HPO4 2- Lewis Structure: 3 Easy Step-by-Step Guide

Hydrogen phosphate or HPO42- is the conjugate base of dihydrogen phosphate having molecular weight 95.98 g/mol. It is a divalent inorganic phosphate anion. It has one hydrogen, one phosphate and four oxygen atoms attached by one double bond and three single bonds around the phosphorous atom.

The Lewis structure of HPO4 2- features a phosphorus (P) atom at the center, bonded to one hydrogen (H) atom and three oxygen (O) atoms. Two of the O atoms are double-bonded to P, each carrying two pairs of non-bonding electrons. The third O atom is single-bonded to P and carries three pairs of non-bonding electrons, and is also bonded to the H atom. This results in a formal charge of -2 on the molecule, with the P atom having a formal charge of +1, one O atom (the one bonded to H) having a -1 charge, and the remaining atoms being neutral.

Let’s highlight the following topics on HPO₄^2-.

How to draw HPO₄ ^2- lewis structure?

To draw the lewis structure of any ion or molecule the following factors should be known perfectly.

1. Determining the number of valence electron: Phosphorus is a group 15 element. Thus, it has five valence electrons, whereas being a group 16 electron, oxygen has 6 electrons in its outer most shell and hydrogen has one valence electron in its 1s orbital.
2. Finding out the bonding electrons:  Total three single (sigma) and one double bond are present in HPO₄^2-. Therefore, the number of electrons involved in bonding are = (3×2) + (2×2) = 10
3. Finding out the nonbonding electrons:  Each of the single bonded oxygen (with phosphorus) has six electrons and the two oxygen atoms (one is attached with phosphorous by double bond and other one is attached by two single bonds with phosphorous and hydrogen atom) has four electrons as nonbonded electron pairs.

HPO₄^2- Lewis structure Shape

Lewis structure shape of any molecule denotes the geometrical structure of any molecule which can be determined by the hybridization of central atom and the lone pairs (if present in central atom). In the following table the changes of geometry with the changing of hybridization are shown-

Therefore, to determine the structure or shape of HPO₄^2-, the hybridization of central atom (phosphorous) should be determined first. The hybridization of phosphorous in this molecule is sp³. Thus, the geometrical structure should be tetrahedral. As phosphorous has no electrons left as nonbonding thus repulsion like lone pair-lone pair, lone pair-bond pair and bond pair-bond pair are not deviating its structure from the ideal geometry.

Otherwise, these repulsive factors the structure as well as bond angle of any molecule from the ideal case.

HPO₄ ^2- lewis structure Formal Charge

Formal charge calculation helps to identify the most stable lewis structure among different possible lewis structures. It also decides the total charge present in the molecule.

• Formal charge = Total number of valance electrons – number of electrons remain as nonbonded – (number of electrons involved in bond formation/2)
• Formal charge of central atom, phosphorous = 5 – 0 – (10/2) = 0
• Formal charge of each of the two oxygen atoms, attached with phosphorous by single bonds = 6 – 6 – (2/2) = -1
• Formal charge of another two oxygen atoms (one is OH oxygen and another oner is attached with phosphorous by double bond) = 6 – 4 – (4/2) = 0

From the calculation of formal charge, it is proved that this molecule has two negative charges on two oxygen atoms but these two negative charges are delocalized over the whole molecule.

HPO₄ ^2- Lewis Structure Angle

Like the geometrical molecular structure, bond angle in any molecule also depends upon the hybridization of central atom. There is a particular bond angle for each of the hybridization.

As the central atom phosphorous is sp³ hybridized in HPO₄^2-, the bond angle should be 109.5⁰. Due to absence of any repulsion involving bond pairs and lone pairs, it shows its actual bond angle without any deviation because phosphorous has no nonbonding electrons left to participate in lone pair-lone pair and lone pair-bond pair repulsion.

HPO₄ ^2- Lewis Structure Octet Rule

Octet rule tells about the valence shell electron configuration of any atom that should match with its nearest noble gas (according to periodic table) valence shell electron configuration.

In HPO₄ ^2-octet rule is not obeyed as phosphorous does not satisfy the octet rule. Phosphorous already contains five electrons in its valence shell (3s²3p⁵). After bond formation, it achieves five more electrons in its valence shell. Therefore, the total number of electrons in the valence shell of phosphorous become 10 which does not match with the valence electron numbers of its nearest noble gas, argon (3s² 3p⁶).

But octet rule is not violated for oxygen atom. It has six electrons and after bonding with neighboring phosphorous and hydrogen atom it achieves two more electrons in its outer most shell which resembles with its nearest noble gas, neon (2s² 2p⁶).

Hydrogen has one electron and after bond formation with oxygen, its valence shell is filled with two electrons like its nearest noble gas helium (1s²).

HPO₄ ^2- Lewis Structure Lone Pairs

Lone pairs are actually those valence electrons which are shown as the electron dot in lewis structure. They are not participating in bond formation with other atoms.

• Nonbonded electron = Total number of valance electron – number of bonded electrons.
• Nonbonding electrons in phosphorous = 5 – 5 = 0
• Nonbonding electrons of each of the two oxygen atoms, attached with phosphorous by single bonds = 6 – 2 = 4 or two pair of lone electrons.
• Nonbonding electrons of another two oxygen atoms (one is OH oxygen and another oner is attached with phosphorous by double bond) = 6 – 0 = 6 or three lone pairs.

Therefore, total numbers of nonbonding electrons in HPO₄^2- = (2×4) + (2×6) = 20

HPO₄ ^2- Valence Electrons

Valance electrons indicate the outer most shell electrons in every atom. The reactivity of valence electrons is the most and they mainly participate in bond formation. The reason behind this greater reactivity of valence electron is due to lesser attraction force of nucleus on the valence shell.

Phosphorous is nitrogen group (group 15) element. It has five electrons in its outer most shell and all the valence electrons are used up for bond formation. Hydrogen has one electron (1s¹) and it is regarded as the valence electron of hydrogen. Oxygen is group 16 element. It has total eight electrons and among these eight electrons, six electrons are in valence shell and are regarded as valence electrons.

Therefore the number of valence electron in HPO₄^2- = 5 + (4×6) + 1 = 30

HPO₄ ^2- Solubility

All phosphates are basically insoluble excluding sodium, potassium and ammonium phosphates in water but some hydrogen phosphates are soluble like disodium dihydrogen phosphate. The solubility of disodium dihydrogen phosphate is 7.7 g/100 ml at 20⁰ C and 11.8 g/100 ml at 25⁰ C which indicated that solubility in water of dihydrogen diphosphate increases with increasing temperature.

Is HPO₄ ^2- polar or nonpolar?

Polarity or dipole moment depends on the following factors-

• Electronegativity difference between the atoms
• Shape of the molecule or orientation of the bonds with respect to each other.

HPO₄^2- is tetrahedral shaped with the bond angle 109.5⁰. There is difference of electronegativity present between oxygen, phosphorous and hydrogen. Thus, the O-P bonds are polar. Due to tetrahedral shape, the bond moments are not cancelled by each other and the anion possesses a permanent dipole moment.

Therefore HPO₄^2- is a polar molecule.

 Is HPO₄ ^2- an acid or base?

Phosphoric acid (H₃PO₄) is an acid. After elimination of two hydrogen atom, HPO₄^2-is formed which is the conjugate base of H₂PO₄^–(formed after elimination of one hydrogen atom from H₃PO₄). Therefore, it can uptake two protons. It can also donate its last proton and form PO₄^3-.

Thus, it acts both as acid and base.

Is HPO₄ ^2- amphiprotic?

Yes, HPO₄^2-is an amphiprotic ion which means it can both accept and donate proton. It is the conjugate base of H₂PO₄^–, and the conjugate acid of PO₄^3-. After accepting two protons, it forms stable phosphoric acid (H₃PO₄).

H₂PO₄^– Lewis Structure

It is basically the conjugate base of phosphoric acid. After eliminating one proton from phosphoric acid, H₂PO₄^– is obtained. In this mono negative molecule, phosphorus is bonded with two OH group by two single bonds and with two oxygen atoms by a single bond and double bond respectively.

Is H₂PO₄^– an acid or base?

It is also an amphiprotic substance because it is the conjugate base of phosphoric acid. On the other hand, it can donate its two proton to form PO₄^3-. Thus, it also acts as Bronsted acid. For this dual behaviour as acid and base it is defined as an amphiprotic substance.

Is H₂PO₄^– stronger than HPO₄^2-?

H₂PO₄^– is stronger than HPO₄^2-because after accepting one hydrogen atom, H₂PO₄^– becomes phosphoric acid. As phosphoric acid is a weak acid and after eliminating one proton it becomes weaker and further donation of proton HPO₄^2-becomes the weakest.

Conclusion

It can be concluded from the above article on HPO₄^2- that it is conjugate base of H₂PO₄^– and it is a tetrahedral anion having bond angle 109.5⁰. It is a polar molecule having a permanent dipole moment.

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Aditi Roy

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I am Aditi Ray, a chemistry SME on this platform. I have completed graduation in Chemistry from the University of Calcutta and post graduation from Techno India University with a specialization in Inorganic Chemistry. I am very happy to be a part of the Lambdageeks family and I would like to explain the subject in a simplistic way.
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