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Chapter 2: Problem 71

Name each of the following compounds: a. CuI f. \(\mathrm{S}_{4} \mathrm{~N}_{4}\) b. \(\mathrm{CuI}_{2}\) g. \(\mathrm{SF}_{4}\) c. CoI \(_{2}\) h. \(\mathrm{NaOCl}\) d. \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) i. \(\mathrm{BaCrO}_{4}\) e. \(\mathrm{NaHCO}_{3}\) j. \(\mathrm{NH}_{4} \mathrm{NO}_{3}\)

Short Answer

Expert verified
a. Copper(I) Iodide f. Tetrasulfur Tetranitride b. Copper(II) Iodide g. Sulfur Tetrafluoride c. Cobalt(II) Iodide h. Sodium Hypochlorite d. Sodium Carbonate i. Barium Chromate e. Sodium Bicarbonate j. Ammonium Nitrate

Step by step solution

01

Compound a: CuI

This is a binary compound. Copper (Cu) is a transition metal, so its oxidation state should be written in Roman numerals after the name. The oxidation state of copper here is +1 because iodine (I) has a -1 charge. This gives us the name Copper(I) Iodide.
02

Compound f: \(\mathrm{S}_{4} \mathrm{N}_{4}\)

This is a binary compound. We use prefixes to show the number of atoms of each element: Tetrasulfur (S4) and Tetranitrogen (N4). The name of this compound is Tetrasulfur Tetranitride.
03

Compound b: \(\mathrm{CuI}_{2}\)

This is a binary compound containing copper and iodine. Copper (Cu) in this compound has an oxidation state of +2 since 2 iodine atoms balance the charge. The name of this compound is Copper(II) Iodide.
04

Compound g: \(\mathrm{SF}_{4}\)

This is a binary compound. The name of this compound is Sulfur Tetrafluoride, following the same naming rules as in Compound f.
05

Compound c: CoI\(_{2}\)

This is a binary compound. Cobalt (Co) is a transition metal, so its oxidation state should be written in Roman numerals after the name. The oxidation state of cobalt here is +2 because 2 iodine atoms balance the charge. This gives us the name Cobalt(II) Iodide.
06

Compound h: \(\mathrm{NaOCl}\)

This compound contains sodium (Na), oxygen (O), and chlorine (Cl). Here, we have the polyatomic ion hypochlorite (OCl-), so the name is Sodium Hypochlorite.
07

Compound d: \(\mathrm{Na}_{2} \mathrm{CO}_{3}\)

This compound is composed of sodium (Na) and the polyatomic ion carbonate (CO3^2-). The name of this compound is Sodium Carbonate.
08

Compound i: \(\mathrm{BaCrO}_{4}\)

This compound consists of barium (Ba) and the polyatomic ion chromate (CrO4^2-). The name of this compound is Barium Chromate.
09

Compound e: \(\mathrm{NaHCO}_{3}\)

This compound is composed of sodium (Na) and the polyatomic ion hydrogen carbonate (HCO3-), which is also known as bicarbonate. The name of this compound is Sodium Bicarbonate.
10

Compound j: \(\mathrm{NH}_{4} \mathrm{NO}_{3}\)

This compound is composed of the ammonium (NH4+) polyatomic ion and the nitrate (NO3-) polyatomic ion. The name for this compound is Ammonium Nitrate.

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