/*! 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 60 A Two elements in the second tra... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A Two elements in the second transition series (Y through Cd) have four unpaired electrons in their \(3+\) ions. What elements fit this description?

Short Answer

Expert verified
Molybdenum (Mo) and Ruthenium (Ru) have four unpaired electrons in their 3+ ions.

Step by step solution

01

Understanding the Problem

We need to identify the elements in the second transition series (Y to Cd) whose 3+ ions have four unpaired electrons in their d orbitals.
02

List Second Transition Series Elements

The elements in the second transition series are Yttrium (Y), Zirconium (Zr), Niobium (Nb), Molybdenum (Mo), Technetium (Tc), Ruthenium (Ru), Rhodium (Rh), Palladium (Pd), Silver (Ag), and Cadmium (Cd).
03

Consider Electron Configuration

The electron configurations affect the number of unpaired electrons. For a 3+ ion, electrons are removed starting from the outermost shell. Typically, electrons are removed from "s" orbitals before "d" orbitals.
04

Calculate Unpaired Electrons

Determine the number of unpaired electrons in the d orbitals for each element's 3+ ion. For example, the electron configuration for Ru is [Kr] 4d^7 5s^1. After losing 3 electrons (2 from 5s and 1 from 4d), the configuration becomes [Kr] 4d^5, which has 4 unpaired electrons.
05

Identify Elements

Upon analysis, Molybdenum (Mo) with [Kr] 4d^5 and Ruthenium (Ru) with [Kr] 4d^5 in their 3+ states both have 4 unpaired electrons in their d orbitals.

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!

Key Concepts

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

Electron Configuration
Electron configuration is an essential concept in understanding the behavior of atoms. It tells us how an atom's electrons are distributed across different atomic orbitals. Each electron in an atom occupies a specific energy level within these orbitals.

In the context of transition metals, electron configurations become particularly interesting because of the involvement of d orbitals. These d orbitals start filling after the s orbital of the next principal energy level. This is why, for example, even though the d orbitals are part of the third energy level, they begin filling after the 4s orbital is filled.

For transition metals like those in the second series (from Yttrium to Cadmium), it's essential to note how electrons in the "s" and "d" orbitals contribute to the electron configuration. When we convert an element to a 3+ ion, we first remove electrons from the outermost "s" orbital before removing electrons from the "d" orbital. This loss alters the electron configuration to reflect remaining unpaired electrons, critical for determining chemical properties.
Unpaired Electrons
In atoms, unpaired electrons refer to electrons in an orbital that do not have a partner with opposite spin. Electrons prefer to be in pairs within an orbital, as paired spins lead to more stable configurations. However, in transition metals, it's common to find unpaired electrons, particularly in their d orbitals.

The number of unpaired electrons significantly influences a metal's magnetic properties. Atoms or ions with unpaired electrons are typically paramagnetic, which means they are attracted to a magnetic field. In our specific case with transition series metals, identifying four unpaired electrons in a 3+ ion helps determine the element.

As seen with elements like Molybdenum and Ruthenium, their respective 3+ ions each have configurations of \([Kr] 4d^5\), leaving four unpaired electrons. This is crucial for understanding their chemical and physical properties, such as their high level of reactivity and specific oxidation states.
Oxidation States
Oxidation states refer to the degree of oxidation (or loss of electrons) of an atom in a chemical compound. Transition metals are known for exhibiting a variety of oxidation states due to their partially filled d orbitals, allowing for different levels of electron loss or gain.

For many transition metals in the second series, like Mo and Ru, the oxidation state can be affected by the removal of electrons when forming ions. The oxidation state is critical in determining the element's role in chemical reactions, particularly redox reactions where electron transfer takes place.

When assessing a metal in a 3+ oxidation state, you are looking at the ion after it has lost three electrons. This step is crucial because it directly modifies the electron configuration and can change the magnetic and chemical properties of the ion. Understanding these oxidation states helps predict how these metals will interact with other elements and compounds.

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

Write the electron configurations for \(P\) and \(C l\) using both spdf notation and orbital box diagrams. Describe the relationship between each atom's electron configuration and its position in the periodic table.

Manganese is found as \(\mathrm{MnO}_{2}\) in deep ocean deposits. (a) Depict the electron configuration of this element using the noble gas notation and an orbital box diagram. (b) Using an orbital box diagram, show the electrons beyond those of the preceding noble gas for \(\mathrm{Mn}^{4+}\) (c) Is \(\mathrm{Mn}^{4+}\) paramagnetic? (d) How many unpaired electrons does the \(\mathrm{Mn}^{4+}\) ion have?

Using orbital box diagrams, depict an electron configuration for each of the following ions: (a) \(\mathrm{Na}^{+}\) (b) \(\mathrm{Al}^{3+},(\mathrm{c}) \mathrm{Ge}^{2+},\) and (d) \(\mathrm{F}^{-}\)

Using spdf and noble gas notations, write electron configurations for atoms of the following elements. (Try to do this by looking at the periodic table but not at Table \(7.3 .\) ) (a) Strontium, Sr. This element is named for a town in Scotland. (b) Zirconium, Zr. The metal is exceptionally resistant to corrosion and so has important industrial applications. Moon rocks show a surprisingly high zirconium content compared with rocks on Earth. (c) Rhodium, Rh. This metal is used in jewelry and in catalysts in industry. (d) Tin, Sn. The metal was used in the ancient world. Alloys of tin (solder, bronze, and pewter) are important.

List the first five orbitals (the five orbitals in an atom at lowest energy) in order of filling, according to the Aufbau principle.

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.