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Match the following $$ \begin{aligned} &\begin{array}{ll} \text { List-I } & \text { List-II } \\ \hline \text { (a) Coloured ion } & \text { (p) } \mathrm{Cu}^{+} \\ \text {(b) } \mu=1.73 \mathrm{~B} \cdot \mathrm{M} & \text { (q) } \mathrm{Cu}^{2+} \\ \text { (c) } \mathrm{d}^{10} \text { configuration } & \text { (r) } \mathrm{Fe}^{2+} \\ \text { (d) } \begin{array}{l} \text { More than } 3 \text { un-paired } \\ \text { electrons } \end{array} & \text { (s) } \mathrm{Mn}^{2+} \\ \hline \end{array}\\\ &\text { - } \end{aligned} $$

Short Answer

Expert verified
(a)-(q), (b)-(q), (c)-(p), (d)-(s).

Step by step solution

01

Analyzing Coloured Ions

The term 'coloured ion' typically refers to ions with unpaired electrons in their d-orbitals, leading to the absorption of visible light. Let's consider the ions: - \( \mathrm{Cu}^{+} \) has a \( \mathrm{d}^{10} \) configuration (all paired electrons), so it does not appear coloured. - \( \mathrm{Cu}^{2+} \) has a \( \mathrm{d}^9 \) configuration, which means it has one unpaired electron, often leading to colour.Thus, (a) pairs with (q) \( \mathrm{Cu}^{2+} \).
02

Understanding Magnetic Moment (\( \mu \))

The magnetic moment is calculated using the formula \( \mu = \sqrt{n(n+2)} \) where \( n \) is the number of unpaired electrons. A magnetic moment of \( 1.73 \mathrm{~B} \cdot \mathrm{M} \) corresponds to one unpaired electron:- \( \mathrm{Cu}^{2+} \) has 1 unpaired electron.Based on this, (b) pairs with (q) \( \mathrm{Cu}^{2+} \).
03

d10 Configuration Matching

The \( \mathrm{d}^{10} \) configuration implies all the d electrons are paired:- \( \mathrm{Cu}^{+} \) has a \( \mathrm{d}^{10} \) configuration.So, (c) pairs with (p) \( \mathrm{Cu}^{+} \).
04

Identifying More Than 3 Unpaired Electrons

Ions with more than 3 unpaired electrons typically have high spin configurations:- \( \mathrm{Mn}^{2+} \) has the \([Ar] 3d^{5}\) configuration with 5 unpaired electrons.Therefore, (d) pairs with (s) \( \mathrm{Mn}^{2+} \).

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Key Concepts

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

Coloured Ions
When we talk about coloured ions in transition metals, we're diving into the magical world of d-orbitals. These orbitals can absorb visible light, a characteristic feature of many transition metals and their compounds.
The colour we observe is due to d-electron excitation. When visible light hits an ion, electrons can absorb some of this energy and jump to higher energy levels. The light that isn't absorbed is what we see, and this gives the ion its unique colour.
For example,
  • \( \mathrm{Cu}^{2+} \) ions, with a \( \mathrm{d}^9 \) configuration, exhibit a lovely blue-green hue. This is because it has one unpaired electron that can undergo these light absorptions.
  • Meanwhile, \( \mathrm{Cu}^{+} \) ions don't show colour due to their \( \mathrm{d}^{10} \) configuration, where all electrons are paired and cannot absorb visible light effectively.
Understanding the colour of transition metal ions is not just about their appearance. It provides insights into their electronic structures and chemical reactions.
Magnetic Moment
Magnetic moment is a measure of the magnetism in an atom or ion, often used to understand the presence and arrangement of electrons in d-orbitals. In the transition metals, magnetism mainly arises due to unpaired electrons.

It's calculated using the formula \( \mu = \sqrt{n(n+2)} \), where \( n \) denotes the number of unpaired electrons. When you have unpaired electrons, their spins can align to create a net magnetic moment.
  • For example, the magnetic moment of \( 1.73 \mathrm{B} \cdot \mathrm{M} \) indicates one unpaired electron, matching well with ions like \( \mathrm{Cu}^{2+} \).
  • This contrasts with \( \mathrm{Cu}^{+} \), which, with all paired electrons, exhibits no considerable magnetism.
This characteristic allows chemists to understand more about the internal configuration of electrons and predict the behavior of substances in magnetic fields.
d-electron Configuration
d-electron configuration tells us how electrons are distributed in the d-orbitals of an atom or ion. In transition metals, these configurations significantly influence their chemical and physical properties, such as colour and magnetism.
For instance:
  • \( \mathrm{Cu}^{+} \)'s \( \mathrm{d}^{10} \) configuration indicates all the electrons in the d-orbitals are paired. This prevents visible light absorption, resulting in a colourless ion.
  • In contrast, \( \mathrm{Cu}^{2+} \) has a \( \mathrm{d}^{9} \) configuration, meaning one electron is unpaired, leading to the colourful properties.
The difference in the electron configurations of similar elements or ions can meaningfully alter their chemical behavior, reactivity, and observable properties.
Unpaired Electrons
Unpaired electrons in the d-orbitals of transition metals are pivotal for understanding their magnetic and optical behaviors. When electrons are unpaired, their spins do not cancel out, allowing these materials to explore interesting physical properties.
Transition metals with several unpaired electrons exhibit higher magnetic moments. For instance:
  • \( \mathrm{Mn}^{2+} \) showcases five unpaired electrons, resulting in a pronounced magnetic moment.
  • Such ions carry significant magnetic properties also seen in their spectrometry reactions.
Furthermore, the presence of these unpaired electrons can affect how transition metals interact with light, rendering them colourful. Comprehending the quantity and arrangement of these unpaired electrons offers insights into the diverse applications and unique characteristics of transition metals.

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Most popular questions from this chapter

\(\mathrm{CuSO}_{4}\) decolourizes on addition of KCN, the product is (a) \(\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]^{2-}\) (b) \(\mathrm{Cu}^{2+}\) gets reduced to form \(\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]^{3-}\) (c) \(\mathrm{Cu}(\mathrm{CN})_{2}\) (d) \(\mathrm{CuCN}\)

Anhydrous ferric chloride is prepared by (a) heating hydrated ferric chloride at a high temperature in a stream of air (b) heating metallic iron in a stream of dry chlorine gas (c) reaction of ferric oxide with hydrochloric acid (d) reaction of metallic iron with hydrochloric acid

The basic character of the transition metal monoxides follow the order (a) \(\mathrm{CrO}>\mathrm{VO}>\mathrm{FeO}>\mathrm{TiO}\) (b) \(\mathrm{TiO}>\mathrm{VO}>\mathrm{CrO}>\mathrm{FeO}\) (c) \(\mathrm{TiO}>\mathrm{FeO}>\mathrm{VO}>\mathrm{CrO}\) (d) \(\mathrm{VO}>\mathrm{CrO}>\mathrm{TiO}>\mathrm{FeO}\)

Which compound is formed when excess of \(\mathrm{KCN}\) is added to an aqueous solution of copper sulphate? (a) \(\mathrm{Cu}(\mathrm{CN})_{2}\) (b) \(\mathrm{K}_{2}\left[\mathrm{Cu}(\mathrm{CN})_{6}\right]\) (c) \(\mathrm{K}\left[\mathrm{Cu}(\mathrm{CN})_{2}\right]\) (d) \(\mathrm{K}_{3}\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]\)

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) \(\mathrm{sp}^{3}, \mathrm{sp}^{3}\)

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