The aqueous solution containing which one of the following ions will be colourless? (a) \(\mathrm{Sc}^{3+}\) (b) \(\mathrm{Fe}^{3+}\) (c) \(\mathrm{Fe}^{2+}\) (d) \(\mathrm{Mn}^{2+}\)

Short Answer

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Sc3+ (a) will form a colorless solution.

Step by step solution

01

Understand the Basics of Color in Ions

The color of ions in aqueous solutions is typically due to electronic transitions within the d-orbitals of transition metals. Ions with a d0 or d10 electronic configuration, where there are no electrons or a full set of electrons in the d-orbitals, are typically colorless because there are no electronic transitions that can absorb visible light.
02

Determine the Electronic Configuration

Examine the electronic configuration of the metal ions to determine if they have a d0 or d10 configuration. Scandium (Sc) has an atomic number of 21, and as a Sc3+ ion, it loses 3 electrons, resulting in an electronic configuration without any d-electrons (d0). Iron (Fe) and Manganese (Mn), on the other hand, have partially filled d-orbitals in the Fe3+, Fe2+, and Mn2+ ions, which can lead to color due to electronic transitions.
03

Identify the Colorless Solution

Based on the electronic configurations, Sc3+ with a d0 configuration is likely to be the ion that forms a colorless solution in water. Fe3+ (d5), Fe2+ (d6), and Mn2+ (d5) have partially filled d-orbitals, and hence, are likely to be colored.

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Key Concepts

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

Transition Metal Ion Colors
When it comes to the enchanting world of colorful solutions, transition metal ions are the star performers. The variety of colors we observe in aqueous solutions of these ions is due to the presence of d-orbitals, which are unique to transition elements. These d-orbitals allow for a fascinating phenomenon known as 'd-d transitions.' If we take a closer look, we can see that ions with incompletely filled d-orbitals usually exhibit colors in solution. Imagine an artist's palette, but instead of paints, we're dabbing with electrons!

For instance, the copper(II) ion (Cu^{2+}) adds a beautiful blue hue to its solution. It's all about what’s happening on a microscopic scale – electrons leaping between energy levels. However, not all transition metals show color. Scandium ions (Sc^{3+}), with their d-orbitals empty (d0), lack the ability to absorb visible light to promote electrons and, therefore, are like invisible ink – colorless in solution.
Electronic Configurations
Electronic configurations are like the personal ID of an atom, telling us how its electrons are arranged. These configurations follow a specific order and can sometimes appear as complex as a social security number. Transition metals, with their d-orbitals, have electronic configurations that directly influence their chemical behavior and properties. In the context of our color conundrum, knowing whether a transition metal ion has an electronic configuration of d0 or d10 is crucial.

An ion with d0 means all available seats in the d-orbital are empty, think of it as a concert hall before the audience arrives – not much is happening. Likewise, a d10 configuration means the d-orbital is at full capacity, like a packed stadium during the finale of a concert – no room for any more action. These 'quiet' or 'full house' scenarios result in the aqueous ions being colorless, as there's no room for electrons to move to higher energy states.
d-Orbital Electronic Transitions
Descending into the world of d-orbital electronic transitions is like diving into a sea of intricate dances between light and electrons. In transition metal ions, electrons can get excited and jump between different levels within the d-orbitals. These transitions aren't random; they need just the right amount of energy, typically from visible light, to make the leap.

As light hits an aqueous solution, certain wavelengths are absorbed by these electrons as they jump up to higher energy states. Once the light is absorbed, the remaining light – which we see – is missing those wavelengths, giving us the perceived color of the solution. Think of it as a spectral selective filter: certain wavelengths go in, some are trapped, and the remaining spectrum that passes through 'paints' the solution with color. If there's no place for an electron to jump to – as with a d0 or d10 configuration – the solution will stay colorless. This makes d-orbital electronic transitions the nucleus of understanding why we see the vivid spectrum of colors in transition metal ion solutions.

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Most popular questions from this chapter

Which one of the following ions is colourless in its aqueous solution? (a) \(\mathrm{T}_{1}^{3+}\) (b) \(\mathrm{Cu}^{2+}\) (c) \(\mathrm{Ni}^{2+}\) (d) \(\mathrm{Zn}^{2+}\)

Two stable ions of first transition series which have the largest number of unpaired electrons and have the highest magnetic moment are: (a) \(\mathrm{Ni}^{2+}\) and \(\mathrm{Co}^{2+}\) (b) \(\mathrm{Cr}^{3+}\) and \(\mathrm{Fe}^{2+}\) (c) \(\mathrm{Fe}^{3+}\) and \(\mathrm{Mn}^{2+}\) (d) \(\mathrm{Ti}^{4+}\) and \(\mathrm{Cr}^{3+}\)

What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid? (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}\) and \(\mathrm{H}_{2} \mathrm{O}\) are formed (b) \(\mathrm{Cr}^{3+}\) and \(\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}\) are formed (c) \(\mathrm{CrO}_{4}{ }^{2-}\) is oxidized to \(+7\) state to \(\mathrm{Cr}\) (d) \(\mathrm{CrO}_{4}{ }^{2-}\) is reduced to \(+3\) state of \(\mathrm{Cr}\)

The soldiers of Napolean's army, while on Alps during freezing winter, suffered a serious problem as regards to the tin buttons of their uniforms. White metallic tin buttons got converted to grey powder. This transformation is related to: (a) An interaction with nitrogen of the air at very low temperatures (b) A change in the crystalline structure of tin (c) An interaction with water vapour contained in the humid air (d) A change in the partial pressure of oxygen in the air.

A complex compound of \(\mathrm{Co}^{3+}\) with molecular formula CoCl \(_{x} \cdot y \mathrm{NH}_{3}\) gives a total of 3 ions when dissolved in water. How many \(\mathrm{Cl}-\) ions satisfy both primary as well as the secondary valencies in this complex? (a) 3 (b) 1 (c) 4 (d) zero

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