Chapter 14, Problem 14.83P

(a)

Interpretation Introduction

Interpretation:

Lewis structure of ${\text{CO, CN}}^{-}{\text{, and C}}_2^{2-}$  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.

Outer valence electrons of Selenium and Florine are six and seven respectively.

Draw Lewis structure of Selenium tetrafluoride:

  $\begin{array}{l}Se\text{ = 1}\times \text{6 = 6}\\ \text{F = 4}\times \text{7 = 28}\\ Total{\text{ e}}^{-}\text{ = 6 + 28 = 34}\\ \text{4 Bonds = }\left(4\times 2\right)=8\\ {\text{Remaining e}}^{-}\text{ = 34}-8=26\end{array}$

The Lewis structure of Selenium tetrafluoride follows as,

After the distribution of electrons, Selenium atom gets a lone pair of electrons.

Selenium has $4$ bond pairs and a lone pair (five electron domains).  Therefore, the molecular geometry of $Se{F}_4$ is see-saw.  However, the bond angle of $Se{F}_4$ is deviated from the ideal bond angle due to the lone pair-bond pair repulsion.

Draw Lewis structure of Selenium hexafluoride:

  $\begin{array}{l}Se\text{ = 1}\times \text{6 = 6}\\ \text{F = 6}\times \text{7 = 42}\\ Total{\text{ e}}^{-}\text{ = 6 + 42 = 48}\\ \text{6 Bonds = }\left(6\times 2\right)=12\\ {\text{Remaining e}}^{-}\text{ = 48}-12=36\end{array}$

Selenium has $6$ bond pairs (six electron domains).

The Lewis structure of Selenium hexafluoride follows as,

(b)

Interpretation Introduction

Interpretation:

Change in orbital hybridization of the central $Se$ during the reaction 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 a molecule can be predicted by knowing its hybridization.

Explanation of Solution

Selenium tetrafluoride and Fluorine reacts to form Selenium hexafluoride.

  $S{F}_4{\text{ + F}}_2\stackrel{}{\to }{\text{ SF}}_6$

The geometry of $S{F}_4$ is see-saw whereas geometry of $S{F}_6$ is octahedral.

In $S{F}_4$, Selenium has $4$ bond pairs and a lone pair (five electron domains).  Hence it requires five orbitals for bonding. Therefore, hybridization is $s{p}^{3}d.$

In $S{F}_6$, Selenium has $6$ bond pairs (six electron domains).  Hence it requires six orbitals for bonding. Therefore, hybridization is $s{p}^3{d}^2.$

During the reaction the hybridization of Selenium changes from $s{p}^{3}dtos{p}^3{d}^2.$