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Chapter 2: Problem 8

Sodium chloride \((\mathrm{NaCl})\) exhibits predominantly ionic bonding. The \(\mathrm{Na}^{+}\)and \(\mathrm{Cl}^{-}\)ions have electron structures that are identical to which two inert gases?

Short Answer

Expert verified
Answer: The Na+ ion shares the same electron configuration as Neon (Ne), and the Cl- ion shares the same electron configuration as Argon (Ar).

Step by step solution

01

Determine the electron configurations of Sodium (Na) and Chlorine (Cl)

The atomic number of Sodium (Na) is 11, so its electron configuration is 1s² 2s² 2p⁶ 3s¹. Chlorine (Cl) has an atomic number of 17, so its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁵.
02

Determine the electron configurations of Na+ and Cl- ions

Sodium (Na) loses one electron to form a positively charged ion (Na+), which means that Na+ has one less electron than the neutral atom. So, the electron configuration of Na+ is 1s² 2s² 2p⁶. Chlorine (Cl) gains one electron to form a negatively charged ion (Cl-), which means that Cl- has one more electron than the neutral atom. The electron configuration of Cl- is 1s² 2s² 2p⁶ 3s² 3p⁶.
03

Find the inert gases with the same electron configurations as Na+ and Cl- ions

Now, let's look at the inert gases in the Periodic Table. The final electron configurations of the first few inert gases are as follows: -Helium (He): 1s² -Neon (Ne): 1s² 2s² 2p⁶ -Argon (Ar): 1s² 2s² 2p⁶ 3s² 3p⁶ Comparing these configurations to those of Na+ and Cl-, we can conclude that the Na+ ion has an electron configuration identical to Neon (Ne), and Cl- ion has an electron configuration identical to Argon (Ar).

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