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Chapter 4: Problem 65

The correct order of number of unpaired electrons in the ion \(\mathrm{Cu}^{2+} \mathrm{Ni}^{2+}, \mathrm{Fe}^{3+}\) and \(\mathrm{Cr}^{3+}\) is (a) \(\mathrm{Cu}^{2+}>\mathrm{Ni}^{2+}>\mathrm{Cr}^{3+}>\mathrm{Fe}^{3+}\) (b) \(\mathrm{Ni}^{2+}>\mathrm{Cu}^{2+}>\mathrm{Fe}^{3+}>\mathrm{Cr}^{3+}\) (c) \(\mathrm{Fe}^{3+}>\mathrm{Cr}^{3+}>\mathrm{Ni}^{2+}>\mathrm{Cu}^{2+}\) (d) \(\mathrm{Fe}^{3+}>\mathrm{Cr}^{3+}>\mathrm{Cu}^{2+}>\mathrm{Ni}^{2+}\)

Short Answer

Expert verified
The correct order is (c) Fe\(^{3+}\) > Cr\(^{3+}\) > Ni\(^{2+}\) > Cu\(^{2+}\).

Step by step solution

01

Determine Electron Configuration for Neutral Atoms

Start by determining the electronic configuration of the neutral atoms for Cu (Copper), Ni (Nickel), Fe (Iron), and Cr (Chromium). - Cu: [Ar] 3d\(^ { 10 }\) 4s\(^1\) - Ni: [Ar] 3d\(^ { 8 }\) 4s\(^2\) - Fe: [Ar] 3d\(^ { 6 }\) 4s\(^2\) - Cr: [Ar] 3d\(^ { 5 }\) 4s\(^1\)
02

Determine Electron Configuration for Ions

Subtract the necessary electrons to determine the electron configuration for the ions: - \(\text{Cu}^{2+}\): Remove 2 electrons from Cu, resulting in [Ar] 3d\(^ { 9 }\)- \(\text{Ni}^{2+}\): Remove 2 electrons from Ni, resulting in [Ar] 3d\(^ { 8 }\)- \(\text{Fe}^{3+}\): Remove 3 electrons from Fe, resulting in [Ar] 3d\(^ { 5 }\)- \(\text{Cr}^{3+}\): Remove 3 electrons from Cr, resulting in [Ar] 3d\(^ { 3 }\)
03

Count Unpaired Electrons

Count the number of unpaired electrons in the d-orbitals for each ion:- \(\text{Cu}^{2+}\): 1 unpaired electron in 3d\(^9\)- \(\text{Ni}^{2+}\): 2 unpaired electrons in 3d\(^8\)- \(\text{Fe}^{3+}\): 5 unpaired electrons in 3d\(^5\)- \(\text{Cr}^{3+}\): 3 unpaired electrons in 3d\(^3\)
04

Order the Ions

Order the ions by the number of unpaired electrons from greatest to least:- \(\text{Fe}^{3+}\)(5 unpaired) > \(\text{Cr}^{3+}\)(3 unpaired) > \(\text{Ni}^{2+}\)(2 unpaired) > \(\text{Cu}^{2+}\)(1 unpaired).

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Key Concepts

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

Ion Formation
When atoms form ions, they are either gaining or losing electrons. This helps them achieve a stable electronic configuration, often similar to the nearest noble gas. Metals, in particular, tend to form positive ions (cations) by losing electrons. Transition metals, such as copper (Cu), nickel (Ni), iron (Fe), and chromium (Cr), are no exception to this.

For instance:
  • Cu becomes \( ext{Cu}^{2+}\) by losing two electrons.
  • Ni transforms into \( ext{Ni}^{2+}\) also by shedding two electrons.
  • Fe turns into \( ext{Fe}^{3+}\) by losing three electrons.
  • Cr becomes \( ext{Cr}^{3+}\) by losing three electrons as well.
The process of ion formation gives these metals a charge, which affects their position and role in various chemical reactions.
Unpaired Electrons
In the study of electron configurations, unpaired electrons play a significant role, especially when considering the magnetic properties of an element or ion. Unpaired electrons are present in partially filled orbitals and contribute to magnetic moments due to their spin.

For the ions \( ext{Cu}^{2+}\), \( ext{Ni}^{2+}\), \( ext{Fe}^{3+}\), and \( ext{Cr}^{3+}\):
  • \( ext{Cu}^{2+}\) has 1 unpaired electron in its 3d orbital.
  • \( ext{Ni}^{2+}\) features 2 unpaired electrons.
  • \( ext{Fe}^{3+}\) includes 5 unpaired electrons, contributing a high magnetic property.
  • \( ext{Cr}^{3+}\) possesses 3 unpaired electrons.
Understanding which electrons remain unpaired and how many there are helps predict an element's or ion's behavior in chemical interactions, including magnetism.
Transition Metals
Transition metals are unique elements often found in the center of the periodic table. They possess partial d subshells, which lead to fascinating properties, such as the formation of colored compounds and multiple oxidation states. These metals often serve as catalysts in various chemical reactions.

Considering elements like Cu, Ni, Fe, and Cr:
  • They can form different ions by losing different numbers of electrons.
  • Their d orbitals in ions can have unpaired electrons that are instrumental in their chemistry.
  • Since their electronic configurations can change, they often show a variety of chemical and physical properties.
The distinct behavior of transition metals makes them essential in industry and biology. Their ability to easily exchange electrons means they catalyze important biological processes, like oxygen transport in hemoglobin where iron plays a crucial role.

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Most popular questions from this chapter

In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen? (a) \(3 \longrightarrow 2\) (b) \(5 \longrightarrow 2\) (c) \(4 \longrightarrow 1\) (d) \(2 \longrightarrow 5\)

The radius of which of the following orbits is same as that of the first Bohr's orbit of hydrogen atom? (a) \(\mathrm{He}^{+}(\mathrm{n}=2)\) (b) \(\mathrm{Li}^{2+}(\mathrm{n}=2)\) (c) \(\mathrm{Li}^{2+}(\mathrm{n}=3)\) (d) \(\mathrm{Be}^{3+}(\mathrm{n}=2)\)

In which orbital is the angular momentum of an electron zero? (a) \(2 \mathrm{p}\) (b) \(2 \mathrm{~s}\) (c) \(3 \mathrm{~d}\) (d) \(4 \mathrm{f}\)

Electron energy of a photon is given as: \(\Delta \mathrm{E} /\) atom \(=3.03 \times 10^{-19} \mathrm{~J}\) atom \(^{-1}\) then, the wavelength of the photon is (a) \(6.56 \mathrm{~nm}\) (b) \(65.6 \mathrm{~nm}\) (c) \(656 \mathrm{~nm}\) (d) \(0.656 \mathrm{~nm}\) Given, \(\mathrm{h}\) (Planck constant) \(=6.63 \times 10^{-34} \mathrm{Js} \mathrm{c}\) (velocity of light \()=3.00 \times 10^{8} \mathrm{~ms}^{-1}\)

Rutherford's experiment, which established the nuclear model of the atom, used a beam of (a) \(\beta\)-particles, which impinged on a metal foil and got absorbed (b) \(\gamma\)-rays, which impinged on a metal foil and ejected electrons (c) Helium atoms, which impinged on a metal foil and got scattered (d) Helium nuclei, which impinged on a metal foil and got scatterd.
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