Explanation Sodium and Chlorine belongs to the same period (Period 3). In Period 3, Sodium lies on the left of Chlorine. Atomic size decreases along the period (from left to right). Hence, atomic size of Sodium will be larger than that of Chlorine. Sodium and Chlorine reacts to forms N a + by losing an electrons and C l − by gaining of one electrons. The attraction by the nucleus gets stronger when the atom loses its valence electron resulting in the decrease in ionic radii. When the number of valence electrons is increased in the anion that increases electron-electron repulsion resulting in the increase in ionic radii. Thus, cations have smaller size comparing to its parent atom whereas anions are larger than its parent atom. So the atomic size of N a + is smaller comparing to C l −

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Chapter 1, Problem 1.23E

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Each element and the ions formed during the reaction between an atom of Sodium and an atom of Chlorine has to be identified and explained.

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Chapter 1, Problem 1.22E

Chapter 1, Problem 1.24E

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Each element and the ions formed during the reaction between an atom of Sodium and an atom of Chlorine has to be identified and explained.

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Interpretation: Each element and the ions formed during the reaction between an atom of Sodium and an atom of Chlorine has to be identified and explained.

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Chapter 1 Solutions

Interpretation:

Each element and the ions formed during the reaction between an atom of Sodium and an atom of Chlorine has to be identified and explained.