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Chapter 18: Problem 2488

which of the following atom has the lowest ionization potential? (A) \({ }^{14}{ }_{7} \mathrm{~N}\) (B) \({ }^{40}{ }_{18} \mathrm{Ar}\) (C) \({ }^{133} 55 \mathrm{Cs}\) (D) \({ }^{16}{ }_{8} \mathrm{O}\)

Short Answer

Expert verified
The atom with the lowest ionization potential is (C) \({ }^{133} 55 \mathrm{Cs}\) (Cesium), as it is farther down and to the left in the periodic table compared to the other options.

Step by step solution

01

Identify each atom's position on the periodic table

To analyze the ionization potential, first, we need to determine the position of each atom in the periodic table. The atomic number (number of protons) will help in identifying the element, and the period and group they belong to. (A) \({ }^{14}{ }_{7} \mathrm{~N}\) : Nitrogen, Period 2, Group 15 (5A) (B) \({ }^{40}{ }_{18} \mathrm{Ar}\) : Argon, Period 3, Group 18 (8A) (C) \({ }^{133} 55 \mathrm{Cs}\) : Cesium, Period 6, Group 1 (1A) (D) \({ }^{16}{ }_{8} \mathrm{O}\) : Oxygen, Period 2, Group 16 (6A)
02

Compare the atoms' ionization potentials based on their position in the periodic table

As mentioned before, ionization potential decreases down a group and increases across a period. We can now compare the atoms according to their position: - Nitrogen and Oxygen are in the same period (Period 2), but Oxygen has a higher ionization potential because it is farther to the right. - Argon is in a higher period (Period 3) than Nitrogen and Oxygen and is in Group 18, which is the farthest right group on the periodic table, thus having a very high ionization potential. - Cesium is in Period 6 and Group 1. It is farther down and to the left on the periodic table, indicating lower ionization potential compared to the other atoms.
03

Determine which atom has the lowest ionization potential

Based on the positions of each atom on the periodic table, we can conclude that Cesium (\({ }^{133} 55 \mathrm{Cs}\)) has the lowest ionization potential among the given options. Thus, the correct answer is (C) \({ }^{133} 55 \mathrm{Cs}\).

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Most popular questions from this chapter

An electron change its Position from orbit \(n=4\) to the orbit \(\mathrm{n}=2\) of an atom the wave length of emitted radiation in the form of \(\mathrm{R}\) (where \(\mathrm{R}\) is Redburg constants) (A) \((16 / 7 \mathrm{R})\) (B) \((16 / \mathrm{R})\) (C) (16 / 3R) (D) \((16 / 5 \mathrm{R})\)

A freshly Prepared radioactive source of half life \(2 \mathrm{~h}\) emits radiation of intensity which is 64 times the Permissible safe level. The minimum time after which is would be possible to work safely with this source is. (A) \(6 \mathrm{~h}\) (B) \(24 \mathrm{~h}\) (C) \(12 \mathrm{~h}\) (D) \(36 \mathrm{~h}\)

which of these is a fusion reaction (A) ${ }^{3}{ }_{1} \mathrm{H}+{ }_{1}^{2} \mathrm{H}={ }_{2}^{4} \mathrm{He}+{ }_{0} \mathrm{n}^{1}$ (B) ${ }^{12}{ }_{7} \mathrm{C} \rightarrow{ }_{6} \mathrm{C}^{12}+\beta^{+}+\mathrm{g}$ (C) ${ }_{92} \mathrm{U}^{238} \rightarrow{ }^{206}{ }_{82} \mathrm{~Pb}+8\left({ }_{2}^{4} \mathrm{He}\right)+6\left(_{1-} \mathrm{e}\right)$ (D) None of these

According to Bohr's theory the radius of electron in an orbit described by Principal quantum number \(\mathrm{n}\) and atomic number \(Z\), is Proportional to. (A) \(Z^{2} n^{2}\) (B) \(\left(\mathrm{n}^{2} / \mathrm{Z}\right)\) (C) \(\left(\mathrm{Z}^{2} / \mathrm{n}\right)\) (D) \(\left(\mathrm{n} / \mathrm{Z}^{2}\right)\)

The ratio of atomic volume of nuclear volume is of the order of (A) \(10^{-15}\) (B) \(10^{-10}\) (C) \(10^{15}\) (D) \(10^{-10}\)
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