/*! 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 each of the following. a. \(\mathrm{Cr}, \mathrm{Cr}^{2+}, \mathrm{Cr}^{3+}\) b. \(\mathrm{Cu}, \mathrm{Cu}^{+}, \mathrm{Cu}^{2+}\) c. \(\mathrm{V}, \mathrm{V}^{2+}, \mathrm{V}^{3+}\)

Short Answer

Expert verified
a. Cr: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5\), Cr虏鈦: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^4\), Cr鲁鈦: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\) b. Cu: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^{10}\), Cu鈦: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10}\), Cu虏鈦: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^9\) c. V: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^3\), V虏鈦: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\), V鲁鈦: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^2\)

Step by step solution

01

a. Electron configurations for Cr, Cr虏鈦, Cr鲁鈦

First, we need to determine the atomic number of Chromium (Cr). Cr has an atomic number of 24, which means it has 24 electrons. Now, let's find the electron configurations for each species: 1. Cr: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5\). This is an exception to the rule when writing electron configurations as the half-filled 3d subshell is more stable than the fully-filled 4s subshell. 2. Cr虏鈦: To form Cr虏鈦, two electrons are removed from Cr. Electrons are removed starting from the 4s subshell. The resulting configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^4\). 3. Cr鲁鈦: To form Cr鲁鈦, three electrons are removed from Cr. Again, after removing two electrons from the 4s subshell, the next electron will be removed from the 3d subshell. The final configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\).
02

b. Electron configurations for Cu, Cu鈦, Cu虏鈦

Copper (Cu) has an atomic number of 29, which means it has 29 electrons. Now, let's find the electron configurations for each species: 1. Cu: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^{10}\). Just like Cr, Cu has a more stable configuration with a completely-filled 3d subshell. 2. Cu鈦: To form Cu鈦, one electron is removed from Cu. The electron is removed from the 4s subshell, resulting in the configuration \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10}\). 3. Cu虏鈦: To form Cu虏鈦, two electrons are removed from Cu. After removing one electron from the 4s subshell, the other electron is removed from the 3d subshell. The final configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^9\).
03

c. Electron configurations for V, V虏鈦, V鲁鈦

Vanadium (V) has an atomic number of 23, which means it has 23 electrons. Let's find the electron configurations for each species: 1. V: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^3\). 2. V虏鈦: To form V虏鈦, two electrons are removed from V. The electrons are removed from the 4s subshell first. The resulting configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\). 3. V鲁鈦: To form V鲁鈦, three electrons are removed from V. After removing two electrons from the 4s subshell, the next electron is removed from the 3d subshell. The final configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^2\).

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.

Chromium Electron Configuration
Understanding the electron configuration of chromium (Cr) is essential as it represents a notable exception to typical electron distribution rules. Chromium has an atomic number of 24, indicating 24 electrons in its neutral state.

The ground-state electron configuration for Cr is expressed as:
  • Chromium (Cr): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5\)
This configuration deviates from the expected \(4s^2 3d^4\). Instead, it occurs because a half-filled \(3d\) subshell (\(3d^5\)) offers additional stability compared to a completely filled \(4s\) subshell. This stability results from the symmetrical distribution of the electrons in the \(3d\) subshell, which provides a 'balance'.

When chromium ions are formed, electrons are removed first from the \(4s\) subshell,:
  • Chromium (II) ion (Cr^{2+}): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^4\)
  • Chromium (III) ion (Cr^{3+}): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\)
This demonstrates that stability can also be achieved through specific electron arrangements within the d-orbital. After reaching the trivalent state, more electrons are removed from the \(3d\), revealing how electron configurations adapt to maintain stability.
Copper Electron Configuration
Copper (Cu), with an atomic number of 29, presents another fascinating case of electron configuration that prioritizes stability. A neutral copper atom's electron configuration is:
  • Copper (Cu): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^{10}\)
This arrangement features a fully filled \(3d\) subshell, providing increased stability, unlike the expected \(4s^2 3d^9\). The electron moves from the \(4s\) to the \(3d\) orbital to achieve a complete \(3d\) block, which is considered lower in energy and more stable.

Copper ions continue to exhibit fascinating electron removal pathways:
  • Copper (I) ion (Cu^{+}): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10}\)
  • Copper (II) ion (Cu^{2+}): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^9\)
The formation of Cu鈦 involves removing an electron from the \(4s\) subshell, maintaining the complete \(3d\) subshell. Continuing with Cu虏鈦, an additional electron is removed from the \(3d\) subshell, although this disrupts its overall symmetry slightly, making it slightly less stable compared to Cu鈦.
Vanadium Electron Configuration
Vanadium (V) is another transition metal where the awareness of electron removal order becomes crucial. With an atomic number of 23, a neutral vanadium atom's electron configuration is:
  • Vanadium (V): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^3\)
Vanadium follows the expected filling order. However, for ions, electron removal starts from the \(4s\) before the \(3d\), aligning with the lower energy level principles.

For its ions:
  • Vanadium (II) ion (V^{2+}): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\)
  • Vanadium (III) ion (V^{3+}): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^2\)
Formation of V虏鈦 involves removing electrons first from the \(4s\) subshell, resulting in leaves the three \(3d\) electrons untouched, which are important for the metal's chemical characteristics. Progressing to V鲁鈦, a single \(3d\) electron is removed next to balance energy further, demonstrating once more how electrons arrange for optimal stability across transition metals.

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

The following statements discuss some coordination compounds. For each coordination compound, give the complex ion and the counterions, the electron configuration of the transition metal, and the geometry of the complex ion. a. \(\mathrm{CoCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) is a compound used in novelty devices that predict rain. b. During the developing process of black-and-white film, silver bromide is removed from photographic film by the fixer. The major component of the fixer is sodium thiosulfate. The equation for the reaction is: \(\begin{aligned} \mathrm{AgBr}(s)+2 \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}(a q) & \longrightarrow \\\& \quad\quad \mathrm{Na}_{3}\left[\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}\right](a q)+\mathrm{NaBr}(a q) \end{aligned}\) c. In the production of printed circuit boards for the electronics industry, a thin layer of copper is laminated onto an insulating plastic board. Next, a circuit pattern made of a chemically resistant polymer is printed on the board. The unwanted copper is removed by chemical etching, and the protective polymer is finally removed by solvents. One etching reaction is: $$\mathrm{Cu}\left(\mathrm{NH}_{3}\right){ }_{4} \mathrm{Cl}_{2}(a q)+4 \mathrm{NH}_{3}(a q)+\mathrm{Cu}(s) \longrightarrow \\\ \mathrm{Cu}\left(\mathrm{NH}_{3}\right){ }_{4} \mathrm{Cl}(a q)$$

Figure \(21.17\) shows that the cis isomer of \(\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}^{+}\) is optically active while the trans isomer is not optically active. Is the same true for \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}^{+} ?\) Explain.

Why are \(\mathrm{CN}^{-}\) and \(\mathrm{CO}\) toxic to humans?

Draw the \(d\) -orbital splitting diagrams for the octahedral complex ions of each of the following. a. \(\mathrm{Fe}^{2+}\) (high and low spin) b. \(\mathrm{Fe}^{3+}\) (high spin) c. \(\mathrm{Ni}^{2+}\)

Henry Taube, 1983 Nobel Prize winner in chemistry, has studied the mechanisms of the oxidation-reduction reactions of transition metal complexes. In one experiment he and his students studied the following reaction: \(\begin{aligned} \mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}(a q) &+\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}^{2+}(a q) \\ & \longrightarrow \mathrm{Cr}(\mathrm{III}) \text { complexes }+\mathrm{Co}(\mathrm{II}) \text { complexes } \end{aligned}\) Chromium(III) and cobalt(III) complexes are substitutionally inert (no exchange of ligands) under conditions of the experiment. Chromium(II) and cobalt(II) complexes can exchange ligands very rapidly. One of the products of the reaction is \(\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}^{2+} .\) Is this consistent with the reaction proceeding through formation of \(\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cr}-\mathrm{Cl}-\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\) as an intermediate? Explain.

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.