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Chapter 4: Problem 7

Recall that the IE of \(\mathrm{H}\) is \(1.31 \mathrm{MJ} / \mathrm{mole}\). If all three electrons in Li were in the first shell at a distance equal to that of hydrogen, which of the following values would be the better estimate of the IE of Li: 3.6 MJ/mole or 0.6 MJ/mole? Explain.

Short Answer

Expert verified
The better estimate for the ionization energy of Li is 3.6 MJ/mole.

Step by step solution

01

Understand the relationship between number of electrons and IE

Notice that the more electrons an atom has in its outer shell, the higher its ionization energy will be. This is because you need more energy to remove additional electrons. Since the ionization energy of hydrogen, which has one electron, is 1.31 MJ/mole, the ionization energy of Li, which has three electrons, should logically just be three times greater than that of H, assuming their distance from the nucleus is equivalent.
02

Calculate the expected IE for Li

Multiply the ionization energy of hydrogen by the number of electrons in Li (which is 3): IE(Li) = 3 * 1.31 MJ/mole = 3.93 MJ/mole.
03

Compare the calculated value with the given options

Comparing the calculated Li IE of 3.93 MJ/mole with the two provided values (3.6 MJ/mole and 0.6 MJ/mole), it is evident that 3.6 MJ/mole is closer to our computed value than 0.6 MJ/mole.
04

Make final assertions

The value of 3.6 MJ/mole would, thus, be a better estimate than 0.6 MJ/mole for the ionization energy of Lithium.

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Most popular questions from this chapter

Is the amount of energy required to remove one of the electrons from the first shell of Li greater than, less than, or equal to the \(\mathrm{IE}_{1}\) for Li? Explain your reasoning.

Predict the relationship between \(\mathrm{IE}_{1}\) and atomic number by making a rough graph of IE \(_{1}\) vs. atomic number. DO NOT PROCEED TO THE NEXT PAGE UNTIL YOU HAVE COMPLETED THIS GRAPH.

The value of the ionization energy of He given in Table 1 is described as being consistent with two electrons in a "shell" approximately the same distance from the nucleus as the one electron in H. Use the Coulombic Potential Energy equation, \(V=\frac{\mathrm{kq}_{1} \mathrm{q}_{2}}{\mathrm{~d}}\) to explain how this conclusion can be reached. Hint: recall the relationship between \(V\) and \(\mathrm{IE}_{1}\).

Using grammatically correct English sentences: a) provide a possible explanation for why \(\mathrm{IE}_{1}\) for \(\mathrm{He}\) is greater than \(\mathrm{IE}_{1}\) for \(\mathrm{H}\). b) provide a possible explanation for why \(\mathrm{IE}_{1}\) for \(\mathrm{Li}\) is less than \(\mathrm{IE}_{1}\) for He.

For atoms with many electrons, not all electrons are at the same distance from the nucleus. In this case, which electron would have the lowest ionization energy: the electron that is closest to the nucleus or the electron that is farthest from the nucleus? Explain.
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