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Chapter 8: Problem 61

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 \(m_{s}=+\frac{1}{2} ?\)

Short Answer

Expert verified
(a) 1 electron. (b) 2 electrons. (c) 10 electrons. (d) 32 electrons. (e) 5 electrons.

Step by step solution

01

(a) n=4, l=2, m_l=1, m_s=+1/2

Only one electron can have these exact quantum numbers because the Pauli Exclusion Principle states that no two electrons in an atom can have the same set of four quantum numbers.
02

(b) n=4, l=2, m_l=1

There can be two electrons with these quantum numbers, it can have one up-spin electron (\(m_s=+1/2\)) and one down-spin electron (\(m_s=-1/2\)).
03

(c) n=4, l=2

For given values of \(n=4\) and \(\ell=2\), \(m_{\ell}\) can range from \(-2\) to \(2\) so there are five possible values. In each orbital, there can be two electrons (one with spin up and one with spin down), hence a total of \(5*2 = 10\) electrons.
04

(d) n=4

For given value of the principal quantum number \(n=4\), there are four subshells \(\ell = 0, 1, 2, 3\). With 1 orbital for \(\ell = 0\), 3 orbitals for \(\ell = 1\), 5 orbitals for \(\ell = 2\), and 7 orbitals for \(\ell = 3\), and each orbital can accommodate 2 electrons, so the total number of electrons is \(2*(1+3+5+7) = 32\).
05

(e) n=4, l=2, m_s=+1/2

Here, for given \(n=4\), \(\ell=2\), and \(m_s=+1/2\), they needs the value of \(m_{\ell}\) which ranges from \(-2\) to \(2\). So there can be \(5\) electrons having \(m_s= +1/2\) in each orbital.

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