Chapter 14, Problem 14.124P

(a)

Interpretation Introduction

Interpretation:

Molecular geometry of $E{F}_5^{-}$ has to be determined.

Concept-Introduction:

Lewis structure

Electron dot structure also known as Lewis dot structure represents the number of valence electrons of an atom or constituent atoms bonded in a molecule.  Each dot corresponds to one electron.

According to VSEPR theory, the geometry is predicted by the minimizing the repulsions between electron-pairs in the bonds and lone-pairs of electrons. The VSEPR theory is summarized in the given table as,

  $\begin{array}{cccc}\text{Electron-pair}& \text{lone-pair}& \text{Electron-pair}\text{geometry}& \text{Molecular}\text{shape}\\ 2& 0& \text{Linear}& \text{Linear}\\ 3& 0& \text{Trigonal}\text{planar}& \text{Trigonal}\text{planar}\\ 2& 1& \text{Trigonal}\text{planar}& \text{Bent}\\ 4& 0& \text{Tetrahedral}& \text{Tetrahedral}\\ 3& 1& \text{Tetrahedral}& \text{Pyramidal}\\ 2& 2& \text{Tetrahedral}& \text{V}-\text{shape}\\ 5& 0& \text{Trigonal}\text{bipyramidal}& \text{Trigonal}\text{bipyramidal}\\ 4& 1& \text{Trigonal}\text{bipyramidal}& \text{See}-\text{saw}\\ 3& 2& \text{Trigonal}\text{bipyramidal}& \text{T}-\text{shape}\\ 2& 3& \text{Trigonal}\text{bipyramidal}& \text{Linear}\\ 6& 0& \text{Octahedral}& \text{Octahedral}\\ 5& 1& \text{Octahedral}& \text{Square}\text{pyramidal}\\ 4& 2& \text{Octahedral}& \text{Square}\text{planar}\end{array}$

Explanation of Solution

The Lewis electron dot structure for given molecules are determined by first drawing the skeletal structure for the given molecules, then the total number of valence electrons for all atoms present in the molecules are determined.

The next step is to subtract the electrons present in the total number of bonds present in the skeletal structure of the molecule with the total valence electrons such that considering each bond contains two electrons with it.

Finally, the electrons which got after subtractions have to be equally distributed considering each atom contains eight electrons in its valence shell.

For Fluorine, outer valence electrons are seven.

Here, the highest oxidation state of Element E is 6+(in structure A). Therefore, the Element E must be a Group 16 element and has six outer valence electrons.

  $\begin{array}{l}E\text{ = 1}\times \text{6 = 6}\\ \text{F = 5}\times \text{7 = 35}\\ \text{Charge =}-\text{1}\\ Total{\text{ e}}^{-}\text{ = 6+35+1 = 42}\\ \text{5 Bonds = }\left(5\times 2\right)=10\\ {\text{Remaining e}}^{-}\text{ = 42}-10=32\end{array}$

After the distribution of electrons, each terminal fluorine atom gets 3 pairs of electrons (30 electrons) and the central E atom gets a lone pair of electrons.

Element E has 5 bond pair and one lone pair (6 electron domains).  Therefore, the molecular geometry of $E{F}_5^{-}$ is square pyramidal.

(b)

Interpretation Introduction

Interpretation:

Hybridization of E n $E{F}_5^{-}$ has to be determined.

Concept-Introduction:

Hybridization is the mixing of valence atomic orbitals to get equivalent hybridized orbitals that having similar characteristics and energy.

Geometry of different types of molecule with respect to the hybridizations are mentioned are mentioned below,

$\begin{array}{l}\begin{array}{cccc}Typeofmolecule& Hybridaization& \begin{array}{l}Atomicorbitals\\ usedforhybridaization\end{array}& Geometry\\ A{X}_2& sp& 1s+1p& Linear\\ A{X}_3,A{X}_{2}B& s{p}^2& 1s+2p& Trigonalplanar\\ A{X}_4,A{X}_{3}B,A{X}_2{B}_2& s{p}^3& 1s+3p& Tetrahedral\\ A{X}_5,A{X}_{4}B,A{X}_3{B}_2,A{X}_2{B}_3& s{p}^{3}d& 1s+3p+1d& Trigonalbipyramidal\\ A{X}_6,A{X}_{5}B,A{X}_4{B}_2& s{p}^3{d}^2& 1s+3p+2d& Octahedral\end{array}\\ A\to Centralatom\\ X\to AtomsbondedtoA\\ B\to NonbondingelectronpairsonA\end{array}$

Explanation of Solution

Element E has 5 bond pair and one lone pair (6 electron domains) $\left(A{X}_{5}B\text{ type}\right)$ .  Hence it requires six orbitals for bonding. Therefore, hybridization of E is $s{p}^3{d}^2.$

(b)

Interpretation Introduction

Interpretation:

Oxidation number of E in $E{F}_5^{-}$ has to be determined.

Concept-Introduction:

Oxidation number: The total number of electrons in an atom after losing or gaining electrons to make a bond with another atom. It indicates the charge of an ion.

Explanation of Solution

Charge of the molecule is 1-.

Oxidation number of Fluoride is 1-

Oxidation number of E is determined as given,

Here, the oxidation number of E is taken as x

  $\begin{array}{l}E{F}_5^{-},\\ \\ \text{x +}\left(5\times \left(1-\right)\right)\text{ = }\left(\text{1}-\right)\\ \\ \text{x = }\left(\text{5 +}\right)\text{+}\left(\text{1}-\right)\\ \\ =\text{ 4+}\end{array}$

Oxidation number of E in $E{F}_5^{-}$ is 4+.