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Chapter 22: Problem 2

Fluorine is able to stabilize elements in very high oxidation states. For each of the elements \(\mathrm{Na}, \mathrm{Mg}, \mathrm{Al}, \mathrm{Si}\) P, S, and Cl, give the formula of the highest-order fluoride that is known to exist. Then, describe the variation in bonding that occurs as we move from left to right across the period.

Short Answer

Expert verified
The highest-order fluorides of given elements which are known to exist are \( \mathrm{NaF}, \mathrm{MgF_2}, \mathrm{AlF_3}, \mathrm{SiF_4}, \mathrm{PF_5}, \mathrm{SF_6}, \mathrm{ClF_7} \). As we move from left to right across the period, the number of bonds that an element forms with Fluorine increases, signifying the increase in available valence electrons for bonding.

Step by step solution

01

Identifying highest order fluorides

The highest-order fluorides that exist for the given elements are \( \mathrm{NaF}, \mathrm{MgF_2}, \mathrm{AlF_3}, \mathrm{SiF_4}, \mathrm{PF_5}, \mathrm{SF_6}, \mathrm{ClF_7} \). This means that Sodium (Na) can form one bond with Fluorine (F), Magnesium (Mg) can form two bonds, Aluminium (Al) can form three, Silicon (Si) can form four, Phosphorus (P) can form five, Sulfur (S) can form six, and Chlorine (Cl) can form seven.
02

Describing Variation in Bonding

As we move from left to right across the period in the Periodic Table, the number of bonds an element can form with Fluorine increases. This is because the number of available valence electrons for bonding increases. Sodium (Na) is in Group 1 and has one valence electron, hence it forms one bond. Magnesium (Mg) is in Group 2 and can form two bonds, and so forth. Silicon (Si) is in Group 4 and forms four bonds, Phosphorus (P) is in Group 5 and forms five bonds, Sulfur (S) is in Group 6 and forms six bonds, and Chlorine (Cl) is in Group 7 and forms seven bonds.

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Most popular questions from this chapter

The solubility of \(\mathrm{Cl}_{2}(\mathrm{g})\) in water is \(6.4 \mathrm{g} \mathrm{L}^{-1}\) at \(25^{\circ} \mathrm{C}\) Some of this chlorine is present as \(\mathrm{Cl}_{2},\) and some is found as HOCl or Cl^- For the hydrolysis reaction $$\begin{array}{c}\mathrm{Cl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(1) \longrightarrow \\ \mathrm{HOCl}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) \\\K_{c}=4.4 \times 10^{-4}\end{array}$$ For a saturated solution of \(\mathrm{Cl}_{2}\) in water, calculate \(\left[\mathrm{Cl}_{2}\right],[\mathrm{HOCl}],\left[\mathrm{H}_{3} \mathrm{O}^{+}\right],\) and \(\left[\mathrm{Cl}^{-}\right]\).

How many grams of \(\mathrm{CaH}_{2}(\mathrm{s})\) are required to generate sufficient \(\mathrm{H}_{2}(\mathrm{g})\) to fill a \(235 \mathrm{L}\) weather observation balloon at \(722 \mathrm{mmHg}\) and \(19.7^{\circ} \mathrm{C} ?\) \(\mathrm{CaH}_{2}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(1) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{aq})+2 \mathrm{H}_{2}(\mathrm{g})\).

All the group 15 elements form trifluorides, but nitrogen is the only group 15 element that does not form a pentafluoride. (a) Suggest a reason why nitrogen does not form a pentafluoride. (b) The observed bond angle in \(\mathrm{NF}_{3}\) is approximately \(102.5^{\circ} \mathrm{C} .\) Use VSEPR theory to rationalize the structure of the \(\mathrm{NF}_{3}\) molecule.

All of the following compounds yield \(\mathrm{O}_{2}(\mathrm{g})\) when heated to about \(1000 \mathrm{K}\) except (a) \(\mathrm{KClO}_{3} ;\) (b) \(\mathrm{KClO}_{4}\) (c) \(\mathrm{N}_{2} \mathrm{O} ;\) (d) \(\mathrm{CaCO}_{3} ;\) (e) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\).

The bond energies of \(\mathrm{Cl}_{2}\) and \(\mathrm{F}_{2}\) are 243 and \(159 \mathrm{kJ} \mathrm{mol}^{-1},\) respectively. Use these data to explain why \(\mathrm{XeF}_{2}\) is a much more stable compound than \(\mathrm{XeCl}_{2} .[\text { Hint: Recall that Xe exists as a monatomic gas. }]\).
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