Q10.91 - Chemistry: Molecular Nature Of Matter And Change

Question

A student isolates a product with the molecular shape shown at right (F is orange).

1. If the species is a neutral compound, can the black sphere represent selenium (Se)?
2. If the species is an anion, can the black sphere represent N?
3. If the black sphere represents Br, what is the charge of the species?

Step-by-Step Solution

Answer

1. Yes, the black sphere can represent Selenium.
2. Yes, the black sphere can indeed stand in for nitrogen.
3. The charge on ${\text{BrF}}_{\text{4}}^{\text{ - }}$is (+1).

Step 1: Subpart (a) The black sphere when the species is a neutral compound.

The given structure is a see-saw having 5 electron group arrangements. So, the compound has 1 lone pair of electrons.

Yes, if the species is neutral, the black sphere represents the selenium (Se).

For example: ${\text{SeF}}_{\text{4}}$

Selenium atom has 6 valence electrons and it forms 4 sigma bonds with 4 fluorine atoms.  The fluorine atom has 7 valence electrons. In the Lewis structure of  ${\text{SeF}}_{\text{4}}$, Se is surrounded by 5 electron groups and the electronic geometry of  ${\text{SeF}}_{\text{4}}$ is trigonal bipyramidal but due to the presence of a lone pair of electrons, it becomes see-saw shaped.

Step 2: Subpart (b) The black sphere when the species is an anion.

Yes, if the species is an anion, the black sphere represents nitrogen.

For example: $\text{Br}{{\text{F}}_{\text{4}}}^{\text{ + }}$

Nitrogen atom has 3 valence electrons and it forms 4 sigma bonds with fluorine atoms. In the Lewis structure of ${\mathbf{Br}}^{\oplus }$  N has a negative charge (negative charge contains a lone pair of electrons) i.e., it’s an anion species and N is surrounded by 4 fluorine atoms so the geometry of    $\text{Br}{{\text{F}}_{\text{4}}}^{\text{ + }}$ is trigonal bipyramidal but due to the presence of lone pair of electrons it becomes see- saw shaped.

Step 3: Subpart (c) When black sphere represents Br , the charge on Bromine.

When the black sphere represents Br, the charge on Br is +1.

For example:  $\text{Br}{{\text{F}}_{\text{4}}}^{\text{ + }}$

Br has 8 valence electrons and ${\mathbf{Br}}^{\mathbf{\oplus }}$has 6 valence electrons from which 4 electrons of Br form 4 sigma bonds with 4 fluorine atoms. In the Lewis structure of $\text{Br}{{\text{F}}_{\text{4}}}^{\text{ + }}$ Br is surrounded by 4 fluorine atoms and has a lone pair of electrons.

So, the shape will be similar to the given structure.