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Chapter 2: Problem 96

In the ground state of element \(115,\) Uup, a. how many electrons have \(n=5\) as one of their quantum numbers? b. how many electrons have \(\ell=3\) as one of their quantum numbers? c. how many electrons have \(m_{\ell}=1\) as one of their quantum numbers? d. how many electrons have \(m_{s}=-\frac{1}{2}\) as one of their quantum numbers?

Short Answer

Expert verified
a. There are 32 electrons with n=5 as one of their quantum numbers. b. There are 28 electrons with ℓ=3 as one of their quantum numbers. c. There are 14 electrons with mℓ=1 as one of their quantum numbers. d. There are 58 electrons with ms=-1/2 as one of their quantum numbers.

Step by step solution

01

Determine the Electronic Configuration of Ununpentium

Ununpentium (Uup) has an atomic number of 115, meaning it has 115 electrons. To determine the electron configuration, follow the Aufbau principle and fill orbitals in order of increasing energy. The electronic configuration for Ununpentium is: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s² 5f¹⁴ 6d¹⁰ 7p³ Now let's answer each question:
02

a. Find electrons with n=5

Count the electrons in the orbitals where the principal quantum number (n) is 5. From the electron configuration: 5s² 5p⁶ 5d¹⁰ 5f¹⁴ 2 + 6 + 10 + 14 = 32 electrons There are 32 electrons with n=5 as one of their quantum numbers.
03

b. Find electrons with ℓ=3

Count the electrons in the orbitals where the azimuthal quantum number (ℓ) is 3. ℓ=3 corresponds to the f orbitals. From the electron configuration: 4f¹⁴ 5f¹⁴ 14 + 14 = 28 electrons There are 28 electrons with ℓ=3 as one of their quantum numbers.
04

c. Find electrons with mℓ=1

Count the electrons in the orbitals where the magnetic quantum number (mℓ) is 1. For s orbitals (ℓ=0), there is no mℓ=1 For p orbitals (ℓ=1), mℓ can be -1, 0, 1 (1 electron mℓ=1 per p orbital ) For d orbitals (ℓ=2), mℓ can be -2, -1, 0, 1, 2 (1 electron mℓ=1 per d orbital) For f orbitals (ℓ=3), mℓ can be -3, -2, -1, 0, 1, 2, 3 (1 electron mℓ=1 per f orbital) From the electron configuration: 6 p orbitals × 1 mℓ=1 electron per p orbital = 6 mℓ=1 electrons 6 d orbitals × 1 mℓ=1 electron per d orbital = 6 mℓ=1 electrons 2 f orbitals × 1 mℓ=1 electron per f orbital = 2 mℓ=1 electrons 6 + 6 + 2 = 14 electrons There are 14 electrons with mℓ=1 as one of their quantum numbers.
05

d. Find electrons with ms=-1/2

Every orbital can hold a maximum of 2 electrons, one with ms=+1/2 and one with ms=-1/2. Since Ununpentium has 115 electrons, half of them will have ms=-1/2. 115 electrons / 2 = 57.5 As electrons are always in pairs, we need to round up to the nearest whole number: 58 electrons There are 58 electrons with ms=-1/2 as one of their quantum numbers.

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Most popular questions from this chapter

What are the possible values for the quantum numbers \(n, \ell\) and \(m_{\ell} ?\)

How many electrons in an atom can have the designation \(1 p\) \(6 d_{x^{2}-y^{2}}, 4 f, 7 p_{y}, 2 s, n=3 ?\)

Does a photon of visible light \((\lambda=400 \text { to } 700 \mathrm{nm}\) ) have sufficient energy to excite an electron in a hydrogen atom from the \(n=1\) to the \(n=5\) energy state? From the \(n=2\) to the \(n=\) 6 energy state?

A carbon-oxygen double bond in a certain organic molecule absorbs radiation that has a frequency of \(6.0 \times 10^{13} \mathrm{s}^{-1}\). a. What is the wavelength of this radiation? b. To what region of the spectrum does this radiation belong? c. What is the energy of this radiation per photon? d. A carbon-oxygen bond in a different molecule absorbs radiation with frequency equal to \(5.4 \times 10^{13} \mathrm{s}^{-1} .\) Is this radiation more or less energetic?

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 \(I_{4}\).
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