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Chapter 2: Problem 97

Give a possible set of values of the four quantum numbers for all the electrons in a boron atom and a nitrogen atom if each is in the ground state.

Short Answer

Expert verified
For a boron atom in the ground state, the set of quantum numbers for all the electrons is: - Electron 1: \(n = 1\), \(l = 0\), \(m_l = 0\), \(m_s = +\frac{1}{2}\) - Electron 2: \(n = 1\), \(l = 0\), \(m_l = 0\), \(m_s = -\frac{1}{2}\) - Electron 3: \(n = 2\), \(l = 0\), \(m_l = 0\), \(m_s = +\frac{1}{2}\) - Electron 4: \(n = 2\), \(l = 0\), \(m_l = 0\), \(m_s = -\frac{1}{2}\) - Electron 5: \(n = 2\), \(l = 1\), \(m_l = -1\), \(m_s = +\frac{1}{2}\) For a nitrogen atom in the ground state, the set of quantum numbers for all the electrons are: - Electron 1: \(n = 1\), \(l = 0\), \(m_l = 0\), \(m_s = +\frac{1}{2}\) - Electron 2: \(n = 1\), \(l = 0\), \(m_l = 0\), \(m_s = -\frac{1}{2}\) - Electron 3: \(n = 2\), \(l = 0\), \(m_l = 0\), \(m_s = +\frac{1}{2}\) - Electron 4: \(n = 2\), \(l = 0\), \(m_l = 0\), \(m_s = -\frac{1}{2}\) - Electron 5: \(n = 2\), \(l = 1\), \(m_l = -1\), \(m_s = +\frac{1}{2}\) - Electron 6: \(n = 2\), \(l = 1\), \(m_l = 0\), \(m_s = +\frac{1}{2}\) - Electron 7: \(n = 2\), \(l = 1\), \(m_l = 1\), \(m_s = +\frac{1}{2}\)

Step by step solution

01

Boron's Electron Configuration and Quantum Numbers

First, we need to determine the electron configuration of boron in the ground state. Boron has 5 electrons, so its electron configuration is 1s², 2s², 2p¹. Now, let's find out the set of quantum numbers for all the electrons. For the 1s² electrons: - Electron 1: \(n = 1\), \(l = 0\), \(m_l = 0\), \(m_s = +\frac{1}{2}\) - Electron 2: \(n = 1\), \(l = 0\), \(m_l = 0\), \(m_s = -\frac{1}{2}\) For the 2s² electrons: - Electron 3: \(n = 2\), \(l = 0\), \(m_l = 0\), \(m_s = +\frac{1}{2}\) - Electron 4: \(n = 2\), \(l = 0\), \(m_l = 0\), \(m_s = -\frac{1}{2}\) For the 2p¹ electron: - Electron 5: \(n = 2\), \(l = 1\), \(m_l = -1\), \(m_s = +\frac{1}{2}\)
02

Nitrogen's Electron Configuration and Quantum Numbers

Next, let's determine the electron configuration of nitrogen in the ground state. Nitrogen has 7 electrons, so its electron configuration is 1s², 2s², 2p³. Let's find the set of quantum numbers for all the electrons. For the 1s² electrons: - Electron 1: \(n = 1\), \(l = 0\), \(m_l = 0\), \(m_s = +\frac{1}{2}\) - Electron 2: \(n = 1\), \(l = 0\), \(m_l = 0\), \(m_s = -\frac{1}{2}\) For the 2s² electrons: - Electron 3: \(n = 2\), \(l = 0\), \(m_l = 0\), \(m_s = +\frac{1}{2}\) - Electron 4: \(n = 2\), \(l = 0\), \(m_l = 0\), \(m_s = -\frac{1}{2}\) For the 2p³ electrons: - Electron 5: \(n = 2\), \(l = 1\), \(m_l = -1\), \(m_s = +\frac{1}{2}\) - Electron 6: \(n = 2\), \(l = 1\), \(m_l = 0\), \(m_s = +\frac{1}{2}\) - Electron 7: \(n = 2\), \(l = 1\), \(m_l = 1\), \(m_s = +\frac{1}{2}\) In conclusion, we have identified the possible sets of values of the four quantum numbers for all the electrons in a boron atom and a nitrogen atom if each is in the ground state.

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Most popular questions from this chapter

Assume that we are in another universe with different physical laws. Electrons in this universe are described by four quantum numbers with meanings similar to those we use. We will call these quantum numbers \(p, q, r,\) and \(s .\) The rules for these quantum numbers are as follows: \(p=1,2,3,4,5, \dots\) \(q\) takes on positive odd integers and \(q \leq p\) \(r\) takes on all even integer values from \(-q\) to \(+q\). (Zero is considered an even number.) \(s=+\frac{1}{2}\) or \(-\frac{1}{2}\) a. Sketch what the first four periods of the periodic table will look like in this universe. b. What are the atomic numbers of the first four elements you would expect to be least reactive? c. Give an example, using elements in the first four rows, of ionic compounds with the formulas XY, XY \(_{2}, X_{2} Y, X Y_{3}\) and \(\mathrm{X}_{2} \mathrm{Y}_{3}\) d. How many electrons can have \(p=4, q=3 ?\) e. How many electrons can have \(p=3, q=0, r=0 ?\) f. How many electrons can have \(p=6 ?\)

The wave function for the \(2 p_{z}\) orbital in the hydrogen atom is $$ \psi_{2 p_{i}}=\frac{1}{4 \sqrt{2 \pi}}\left(\frac{Z}{a_{0}}\right)^{3 / 2} \sigma \mathrm{e}^{-\alpha / 2} \cos \theta $$ 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 \(\psi_{2 p^{2}}\) at \(r=a_{0}\) for \(\theta=0^{\circ}\left(z \text { axis) and for } \theta=90^{\circ}\right.\) (xy plane).

One of the emission spectral lines for \(\mathrm{Be}^{3+}\) has a wavelength of \(253.4 \mathrm{nm}\) for an electronic transition that begins in the state with \(n=5 .\) What is the principal quantum number of the lower-energy state corresponding to this emission? (Hint: The Bohr model can be applied to one- electron ions. Don't forget the \(Z\) factor: \(Z=\) nuclear charge \(=\) atomic number.

Arrange the following groups of atoms in order of increasing size. a. \(\mathrm{Te}, \mathrm{S}, \mathrm{Se}\) b. \(\mathrm{K}, \mathrm{Br}, \mathrm{Ni}\) c. \(\mathrm{Ba}, \mathrm{Si}, \mathrm{F}\)

Order the atoms in each of the following sets from the least negative electron affinity to the most. a. \(\mathrm{S}, \mathrm{Se}\) b. \(\mathrm{F}, \mathrm{Cl}, \mathrm{Br}, \mathrm{I}\)
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