The complex ion \(\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) has an absorption maximum at around 800 \(\mathrm{nm}\) . When four ammonias replace water, $\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}^{2+},$ the absorption maximum shifts to around 600 \(\mathrm{nm} .\) What do these results signify in terms of the relative field splittings of \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\) ? Explain.

Short Answer

Expert verified
In conclusion, the absorption maximum of Cu(NH3)4(H2O)2^2+ at around 600 nm compared to 800 nm for Cu(H2O)6^2+ signifies that the field splitting of NH3 is greater than that of H2O. This is because the lower wavelength indicates a larger energy difference between the d-orbitals in Cu(NH3)4(H2O)2^2+, meaning NH3 creates a stronger crystal field around the central Cu ion than H2O.

Step by step solution

01

Understanding Crystal Field Theory (CFT)

Crystal Field Theory (CFT) is a model that helps in understanding the electronic structure of complex ions. It describes the interaction of the central metal ion and the ligands surrounding it. The interaction between the central metal ion and the ligands causes the energy levels of d-orbitals to split into different energy levels, which is known as the field splitting.
02

Relating absorption wavelength to field splitting#:tag_content# When a complex absorbs light, an electron is excited from lower energy d-orbitals to higher energy d-orbitals. The energy difference between these orbitals is directly related to the absorption wavelength. The smaller the energy difference is, the more significant absorption wavelength. Mathematically, this can be represented as: \[ ΔE = h \nu = \dfrac{hc}{λ} \] where ΔE is the energy difference between the d-orbitals, h is the Planck's constant, ν is the frequency of absorbed light, c is the speed of light, and λ is the wavelength of absorbed light.

Step 3: Observing the absorption wavelengths of given complex ions#:tag_content# We have two complex ions: 1. Cu(H2O)6^2+ with an absorption maximum at around 800 nm 2. Cu(NH3)4(H2O)2^2+ with an absorption maximum at around 600 nm Since the absorption maximum is inversely proportional to the energy difference, we can conclude that the energy difference between the d-orbitals in Cu(NH3)4(H2O)2^2+ is larger than that of Cu(H2O)6^2+.
03

Interpreting the results in terms of field splitting

As we have concluded that the energy difference between the d-orbitals in Cu(NH3)4(H2O)2^2+ is larger than that of Cu(H2O)6^2+, it means that the field splitting due to NH3 ligands is more significant than that of H2O ligands. Therefore, NH3 creates a stronger crystal field around the central Cu ion than H2O. In conclusion, these results signify that the field splitting of NH3 is greater than that of H2O, as the absorption maximum shifts to lower wavelength (600 nm) when four NH3 ligands replace water (with an absorption maximum at around 800 nm) in the complex ion.

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!

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

Which is more likely to be paramagnetic, \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\) or \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) ? Explain.

Carbon monoxide is toxic because it binds more strongly to iron in hemoglobin \((\mathrm{Hb})\) than does \(\mathrm{O}_{2} .\) Consider the following reactions and approximate standard free energy changes: $$\begin{aligned} \mathrm{Hb}+\mathrm{O}_{2} \longrightarrow \mathrm{HbO}_{2} & \Delta G^{\circ}=-70 \mathrm{kJ} \\ \mathrm{Hb}+\mathrm{CO} \longrightarrow \mathrm{HbCO} & \Delta G^{\circ}=-80 \mathrm{kJ} \end{aligned}$$ Using these data, estimate the equilibrium constant value at $25^{\circ} \mathrm{C}$ for the following reaction: $$\mathrm{HbO}_{2}(a q)+\mathrm{CO}(g) \rightleftharpoons \mathrm{HbCO}(a q)+\mathrm{O}_{2}(g)$$

The compound $\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6} \mathrm{Cl}_{2}$ is green, whereas \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{Cl}_{2}\) is violet. Predict the predominant color of light absorbed by each compound. Which compound absorbs light with the shorter wavelength? Predict in which compound \(\Delta\) is greater and whether \(\mathrm{H}_{2} \mathrm{O}\) or \(\mathrm{NH}_{3}\) is the stronger field ligand. Do your conclusions agree with the spectrochemical series?

When aqueous KI is added gradually to mercury(II) nitrate, an orange precipitate forms. Continued addition of KI causes the precipitate to dissolve. Write balanced equations to explain these observations. $\left(\text {Hint} : \mathrm{Hg}^{2+} \text { reacts with } \mathrm{I}^{-} \text { to form } \mathrm{HgI}_{4}^{2-} .\right)\( Would you expect \)\mathrm{Hg} \mathrm{I}_{4}^{2-}$ to form colored solutions? Explain.

Name the following coordination compounds. a. $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Br}\right] \mathrm{Br}_{2}$ b. \(\mathrm{Na}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]\) c. $\left[\mathrm{Fe}\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)_{2}\left(\mathrm{NO}_{2}\right)_{2}\right] \mathrm{Cl}$ d. $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{2}\right]\left[\mathrm{PtI}_{4}\right]$

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