Question

Number of double bond present in the Lewis structure of $PF_6^ -$ is :

Answer 1

For hexafluorophosphate, [ $P{F_6}$ ]-, there are six bonded groups and so no lone pairs. This anion is useful in synthesis as it also aids the crystallisation of bulky cations by having a suitable size match for the cation. The ion's negative charge is distributed through all seven atoms.

Complete answer:
The valence shell electron pair repulsion (VSEPR) theory is a mathematical model for predicting 3-D molecular geometry based on the number of valence shell electron bond pairs between atoms in a molecule or ion. As long as the central atom is not a metal, it can predict the form of almost all compounds with a central atom.
In this compound the central atom is phosphorus. We know that initially the valence electron on the central atom is $5$ . The $6$ Fluoride atom contributes one electron each. Hence, we add one negative charge to P. So the total valence electron is $12$ .
Dividing $12$ by $2$ , we get $6$ . Hence, the geometry formed in $6$ electron pairs is Octahedral geometry. Therefore, The VSEPR model for ${\left[ {P{F_6}} \right]^ - }$ is Octahedral.

In this figure, we can conclude that there are no double bonds in the compound. Hence, the No. of double bonds present in the Lewis structure of $PF_6^ -$ is $0$ .

Note:
Hexafluorophosphate is most commonly used as the lithium salt, lithium hexafluorophosphate. This salt is a natural electrolyte in industrial secondary batteries, such as lithium-ion cells, when combined with dimethyl carbonate.