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\(\mathrm{SF}_{6}, \mathrm{ClF}_{5}\), and \(\mathrm{XeF}_{4}\) are three compounds whose central atoms do not follow the octet rule. Draw Lewis structures for these compounds.

Short Answer

Expert verified
The Lewis structures for the given compounds are as follows: \(\mathrm{SF}_{6}\): ``` F F \ / --S------ / \ F F ``` \(\mathrm{ClF}_{5}\): ``` F F \ / --Cl----- / \ F F \ F ``` \(\mathrm{XeF}_{4}\): ``` F F \ / --Xe----- / \ F F ```

Step by step solution

01

Total number of valence electrons

First, we need to determine the total number of valence electrons for each compound. \(\mathrm{SF}_{6}:\) Sulfur has 6 valence electrons, and each fluorine has 7 valence electrons. Since there are 6 fluorine atoms, the total number of valence electrons for this compound is \(6 + (6 \times 7) = 48\). \(\mathrm{ClF}_{5}:\) Chlorine has 7 valence electrons, and each fluorine has 7 valence electrons. Since there are 5 fluorine atoms, the total number of valence electrons for this compound is \(7 + (5 \times 7) = 42\). \(\mathrm{XeF}_{4}:\) Xenon has 8 valence electrons, and each fluorine has 7 valence electrons. Since there are 4 fluorine atoms, the total number of valence electrons for this compound is \(8 + (4 \times 7) = 36\).
02

Determine the central atom

In each compound, the central atom will be the one with the least electronegative element. In all three cases, the central atom is given: \(\mathrm{SF}_{6}:\) Sulfur is the central atom. \(\mathrm{ClF}_{5}:\) Chlorine is the central atom. \(\mathrm{XeF}_{4}:\) Xenon is the central atom.
03

Complete the octets of the surrounding atoms

Now we will add single bonds between the central atom and the surrounding fluorine atoms. Then, we will complete the octets of the fluorine atoms with lone pairs. \(\mathrm{SF}_{6}:\) Add 6 single bonds between sulfur and the 6 fluorine atoms and complete the fluorine octets. This uses up 48 electrons, so there are no remaining electrons. \(\mathrm{ClF}_{5}:\) Add 5 single bonds between chlorine and the 5 fluorine atoms and complete the fluorine octets. This uses up 40 electrons, so there are 2 remaining electrons. \(\mathrm{XeF}_{4}:\) Add 4 single bonds between xenon and the 4 fluorine atoms and complete the fluorine octets. This uses up 32 electrons, so there are 4 remaining electrons.
04

Place remaining electrons on the central atom

Finally, we will place any remaining electrons on the central atom. \(\mathrm{SF}_{6}:\) No remaining electrons. \(\mathrm{ClF}_{5}:\) Place the 2 remaining electrons on chlorine as a lone pair. \(\mathrm{XeF}_{4}:\) Place the 4 remaining electrons on xenon as two lone pairs. The Lewis structures for these compounds are: \(\mathrm{SF}_{6}:\) ``` F F \ / --S------ / \ F F ``` \(\mathrm{ClF}_{5}:\) ``` F F \ / --Cl----- / \ F F \ F ``` \(\mathrm{XeF}_{4}:\) ``` F F \ / --Xe----- / \ F F ```

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Most popular questions from this chapter

A polyatomic ion is composed of \(\mathrm{C}, \mathrm{N}\), and an unknown element \(\mathrm{X} .\) The skeletal Lewis structure of this polyatomic ion is \([\mathrm{X}-\mathrm{C}-\mathrm{N}]^{-} .\) The ion \(\mathrm{X}^{2-}\) has an electron configuration of \([\mathrm{Ar}] 4 s^{2} 3 d^{10} 4 p^{6} .\) What is element \(\mathrm{X} ?\) Knowing the identity of \(\mathrm{X}\), complete the Lewis structure of the polyatomic ion, including all important resonance structures.

List all the possible bonds that can occur between the elements \(\mathrm{P}, \mathrm{Cs}, \mathrm{O}\), and \(\mathrm{H}\). Predict the type of bond (ionic, covalent, or polar covalent) one would expect to form for each bond.

Predict the empirical formulas of the ionic compounds formed from the following pairs of elements. Name each compound. a. \(\mathrm{Li}\) and \(\mathrm{N}\) c. \(\mathrm{Rb}\) and \(\mathrm{Cl}\) b. Ga and \(\mathrm{O}\) d. \(\mathrm{Ba}\) and \(\mathrm{S}\)

Two different compounds have the formula \(\mathrm{XeF}_{2} \mathrm{Cl}_{2} .\) Write Lewis structures for these two compounds, and describe how measurement of dipole moments might be used to distinguish between them.

Write Lewis structures for \(\mathrm{CO}_{3}^{2-}, \mathrm{HCO}_{3}^{-}\), and \(\mathrm{H}_{2} \mathrm{CO}_{3}\). When acid is added to an aqueous solution containing carbonate or bicarbonate ions, carbon dioxide gas is formed. We generally say that carbonic acid \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right)\) is unstable. Use bond energies to estimate \(\Delta H\) for the reaction (in the gas phase) $$ \mathrm{H}_{2} \mathrm{CO}_{3} \longrightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} $$ Specify a possible cause for the instability of carbonic acid.

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