fundamental chemistry, but also for practical applications, such as materials science and nuclear chemistry. Knowing stable elements and their properties enables chemists to predict and explain various chemical reactions and why some elements readily combine while others do not.
Element Order Discrepancy Explained
The element order discrepancy is observed when the atomic weights of certain pairs of elements do not align with the increasing order of their atomic numbers. This can seem counterintuitive since one might expect these two values to correlate directly. However, this is where the nuanced nature of atomic structure plays a role.

Several factors contribute to this discrepancy, including variations in the number of neutrons and the distribution of naturally occurring isotopes. Isotopes are atoms of the same element that have different numbers of neutrons, causing them to have different atomic weights. Naturally occurring isotopes have different abundances, which affect the average atomic weight of an element. In some cases, an element with a higher atomic number might have a lower atomic weight than its predecessor due to these isotopic abundances.

For example, cobalt with an atomic number of 27 actually has a slightly larger atomic weight than nickel with an atomic number of 28 due to the relative isotopic abundances. When evaluating elements based on these criteria, students are encouraged to be attentive to the precise definition of atomic weight as an average that includes isotopic composition, which clarifies why these order discrepancies occur.

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Most popular questions from this chapter

Detailed calculations show that the value of \(Z_{\text {eff }}\) for the outermost electrons in \(\mathrm{Na}\) and \(\mathrm{K}\) atoms is \(2.51+\) and \(3.49+\), respectively. (a) What value do you estimate for \(Z_{\text {eff }}\) experienced by the outermost electron in both \(\mathrm{Na}\) and \(\mathrm{K}\) by assuming core electrons contribute \(1.00\) and valence electrons contribute \(0.00\) to the screening constant? (b) What values do you estimate for \(Z_{\text {eff }}\) using Slater's rules? (c) Which approach gives a more accurate estimate of \(Z_{\text {eff ? }}\) ? (d) Does either method of approximation account for the gradual increase in \(Z_{\text {eff }}\) that occurs upon moving down a group? (e) Predict \(Z_{\text {eff }}\) for the outermost electrons in the \(\mathrm{Rb}\) atom based on the calculations for \(\mathrm{Na}\) and \(\mathrm{K}\).

It is possible to define metallic character as we do in this book and base it on the reactivity of the element and the ease with which it loses electrons. Alternatively, one could measure how well electricity is conducted by each of the elements to determine how "metallic" the elements are. On the basis of conductivity, there is not much of a trend in the periodic table: Silver is the most conductive metal, and manganese the least. Look up the first ionization energies of silver and manganese; which of these two elements would you call more metallic based on the way we define it in this book?

The electron affinities, in \(\mathrm{kJ} / \mathrm{mol}\), for the group \(1 \mathrm{~B}\) and group 2B metals are as follows: (a) Why are the electron affinities of the group \(2 B\) elements greater than zero? (b) Why do the electron affinities of the group \(1 \mathrm{~B}\) elements become more negative as we move down the group? [Hint: Examine the trends in the electron affinities of other groups as we proceed down the periodic table.]

(a) If the core electrons were totally effective at screening the valence electrons and the valence electrons provided no screening for each other, what would be the effective nuclear charge acting on the \(3 s\) and \(3 p\) valence electrons in \(\mathrm{P}\) ? (b) Repeat these calculations using Slater's rules. (c) Detailed calculations indicate that the effective nuclear charge is \(5.6+\) for the \(3 s\) electrons and \(4.9+\) for the \(3 p\) electrons. Why are the values for the \(3 s\) and \(3 p\) electrons different? (d) If you remove a single electron from a P atom, which orbital will it come from?

Based on their positions in the periodic table, predict which atom of the following pairs will have the smaller first ionization energy: (a) \(\mathrm{Cl}, \mathrm{Ar}\); (b) Be, Ca; (c) K, Co; (d) S, Ge; (e) Sn, Te.

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