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Chapter 22: Problem 82

In \(1988,\) G. J. Schrobilgen, professor of chemistry at McMaster University in Canada, reported the synthesis of an ionic compound, \([\mathrm{HCNKrF}]\left[\mathrm{AsF}_{6}\right],\) which consists of \(\mathrm{HCNKr} \mathrm{F}^{+}\) and \(\mathrm{AsF}_{6}^{-}\) ions. In the \(\mathrm{HCNKr} \mathrm{F}^{+}\) ion, the krypton is covalently bonded to both fluorine and nitrogen. Draw Lewis structures for these ions, and estimate the bond angles.

Short Answer

Expert verified
The Lewis structure of \(\mathrm{HCNKrF}^{+}\) has SP2 and SP hybridization on carbon and krypton, respectively, with bond angles of 120 and 180 degrees. The Lewis structure for \(\mathrm{AsF}_{6}^{-}\) has SP3d2 hybridization on arsenic, with a bond angle of 90 degrees.

Step by step solution

01

Draw Lewis Structures

First, we draw Lewis structures for both ions. For \(\mathrm{HCNKrF}^{+}\):- Hydrogen and fluorine each have 1 and 7 valence electrons, respectively.- Carbon and nitrogen have 4 and 5 valence electrons, respectively.- Their total count gives us \(1+4+5+7 = 17\), but we subtract 1 because it's a positive ion, giving us 16 electrons.- Hydrogen is bonded to carbon, carbon to nitrogen, and nitrogen to krypton and fluorine.For \(\mathrm{AsF}_{6}^{-}\):- Arsenic has 5 valence electrons and each fluorine has 7.- Their total gives us \(5+6*7 = 47\), but we add 1 because it's a negative ion, giving us 48 electrons.- Arsenic is centrally located and is bonded to six fluorine atoms.
02

Predict Bond Angles

Next, we estimate the bond angles consistent with the VSEPR theory. For \(\mathrm{HCNKrF}^{+}\):- The carbon is bonded to three other atoms, so it adopts an SP2 hybridization and, thus, a trigonal planar geometry with bond angles of 120 degrees.- Krypton is bonded to two other atoms, showing SP hybridization and a linear shape with bond angles of 180 degrees between the nitrogen, krypton, and fluorine atoms.For \(\mathrm{AsF}_{6}^{-}\):- The central arsenic atom is bonded to six other atoms, so it adopts an SP3d2 hybridization and, therefore, an octahedral shape. The bond angles in an octahedral shape are 90 degrees.

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Most popular questions from this chapter

Draw plausible Lewis structures for (a) dimethylhydrazine, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NNH}_{2}\) (b) nitryl chloride, \(\mathrm{ClNO}_{2}\) (Central atom is \(\mathrm{N}\) ) (c) phosphorous acid, a diprotic acid with the empirical formula \(\mathrm{H}_{3} \mathrm{PO}_{3}\).

Write balanced equations for the following important commercial reactions involving nitrogen and its compounds. (a) the principal artificial method of fixing atmospheric \(\mathrm{N}_{2}\) (b) oxidation of ammonia to \(\mathrm{NO}\) (c) preparation of nitric acid from \(\mathrm{NO}\).

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