/*! 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 24 Write electron configurations fo... [FREE SOLUTION] | 91影视

91影视

Write electron configurations for the following transition metal ions. a. \(\mathrm{Sc}^{3+}\) b. \(\mathrm{Ru}^{2+}, \mathrm{Ru}^{3+}\) c. \(\operatorname{Ir}^{+}, \operatorname{Ir}^{3+}\) d. \(M n^{2+}\)

Short Answer

Expert verified
The electron configurations for the given transition metal ions are: a. Sc鲁鈦: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^0\) b. Ru虏鈦: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^4\) b. Ru鲁鈦: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^3\) c. Ir鹿鈦: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^8\) c. Ir鲁鈦: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 4f^{14} 5d^5\) d. Mn虏鈦: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\)

Step by step solution

01

Reminder: Aufbau Principle and Orbital Notations

Keep in mind the Aufbau Principle, which states that electrons fill orbitals in order of increasing energy, starting from the lowest energy orbitals. The typical order is: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, etc.
02

a. Sc鲁鈦 Electron Configuration

Step 1: Write the electron configuration for neutral Scandium (Sc) Sc (atomic number 21) has the electron configuration: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1\) Step 2: Remove 3 electrons from the highest energy orbitals for Sc鲁鈦 Since we have a 3+ ion, we need to remove 3 electrons. We take the electrons from the highest energy orbitals, which are 4s and 3d in this case. Electron configuration for Sc鲁鈦 is: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^0\)
03

b. Ru虏鈦 and Ru鲁鈦 Electron Configurations

Step 1: Write the electron configuration for neutral Ruthenium (Ru) Ru (atomic number 44) has the electron configuration: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^6\) Step 2: Remove 2 electrons for Ru虏鈦 Electron configuration for Ru虏鈦 is: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^4\) Step 3: Remove an additional electron for Ru鲁鈦 Electron configuration for Ru鲁鈦 is: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^3\)
04

c. Ir鹿鈦 and Ir鲁鈦 Electron Configurations

Step 1: Write the electron configuration for neutral Iridium (Ir) Ir (atomic number 77) has the electron configuration: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^7\) Step 2: Add 1 electron for Ir鹿鈦 Electron configuration for Ir鹿鈦 is: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^8\) Step 3: Remove 3 electrons for Ir鲁鈦 Electron configuration for Ir鲁鈦 is: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 4f^{14} 5d^5\)
05

d. Mn虏鈦 Electron Configuration

Step 1: Write the electron configuration for neutral Manganese (Mn) Mn (atomic number 25) has the electron configuration: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5\) Step 2: Remove 2 electrons for Mn虏鈦 Electron configuration for Mn虏鈦 is: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\)

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 91影视!

Key Concepts

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

Transition Metal Ions
Transition metal ions are unique because they can form different ions by losing various numbers of electrons. This variability in their ion forms contributes to their diverse chemistry characteristics. Transition metals can display several oxidation states because of their ability to donate electrons from both their outermost shell and their d-subshells. This quality differs from the main group elements, which often have a maximum of two oxidation states. For example, Ruthenium (Ru) can form ions like Ru虏鈦 and Ru鲁鈦 by losing electrons from its 5s and 4d orbitals. Similarly, Iridium (Ir) can form Ir鈦 and Ir鲁鈦 ions. This ability creates a rich set of chemistry rules that cannot easily be applied to other elements. Understanding these transition metal ions is crucial as it provides insights into why they are widely used in industrial processes, catalysts, and even biological systems.
Aufbau Principle
The Aufbau Principle is a fundamental concept used to determine an atom鈥檚 electron configuration. It derives from the German term "Aufbau," which means "building up," describing how electrons populate an atom鈥檚 orbitals. Electrons fill orbitals starting with the lowest available energy level before moving to higher levels. - For instance, they will complete the 1s orbital before moving to 2s. - The common filling order is: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, and so on. As you advance along the periodic table, the electron configuration becomes progressively complex, but the Aufbau Principle supports a systematic approach to determining the order of electron filling, assisting in understanding the chemical properties of elements.
Orbital Notations
Orbital notation is a system used to represent the distribution of electrons in an atom鈥檚 electron cloud. It aids in visualizing how electrons fill sublevels around the nucleus and is often used alongside the electron configuration. - For example, the orbital notation can be written using arrows and lines (or boxes) where each line or box represents an orbital, and arrows indicate electrons with opposite spins. - A neutral Scandium (Sc) atom electons are filled as follows: 鈫戔啌 鈫戔啌 鈫戔啌 鈫戔啌 鈫戔啌 鈫戔啌 鈫戔啌 鈫戔啌 鈫 showing the 1s, 2s, 2p, 3s, 3p, 4s, and 3d orbitals respectively. Orbital notation helps to visualize unpaired electrons and identify possible magnetic properties and chemical reactivity, making it a valuable tool when studying electron configurations.
Electron Removal from Orbitals
To transform a neutral atom into a positive ion, electrons are removed, typically starting from the outermost shell. In transition metals, however, electrons are first removed from the s-orbital before the d-orbital due to their energy levels. - For example, when forming the Sc鲁鈦 ion, you would first remove electrons from the 4s orbital and then from the 3d orbital, going from Sc ( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1 ) to Sc鲁鈦 ( 1s^2 2s^2 2p^6 3s^2 3p^6 3d^0 ) . - This process emphasizes the unique electron configuration priorities and behavior in transition metals, which influences their extensive chemical versatility.

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

Ammonia and potassium iodide solutions are added to an aqueous solution of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} .\) A solid is isolated (compound A), and the following data are collected: i. When 0.105 g of compound A was strongly heated in excess \(\mathrm{O}_{2}, 0.0203 \mathrm{g} \mathrm{CrO}_{3}\) was formed. ii. In a second experiment it took 32.93 \(\mathrm{mL}\) of 0.100\(M \mathrm{HCl}\) to titrate completely the \(\mathrm{NH}_{3}\) present in \(0.341 \mathrm{g} \mathrm{com}-\) pound A. iii. Compound A was found to contain 73.53\(\%\) iodine by mass. iv. The freezing point of water was lowered by \(0.64^{\circ} \mathrm{C}\) when 0.601 \(\mathrm{g}\) compound A was dissolved in 10.00 \(\mathrm{g} \mathrm{H}_{2} \mathrm{O}\left(K_{\mathrm{f}}=\right.\) \(1.86^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol} )\) What is the formula of the compound? What is the structure of the complex ion present? (Hints: \(\mathrm{Cr}^{3+}\) is expected to be six-coordinate, with \(\mathrm{NH}_{3}\) and possibly I- - as ligands. The I- ions will be the counterions if needed.)

Oxalic acid is often used to remove rust stains. What properties of oxalic acid allow it to do this?

Draw geometrical isomers of each of the following complex ions. a. \(\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}-\) b. \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{2}^{2+}\) c. \(\operatorname{Ir}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\) d. \(\mathrm{Cr}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2} \mathrm{I}_{2}^{+}\)

Use standard reduction potentials to calculate \(\mathscr{C}^{\circ}, \Delta G^{\circ},\) and \(K\) (at 298 K) for the reaction that is used in production of gold: $$2 \mathrm{Au}(\mathrm{CN})_{2}-(a q)+\mathrm{Zn}(s) \longrightarrow 2 \mathrm{Au}(s)+\mathrm{Zn}(\mathrm{CN})_{4}^{2-}(a q)$$ The relevant half-reactions are $$\begin{aligned} \operatorname{Au}(\mathrm{CN})_{2}^{-}+\mathrm{e}^{-} \longrightarrow \mathrm{Au}+2 \mathrm{CN}^{-} & \mathscr{C}^{\circ}=-0.60 \mathrm{V} \\ \mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}+4 \mathrm{CN}^{-} & \mathscr{C}^{\circ}=-1.26 \mathrm{V} \end{aligned}$$

Silver is sometimes found in nature as large nuggets; more often it is found mixed with other metals and their ores. Cyanide ion is often used to extract the silver by the following reaction that occurs in basic solution: $$\mathrm{Ag}(s)+\mathrm{CN}^{-}(a q)+\mathrm{O}_{2}(g) \stackrel{\mathrm{Basic}}{\longrightarrow} \mathrm{Ag}(\mathrm{CN})_{2}^{-}(a q)$$ Balance this equation by using the half-reaction method.

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.