PF_{6}^{–} or hexafluorophosphorus is the anionic form of the pentafluorophosphorus having a molecular weight of 144.96 g/mol. Let us discuss more PF_{6}^{–}.

**PF _{6}^{–} is the octahedral molecule where P also expands its octet and geometry. The stable valency of P is +3 and +5 but here the valency of P is +6, so P expands its valency by violating the octet. The mode of hybridization of the central atom is sp^{3}d. so, here d orbital of P is involved in the bond formation.**

The availability of energetically d orbital P can easily increase its coordination number and violates its octet in PF_{6}^{–} molecule. Now we try to explain the lewis structure, hybridization, octet rule, valence electrons, and polarity of PF_{6}^{–} in the following part of the article with proper explanation.

## 1. How to draw PF_{6}^{–} lewis structure

Lewis structure of PF_{6}^{–} can give us a clear picture of the bonding electrons, valence electrons, etc. so we try to draw PF_{6}^{–} lewis structure in few many steps.

### Counting the valence electrons

**The first step for drawing the lewis structure is counting the total valence electrons for the respective molecule. The number of total valence electrons for the PF _{6}^{–} is 44, and those numbers are the summation of valence electrons of P and five F atoms. For the negative charge, we add one more electron.**

### Choosing the central atom

**The 2 ^{nd} and other important step is selecting the central atom for a molecule while drawing the lewis structure. We should choose the central atom based on its size and electropositivity. Accounting for those conditions P is chosen as the central atom because it is electropositive and larger in size than F.**

### Satisfying the octet

**Every atom after bond formation obeys the octet rule by fulfilling its valence orbital via a suitable number of electrons. Here P violates its octet by accepting more electrons and this is happening because it has a vacant d orbital, due to the presence of excess electron negative charge present over it.**

### Satisfying the valency

**The electrons required for a complete octet 7*8 = 56, but the valence electrons are 48, so the remaining number of electrons accumulated by ½(56-48) = 4 bonds where the pentavalency of P exceeds and accepts a negative charge. Each F is satisfied by its monovalency. For this reason, P is violated octet.**

### Assign the lone pairs

**After successful bond formation, we assign the remaining non-bonded electrons to their respective atoms. Here each F has seven valence electrons and among them, only one is involved in the bond formation, so the remaining six electrons exist as three pairs of lone pairs, but P is lack lone pairs here.**

## 2. PF_{6}^{–} valence electrons

Valence electrons are the electrons present in the outermost orbital and are involved in bond formation. Let us count the valence electrons for PF_{6}^{–}.

**The total valence electrons for the PF _{6}^{–} are 48 which are the contribution from the one P and six F atoms. Total valence electrons for a molecule are always a summation of the valence electrons of the substituent atoms. P and F have five and seven electrons respectively by their electronic configuration.**

- Now we count the total valence electrons for the PF
_{6}^{–}in detail - The valence electrons for the P are 5
- The valence electrons for the F are 7
- So, in the one P and six F atoms are present along with one negative charge
- For negative charge, one electron added extra
- So, the total valence electrons for the PF
_{6}^{–}are, 5+(7*6) + 1 = 48

## 3. PF_{6}^{–} lewis structure octet rule

Octet is the completion of the valence orbital after bond formation by accepting a suitable number of electrons. Let us see if PF_{6}^{–} follows octet or not.

**In the PF _{6}^{–} molecule P not follows the octet here, it expands its valency and violates the octet. From the octet calculation, it requires 56 electrons but the valence electrons are 48, so the remaining 8 electrons are accumulated by 8/2 = 4 bonds, but here P makes six bonds which is a violation of octet.**

From the electronic configuration [Ne]3s^{2}3p^{3}, it is confirmed that P has five electrons in its valence shell so it can accept three more electrons to complete the octet but in the PF_{6}^{–}, it makes six bonds involving its empty d orbital and expand its octet where it shares 12 electrons and violates the rule of the octet.

## 4. PF_{6}^{–} lewis structure lone pairs

Lone pairs are non-bonded valence electrons present in the outermost orbital after bond formation. Let us count the lone pairs of the PF_{6}^{–}.

**The total number of lone pairs are 36**, **only F contains lone pairs, P is lack lone pairs here. Each F makes one single bond and the other six electrons exist as lone pairs over them because F has seven valence electrons. But all five valence electrons of P are involved in the bond formation.**

- Let us count the total lone pairs over the PF
_{6}^{–}in detail by the formula lone pairs = valence electrons – bonded electrons. - Lone pairs over P atom here, 5-6 = -1(negative sign indicates it needs one more electron)
- Lone pairs over each F atom, 7-1 = 6 (three pairs of lone pairs)
- So, the total lone pairs over PF
_{6}^{–}are 6*6 = 36.

## 5. PF_{6}^{–} lewis structure shape

The molecular shape is arranged by the constituent atoms to give a proper geometry to the molecule by arranging them. Let us know the shape of the PF_{6}^{–}.

**The lewis structure shape of the PF _{6}^{–} molecule is octahedral which is confirmed by the following table.**

Molecular Formula | No. of bond pairs | No. of lone pairs | Shape | Geometry |

AX | 1 | 0 | Linear | Linear |

AX_{2} | 2 | 0 | Linear | Linear |

AXE | 1 | 1 | Linear | Linear |

AX_{3} | 3 | 0 | Trigonalplanar | TrigonalPlanar |

AX_{2}E | 2 | 1 | Bent | TrigonalPlanar |

AXE_{2} | 1 | 2 | Linear | TrigonalPlanar |

AX_{4} | 4 | 0 | Tetrahedral | Tetrahedral |

AX_{3}E | 3 | 1 | Trigonalpyramidal | Tetrahedral |

AX_{2}E_{2} | 2 | 2 | Bent | Tetrahedral |

AXE_{3} | 1 | 3 | Linear | Tetrahedral |

AX_{5} | 5 | 0 | trigonal bipyramidal | trigonal bipyramidal |

AX_{4}E | 4 | 1 | seesaw | trigonal bipyramidal |

AX_{3}E_{2} | 3 | 2 | t-shaped | trigonal bipyramidal |

AX_{2}E_{3} | 2 | 3 | linear | trigonal bipyramidal |

AX_{6} | 6 | 0 | octahedral | octahedral |

AX_{5}E | 5 | 1 | square pyramidal | octahedral |

AX_{4}E_{2} | 4 | 2 | square pyramidal | octahedral |

**VSEPR Table**

So, from the above table, it is evident that PF_{6}^{–} is an AX_{6} type of molecular, and for VSEPR (Valence Shell Electrons Pair Repulsion) theory the ideal geometry for the AX_{6} molecule is octahedral and here no lone pairs are present over central P atom, so no need for change the geometry or shape.

## 6. PF_{6}^{–} lewis structure angle

The structure angle in that angle is made by the constituent atoms for the proper orientation of the molecule. Let us calculate the bond angle for PF_{6}^{–}.

**The bond angle for F-P-F in the PF _{6}^{–} the molecule is 90^{0}, a s it is an octahedral molecule, where the central tom is surrounded by six equivalent atoms and the bond angle for the octahedral geometry is perfectly 90^{0}. There is no deviation factor is present so the bond angle can be perfectly ideal.**

- The bond angle of PF
_{6}^{–}can be calculated by the hybridization value. - The bond angle formula according to Bent’s rule is COSθ = (p-1)/p.
- The central atom P is sp
^{3}dhybridized, so the p character here is 3/6 = ½ - So, the bond angle is, COSθ = [{1-(1/2)} / (1/2)] = -1
- Θ = COS
^{-1}(-1) = 180^{0}, 180^{0}is not allowed and we deducted 90^{0}from it, - So, the bond angle is 180
^{0}– 90^{0}= 90^{0}

## 7. PF_{6}^{–} lewis structure formal charge

By the concept of formal charge, we can predict the amount of charge present within the molecule or particular atom. let us calculate the formal charge of the PF_{6}^{–} molecule.

**The formal charge of PF _{6}^{–} is -1 because there is already a charge present within the molecule so, the amount of charge by the anions is not fully satisfied by the charge quantity by the cations. Here several anions are greater than the valency of the cation. That’s why a negative charge is added.**

- Let us calculate the formal charge of PF
_{6}^{–}molecule by the formula, F.C. = N_{v}– N_{l.p.}-1/2 N_{b.p} - Formal charge present over the P atom is, 5-0-(12/2) = -1
- Formal charge present over each F atom is, 7-6-(2/2) = 0
- So, the overall formal charge presents over the PF
_{6}^{–}the molecule is -1 which is actually from the P site because it has more electrons than its valency.

## 8. PF_{6}^{–} hybridization

Hybridization is the mixing of the atomic orbitals of different energy to produce a new hybrid orbital of equivalent energy. Now we explore the hybridization of the PF_{6}^{–}.

**The Hybridization of the central P in PF _{6}^{–} is sp^{3}d which can be confirmed by the following table.**

Structure | Hybridization value | State of hybridization of central atom | Bond angle |

1.Linear | 2 | sp /sd / pd | 180^{0} |

2.Planner trigonal | 3 | sp^{2 } | 120^{0} |

3.Tetrahedral | 4 | sd^{3}/ sp^{3} | 109.5^{0} |

4.Trigonal bipyramidal | 5 | sp^{3}d/dsp^{3} | 90^{0} (axial), 120^{0}(equatorial) |

5.Octahedral | 6 | sp^{3}d^{2}/ d^{2}sp^{3} | 90^{0} |

6.Pentagonal bipyramidal | 7 | sp^{3}d^{3}/d^{3}sp^{3} | 90^{0},72^{0} |

**Hybridization Table**

- We can calculate the hybridization by the convention formula, H = 0.5(V+M-C+A),
- So, the hybridization of central P is, ½(5+6+0+1) = 6 (sp
^{3}d) - One s orbital, three p orbitals, and one d orbital of P is involved in the hybridization.
- The lone pairs of F are not involved in the hybridization.

## 9. Is PF_{6}^{–} polar or nonpolar?

A molecule is said to be polar if there is a presence of a permanent dipole- moment. Let us check whether PF_{6}^{–} is polar or not.

**PF _{6}^{–} is non-polar because it has a symmetrical octahedral geometry, so the directions of dipole-moment from P to F are equal and opposite in direction to each other. So, they cancel out very easily and for this reason, the value of dipole-moment for the PF_{6}^{–} is zero which makes the molecule non-polar.**

## 10. Is PF_{6}^{–} an electrolyte?

A substance is called an electrolyte when it is soluble in water and gets ionized and carries electricity through the solution. Let us see whether PF_{6}^{–} is an electrolyte or not.

**PF _{6}^{–} is an electrolyte because a negative charge is already present over the molecule, so it is an already charged particle and when it gets dissolved in the water it can carry electricity. Also, in the aqueous solution, it ionized fully to form highly electronegative fluoride ions and carries electricity.**

## 11. Is PF_{6}^{–} soluble in water?

When a molecule breaks its bond and gets dissolved in water by partial or full hydrolysis then it can be soluble in water. Let us see if PF_{6}^{–} is soluble in water or not

**PF _{6}^{–} is soluble in water because it can form H-bonding wither the water molecule. The size of F is very small and the electronegativity is very high so it is compatible with H-bonding. Although there is more hydrophobic part is present which makes the low solubility of the molecule in water.**

## 12. Is PF_{6}^{–} ionic or covalent?

According to Fajan’s rule, no molecule has 100% ionic or covalent character it is vice versa based on polarizability. Let us see whether PF_{6}^{–} is covalent or ionic.

**PF _{6}^{–} is a purely covalent molecule, made by the hybridization of the atomic orbitals of P. the bond between P and F is equally shared by the electrons of those atoms. Unlike ionic molecules, the bond is not polar and has no unsymmetrical distribution of electrons.**

The polarizing power of cation is very poor (low ionic potential – charge density low and size is larger), again the size of the anion is smaller so it cannot get polarized. On the other hand, the polarizability of the anion is also low.so, it has the least ionic character.

#### Conclusion

PF_{6}^{–} can acts as a strong field ligand in organometallic chemistry, because it can donate the electron density and also accept the electron density to the vacant orbital of P. even six electronegative F atoms draw the electron density from the P toward themselves so the orbital energy of P decreases and can easily gets bonded.