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Chapter 4: Problem 4

Which set has same number of unpaired electrons? (a) \(\mathrm{Fe}^{2+}, \mathrm{Mn}^{+2}\) (b) \(\mathrm{Fe}^{3+}, \mathrm{Mn}^{+2}\) (c) \(\mathrm{Cr}^{+3}, \mathrm{Ni}^{2+}\) (d) \(\mathrm{Zn}^{2+} \mathrm{Cu}^{2+}\)

Short Answer

Expert verified
The set \((\mathrm{Fe}^{3+}, \mathrm{Mn}^{2+})\) has the same number of unpaired electrons, which is 5.

Step by step solution

01

Determine Electron Configuration

Identify the electron configuration for the given ions:- \( \mathrm{Fe}^{2+} \): Atomic number of Fe is 26. The neutral Fe electron configuration is \([\mathrm{Ar}] 3d^6 4s^2\). The \( \mathrm{Fe}^{2+} \) ion loses 2 electrons from the \( 4s \) and \( 3d \) orbitals, resulting in \([\mathrm{Ar}] 3d^6\).- \( \mathrm{Mn}^{2+} \): Atomic number of Mn is 25. The neutral Mn electron configuration is \([\mathrm{Ar}] 3d^5 4s^2\). The \( \mathrm{Mn}^{2+} \) ion loses 2 electrons, resulting in \([\mathrm{Ar}] 3d^5\).- \( \mathrm{Fe}^{3+} \): Remove one more electron from \( \mathrm{Fe}^{2+} \)'s \( 3d \) orbital, resulting in \([\mathrm{Ar}] 3d^5\).- \( \mathrm{Cr}^{3+} \): Atomic number of Cr is 24. For Cr \([\mathrm{Ar}] 3d^5 4s^1\); removing 3 electrons gives \([\mathrm{Ar}] 3d^3\).- \( \mathrm{Ni}^{2+} \): Atomic number of Ni is 28. For neutral Ni \([\mathrm{Ar}] 3d^8 4s^2\), and so \( \mathrm{Ni}^{2+} \) ion configuration is \([\mathrm{Ar}] 3d^8\).- \( \mathrm{Zn}^{2+} \): Atomic number of Zn is 30. Zn's configuration is \([\mathrm{Ar}] 3d^{10} 4s^2\), so \( \mathrm{Zn}^{2+} \) is \([\mathrm{Ar}] 3d^{10}\).- \( \mathrm{Cu}^{2+} \): Atomic number of Cu is 29. For neutral Cu \([\mathrm{Ar}] 3d^{10} 4s^1\), \( \mathrm{Cu}^{2+} \) becomes \([\mathrm{Ar}] 3d^9\).
02

Count Unpaired Electrons

Count the unpaired electrons in the electronic configuration of each ion:- \( \mathrm{Fe}^{2+} \): \( 3d^6 \) has 4 unpaired electrons.- \( \mathrm{Mn}^{2+} \): \( 3d^5 \) has 5 unpaired electrons.- \( \mathrm{Fe}^{3+} \): \( 3d^5 \) has 5 unpaired electrons.- \( \mathrm{Cr}^{3+} \): \( 3d^3 \) has 3 unpaired electrons.- \( \mathrm{Ni}^{2+} \): \( 3d^8 \) has 2 unpaired electrons.- \( \mathrm{Zn}^{2+} \): \( 3d^{10} \) has 0 unpaired electrons.- \( \mathrm{Cu}^{2+} \): \( 3d^9 \) has 1 unpaired electron.
03

Identify Matching Sets

Compare the number of unpaired electrons in each pair from the options given:- Option (a) \( \mathrm{Fe}^{2+}, \mathrm{Mn}^{2+} \): 4 and 5 unpaired electrons.- Option (b) \( \mathrm{Fe}^{3+}, \mathrm{Mn}^{2+} \): Both have 5 unpaired electrons.- Option (c) \( \mathrm{Cr}^{3+}, \mathrm{Ni}^{2+} \): 3 and 2 unpaired electrons.- Option (d) \( \mathrm{Zn}^{2+}, \mathrm{Cu}^{2+} \): 0 and 1 unpaired electron.Option (b), which consists of \( \mathrm{Fe}^{3+} \) and \( \mathrm{Mn}^{2+} \), has the same number of unpaired electrons.

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Key Concepts

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

Electron Configuration
Electron configuration refers to the distribution of electrons in an atom or ion across various orbitals. This arrangement is crucial for understanding the chemical properties and reactions of an element. Each electron occupies the lowest energy orbital available, following the Aufbau principle.
To determine the electron configuration, start from the lowest energy level, filling each orbital until all electrons are placed.
  • Use the sequence: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, and so on.
  • Remember that the 'd' subshell in transition metals introduces complexity because its electrons are filled after the 's' orbital of the next principal energy level.
For example, iron (Fe) has an atomic number of 26, leading to a neutral electron configuration of \[ [\text{Ar}] 3d^6 4s^2 \]. However, ion formation changes this configuration by removing electrons, usually starting from the outermost 's' orbital.
Transition Metals
Transition metals are elements that have partially filled d orbitals. They represent a unique category in the periodic table, exhibiting notable properties such as variable oxidation states and the ability to form colored compounds. These metals are located in groups 3 through 12 on the periodic table.
  • The d orbital filling is responsible for the distinct characteristics of transition metals.
  • Unlike s and p block elements, transition metals can have unfilled inner d orbitals even after forming certain ions.
For instance, manganese (Mn), which has the electron configuration \[ [\text{Ar}] 3d^5 4s^2 \] in its neutral form, loses two electrons to form \[ [\text{Ar}] 3d^5 \]. This results in an ion with unique properties compared to other blocks.
Ion Formation
Ion formation occurs when atoms gain or lose electrons to achieve a more stable electron configuration, often resembling a noble gas. This process is essential for the formation of compounds and determines the chemical behavior of elements. In transition metals, it primarily involves changes in the d and sometimes s orbitals.
  • Commonly, transition metals lose electrons from the s orbital before the d orbital, even though the d subshell is energetically lower once filled.
  • The number of electrons lost varies, leading to multiple oxidation states.
Consider iron (Fe), which can lose two electrons to form \[ \text{Fe}^{2+} \] with the configuration \[ [\text{Ar}] 3d^6 \], or three electrons to form \[ \text{Fe}^{3+} \] with \[ [\text{Ar}] 3d^5 \], impacting its reactivity and bonding characteristics.
Electron Orbitals
Electron orbitals are regions around an atom's nucleus where electrons are likely to be found. These orbitals define the shape and energy levels of the electron cloud, influencing the atom's interaction with other atoms.
  • Each orbital can hold a maximum of two electrons with opposite spins, as stated by Pauli's exclusion principle.
  • The main types of orbitals are s, p, d, and f, each with unique shapes and energy levels.
For transition metals, d orbitals play a crucial role. Unpaired electrons in these d orbitals lead to magnetism and affect compound coloration. For example, chromium (Cr) forms a \[ \text{Cr}^{3+} \] ion with three unpaired electrons in its d orbitals, significantly influencing its chemical properties and use in industrial applications.

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Most popular questions from this chapter

The frequency of radiation emitted when the electron falls from \(\mathrm{n}=4\) to \(\mathrm{n}=1\) in a hydrogen atom will be (Given ionization energy of \(\mathrm{H}=2.18 \times 10^{18} \mathrm{~J}\) atom \(^{-1}\) and \(h=6.625 \times 10^{-34} \mathrm{Js}\) ) (a) \(1.54 \times 10^{15} \mathrm{~s}^{-1}\) (b) \(1.03 \times 10^{15} \mathrm{~s}^{-1}\) (c) \(3.08 \times 10^{15} \mathrm{~s}^{-1}\) (d) \(2.00 \times 10^{15} \mathrm{~s}^{-1}\)

For the electronic transition from \(\mathrm{n}=2 \rightarrow \mathrm{n}=1\) which of the following will produce shortest wave length? (a) \(\mathrm{Li}^{2+}\) ion (b) D atom (c) He' ion (d) \(\mathrm{H}\) atom

What is the wavelength of the radiation emitted produced in a line in the Lyman series when an electron falls from fourth stationary state in hydrogen atom? \(\left(\mathrm{R}_{\mathrm{H}}=1.1 \times 10^{7} \mathrm{~m}^{-1}\right)\) (a) \(96.97 \mathrm{~nm}\) (b) \(969.7 \mathrm{~nm}\) (c) \(9.697 \mathrm{~nm}\) (d) none

The charge cloud of a single electron in a \(2 \mathrm{p}_{\mathrm{x}}\) atomic orbital has two lobes of electron density. This means (a) there is a high probability of locating the electron in a \(2 \mathrm{p}_{x}\) atomic orbital at values of \(\mathrm{x}>0\) (b) there is a great probability of finding a p electron right at the nucleus (c) there is a high probability of locating it values of \(x<0\) but no probability at alloy locating if any where in the yz plane along which \(\mathrm{x}=0\). (d) both (a) and (c)

Which of the following statement(s) are correct? (1) the electronic configuration of \(\mathrm{Cr}\) is \([\mathrm{Ar}] 3 \mathrm{~d}^{5} 4 \mathrm{~s}^{1}\) (atomic number of \(\mathrm{Cr}=24\) ) (2) the magnetic quantum number may have a negative value (3) in silver atom, 23 electrons have a spin of one type and 24 of the opposite type (atomic number of \(\mathrm{Ag}=47\) )
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