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Chapter 3: Problem 14

Why do we call \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) barium nitrate, but we call \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}\) iron(II) nitrate?

Short Answer

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We call \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) barium nitrate because barium has a fixed oxidation state of +2 and there is no ambiguity in the charge. In contrast, we call \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}\) iron(II) nitrate because iron can form multiple oxidation states, and the (II) in the name specifies that it has a +2 charge in this particular compound.

Step by step solution

01

Understanding Ionic Compounds

Barium nitrate and iron(II) nitrate are ionic compounds, which are formed between metal and non-metal ions, specifically a positive ion (cation) and a negative ion (anion). In these compounds, the metal (cation) is written first and the non-metal (anion) is written second.
02

Barium Nitrate

Barium nitrate has the chemical formula \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\). In this compound, Ba is the symbol for barium and it is an alkaline earth metal. The (NO3) group represents the nitrate ion with a -1 charge. Since barium has an oxidation state of +2 and always forms 2+ cations, there is no ambiguity in the charge, and we simply call it barium nitrate.
03

Iron(II) Nitrate

Iron(II) nitrate has the chemical formula \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}\). In this compound, Fe is the symbol for iron, which is a transition metal and can form different oxidation states. The nitrate ion (NO3) has a -1 charge. Since iron can have multiple positive charges as cations (+2 and +3 are common), we need to specify which iron cation is present in the compound. In this case, Fe has a +2 charge in the compound as there are two nitrate ions present, so we call it iron(II) nitrate to indicate the oxidation state or charge of the iron.
04

Summary

To recap, we call \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) barium nitrate because barium has a fixed oxidation state of +2 and there is no ambiguity in the charge. In contrast, we call \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}\) iron(II) nitrate because iron can form multiple oxidation states, and the (II) in the name specifies that it has a +2 charge in this particular compound.

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Most popular questions from this chapter

Consider the following bond lengths: \(\begin{array}{cccccc}\mathbf{C}-\mathbf{O} & 143 \mathrm{pm} & \mathbf{C}=\mathbf{O} & 123 \mathrm{pm} & \mathbf{C} \equiv \mathbf{O} & 109 \mathrm{pm}\end{array}\) In the \(\mathrm{CO}_{3}^{2-}\) ion, all three \(\mathrm{C}-\mathrm{O}\) bonds have identical bond lengths of \(136 \mathrm{pm}\). Why?

Without using Fig. \(3-4,\) predict which bond in each of the following groups will be the most polar. a. \(\mathrm{C}-\mathrm{H}, \mathrm{Si}-\mathrm{H}, \mathrm{Sn}-\mathrm{H}\) \(\mathbf{b .}\) Al- \(\mathbf{B r}, \mathbf{G a}-\mathbf{B r}, \operatorname{In}-\mathbf{B r}, \mathbf{T}-\mathbf{B r}\) c. \(C-O\) or \(S i-O\) d. \(\mathrm{O}-\mathrm{F}\) or \(\mathrm{O}-\mathrm{Cl}\)

Write electron configurations for the most stable ion formed by each of the elements Te, \(\mathrm{Cl}, \mathrm{Sr},\) and \(\mathrm{Li}\) (when in stable ionic compounds).

Lewis structures can be used to understand why some molecules react in certain ways. Write the Lewis structures for the reactants and products in the reactions described below. a. Nitrogen dioxide dimerizes to produce dinitrogen tetroxide. b. Boron trihydride accepts a pair of electrons from ammonia, forming \(\mathrm{BH}_{3} \mathrm{NH}_{3}\) Give a possible explanation for why these two reactions occur.

Nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) has three possible Lewis structures: $$\therefore N=N=O^{\cdot} \leftrightarrow: N \equiv N-\vec{O}: \longleftrightarrow: N-N \equiv 0$$ Given the following bond lengths, $$\begin{aligned} &\mathrm{N}-\mathrm{N} \quad 167 \mathrm{pm} \quad \mathrm{N}=\mathrm{O} \quad 115 \mathrm{pm}\\\ &\mathrm{N}=\mathrm{N} \quad 120 \mathrm{pm} \quad \mathrm{N}-\mathrm{O} \quad 147 \mathrm{pm}\\\ &\mathrm{N} \equiv \mathrm{N} \quad 110 \mathrm{pm} \end{aligned}$$ rationalize the observations that the \(\mathrm{N}-\mathrm{N}\) bond length in \(\mathrm{N}_{2} \mathrm{O}\) is \(112 \mathrm{pm}\) and that the \(\mathrm{N}-\mathrm{O}\) bond length is \(119 \mathrm{pm}\). Assign formal charges to the resonance structures for \(\mathrm{N}_{2} \mathrm{O}\). Can you eliminate any of the resonance structures on the basis of formal charges? Is this consistent with observation?
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