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Consider the ground state of arsenic, As. How many electrons have \(\ell=1\) as one of their quantum numbers? How many electrons have \(m_{\ell}=0 ?\) How many electrons have \(m_{\ell}=+1 ?\)

Short Answer

Expert verified
In the ground state of arsenic, 15 electrons have \(\ell=1\) as one of their quantum numbers, 6 electrons have \(m_{\ell}=0\), and 6 electrons have \(m_{\ell}=+1\).

Step by step solution

01

Find the electron configuration of arsenic

To find the electron configuration of arsenic, we need to know its atomic number. Arsenic has an atomic number of 33, which means it has 33 electrons in its ground state. We can now fill the orbitals in ascending order of energy, following the Aufbau principle. The electron configuration of arsenic is: \[1s^22s^22p^63s^23p^64s^23d^{10}4p^3\]
02

Count electrons with \(\ell=1\)

Now, we have to find the number of electrons that have \(\ell =1\). The value \(\ell=1\) corresponds to the p-orbitals. In the electron configuration of arsenic, we have: 2p: 6 electrons 3p: 6 electrons 4p: 3 electrons Total number of electrons with \(\ell=1\) is: \(6+6+3 = 15\)
03

Count electrons with \(m_{\ell}=0\)

Next, we need to find the number of electrons with \(m_{\ell}=0\). The value \(m_\ell=0\) corresponds to those orbitals which are aligned along the z-axis. To find them, we'll look at the electron configuration. In p-orbitals with \(\ell=1\), \(m_\ell\) can be -1, 0, or +1. Since each p-orbital can hold 6 electrons, there will be 2 electrons in each \(m_\ell\) state. So, there are 2 electrons with \(m_\ell=0\) in each of the following orbitals: 2p, 3p, and 4p. Total number of electrons with \(m_{\ell}=0\): \(2+2+2 = 6\)
04

Count electrons with \(m_{\ell}=+1\)

Lastly, we need to find the number of electrons with \(m_{\ell}=+1\). As mentioned earlier, in p-orbitals with \(\ell=1\), \(m_\ell\) can be -1, 0, or +1. Similar to the previous step, there will be 2 electrons with \(m_\ell=+1\) in each of the following orbitals: 2p, 3p, and 4p. Total number of electrons with \(m_{\ell}=+1\): \(2+2+2 = 6\) So, in the ground state of arsenic: - 15 electrons have \(\ell=1\) as one of their quantum numbers. - 6 electrons have \(m_{\ell}=0\). - 6 electrons have \(m_{\ell}=+1\).

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Most popular questions from this chapter

Valence electrons are those electrons in the outermost principal quantum level (highest n level) of an atom in its ground state. Groups 1A to 8A have from 1 to 8 valence electrons. For each group of the representative elements (1A–8A), give the number of valence electrons, the general valence electron configuration, a sample element in that group, and the specific valence electron configuration for that element.

Complete and balance the equations for the following reactions. a. \(\operatorname{Li}(s)+\mathrm{N}_{2}(g) \rightarrow\) b. \(\operatorname{Rb}(s)+\mathrm{S}(s) \rightarrow\)

An electron is excited from the \(n=1\) ground state to the \(n=\) 3 state in a hydrogen atom. Which of the following statements are true? Correct the false statements to make them true. a. It takes more energy to ionize (completely remove) the electron from \(n=3\) than from the ground state. b. The electron is farther from the nucleus on average in the \(n=3\) state than in the \(n=1\) state. c. The wavelength of light emitted if the electron drops from \(n=3\) to \(n=2\) will be shorter than the wavelength of light emitted if the electron falls from \(n=3\) to \(n=1 .\) d. The wavelength of light emitted when the electron returns to the ground state from \(n=3\) will be the same as the wavelength of light absorbed to go from \(n=1\) to \(n=3\) e. For \(n=3,\) the electron is in the first excited state.

Calculate the de Broglie wavelength for each of the following. a. an electron with a velocity 10.% of the speed of light b. a tennis ball ( 55 g) served at 35 \(\mathrm{m} / \mathrm{s}(\sim 80 \mathrm{mi} / \mathrm{h})\)

Assume that a hydrogen atom's electron has been excited to the \(n=6\) level. How many different wavelengths of light can be emitted as this excited atom loses energy?

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