Which \(\mathrm{M}^{3+}\) ions \((\mathrm{M}=\) metal) are predicted to have the following ground-state electron configurations: (a) \(\mid \mathrm{Ar}] 3 d^{\prime} ;\) (b) \(\left.\mid \mathrm{Ar}\right] 3 d^{3} ;\) (c) \([K r] 4 d^{3} ;\) (d) [Kr]4d?

Short Answer

Expert verified
The \texttt{\text{\(M^{3+}\) ions with the given ground-state electron configurations are as follows: (a) \(Co^{3+}\), (b) \(Cr^{3+}\), \((c) \)Mo^{3+}\(, (d) \)Ru^{3+}$.}}

Step by step solution

01

Identifying Metal Ion for Configuration (a)

For the ground-state electron configuration \texttt{\text{(a) \([Ar] 3d^6\)}}, we start with the noble gas argon, which has a complete shell of 18 electrons. Adding six electrons to the 3d orbital gives us a configuration of 24 electrons. This represents a metal that has lost three electrons (as \texttt{\text{\(M^{3+}\)}}) to become a +3 cation. The neutral metal would then have 27 electrons, which corresponds to the element cobalt (Co). As a +3 ion, Co would have the configuration \texttt{\text{\([Ar] 3d^6\)}}.
02

Identifying Metal Ion for Configuration (b)

For the ground-state electron configuration \texttt{\text{(b) \([Ar] 3d^3\)}}, we also begin with argon's electron configuration. By adding three electrons to the 3d orbital, the ion will have a total of 21 electrons. As a +3 ion, the neutral atom should have 24 electrons, indicating that the metal is chromium (Cr). The chromium ion \texttt{\text{\(Cr^{3+}\)}} has the electron configuration \texttt{\text{\([Ar] 3d^3\)}}.
03

Identifying Metal Ion for Configuration (c)

For the ground-state electron configuration \texttt{\text{(c) \([Kr] 4d^3\)}}, we begin with krypton, which has 36 electrons in its neutral state. Upon addition of three electrons to the 4d orbital, our ion's configuration shows 39 electrons. For the neutral atom to form a +3 ion, its atomic number would be 42, which is the element molybdenum (Mo). Molybdenum in its +3 oxidation state would have the electron configuration \texttt{\text{\([Kr] 4d^3\)}}.
04

Identifying Metal Ion for Configuration (d)

The ground-state electron configuration for \texttt{\text{(d) \([Kr] 4d^5\)}} starts with krypton at 36 electrons in the neutral atom. Adding five electrons to the 4d orbital gives 41 electrons in total. The neutral atom must then have 44 electrons to become a +3 ion, which is the element ruthenium (Ru). Therefore, ruthenium in the +3 state has the electron configuration \texttt{\text{\([Kr] 4d^5\)}}.

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 Vaia!

Key Concepts

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

Ground-State Electron Configuration
Understanding ground-state electron configurations is critical for students tackling chemistry. A ground-state electron configuration refers to the arrangement of electrons around the nucleus of an atom in the most stable, lowest energy configuration. It's like finding the best seating arrangement in a theater where everyone can see the stage without moving around too much!

Here's a real-life example. Imagine a metal ion as your favorite band, with the different energy levels being sections in the concert hall. The ground state would be the setup where most attendees are closer to the stage (lower energy levels), and each section is filled orderly before moving outwards. The same principle applies to electrons in an atom.

Electron Shell Filling Order

Electrons fill 'seats' or orbitals from low to high energy levels following certain rules—like the aufbau principle—starting from the 1s orbital up to the highest available within that atom's energy levels.

Hund's Rule & The Pauli Exclusion Principle

These rules govern how electrons are distributed among orbitals. Hund's Rule tells us to spread electrons out in the same subshell before doubling up, while Pauli's principle states that no two electrons can have the same set of quantum numbers, essentially meaning no identical twins in the same seat!

When we're dealing with metal ions, the band (metal ion) loses some fans (electrons), especially those furthest from the stage (highest energy levels), leading to a 'positive vibe' or cation with a positive charge.
Metal Cations
Now, let's groove onto the concept of metal cations. A metal cation is an atom that has more protons than electrons, thus carrying a positive charge. Think of it as a generous person giving out free tickets (electrons) to a concert (chemical reaction). The metal becomes positively charged because it now has fewer electron 'fans' than it had 'security guards' (protons).

In our textbook example, metal ions lose electrons to achieve a lower, more stable energy state. Imagine throwing a concert and realizing you need a smaller venue. You might have to turn away some fans (lose some electrons) to fit the new 'venue size', resulting in a smaller, positively charged 'fan club' (metal cation).

Formation of Metal Cations

This act of giving up electrons is known as ionization. The metals we've discussed, like cobalt or chromium, are like rockstars who've just electrified the audience, and in their energized state, they 'jump' into forming a cation, which is a plus for everyone involved. It's all about achieving stability, much like how a sold-out concert feels just right with a packed crowd.
Oxidation States
Finally, the concept of oxidation states sets the stage for understanding the charge of our metal cation rockstars. The oxidation state, often thought of as the 'mood' of the atom, tells us how many electrons an atom has gained or lost. In essence, it's like balancing a budget; you want to know if you're in the red or in the black financially.

For example, a metal ion with a +3 oxidation state has lost three electrons - it’s as if the band paid for three concert goers to see the show (but in a chemical sense, of course). It's a way of keeping track of electron flow in reactions, much like a ticket stub shows who has entered the concert venue.

Conservation of Charge

In any chemical reaction, the total charge before and after must be the same - it's like making sure nobody sneaks in or out of the concert. When an atom like cobalt loses three electrons, we say it has an oxidation state of +3. It's a way to signal that the metal is ready to mingle, chemically speaking, and form different compounds – maybe even a rock group with other elements!

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

2.14 For each of the following ground-state ions, predict the type of orbital \((1 s, 2 p, 3 d, 4 f\), and so on) from which an electron will need to be removed to form the ions of one greater positive charge: (a) \(\mathrm{Ti}^{2+}\); (b) \(\ln ^{+} ;\)(c) \(\mathrm{Te}^{2-} ;\) (d) \(\mathrm{Ag}^{+} .\)

Write the complete Lewis structure for (a) formaldehyde, HCHO, which as its aqueous solution "formalin" is used to preserve biological specimens; (b) methanol, \(\mathrm{CH}_{3} \mathrm{OH}\), the toxic compound also called wood alcohol; (c) glycine, \(\mathrm{CH}_{2}\left(\mathrm{NH}_{2}\right) \mathrm{COOH}\), the simplest of the amino acids, the building blocks of protcins.

\(2.6\) Give the number of valence electrons for each of the following elements: (a) As; (b) \(\mathrm{V}_{\text {; (c) } \mathrm{Cl} \text {; (d) } \mathrm{Ba} \text {. }\)

Which of the following species are radicals: (a) \(\mathrm{NO}\); (b) \(\mathrm{CH}_{3}^{*}\); (c) \(\mathrm{BF}_{4}\); (d) \(\mathrm{BrO}\) ?

An important principle in chemistry is the isolobal analogy. This very simple principle states that chemical fragments with similar valence orbital structures can replace each other in molecules. For example, \(\cdot \mathrm{C}-\mathrm{H}\) and \(\cdot \mathrm{S}-\mathrm{H}\) are isolobal fragments, cach having three electrons with which to form bonds besides the bond to \(\mathrm{H}\). An isolobal series of molecules would be HCCH, HCSiH, and HSiSiH. Similarly, a lone pair of clectrons can be used to replace a bond so that - \(\mathrm{N}\) : is isolobal with ' \(\mathrm{C}-\mathrm{H}\), with the lone pair taking the place of the \(\mathrm{C}-\mathrm{H}\) bond. The isolobal set here is HOCH, HCN, and NN. (a) Draw the Lewis structures for the molecules HCCH, HCSiH, HSiSiH, \(\mathrm{HCN}\), and NN. (b) Using the isolobal principle, draw Lewis structures for molecules based on the structure of benzenc, \(\mathrm{C}_{6} \mathrm{H}_{6}\), in which one or more \(\mathrm{CH}\) groups are replaced with \(\mathrm{N}\) atoms.

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.

Sign-up for free