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Chapter 8: Problem 75

On the basis of the periodic table and rules for electron configurations, indicate the number of (a) \(2 p\) electrons in \(\mathrm{N} ;\) (b) \(4 \mathrm{s}\) electrons in \(\mathrm{Rb} ;\) (c) \(4 \mathrm{d}\) electrons in As; (d) \(4 f\) electrons in \(\mathrm{Au} ;\) (e) unpaired electrons in \(\mathrm{Pb} ;\) (f) elements in group 14 of the periodic table; (g) elements in the sixth period of the periodic table.

Short Answer

Expert verified
a) Nitrogen has 3 electrons in the \(2p\) orbital. b) Rubidium has 2 electrons in the \(4s\) orbital. c) Arsenic has no \(4d\) electrons. d) Gold has 14 electrons in the \(4f\) orbital. e) Lead has 2 unpaired electrons. f) There are 5 elements in group 14 of the periodic table. g) There are 32 elements in the sixth period of the periodic table.

Step by step solution

01

Understanding Electron Configuration

The electron configuration of an element in the periodic table is determined by its atomic number. For example, for Nitrogen (\(N\)), which has an atomic number of 7, the electron configuration is \(1s^2 2s^2 2p^3\). This means there are 2 electrons in the \(1s\) orbital, 2 in the \(2s\) and 3 in the \(2p\) orbital.
02

Determining Number of 2p Electrons in Nitrogen

From the electron configuration of Nitrogen, \(1s^2 2s^2 2p^3\), we can see that the \(2p\) subshell has 3 electrons.
03

Determining Number of 4s Electrons in Rubidium

Rubidium (\(Rb\)) is number 37 on the periodic table. Its electron configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^1\). Note how there are 2 electrons in the \(4s\) subshell.
04

Determining Number of 4d Electrons in Arsenic

Arsenic (\(As\)) is number 33 on the periodic table. Its electron configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^3\). It doesn't have \(4d\) electrons.
05

Determining Number of 4f Electrons in Gold

Gold (\(Au\)) is number 79 on the periodic table. Its electron configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^1 4f^{14} 5d^{10}\). From this, we can see that there are 14 electrons in the \(4f\) subshell.
06

Determining Number of Unpaired Electrons in Lead

Lead (\(Pb\)) is number 82 on the periodic table. Its electron configuration ends in \(6p^2\) which means there are 2 unpaired electrons in the outermost \(p\) subshell.
07

Determining Elements in Group 14 of the Periodic Table

Elements in group 14 of the periodic table are Carbon (\(C\)), Silicon (\(Si\)), Germanium (\(Ge\)), Tin (\(Sn\)), and Lead (\(Pb\)).
08

Determining Elements in the Sixth Period of the Periodic Table

The sixth period of the periodic table has 32 elements, starting from Cesium (\(Cs\)) up to Radon (\(Rn\)).

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Most popular questions from this chapter

How long does it take light from the sun, 93 million miles away, to reach Earth?

Without doing detailed calculations, determine which of the following wavelengths represents light of the highest frequency: (a) \(6.7 \times 10^{-4} \mathrm{cm} ;\) (b) \(1.23 \mathrm{mm}\) (c) \(80 \mathrm{nm} ;\) (d) \(6.72 \mu \mathrm{m}\)

Based on the relationship between electron configurations and the periodic table, give the number of (a) outer-shell electrons in an atom of \(\mathrm{Sb} ;\) (b) electrons in the fourth principal electronic shell of \(\mathrm{Pt} ;\) (c) elements whose atoms have six outer-shell electrons; (d) unpaired electrons in an atom of Te; (e) transition elements in the sixth period.

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

An electron in a one-dimensional box requires a wavelength of \(618 \mathrm{nm}\) to excite an electron from the \(n=2\) level to the \(n=4\) level. Calculate the length of the box.
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