/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 Silver and rubidium both form \(... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Silver and rubidium both form \(+1\) ions, but silver is far less reactive. Suggest an explanation, taking into account the ground-state electron configurations of these elements and their atomic radii.

Short Answer

Expert verified
In conclusion, Rubidium's reactivity is higher than Silver's due to its larger atomic radius (248 pm vs 144 pm), making its outermost electron easier to remove. Additionally, Silver has more inner electron shells providing shielding to the outer electrons, reducing the effective nuclear charge. Therefore, Rubidium's increased reactivity is primarily a result of its larger atomic radius and weaker effective nuclear charge compared to Silver.

Step by step solution

01

1. Identify the electron configuration of silver and rubidium

To find the electron configuration of each element, we must look at the periodic table and determine their atomic numbers (number of protons). The atomic number also represents the number of electrons in a neutral atom. For silver (Ag), the atomic number is 47. Its electron configuration is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^9\] For rubidium (Rb), the atomic number is 37. Its electron configuration is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^1\]
02

2. Compare the atomic radii of silver and rubidium

The atomic radius is a measure of the size of an atom. Larger atomic radii imply that the electrons are further away from the nucleus. This can be important when looking at the reactivity of elements. In general, elements with larger atomic radii are more reactive since their outer electrons are less tightly bound by the nucleus. Rubidium has an atomic radius of 248 pm (picometers). Silver has an atomic radius of 144 pm.
03

3. Explain the difference in reactivity

When comparing their electron configurations, we can see that both elements contain a single electron in their outermost s-orbital (5s^1 for Rb and 5s^2 4d^9 for Ag). This is essential for forming a +1 ion, as the loss of an outer electron results in a stable, positively charged ion. However, the atomic radius of Rubidium is larger than that of Silver (248 pm vs 144 pm). This means that the outermost electron in Rubidium is further away from the nucleus, making it less tightly bound and easier to remove. That partly explains why Rubidium is more reactive than Silver. Another point worth considering is the shielding effect of the inner electron shells. Silver has more inner electron shells (full 4d orbitals) that shield the outer electrons from the positive charge of the nucleus. This means that Silver's outermost electron feels a smaller net positive charge. In conclusion, with a larger atomic radius and a weaker effective nuclear charge, Rubidium's outermost electron is more easily removed, making the element more reactive than Silver. This factor, combined with the shielding provided by additional inner electron shells in Silver, helps to explain the observed difference in reactivity between these two elements.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) What is the trend in first ionization energies as one proceeds down the group 7A elements? Explain how this trend relates to the variation in atomic radii. (b) What is the trend in first ionization energies as one moves across the fourth period from \(\mathrm{K}\) to \(\mathrm{Kr}\) ? How does this trend compare with the trend in atomic radii?

Using only the periodic table, arrange each set of atoms in order from largest to smallest: (a) K, Li, Cs; (b) Pb, Sn, Si; (c) F, O, N.

(a) Why does Li have a larger first ionization energy than \(\mathrm{Na}\) ? (b) The difference between the third and fourth ionization energies of scandium is much larger than that of titanium. Why? (c) Why does Li have a much larger second ionization energy than Be?

Little is known about the properties of astatine, At, because of its rarity and high radioactivity. Nevertheless, it is possible for us to make many predictions about its properties. (a) Do you expect the element to be a gas, liquid, or solid at room temperature? Explain. (b) Would you expect At to be a metal, nonmetal, or metalloid? Explain. (c) What is the chemical formula of the compound it forms with \(\mathrm{Na}\) ?

When magnesium metal is burned in air (Figure 3.6), two products are produced. One is magnesium oxide, \(\mathrm{MgO}\). The other is the product of the reaction of \(\mathrm{Mg}\) with molecular nitrogen, magnesium nitride. When water is added to magnesium nitride, it reacts to form magnesium oxide and ammonia gas. (a) Based on the charge of the nitride ion (Table 2.5), predict the formula of magnesium nitride. (b) Write a balanced equation for the reaction of magnesium nitride with water. What is the driving force for this reaction? (c) In an experiment, a piece of magnesium ribbon is burned in air in a crucible. The mass of the mixture of \(\mathrm{MgO}\) and magnesium nitride after burning is \(0.470 \mathrm{~g}\). Water is added to the crucible, further reaction occurs, and the crucible is heated to dryness until the final product is \(0.486 \mathrm{~g}\) of \(\mathrm{MgO}\). What was the mass percentage of magnesium nitride in the mixture obtained after the initial burning? (d) Magnesium nitride can also be formed by reaction of the metal with ammonia at high temperature. Write a balanced equation for this reaction. If a 6.3-g Mg ribbon reacts with \(2.57 \mathrm{~g} \mathrm{NH}_{3}(g)\) and the reaction goes to completion, which component is the limiting reactant? What mass of \(\mathrm{H}_{2}(g)\) is formed in the reaction? (e) The standard enthalpy of formation of solid magnesium nitride is \(-461.08 \mathrm{~kJ} / \mathrm{mol}\). Calculate the standard enthalpy change for the reaction between magnesium metal and ammonia gas.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.