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Chapter 6: Problem 66

For each element, count the number of valence electrons, core electrons, and unpaired electrons in the ground state: (a) carbon, (b) phosphorus, (c) neon.

Short Answer

Expert verified
For the given elements in their ground state: (a) Carbon has 4 valence electrons, 2 core electrons, and 2 unpaired electrons; (b) Phosphorus has 5 valence electrons, 10 core electrons, and 3 unpaired electrons; (c) Neon has 8 valence electrons, 2 core electrons, and 0 unpaired electrons.

Step by step solution

01

(a) Carbon

Step 1: Determine the electron configuration Carbon has an atomic number of 6, meaning it has 6 electrons in its neutral state. The electron configuration is given by: 1s虏 2s虏 2p虏 Step 2: Identify valence and core electrons Valence electrons are the electrons in the outermost shell of an atom. In this case, carbon has 4 valence electrons (2s虏 and 2p虏). The remaining 2 electrons (1s虏) are core electrons. Step 3: Determine unpaired electrons Unpaired electrons are electrons without a full set of paired electrons. Here, carbon has 2 unpaired electrons (2p虏). So, for carbon: - Valence electrons: 4 - Core electrons: 2 - Unpaired electrons: 2
02

(b) Phosphorus

Step 1: Determine the electron configuration Phosphorus has an atomic number of 15, meaning it has 15 electrons in its neutral state. The electron configuration is given by: 1s虏 2s虏 2p鈦� 3s虏 3p鲁 Step 2: Identify valence and core electrons Phosphorus has 5 valence electrons (3s虏 and 3p鲁). The remaining 10 electrons (1s虏, 2s虏, and 2p鈦�) are core electrons. Step 3: Determine unpaired electrons Unpaired electrons are electrons without a full set of paired electrons. Here, phosphorus has 3 unpaired electrons (3p鲁). So, for phosphorus: - Valence electrons: 5 - Core electrons: 10 - Unpaired electrons: 3
03

(c) Neon

Step 1: Determine the electron configuration Neon has an atomic number of 10, meaning it has 10 electrons in its neutral state. The electron configuration is given by: 1s虏 2s虏 2p鈦� Step 2: Identify valence and core electrons Neon has 8 valence electrons (2s虏 and 2p鈦�). The 2 remaining electrons (1s虏) are core electrons. Step 3: Determine unpaired electrons Since neon has a full set of paired electrons in the 2s and 2p orbitals, it has 0 unpaired electrons. So, for neon: - Valence electrons: 8 - Core electrons: 2 - Unpaired electrons: 0

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Most popular questions from this chapter

Neutron diffraction is an important technique for determining the structures of molecules. Calculate the velocity of a neutron needed to achieve a wavelength of \(0.955 \AA\). (Refer to the inside cover for the mass of the neutron).

(a) What experimental evidence is there for the electron having a "spin"? (b) Draw an energy-level diagram that shows the relative energetic positions of a \(1 s\) orbital and a 2 s orbital. Put two electrons in the 1 s orbital. (c) Draw an arrow showing the excitation of an electron from the 1s to the 2 s orbital.

A diode laser emits at a wavelength of \(987 \mathrm{~nm}\). (a) In what portion of the electromagnetic spectrum is this radiation found? (b) All of its output energy is absorbed in a detector that measures a total energy of \(0.52 \mathrm{~J}\) over a period of \(32 \mathrm{~s}\). How many photons per second are being emitted by the laser?

(a) For \(n=4\), what are the possible values of \(l ?\) (b) For \(l=2\), what are the possible values of \(m_{l} ?(c)\) If \(m_{l}\) is 2, what are the possible values for \(l ?\)

(a) Using Equation \(6.5\), calculate the energy of an electron in the hydrogen atom when \(n=2\) and when \(n=6\). Calculate the wavelength of the radiation released when an electron moves from \(n=6\) to \(n=2\). Is this line in the visible region of the electromagnetic spectrum? If so, what color is it? (b) Calculate the energies of an electron in the hydrogen atom for \(n=1\) and for \(n=(\infty)\). How much energy does it require to move the electron out of the atom completely (from \(n=1\) to \(n=\infty\) ), according to Bohr? Put your answer in \(\mathrm{kJ} / \mathrm{mol}\). (c) The energy for the process \(\mathrm{H}+\) energy \(\rightarrow \mathrm{H}^{+}+\mathrm{e}^{-}\) is called the ionization energy of hydrogen. The experimentally determined value for the ionization energy of hydrogen is \(1310 \mathrm{~kJ} / \mathrm{mol}\). How does this compare to your calculation?
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