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Chapter 7: Problem 79

How many orbitals in an atom can have the designation \(5 p\), \(3 d_{z^{2}}, 4 d, n=5, n=4 ?\)

Short Answer

Expert verified
The total number of orbitals for the given designations is 50, calculated as follows: 3 (5p) + 1 (3d_{z^2}) + 5 (4d) + 25 (n=5) + 16 (n=4).

Step by step solution

01

1. Find the number of 5p orbitals

In the 5p designation, the principal quantum number n = 5 and the angular momentum quantum number l = 1 (for p orbitals). The magnetic quantum number (m_l) can have values in the range of -l to +l, in this case, from -1 to +1. Thus, for 5p orbitals, there are three possible values of m_l: -1, 0, and +1. This means there are three 5p orbitals.
02

2. Find the number of 3dz^2 orbitals

For 3d_{z^2}, the principal quantum number n = 3 and the angular momentum quantum number l = 2 (for d orbitals). Since the magnetic quantum number (m_l) is specifically designated as being z^2, that means there's just one orbital corresponding to this case, the dz^2 orbital.
03

3. Find the number of 4d orbitals

In the case of 4d, the principal quantum number n = 4 and the angular momentum quantum number l = 2 (for d orbitals). The magnetic quantum number (m_l) can have values in the range of -l to +l, in this case, from -2 to +2. Thus, for 4d orbitals, there are five possible values of m_l: -2, -1, 0, +1, and +2. This means there are five 4d orbitals.
04

4. Find the number of orbitals for n=5

For n = 5, the allowed values of the angular momentum quantum number l are 0, 1, 2, 3, and 4. For each value of l, there will be 2l+1 possible values for m_l. To find the total number of orbitals for n=5, we can sum the number of orbitals for each value of l. Orbitals for n = 5 = (2×0+1) + (2×1+1) + (2×2+1) + (2×3+1) + (2×4+1) = 1+3+5+7+9 = 25
05

5. Find the number of orbitals for n=4

For n = 4, the angular momentum quantum number l can have values from 0 to 3. Following the same formula as for n=5, we calculate the total number of orbitals for n=4: Orbitals for n = 4 = (2×0+1) + (2×1+1) + (2×2+1) + (2×3+1) = 1+3+5+7 = 16
06

6. Calculate the total number of orbitals

Now we can combine the number of orbitals for each designation to find the total number of orbitals for the given exercise: Total orbitals = 3 (5p) + 1 (3d_{z^2}) + 5 (4d) + 25 (n=5) + 16 (n=4) = 50 So, there are 50 orbitals in total for the given designations.

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Most popular questions from this chapter

How many unpaired electrons are present in each of the firstrow transition metals in the ground state?

The wave function for the \(2 p_{z}\) orbital in the hydrogen atom is $$ \psi_{2 p_{i}}=\frac{1}{4 \sqrt{2 \pi}}\left(\frac{Z}{a_{0}}\right)^{3 / 2} \sigma \mathrm{e}^{-\sigma / 2} \cos \theta $$ where \(a_{0}\) is the value for the radius of the first Bohr orbit in meters \(\left(5.29 \times 10^{-11}\right), \sigma\) is \(Z\left(r / a_{0}\right), r\) is the value for the distance from the nucleus in meters, and \(\theta\) is an angle. Calculate the value of \(\psi_{2 p_{z}}^{2}\) at \(r=a_{0}\) for \(\theta=0^{\circ}(z\) axis \()\) and for \(\theta=90^{\circ}\) \((x y\) plane).

Which of the following orbital designations are incorrect: \(1 s\), \(1 p, 7 d, 9 s, 3 f, 4 f, 2 d ?\)

Consider the following ionization energies for aluminum: $$ \begin{aligned} \mathrm{Al}(g) \longrightarrow \mathrm{Al}^{+}(g)+\mathrm{e}^{-} & I_{1}=580 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Al}^{+}(g) \longrightarrow \mathrm{Al}^{2+}(g)+\mathrm{e}^{-} & I_{2}=1815 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Al}^{2+}(g) \longrightarrow \mathrm{Al}^{3+}(g)+\mathrm{e}^{-} & I_{3}=2740 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Al}^{3+}(g) \longrightarrow \mathrm{Al}^{4+}(g)+\mathrm{e}^{-} & I_{4}=11,600 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ a. Account for the trend in the values of the ionization energies. b. Explain the large increase between \(I_{3}\) and \(I_{4}\).

Which of the following statements is(are) true? a. The \(2 s\) orbital in the hydrogen atom is larger than the \(3 s\) orbital also in the hydrogen atom. b. The Bohr model of the hydrogen atom has been found to be incorrect. c. The hydrogen atom has quantized energy levels. d. An orbital is the same as a Bohr orbit. e. The third energy level has three sublevels, the \(s, p\), and \(d\) sublevels.
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