Lewis Structure, Hybridization, Polarization, and Molecular Geometry of Compound PO43

Did you know that phosphate ions (tricalcium phosphate) make up the majority of our bones? It is found not only in our bones but also in our teeth. That seems very intriguing, doesn’t it?

It is vital not only for people but also for plants. Therefore, it is essential that we study about this molecule and determine what makes it so unique.

Phosphate, represented by the formula PO43-, is a trivalent inorganic anion produced from phosphoric acid. It is composed of a variety of salts containing the phosphate ion, the dihydrogen phosphate ion, and numerous cations (positive charge) such as calcium, sodium, etc. It is found naturally in food, water, and humans (bones, teeth, and genes).

The United States, China, and Morocco are the main producers of phosphate due to the highest concentration of phosphate rocks (minerals) in these nations.

Now that you understand its significance, let’s examine PO43- in depth and learn its Lewis structure, Hybridization, form, and polarity.

Lewis Structure of PO43

Before we begin to sketch the Lewis structure of PO43- step by step, let’s examine a general representation of the Lewis structure.

To depict the Lewis structure of any molecule, valence electrons are always utilised (number of electrons in the outer shell of each atom).

The total charge on the phosphate ion is -3, and the oxidation state of the phosphorus atom is +5.

P forms single bonds with 3 oxygen atoms and a double bond with one oxygen atom in the Lewis structure of PO43-. These oxygen atoms are negatively charged.

Let us now examine the steps necessary to draw a Lewis structure:

  1. determining the molecule’s total amount of valence electrons.
  2. locating the molecule’s core atom
  3. arranging electrons as lone pairs in order to make bonds with each atom.

The configuration of the remaining valence electrons so that each atom’s octet is completed.

  1. Transforming possible lone pairs into double or triple bonds to increase the stability of the molecule. In addition, we ensure that the formal charge of each atom is as low as possible.

Each atom in the optimal Lewis structure of any molecule will have a formal charge of zero. Below is the formula for calculating formal charge: –

Drawing the lewis structure of PO43-

Count the total number of valence electrons in the PO43- molecule. P is a member of group 5, hence it contains five valence electrons. O is a member of group 6, hence it has 64 = 24 valence electrons (4 atoms of O).

Now the entire atom has a charge of -3, which indicates the presence of 3 more valence electrons. The total number of valence electrons is therefore 5 plus 24 plus 3 = 32 valence electrons.

The atom with the highest valence factor and the greatest number of bonding sites is known as the central atom. According to this argument, the core atom of PO43- is P.

Now, we begin to arrange these electrons as lone pairs on each atom, simulating a chemical connection. For this, eight valence electrons are sacrificed.

Now, the remaining 24 valence electrons are arranged around each atom to complete its octet. Currently, all 32 valence electrons have been consumed.

Now, we examine the formal charge of each PO43- atom. All the O atoms have a -1 charge, while P has a +1 charge. Now, this appears acceptable.

But is this the lowest feasible official fee? Keep in mind that the charges on each atom of a molecule make the molecule more unstable. Therefore, the optimal Lewis structure should have each atom with a charge of zero.

P can store more than 8 valence electrons in PO43- (period 5). Therefore, if P forms a double bond with one of the oxygen atoms, both P and the doubly-bonded O atom will have a charge of zero. The remaining 3 O atoms will have a -1 charge, giving rise to the molecule’s -3 charge.

We cannot induce more double bonds with O because P will gain a -1 charge, rendering the molecule extremely unstable.

Thus, the optimal Lewis structure for PO43- is obtained.

PO43- Hybridization

Let us now examine the molecule’s hybridization, which will provide us with more information about the nature of their chemical bonds.

Sp3 is the hybridization product of PO43-.

This can be comprehended in two ways:-

Method 1 entails summing the amount of bonds and the lone pair of the central atom to determine the theory aspect, hybridization of PO43. The value of Hybridization (H) is determined by the following factors:

If H = 2, then it is a sp hybrid.

If H=3, sp2 hybridization has occurred.

H=4 indicates sp3 hybridization.

H=5 indicates sp3d hybridization.

And H=6 indicates that sp3d2 has been hybridised.

We know that P is the core atom in PO43-. It forms four sigma bonds with each oxygen atom (double and triple bonds are regarded to be one sigma bond) and has no lone pairs. Therefore, the total value of H is 4, which hybridises the Sp3 molecule.

Next, we will discuss the formula component of hybridization. This is beneficial if you cannot recall the theory.

The following formula can be used to determine the hybridization of any molecule:

H= 1/2[V+M-C+A]

H = Molecular hybridization

V = Valence electrons of the core atom.

Charge on a cation or electropositive atom.

A equals the charge on an anion or a more electropositive atom.

Now let’s examine PO43-,

P being the core atom, V = 5 (valence electrons)

Due to the fact that Oxygen (O) is a divalent atom, there are no monovalent atoms linked to the centre atom, hence M = 0.

C = 0 (charge on P)

A= -3 (Combined charge of O atoms)

Hence,

H=1/2[5+3]

H = 4, which indicates that PO43- is sp3 hybridised

Consequently, the hybridization of PO43- can be determined using these two techniques.

PO43- Molecular Geometry

Next, it is crucial to identify the molecular geometry of PO43-.

A molecule’s molecular geometry determines its shape and bond angles.

PO43-‘s molecular geometry is tetrahedral. The angle between each atomic link is 109.5 degrees.

Any molecule’s molecular geometry can be determined via the symbol AXN. A represents the quantity of central atoms.

The number of atoms that are chemically connected to the centre atom is denoted by X. Lastly, N represents the lone pairs on the centre atom ( the nonbonding electrons).

P is the solitary central atom in PO43-, hence A equals one. As there are four O atoms linked to the centre atom, X equals four. As all electrons are involved in bonds and there are no lone pairs, N will be equal to zero. Therefore, using the formula, the shape of PO43- is AX4.

If we look up this notation in the following VSEPR chart, we see that PO43- possesses tetrahedral geometry.

PO43- Polarization

The nonpolar compound PO43-.

Polarity in a molecule results mostly from differences in electronegativity. Typically, polar molecules have an asymmetrical structure, and their net dipole moments are never zero.

Since the difference in electronegativity between P (2.1) and O (3.5) is larger than 0.5, the P-O bond is polar. However, is the PO43- molecule polar?

First, we sketch the Lewis structure, followed by the dipole moments of each link. Due to the symmetrical structure of PO43-, the dipole moments cancel out, rendering the molecule non-polar.

In order to validate this, we can also examine its molecular geometry. PO43- is a non-polar molecule due to its tetrahedral form with each dipole moment pointing in the opposite direction, resulting in a net dipole moment of zero.

Conclusion

PO43- has a molar mass of 94.97 g/mol. It is widely used as a polishing and conditioning component in toothpaste. It is also utilised in a variety of cleaning chemicals and pesticides, making it highly effective at removing dirt.

We now have a complete understanding of the PO43- molecule, including its Lewis structure, hybridization, molecular geometry, and polarity.

Now you have a more foundational understanding of the P043- molecule. If you have any questions on the aforementioned points, feel free to contact me. Study Hard!

Read more: Ionic or covalent is water?

Misha Khatri
Misha Khatri is an emeritus professor in the University of Notre Dame's Department of Chemistry and Biochemistry. He graduated from Northern Illinois University with a BSc in Chemistry and Mathematics and a PhD in Physical Analytical Chemistry from the University of Utah.

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