Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The aqueous solution of the following salts will be coloured in the case of (a) \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (b) \(\mathrm{LiNO}_{3}\) (c) \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\) (d) potash alum

Short Answer

Expert verified
The aqueous solution of \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\) is colored.

Step by step solution

01

Analyze Zn(NO₃)₂

Zinc nitrate, \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\), is a compound of zinc. Zinc ions are \(\mathrm{Zn}^{2+}\), which typically form colorless solutions as they do not have unpaired d-electrons in their electronic configuration. Hence, the solution of zinc nitrate is usually colorless.
02

Analyze LiNO₃

Lithium nitrate, \(\mathrm{LiNO}_{3}\), is a compound of lithium. Lithium ions are \(\mathrm{Li}^{+}\), and like zinc, they do not have unpaired d-electrons and form colorless solutions. Therefore, the lithium nitrate solution will also be colorless.
03

Analyze Co(NO₃)₂

Cobalt nitrate, \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\), contains \(\mathrm{Co}^{2+}\) ions. Cobalt ions have unpaired d-electrons, which allow for d-d electron transitions. These transitions result in absorbing certain wavelengths of light and emitting others, making the solution appear colored. The solution of cobalt nitrate is usually pink.
04

Analyze Potash Alum

Potash alum typically contains potassium, aluminum, sulfate ions, and water of crystallization, \(\mathrm{KAl(SO}_{4})_{2} \, \cdot \, 12\mathrm{H}_2\mathrm{O}\). The ions from potash alum including \(\mathrm{K}^{+}\) and \(\mathrm{Al}^{3+}\) do not have unpaired d-electrons, leading to a colorless solution. Thus, potash alum solutions are generally colorless.

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.

Transition Metal Ions
Transition metal ions play an important role in determining the color of salt solutions. These ions can often cause the aqueous solution to display vibrant colors due to their electronic properties. Transition metals are elements found in the d-block of the periodic table. They are characterized by having partially filled d-orbitals, which are the key to their colorful nature.
Most transition metal ions have one or more unpaired electrons in their d-orbitals. This is crucial because these unpaired electrons enable them to undergo specific electronic transitions that absorb certain wavelengths of light. This absorbed light is what imparts a color to the solution. Thus, if a salt solution contains transition metal ions, it is quite likely to appear colored.
  • Example: Cobalt nitrate, \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\), contains \(\mathrm{Co}^{2+}\) ions with unpaired d-electrons, resulting in a pink solution.
  • Contrast: Non-transition metals like zinc and lithium do not form colored solutions because their ions lack unpaired d-electrons.
Electronic Configuration
The electronic configuration of an ion significantly influences the color it expresses in solution. Electronic configuration refers to the arrangement of electrons in an atom or ion. For transition metals, electrons fill the d-orbitals, the energy levels where electrons have the highest probability of being found.
When these d-orbitals are partly filled, it leads to unpaired d-electrons, which are crucial for color formation due to d-d transitions. For example, in cobalt with an electronic configuration of \( [\mathrm{Ar}] \, 3d^7\) for \( \mathrm{Co}^{2+}\), the presence of unpaired electrons in the d-orbitals allows this ion to absorb specific wavelengths of light and appear colored.
In contrast, ions such as \( \mathrm{Zn}^{2+} \, \text{(with no unpaired d-electrons)}\) have an electronic configuration of \( [\mathrm{Ar}] \, 3d^{10}\), which results in a colorless solution.
  • This difference explains why some metal ions result in colored solutions while others do not.
d-d Electron Transitions
A fascinating phenomenon that explains the color of transition metal ion solutions is the occurrence of d-d electron transitions. These transitions happen when an electron jumps between d-orbitals within the same shell, typically involving the absorption of light.
Due to the arrangement and energy levels of d-orbitals, these transitions require specific energies that correspond to particular wavelengths of visible light. When an electron absorbs light of a particular wavelength, it moves to a higher energy d-orbital. The light that is not absorbed is what our eyes perceive as the color of the solution.
  • An example is the \( \mathrm{Co}^{2+}\) in cobalt nitrate, where unpaired d-electrons transition between orbitals, absorbing specific light wavelengths and giving the solution a pink color.
  • Conversely, ions lacking unpaired d-electrons, like \( \mathrm{Li}^{+}\), do not undergo these transitions and remain colorless.

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

Match the following: List I (n = un-paired e \(\mathbf{~}^{-}\)) \(\quad\) List II (Meg. moment) 1\. \(\mathrm{Sc}^{3+}=0\) (i) 5.92 B.M. 2\. \(\mathrm{V}^{2+}=3\) (ii) \(1.73\) B.M. 3\. \(\mathrm{Fe}^{3+}=5\) (iii) zero 4\. \(\mathrm{Cu}^{2+}=1\) (iv) \(3.87\) B.M. The correct matching is: \(\begin{array}{llll}1 & 2 & 3 & 4\end{array}\) (a) (ii) (i) (iii) (iv) (b) (iii) (iv) (ii) (i)

The d-orbitals participating in hybridization of central metal atom may be from the outermost shell or the penultimate shell. This depends on the nature of metal and the nature of ligand. The complexes involving the inner \(\mathrm{d}\) level (inner orbital complexes) result when the ligand is a powerful or strong ligand resulting in diamagnetic or low spin complexes. A weak ligand usually results in the formation of outer orbital complex or high spin complex. The hybridization of \(\mathrm{Cu}\) in \(\left[\mathrm{CuCl}_{4}\right]^{2-}\) and \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) are respectively. (a) \(\mathrm{sp}^{3}, \mathrm{dsp}^{2}\) (b) \(\mathrm{dsp}^{2}, \mathrm{dsp}^{2}\) (c) \(\mathrm{dsp}^{2}, \mathrm{sp}^{3}\) (d) \(\mathrm{sp}^{3}, \mathrm{sp}^{3}\)

What is the current order of spin only magnetic moment (in B.M.) of \(\mathrm{Mn}^{2+}, \mathrm{Cr}^{2+}\) and \(\mathrm{V}^{2+}\) ? (a) \(\mathrm{Mn}^{2+}>\mathrm{V}^{2+}>\mathrm{Cr}^{2+}\) (b) \(\mathrm{V}^{2+}>\mathrm{Cr}^{2+}>\mathrm{Mn}^{2+}\) (c) \(\mathrm{Mn}^{2+}>\mathrm{Cr}^{2+}>\mathrm{V}^{2+}\) (d) \(\mathrm{Cr}^{2+}>\mathrm{V}^{2+}>\mathrm{Mn}^{2+}\)

Sodium thiosulphate is used in photography because of its (a) reaction with light (b) oxidizing behaviour (c) reducing behaviour (d) complex forming behaviour

The number of unpaired electrons present in \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) and \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) are, respectively (a) 0,0 (b) 0,4 (c) 1,2 (d) 2,4Among the following four sets, the one having the same geometry and same magnetic property for both complexes is (a) \(\left[\mathrm{Co}\left(\mathrm{CN}_{6}\right)\right]^{3-}\) and \(\left[\mathrm{CoF}_{6}\right]^{3-}\) (b) \(\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}\) and \(\left[\mathrm{MnF}_{6}\right]^{3-}\) (c) \([\mathrm{Cr}(\mathrm{CN})]^{3-}\) and \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) (d) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) and \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\)

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