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Would you expect the reaction of \(\mathrm{Ge}(\mathrm{s})\) and \(\mathrm{F}_{2}(\mathrm{g})\) to yield GeF \(_{2},\) with germanium in the +2 oxidation state, or GeF \(_{4},\) with germanium in the +4 oxidation state?

Short Answer

Expert verified
It is expected that the reaction of Ge and F2 will yield GeF4, with germanium in the +4 oxidation state.

Step by step solution

01

Understand the nature of the elements

Germanium (Ge), is in the carbon group of the periodic table, and typically exhibits a +4 oxidation state due to the four outer shell electrons available for bonding. Fluorine (F), in Group 17 (Halogens) is the most electronegative element, always has an oxidation state of -1 when it forms compounds.
02

Predict the resulting compound

Due to the high electronegativity of fluorine, it will draw electrons towards itself, causing other elements to exhibit their highest oxidation states. Therefore, when germanium reacts with fluorine, it is more likely to achieve a +4 oxidation state, forming GeF4.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Germanium Compounds
Germanium is a fascinating element found in the carbon group of the periodic table, often forming compounds in which it exhibits multiple oxidation states. When we discuss germanium compounds, we typically focus on its preferred oxidation state, which is +4. This relates to germanium’s electron configuration and its position in the periodic table – in the same group as carbon and silicon, which also tend to form compounds in which they have four bonds.

Germanium can form various types of compounds including germanides, germinates, and germanium halides. One interesting aspect of germanium is its ability to form stable organogermanium compounds, similar to organosilicon chemistry. In the exercise in focus, we deal with a germanium halide, specifically hypothesizing about its reaction with fluorine. The correct prediction requires understanding germanium’s behavior in chemical reactions, where it can exhibit different oxidation states, usually +2 or +4, depending on the reactants and conditions of the reaction.

How do we determine which oxidation state germanium will adopt in a compound? It often comes down to the nature of the other element it reacts with. For instance, with less electronegative elements, germanium might exhibit lower oxidation states. However, when reacting with highly electronegative elements, like fluorine, germanium is more likely to adopt its maximum oxidation state, resulting in compounds like GeF4.
Fluorine Reactivity and Its Impact on Compounds
Fluorine is a member of the halogen family and is recognized as the most electronegative element on the periodic table. Due to this property, it has a powerful ability to attract electrons and thus, is generally found in the -1 oxidation state in its compounds.

The reactivity of fluorine is significant when predicting the outcomes of chemical reactions. In forming compounds, fluorine's high electronegativity will often oxidize other elements to their highest possible states because it pulls electrons towards itself so effectively. This is a key concept to understand when exploring why certain compounds form over others in reactions involving fluorine.

When paired with a metal or metalloid like germanium, the highly reactive nature of fluorine favors the formation of compounds where the other element is in a higher oxidation state. This means that in a reaction between germanium and fluorine, it is unlikely for GeF2 to form because this does not showcase germanium in its highest common oxidation state. Instead, the reaction would yield GeF4, which is consistent with fluorine's tendency to stabilize other elements in their highest oxidation states.
Predicting Chemical Reactions
Predicting chemical reactions involves understanding the properties of the elements involved, their common oxidation states, and how they react with each other. For students, it can sometimes be challenging to predict the products of a reaction simply by looking at the reactants. However, by following a few guiding principles, we can often make accurate predictions.

In the exercise at hand, two of these principles are critical: knowledge of the elements’ common oxidation states and understanding that elements tend to form compounds in which they have a stable electron configuration. For example, germanium, which commonly has a +4 oxidation state, will likely achieve a stable compound configuration when reacted with the highly electronegative fluorine.

Additionally, considering the periodic trends and the rules of electronegativity can aid in predicting the outcome of reactions involving nonmetals. The more electronegative an element like fluorine is, the more it will affect the oxidation states of the other elements in the compounds it forms. Thus, if we apply this thinking to the reaction between germanium and fluorine, we can predict that the product will be GeF4, as it allows germanium to achieve the more stable +4 oxidation state, consistent with fluorine's reactivity and the periodic trends.

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Most popular questions from this chapter

When a \(0.200 \mathrm{g}\) sample of \(\mathrm{Mg}\) is heated in air, \(0.315 \mathrm{g}\) of product is obtained. Assume that all the Mg appears in the product. (a) If the product were pure \(\mathrm{MgO}\), what mass should have been obtained? (b) Show that the 0.315 g product could be a mixture of \(\mathrm{Mg} \mathrm{O}\) and \(\mathrm{Mg}_{3} \mathrm{N}_{2}.\) (c) What is the mass percent of \(\mathrm{MgO}\) in the \(\mathrm{MgO}-\mathrm{Mg}_{3} \mathrm{N}_{2}\) mixed product?

Would you expect the reaction of \(\mathrm{Pb}(\mathrm{s})\) and \(\mathrm{Cl}_{2}(\mathrm{g})\) to yield \(\mathrm{PbCl}_{2}\) or \(\mathrm{PbCl}_{4} ?\)

The melting point of \(\mathrm{NaCl}(\mathrm{s})\) is \(801^{\circ} \mathrm{C},\) much higher than that of \(\mathrm{NaOH}\left(322^{\circ} \mathrm{C}\right) .\) More energy is consumed to melt and maintain molten NaCl than NaOH. Yet the preferred commercial process for the production of sodium is electrolysis of \(\mathrm{NaCl}(\mathrm{l})\) rather than \(\mathrm{NaOH}(1)\) Give a reason or reasons for this discrepancy.

A chemical dictionary gives the following descriptions of the production of some compounds. Write plausible chemical equations based on these descriptions. (a) lead(II) carbonate: adding a solution of sodium bicarbonate to a solution of lead nitrate. (b) lithium carbonate: reaction of lithium oxide with ammonium carbonate solution. (c) hydrogen peroxide: by the action of dilute sulfuric acid on barium peroxide. (d) lead(IV) oxide: action of an alkaline solution of calcium hypochlorite on lead(II) oxide.

The Gibbs energies of formation, \(\Delta G_{\mathrm{f}}^{\circ},\) for \(\mathrm{Na}_{2} \mathrm{O}(\mathrm{s})\) and \(\mathrm{Na}_{2} \mathrm{O}_{2}(\mathrm{s})\) are \(-379.09 \mathrm{kJ} \mathrm{mol}^{-1}\) and \(-449.63 \mathrm{kJ} \mathrm{mol}^{-1}\) respectively, at 298 K. Calculate the equilibrium constant for the reaction below at \(298 \mathrm{K} .\) Is \(\mathrm{Na}_{2} \mathrm{O}_{2}(\mathrm{s})\) thermodynamically stable with respect to \(\mathrm{Na}_{2} \mathrm{O}(\mathrm{s})\) and \(\mathrm{O}_{2}(\mathrm{g})\) at \(298 \mathrm{K} ?\) $$ \mathrm{Na}_{2} \mathrm{O}_{2}(\mathrm{s}) \longrightarrow \mathrm{Na}_{2} \mathrm{O}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) $$

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