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Chapter 7: Problem 132

The first-row transition metals from chromium through zinc all have some biologic function in the human body. How many unpaired electrons are present in each of these first-row transition metals in the ground state?

Short Answer

Expert verified
The number of unpaired electrons in the ground state for the first-row transition metals from chromium to zinc are: Cr (5), Mn (5), Fe (4), Co (3), Ni (2), Cu (1), and Zn (0).

Step by step solution

01

Identify the atomic number of the elements

The elements are chromium (Cr), manganese (Mn), iron (Fe), cobalt (Co), nickel (Ni), copper (Cu), and zinc (Zn). Their atomic numbers are: Cr (24), Mn (25), Fe (26), Co (27), Ni (28), Cu (29), Zn (30).
02

Write the electronic configuration

Electronic configurations of these elements in the ground state are: - Cr (\(Z = 24\)): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1\) - Mn (\(Z = 25\)): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^2\) - Fe (\(Z = 26\)): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2\) - Co (\(Z = 27\)): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^7 4s^2\) - Ni (\(Z = 28\)): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^8 4s^2\) - Cu (\(Z = 29\)): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1\) - Zn (\(Z = 30\)): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2\)
03

Determine the number of unpaired electrons

To find the number of unpaired electrons, we need to count the number of electrons in the partially filled orbitals: - Cr: 3d orbital has 5 unpaired electrons. - Mn: 3d orbital has 5 unpaired electrons. - Fe: 3d orbital has 4 unpaired electrons. - Co: 3d orbital has 3 unpaired electrons. - Ni: 3d orbital has 2 unpaired electrons. - Cu: 3d orbital is fully filled and only the 4s orbital has 1 unpaired electron. - Zn: All orbitals are fully filled, so there are no unpaired electrons. So, the number of unpaired electrons in the ground state for the first-row transition metals from chromium to zinc are 5, 5, 4, 3, 2, 1, and 0, respectively.

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Most popular questions from this chapter

The bright yellow light emitted by a sodium vapor lamp consists of two emission lines at \(589.0\) and \(589.6 \mathrm{~nm}\). What are the frequency and the energy of a photon of light at each of these wavelengths? What are the energies in \(\mathrm{kJ} / \mathrm{mol}\) ?

What are the possible values for the quantum numbers \(n, \ell\), and \(m_{\ell} ?\)

An excited hydrogen atom with an electron in the \(n=5\) state emits light having a frequency of \(6.90 \times 10^{14} \mathrm{~s}^{-1}\). Determine the principal quantum level for the final state in this electronic transition.

Are the following statements true for the hydrogen atom only, true for all atoms, or not true for any atoms? a. The principal quantum number completely determines the energy of a given electron. b. The angular momentum quantum number, \(\ell\), determines the shapes of the atomic orbitals. c. The magnetic quantum number, \(m_{\ell}\), determines the direction that the atomic orbitals point in space.

Consider the following ionization energies for aluminum: $$ \begin{aligned} \mathrm{Al}(g) \longrightarrow \mathrm{Al}^{+}(g)+\mathrm{e}^{-} & I_{1}=580 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Al}^{+}(g) \longrightarrow \mathrm{Al}^{2+}(g)+\mathrm{e}^{-} & I_{2}=1815 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ $$ \begin{array}{ll} \mathrm{Al}^{2+}(g) \longrightarrow \mathrm{Al}^{3+}(g)+\mathrm{e}^{-} & I_{3}=2740 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Al}^{3+}(g) \longrightarrow \mathrm{Al}^{4+}(g)+\mathrm{e}^{-} & I_{4}=11,600 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ a. Account for the trend in the values of the ionization energies. b. Explain the large increase between \(I_{3}\) and \(I_{4}\). c. Which one of the four ions has the greatest electron affinity? Explain. d. List the four aluminum ions given in order of increasing size, and explain your ordering. (Hint: Remember that most of the size of an atom or ion is due to its electrons.)
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