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Chapter 9: Problem 91

Refer only to the periodic table on the inside front cover and indicate which of the atoms, \(\mathrm{Bi}, \mathrm{S}, \mathrm{Ba}, \mathrm{As}\) and \(\mathrm{Ca},\) (a) is most metallic; (b) is most nonmetallic; (c) has the intermediate value when the five are arranged in order of increasing first ionization energy.

Short Answer

Expert verified
The most metallic element is \(\mathrm{Ba}\), the most nonmetallic element is \(\mathrm{S}\), and the element with the intermediate first ionization energy is \(\mathrm{Ca}\).

Step by step solution

01

Finding the most metallic element

Identify the positions of \(\mathrm{Bi}, \mathrm{S}, \mathrm{Ba}, \mathrm{As}\) and \(\mathrm{Ca}\) on the periodic table. The most metallic element is the one located closest to the left and bottom. The element \(\mathrm{Ba}\) is the farthest to the left and bottom.
02

Finding the most nonmetallic element

By the same logic, the most nonmetallic element is the one located furthest to the right and top. Here, this element is \(\mathrm{S}\).
03

Finding the intermediate ionization energy

Knowing that ionization energy increases from left to right and bottom to top, arrange the given elements in ascending order of their ionization energy: \(\mathrm{Ba}, \mathrm{Bi}, \mathrm{Ca}, \mathrm{As}, \mathrm{S}\). The element with intermediate ionization energy is \(\mathrm{Ca}\) as it falls in the middle of this arrangement.

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Most popular questions from this chapter

Is it possible for two different atoms to be isoelectronic? two different cations? two different anions? a cation and an anion? Explain.

Refer only to the periodic table on the inside front cover, and arrange the following ionization energies in order of increasing value: \(I_{1}\) for \(\mathrm{F} ; I_{2}\) for \(\mathrm{Ba}\) \(I_{3}\) for \(\mathrm{Sc} ; I_{2}\) for \(\mathrm{Na} ; I_{3}\) for \(\mathrm{Mg}\). Explain the basis of any uncertainties.

For the following groups of elements, select the one that has the property noted: (a) the largest atom: \(\mathrm{Mg}, \mathrm{Mn}, \mathrm{Mo}, \mathrm{Ba}, \mathrm{Bi}, \mathrm{Br}.\) (b) the lowest first ionization energy: \(\mathrm{B}, \mathrm{Sr}, \mathrm{Al}, \mathrm{Br}\) \(\mathrm{Mg}_{\ell} \mathrm{Pb}.\) (c) the most negative electron affinity: \(\mathrm{As}, \mathrm{B}, \mathrm{Cl}\) \(\mathrm{K}, \mathrm{Mg}, \mathrm{S}.\) (d) the largest number of unpaired electrons: \(\mathrm{F}, \mathrm{N}, \mathrm{S}^{2-}, \mathrm{Mg}^{2+}, \mathrm{Sc}^{3+}, \mathrm{Ti}^{3+}.\)

Arrange the following in expected order of increasing radius: $\mathrm{Br}, \mathrm{Li}^{+}, \mathrm{Se}, \mathrm{I}^{-} .$ Explain your answer.

Consider a nitrogen atom in the ground state and comment on whether the following statements are true or false. (a) \(\mathrm{Z}_{\text {eff }}\) for an electron in a 2 s orbital is greater than that for the 1 s orbital. (b) The \(Z_{\text {eff for the } 2 p \text { and } 2 s \text { orbitals is the same. }}\) (c) More energy is required to remove an electron from a 2 s orbital than from the \(2 p\) orbital. (d) The 2 s electron is less shielded than the \(2 p\) electron.
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