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Chapter 7: Problem 63

Chlorine reacts with oxygen to form \(\mathrm{Cl}_{2} \mathrm{O}_{7} .\) (a) What is the name of this product (see Table 2.6)? (b) Write a balanced equation for the formation of \(\mathrm{Cl}_{2} \mathrm{O}_{7}(l)\) from the elements. (c) Would you expect \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) to be more reactive toward \(\mathrm{H}^{+}(a q)\) or \(\mathrm{OH}^{-}(a q) ?(\mathbf{d})\) If the oxygen in \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) is considered to have the -2 oxidation state, what is the oxidation state of the \(\mathrm{Cl}\) ? What is the electron configuration of \(\mathrm{Cl}\) in this oxidation state?

Short Answer

Expert verified
The product \(\mathrm{Cl}_{2}\mathrm{O}_{7}\) is called Dichlorine heptoxide. The balanced equation for its formation is \(\mathrm{Cl}_2(g) + \frac{7}{2}\mathrm{O}_2(g) \rightarrow \mathrm{Cl}_{2}\mathrm{O}_{7}(l)\). It would be more reactive towards \(\mathrm{H}^{+}(aq)\) ions. The oxidation state of Cl in the compound is +7, and its electron configuration in this state is \([\mathrm{Ne}]\).

Step by step solution

01

(a) Name of the compound

From Table 2.6, the name of the compound \(\mathrm{Cl}_{2}\mathrm{O}_{7}\) is Dichlorine heptoxide.
02

(b) Balanced equation for the formation of Cl2O7

To write a balanced equation for the formation of \(\mathrm{Cl}_{2}\mathrm{O}_{7}(l)\) from the elements, we'll start with: Chlorine (gas): \(\mathrm{Cl}_2\) Oxygen (gas): \(\mathrm{O}_2\) The formation of the compound will then be: $$\mathrm{Cl}_2(g) + \mathrm{O}_2(g) \rightarrow \mathrm{Cl}_{2}\mathrm{O}_{7}(l)$$ Balancing the equation, we get: $$\mathrm{Cl}_2(g) + \frac{7}{2}\mathrm{O}_2(g) \rightarrow \mathrm{Cl}_{2}\mathrm{O}_{7}(l)$$
03

(c) Reactivity towards H+ or OH- ions

Given that \(\mathrm{Cl}_{2}\mathrm{O}_{7}\) is a compound with a high oxidation state of chlorine (which we will determine in part (d)), it can be expected that it will be a powerful oxidizing agent. Therefore, it will have a higher reactivity towards \(\mathrm{H}^{+}(aq)\) rather than \(\mathrm{OH}^{-}(aq)\) ions, as the H+ ions can easily reduce the high oxidation state of Cl in \(\mathrm{Cl}_{2}\mathrm{O}_{7}\).
04

(d) Oxidation state and electron configuration of Cl

As given, the oxygen in \(\mathrm{Cl}_{2}\mathrm{O}_{7}\) has an oxidation state of -2. To find the oxidation state of Cl, let's use the formula: $$2x + 7(-2) = 0$$ Where x represents the oxidation state of Cl. Solving for x, we get: $$x = +7$$ So, the oxidation state of Cl is +7 in this compound. The electron configuration of a neutral chlorine atom is \([\mathrm{Ne}]3s^23p^5\). To achieve an oxidation state of +7, chlorine has to lose all 5 valence electrons in the 3p orbital and 2 valence electrons in the 3s orbital. Thus, the electron configuration of chlorine in the +7 oxidation state will be the same as that of a neon atom: $$[\mathrm{Ne}]$$

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Most popular questions from this chapter

(a) The five most abundant elements in the Earth's crust are $\mathrm{O}, \mathrm{Si}, \mathrm{Al}, \mathrm{Fe},\( and Ca. Referring to Figure \)7.1,$ are any of these elements among those known before \(1700 ?\) If so which ones? (b) Seven of the nine elements known since ancient times are metals. Referring to Table \(4.5,\) are these metals mostly found at the bottom or top of the activity series?

(a) Why does xenon react with fluorine, whereas neon does not? (b) Using appropriate reference sources, look up the bond lengths of Xe-F bonds in several molecules. How do these numbers compare to the bond lengths calculated from the atomic radii of the elements?

Potassium superoxide, \(\mathrm{KO}_{2},\) is often used in oxygen masks (such as those used by firefighters) because \(\mathrm{KO}_{2}\) reacts with \(\mathrm{CO}_{2}\) to release molecular oxygen. Experiments indicate that 2 mol of \(\mathrm{KO}_{2}(s)\) react with each mole of \(\mathrm{CO}_{2}(g) .\) (a) The products of the reaction are \(\mathrm{K}_{2} \mathrm{CO}_{3}(s)\) and \(\mathrm{O}_{2}(g) .\) Write a balanced equation for the reaction between \(\mathrm{KO}_{2}(s)\) and \(\mathrm{CO}_{2}(g) .(\mathbf{b})\) Indicate the oxidation number for each atom involved in the reaction in part (a). What elements are being oxidized and reduced? (c) What mass of \(\mathrm{KO}_{2}(s)\) is needed to consume \(18.0 \mathrm{~g} \mathrm{CO}_{2}(g)\) ? What mass of \(\mathrm{O}_{2}(g)\) is produced during this reaction?

(a) If the core electrons were totally effective at screening the valence electrons and the valence electrons provided no screening for each other, what would be the effective nuclear charge acting on the \(3 s\) and \(3 p\) valence electrons in P? (b) Repeat these calculations using Slater's rules. (c) Detailed calculations indicate that the effective nuclear charge is \(5.6+\) for the \(3 s\) electrons and \(4.9+\) for the \(3 p\) electrons. Why are the values for the \(3 s\) and \(3 p\) electrons different? (d) If you remove a single electron from a Patom, which orbital will it come from?

Hydrogen is an unusual element because it behaves in some ways like the alkali metal elements and in other ways like nonmetals. Its properties can be explained in part by its electron configuration and by the values for its ionization energy and electron affinity. (a) Explain why the electron affinity of hydrogen is much closer to the values for the alkali elements than for the halogens. (b) Is the following statement true? "Hydrogen has the smallest bonding atomic radius of any element that forms chemical compounds." If not, correct it. If it is, explain in terms of electron configurations. (c) Explain why the ionization energy of hydrogen is closer to the values for the halogens than for the alkali metals. (d) The hydride ion is \(\mathrm{H}^{-}\). Write out the process corresponding to the first ionization energy of the hydride ion. (e) How does the process in part (d) compare to the process for the electron affinity of a neutral hydrogen atom?
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