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Chapter 8: Problem 59

What type of orbital (i.e., \(3 s, 4 p, \ldots)\) is designated by these quantum numbers? (a) \(n=5, \ell=1, m_{\ell}=0\) (b) \(n=4, \ell=2, m_{\ell}=-2\) (c) \(n=2, \ell=0, m_{\ell}=0\)

Short Answer

Expert verified
a) The quantum numbers (\(n=5, \ell=1, m_{\ell}=0\)) represent a 5p orbital. b) The quantum numbers (\(n=4, \ell=2, m_{\ell}=-2\)) represent a 4d orbital. c) The quantum numbers (\(n=2, \ell=0, m_{\ell}=0\)) represent a 2s orbital.

Step by step solution

01

- Analyze the Quantum Numbers for Part (a)

For the first set of quantum numbers (\(n=5, \ell=1, m_{\ell}=0\)), the principal quantum number, \(n=5\), signifies that the electron is in the 5th energy level. The azimuthal quantum number, \(\ell=1\), corresponds to a p-orbital. The magnetic quantum number, \(m_{\ell}=0\), signifies that this orbital is aligned along one of the coordinate axes.
02

- Analyze the Quantum Numbers for Part (b)

For the second set of quantum numbers (\(n=4, \ell=2, m_{\ell}=-2\)), the principal quantum number, \(n=4\), signifies that the electron is in the 4th energy level. The azimuthal quantum number, \(\ell=2\), corresponds to a d-orbital. The magnetic quantum number, \(m_{\ell}=-2\), signifies that this orbital is specifically aligned along one direction in 3D space.
03

- Analyze the Quantum Numbers for Part (c)

For the third set of quantum numbers (\(n=2, \ell=0, m_{\ell}=0\)), the principal quantum number, \(n=2\), signifies that the electron is in the 2nd energy level. The azimuthal quantum number, \(\ell=0\), corresponds to an s-orbital. The magnetic quantum number, \(m_{\ell}=0\), signifies that this orbital is spherically symmetric around the nucleus.

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Most popular questions from this chapter

Construct a concept map for the configurations of multielectron atoms.

Concerning the electrons in the shells, subshells, and orbitals of an atom, how many can have (a) \(n=4, \ell=2, m_{\ell}=1,\) and \(m_{s}=+\frac{1}{2} ?\) (b) \(n=4, \ell=2,\) and \(m_{\ell}=1 ?\) (c) \(n=4\) and \(\ell=2 ?\) (d) \(n=4 ?\) (e) \(n=4, \ell=2,\) and \(m_{s}=+\frac{1}{2} ?\)

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Show that the probability of finding a \(3 d_{x z}\) electron in the \(x y\) plane is zero.

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