lewis structure

Cyanate ion has the chemical formula OCN-. It is also known as isocyanates. It is an ambidentate ligand which forms complexes.

Cyanate is a colorless liquid substance with small odor. If treated with water it creates toxic fumes and also if heated at high temperature until decomposition can produce cyanide and nitiric oxides fumes which are toxic to health. In this editorial we are learning about OCN- lewis structure and its detailed facts.

How to draw lewis structure for OCN-?

Some points to be remember while drawing any lewis structure

1. Structure’s total valence electrons calculation
2. Element needs lowest electronegativity to occupy central position
3. Connecting all elements with bonding

Chemical formula for cyanate ion is OCN-.

Molecular weight of OCN- is 42.017 g mol^-1.

Molecular geometry of OCN- is linear in shape.

OCN- has sp hybridization.

OCN- is polar in nature.

The isomer of less stable fulminate anaion is the cyanate. It also has various salt forms like ammonium cyanate. Cyanate is an anion consisting of three different elements i.e. oxygen, carbon and nitrogen. Most of cyanate compounds toxic to inhale, flammable and also can cause irritation to eyes, skin and mucous membranes.

It is a bridging and ambidentate ligand (ambidentate means it can coordinate with either of its two different elements).

In OCN- lewis structure, there is a single covalent bond between carbon and oxygen atom and a triple bond within oxygen and nitrogen atoms. The nitrogen atom in OCN- lewis structure bears a negative charge on it.

OCN- valence electrons

• OCN- Structure’s total valence electrons calculation

Let us calculate the total valence electrons on OCN- ion. As we know that the OCN- ion involves three elements oxygen, carbon and nitrogen, we have to check the position of group of these three elements in the periodic table. The oxygen atom belongs to 16^th group, carbon atom belongs to 14^th group and the nitrogen atom belongs to 15^th group of the periodic table.

Therefore, O atom contains 6 valence electrons, C atom contains 4 valence electrons and N atom contains 5 valence electrons in their valence shell. As OCN- ion have one O atom, one C atom and one N atom, let us first calculate the total valence electrons for OCN- ion.

Oxygen atom valence electrons = 6 x 1 (O) = 6

Carbon atom valence electrons = 4 x 1 (C) = 4

Nitrogen atom valence electrons = 5 x 1 (N) = 5

Now, add extra one electron for negative (-) charge present on OCN- ion

So, OCN- ion total valence electrons = 6 (O) + 4 (C) + 5 (N) + 1 (-) = 16

Therefore, total valence electrons on OCN- lewis structure is sixteen.

Let us find the total electron pairs on OCN- lewis structure by dividing total valence electrons by two.

OCN- total electron pairs = OCN- total valence electrons / 2 = 16/2 = 8

Hence, there are total eight electron pairs on OCN- lewis structure.

• Element needs lowest electronegativity to occupy central position

The atom with least electronegativity will occupy the central position of the OCN- lewis structure. O atom has electronegativity 3.44, C atom has electronegativity 2.55 and nitrogen atom has electronegativity 3.04. So, carbon atom is least electronegative from all the three elements. Then carbon atom will occupy the central position of the OCN- lewis structure.

• Connecting all elements with bonding

Now, connect all the three elements O, C and N with each other by linking with a single covalent bond between them. There is the involvement of two valence electrons in each single covalent bond.

OCN- lewis structure octet rule

We have total 16 valence electrons in OCN- lewis structure, out of which four electrons involved in bonding between O-C and C-N i.e. two bond pairs. Now we are left with more 12 valence electrons for distribution on OCN- lewis structure to complete the octet of all the three elements present in the ion.

To complete the octet we have to put the eight electrons in the atoms valence shell.  Now we have to complete the octet of outer elements i.e. oxygen and nitrogen atom.

The above structure shows the remaining electrons get shared on O an N, there are six electrons on oxygen and six electrons on nitrogen atom get distributed. So we can see that the O and N atoms has complete octet i.e. two bond pairs and three lone pairs. Similarly, N atom also has teo bond electrons and three lone electrons.

Thus, the O and N atom has complete octet with eight electrons but the C atom has only four electrons i.e. bond pair electrons.

So, carbon atom is not satisfied as its octet is not complete and the structure is unstable. Then we have to move the electrons from nitrogen and oxygen atoms to complete the octet of central carbon atom and also to form double or triple bonds to get a stable OCN- lewis structure which we see further in the explanation of resonance structure.

OCN- lewis structure formal charges

The atoms in any lewis structure having small formal charge give more stable structure. There is a formula to count formal charges on atoms in a lewis structure as follows.

Formal charge = (valence electrons – non-bonding electrons – ½ bonding electrons)

Let we count the formal charges on OCN- lewis structure. So, first we count the formal charge for oxygen atom.

Oxygen atom: Oxygen’s valence electrons = 06

                       Oxygen’s non-bonding electrons = 06

                       Oxygen’s bonding electrons = 02

Formal charge on oxygen = (6 – 6 – 2/2) = -1

Carbon atom: Carbon’s Valence electrons = 04

                       Carbon’s Non-bonding electrons = 00

                       Carbon’s Bonding electrons =04

Formal charge on carbon = (4 – 0 – 4/2) = +2

Nitrogen atom:Nitrogen’s Valence electron = 05

                        Nitrogen’s Non-bonding electron = 06

                        Nitrogen’s Bonding electrons =02

Formal charge on nitrogen = (05 – 06 – 2/2) = -2

Hence, the oxygen atom has -1 formal charge, carbon atom has +2 formal charge and nitrogen atom has -2 formal charge in OCN- lewis structure.

OCN- lewis structure lone pairs

In OCN- lewis structure, after the completion of both the outer elements i.e. O and N there are total twelve non-bonding electrons on both the atoms. It means total six lone electron pairs are present on OCN- lewis structure in unstable form.

Thus, the oxygen atom has three lone electron pairs, carbon atom has zero lone electron pairs and nitrogen atom has three lone electron pairs on OCN- lewis structure.

OCN- lewis structure resonance

OCN- lewis structure shows three non-equivalent resonance structures with the formation of multiple (double / triple) bonds by transfer of electrons within the ion.

In OCN- resonance structure, the charge distributions and bonding’s are vary and having energy levels differ from each other. Some resonance structure shows more stability than other resonance structures. As per the resonance structure rules, if the structure has less energy than it is more stable in nature.

We have already seen in the formal charges topic that the OCN- lewis structure basically consisting three charges -1 on O, +2 on C and -2 on N. Now we have to convert the lone electron pairs of N atom rather than O atom, as O is more electronegative than N atom. Nitrogen being less electronegative can provide more valence electrons for sharing.

To get the more stable resonance structure, the negative charge should be on the element with more electronegativity. So, in the above structure oxygen having negative charge on one resonance structure, which is more stable as oxygen more electronegative than C and N.

We have to convert the nitrogen atoms lone electron to bond pair to make C-N bond within C and N atom. As there are more charges on atoms so we have to convert more lone electron pairs of N atom into bond pair to get a triple bond between C and N, creating negative charge on O atom to give more stable structure.

Now, in the above more stable structure we can see that carbon atom now has eight electrons in four bond pairs i.e. carbon atom is satisfied with complete octet. As well the negative charge is on O atom as it is more electronegative and shows good lewis structure.

So, finally we conclude that, OCN- lewis structure shows three resonances after minimization of charges on it. One resonance structure shows single bond between O-C and triple bond between C-N, second resonance structure shows triple bond between O-C and single bond between C-N and the third resonance structure shows double bonds between O-C and C-N.

OCN- lewis structure shape

As we already seen in above resonance structure explanation, that the O atom bears -1 negative charge as it is more electronegative. Also O and C have a single covalent bond and C and N atom has triple bond showing a stable form of structure. So, there are zero lone electron pairs on central carbon atom. The VSEPR theory generic formula for OCN- ion is AX2.

As per the above structure and its valence electrons, the OCN- atoms arrangement looks planar; also there are no lone electron pairs present and also the geometry s not in bent shape. Then according to VSEPR theory The OCN- lewis structure is linear in shape.

OCN- Hybridization

Hybridization of any molecule can be determined by the electron density present on the atom. In the above OCN- lewis structure we discussed, central carbon atom is creating bonds within oxygen and nitrogen atom, now let us discuss about its hybridization.

There is a triple bond between carbon and nitrogen which shows a one electron density, also carbon has a single covalent bond with oxygen atom shows second electron density. As there are two electron densities for carbon atom in OCN- lewis structure, the two hybrid orbitals can form by carbon atom to form bonding with oxygen and nitrogen atoms.

Therefore, the carbon atom can form one ‘s’ hybrid orbital and one ‘p’ hybrid orbital during formation of single bona and triple bond with oxygen and nitrogen atom. Thus, the central carbon atom has ‘sp’ hybridization in OCN- lewis structure.

OCN- polar or nonpolar

Some points to be noted, while considering any molecule is polar or non-polar in nature as follows.

• If two atoms have more electronegativity difference then tha molecule is polar in nature.
• Symmetry or assymetrical molecular shape, if symmetrical then non-polar and if assymetrical then it is polar molecule.
• Dipole moment creates due to more electronegative central atoms.

Let us discuss the polar or non-polar nature of OCN- lewis structure. Electronegativity of C atom is 2.55, electronegativity of O atom is 3.44 and electronegativity of N atom is 3.04. The differences of electronegativities of all three elements are clearly seen.

In case of OCN- lewis structure, carbon and oxygen (C-O) and carbon and nitrogen (C-N) bond creates dipole on it due to their electronegativity differences. Basically we know that, the molecule is polar when it has one or more lone electron pairs on central element and having different outer elements.

Thus, central carbon atom of OCN- ion has no lone electron pairs and also OCN- ion have assymetrical geometry. So, OCN- ion is polar in nature.

OCN- lewis structure bond angle

OCN- lewis structure has three elements arranged in a single horizontal plane and having assymetrical distribution of electrons which shows linear shape of OCN- ion.

So, the OCN- lewis structure has 180 degree bond angle.

OCN- lewis structure electron geometry

We have already discussed about the leis structure, formal charges, valence electrons and resonance structure of OCN- ion. From which we can see that there are 16 total valence electrons are present on OCN- ion.

From these 16 valence electrons we can create two double bonds between O-C and C-N or we can create one triple bond between either O-C or C-N. But the stable form of OCN- resonance structure show a single bond between O and C atom and a triple bond between C and N atom with negative charge on O atom. Also there are three lone pairs on O atom and two lone pairs on N atom due to which this ion is termed as ambidentate ligand.

So, OCN- ion shows tetrahedral electron pair geometry.

OCN- Uses

As the cyanate ion (OCN-) is toxic in nature i.e. it is flammable, also cause irritations to human skin, eyes and mucous membrane. We have not seen the cyanate ion in our general use of day to day life. It could be used in only high level chemical factories and industries.

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The phosphate ion (PO4 3-) has a central phosphorus (P) atom with 5 valence electrons, surrounded by four oxygen (O) atoms, each with 6 valence electrons. The negative charge adds 3 extra electrons, totaling 32 valence electrons. The Lewis structure shows P at the center with four single P-O bonds and a double bond to one oxygen, and three oxygen atoms bearing a negative charge. The structure has four lone pairs on the terminal oxygen atoms. The ion adopts a tetrahedral geometry with bond angles of approximately 109.5°. This structure is key to understanding the ion’s role in biological systems and its reactivity in forming salts and esters.

How to Draw PO4 3- Lewis Structure

Drawing the Lewis structure of a molecule or ion helps us understand its chemical bonding and molecular geometry. In this section, we will explore the step-by-step process of drawing the Lewis structure for the phosphate ion (PO4 3-). Let’s dive in!

A. Counting the Total Number of Valence Electrons

The first step in drawing the Lewis structure of PO4 3- is to count the total number of valence electrons present in the molecule. Valence electrons are the outermost electrons of an atom and are involved in chemical bonding.

To count the valence electrons in PO4 3-, we need to consider the valence electrons of each atom in the ion. Phosphorus (P) is in Group 5A, so it has 5 valence electrons. Oxygen (O) is in Group 6A, so each oxygen atom has 6 valence electrons. Since there are 4 oxygen atoms in PO4 3-, the total number of valence electrons is:

(1 × 5) + (4 × 6) + 3 (negative charge) = 32 valence electrons

Total valence electron pair (PO4 3-) = sigma (σ) bonds + pi (π) bonds + valence shell lone pair electrons

To count total valence electron pair we have to divide total number of valence electrons by 2.

Hence, total valence electron pair for PO4 3- = 32/2 = 16

B. Determining the Central Atom

The next step is to determine the central atom in the PO4 3- ion. The central atom is usually the least electronegative atom, which is phosphorus in this case. Phosphorus can form multiple bonds and is capable of accommodating more than one atom around it.

PO4 3- structure showing Phosphorous at central position

C. Placing the Remaining Atoms

After identifying the central atom, we need to place the remaining atoms around it. In the case of PO4 3-, we have 4 oxygen atoms. Since oxygen needs 2 electrons to complete its octet, we can form a single bond between each oxygen atom and the central phosphorus atom. This will account for 8 of the 32 valence electrons.

PO4 3- structure showing bonding between P and O

D. Distributing the Remaining Electrons

Next, we distribute the remaining electrons to satisfy the octet rule for each atom. The octet rule states that atoms tend to gain, lose, or share electrons to achieve a stable configuration with 8 electrons in their outermost shell.

In the case of PO4 3-, we have 24 valence electrons remaining after placing the oxygen atoms. We distribute these electrons as lone pairs on the oxygen atoms. Each oxygen atom can accommodate 3 lone pairs, which accounts for 24 electrons.

E. Checking the Octet Rule

Finally, we check if all the atoms in the Lewis structure of PO4 3- have achieved an octet or a stable configuration. In this case, the central phosphorus atom has 10 electrons (6 from the single bonds and 4 from the lone pairs), which exceeds the octet. However, this is acceptable for elements beyond the second period in the periodic table.

The Lewis structure of PO4 3- is represented as follows:

PO4 3- lewis structure showing complete octet of P and O atoms

In conclusion, the step-by-step process of drawing the Lewis structure for the phosphate ion (PO4 3-) involves counting the total number of valence electrons, determining the central atom, placing the remaining atoms, distributing the remaining electrons, and checking the octet rule. By following these steps, we can gain insights into the chemical bonding and molecular geometry of the PO4 3- ion.

The Role of Formal Charge in PO4 3- Lewis Structure

Calculating the Formal Charge for PO4 3-

Let’s now focus on the specific case of the phosphate ion (PO4 3-). To calculate the formal charge for each atom in the PO4 3- ion, we need to consider the valence electrons and the number of electrons assigned to each atom.

The phosphate ion consists of one central phosphorus atom bonded to four oxygen atoms. Phosphorus, being in Group 5A of the periodic table, has five valence electrons. Each oxygen atom, being in Group 6A, has six valence electrons.

To calculate the formal charge for each atom, we use the formula:

Formal Charge = Valence Electrons – (Number of Lone Pair Electrons + 0.5 * Number of Bonded Electrons)

For the central phosphorus atom in PO4 3-, the formal charge can be calculated as follows:

Formal Charge of Phosphorus = 5 – (0 + 0.5 * 8) = 5 – 4 = +1

For each oxygen atom, the formal charge can be calculated as follows:

Formal Charge of Oxygen = 6 – (6 + 0.5 * 2) = 6 – 7 = -1

The Impact of Formal Charge on the Stability of PO4 3-

The formal charges in the PO4 3- ion help us understand its stability and the distribution of electrons within the molecule. In the Lewis structure of PO4 3-, the formal charges on each atom contribute to the overall charge of the ion, which is -3.

The formal charge of +1 on the central phosphorus atom indicates that it has one fewer electron than its neutral state. This positive charge is balanced by the three negative charges on the oxygen atoms, resulting in a net charge of -3 for the entire ion.

The stability of the PO4 3- ion is influenced by the formal charges on each atom. In general, a Lewis structure with formal charges closer to zero is more stable. In the case of PO4 3-, the formal charges of +1 on phosphorus and -1 on each oxygen atom are relatively close to zero, indicating a stable arrangement.

It’s important to note that formal charges are a simplified representation of electron distribution and do not necessarily reflect the true nature of electron density in a molecule. However, they provide a useful tool for predicting molecular stability and guiding the construction of Lewis structures.

In conclusion, formal charge is a valuable concept in understanding the stability and arrangement of atoms in a molecule. By calculating the formal charges in the PO4 3- ion, we can determine the most stable Lewis structure and gain insights into the distribution of electrons within the molecule.

PO4 3- lewis structure showing formal charges on P and O atoms

PO4 3- Lewis structure lone pairs

Sixteen total valence pair electrons are present on PO4 3- structure. Central P atom has four bonds linked to four O atoms in PO4 3- structure. So after four P-O bonds there are only 12 electron pairs get left in PO4 3- ion.

As we already see that oxygen comes under 2^nd period of the periodic table and it do not have more the eight electrons in its last outer shell orbital. So, let we mark lone pair electrons on four outer O atoms. Central P atom has no lone electron pair as it already has 4 bond pairs with 4 O atoms.

So, the outer four O atoms of PO4 3- lewis structure has total 12 lone electron pairs, three pairs on each single O atom.

PO4 3- Lewis structure shape

In PO4 3- lewis structure, the middle phosphorous atom has four bonds with four oxygen atoms. All the P and 4 O elements are settled in a possible geometric arrangement with each other that can they minimize repulsion within bond pairs of each other.

According to VSEPR theory, any structure or molecule accepts the geometry that minimizes repulsion. Means if there are no lone pair electrons available on any lewis structure then there is no repulsion between bond pairs. The generic formula under VSEPR theory is AX4 which is suitable for PO4 3- lewis structure.

AX4:- A = no. of central atoms

          X = no.of outer bonded atoms

Therefore, according to molecular geometry of PO4 3- ion, the shape of PO4 3- lewis structure is tetrahedral.

PO4 3- Hybridization

Any structures or molecules hybridization is based on its steric number. PO4 3- lewis structures hybridization can be determine by finding the steric number of its central atom phosphorous.

Steric number is the sum of total number of bonded atoms linked (attached) with central atom and lone pair of electrons present on it.

Steric number of PO4 3- = (no. of elements or atoms bonded with phosphorous + phosphorous atom lone electrons pair)

As we see the PO4 3- lewis structure, central phosphorous atom attached with four oxygen atom i.e. four bonds with four oxygen atoms and have no lone electron pairs on phosphorous atoms.

So, PO4 3- steric number = 4 + 0 = 4

In PO4 3- lewis structure, central P atom has a steric number four after calculation, so it has a Sp³ hybridization of phosphorous in PO4 3- ion.

The Resonance of PO4 3- Lewis Structure

The Resonance Structures of PO4 3-

Let’s take a closer look at the resonance structures of the phosphate ion (PO4 3-). The phosphate ion consists of a central phosphorus atom bonded to four oxygen atoms. The Lewis structure of PO4 3- can be represented by multiple resonance structures due to the delocalization of electrons.

In the first resonance structure, the central phosphorus atom forms single bonds with three oxygen atoms and a double bond with one oxygen atom. This arrangement satisfies the octet rule for all atoms involved.

In the second resonance structure, the double bond is shifted to a different oxygen atom. This results in a different distribution of electron density within the molecule.

The third resonance structure involves the double bond being shifted to yet another oxygen atom. Again, this leads to a different electron distribution.

The presence of multiple resonance structures indicates that the actual structure of the phosphate ion is a hybrid of these different arrangements. It is important to note that the actual structure does not oscillate between these resonance structures, but rather represents an average of their characteristics.

PO4 3-

Lewis structure showing two resonance structures

The Importance of Resonance in PO4 3- Lewis Structure

The concept of resonance is crucial in understanding the properties and behavior of the phosphate ion. The delocalization of electrons in the resonance structures of PO4 3- contributes to its stability and reactivity.

The presence of multiple resonance structures allows for the distribution of negative charge over the entire molecule, reducing the electron density on any one oxygen atom. This distribution of charge makes the phosphate ion more stable compared to a hypothetical structure with localized charges.

Resonance also influences the reactivity of the phosphate ion. The delocalization of electrons makes it easier for the phosphate ion to participate in chemical reactions. For example, it can readily form covalent bonds with other atoms or ions, contributing to the formation of various phosphate-containing compounds.

In summary, the resonance structures of the phosphate ion (PO4 3-) play a significant role in determining its stability and reactivity. By considering the different resonance structures, chemists can gain insights into the electronic structure and behavior of this important polyatomic ion.

PO4 3- polar or nonpolar

The main reason behind polarity occurs in any molecule or ion is its electronegativity difference. There is a non-zero dipole moment and have asymmetrical shape in polar molecules.

The electronegativity of P atom is 2.19 and electronegativity of O atom is 3.44 and its difference is 1.25 which is greater than Pauling’s rule value 0.5 making polar P-O bond.

But if we draw a dipole moments in PO4 3- lewis structure for each bond. As the PO4 3- molecule has a symmetrical structure, due to which the dipole moments cancel one another making PO4 3- a non-polar molecule.

For the conformation of non-polar nature of PO4 3- ion, we can see its molecular geometry as it has a tetrahedral shape and its each dipole moment goes in opposite direction and has the zero net dipole moment which confirms the non-polar nature of PO4 3- molecule.

PO4 3- Lewis structure bond angle

The molecular geometry of PO4 3- ion shows tetrahedral shape with VSEPR generic formula AX4, it shows that in the PO4 3- lewis structure the central P atom attached to outer four O atoms.

So, the PO4 3- Lewis structure has the 109.5 degree bond angle.

The Real-world Applications of PO4 3- Lewis Structure

A. PO4 3- in Biological Systems

Phosphate ions (PO4 3-) play a crucial role in various biological systems, serving as a fundamental component of DNA, RNA, and ATP (adenosine triphosphate). These molecules are essential for the storage and transfer of genetic information, as well as for energy metabolism within cells. The Lewis structure of PO4 3- helps us understand the chemical bonding and molecular geometry of phosphate ions in biological processes.

In DNA and RNA, phosphate groups form the backbone of the nucleic acid chains. The Lewis dot structure of PO4 3- reveals that the central phosphorus atom is bonded to four oxygen atoms, with three of them carrying a negative charge. This arrangement allows for the formation of strong covalent bonds between the phosphorus atom and the oxygen atoms, providing stability to the DNA and RNA molecules.

Furthermore, the Lewis structure of PO4 3- shows that the phosphorus atom has a lone pair of electrons, which can participate in hydrogen bonding with other molecules. This interaction is vital for the proper functioning of DNA and RNA, as it allows for the recognition and binding of complementary nucleotide sequences during processes like replication and transcription.

B. PO4 3- in Environmental Chemistry

Phosphate ions (PO4 3-) also have significant implications in environmental chemistry, particularly in the context of nutrient cycling and water quality. The presence of excessive phosphates in water bodies can lead to eutrophication, a process characterized by the excessive growth of algae and other aquatic plants. This overgrowth depletes oxygen levels in the water, leading to the death of fish and other organisms.

Understanding the Lewis structure of PO4 3- helps us comprehend the behavior of phosphate ions in aquatic environments. The negative charge on the phosphate ion attracts positively charged ions like calcium and magnesium, resulting in the formation of insoluble compounds such as calcium phosphate and magnesium phosphate. These compounds can precipitate out of the water, reducing the availability of phosphates for algal growth and mitigating the effects of eutrophication.

Moreover, the Lewis dot structure of PO4 3- also aids in the design and development of phosphate removal technologies. By understanding the chemical bonding and molecular geometry of phosphate ions, scientists can develop effective methods to remove phosphates from wastewater and agricultural runoff, thereby minimizing the impact of eutrophication on aquatic ecosystems.

C. PO4 3- in Industrial Applications

The Lewis structure of PO4 3- finds applications in various industrial processes, particularly in the production of fertilizers and detergents. Phosphate compounds are essential components of fertilizers, providing the necessary nutrients for plant growth. The understanding of the chemical bonding and molecular geometry of phosphate ions helps in the formulation of effective fertilizers that deliver phosphates in a readily available form for plants.

In the manufacturing of detergents, phosphate compounds are used as water softeners and builders. The Lewis dot structure of PO4 3- reveals that the negative charge on the phosphate ion allows it to bind with calcium and magnesium ions present in hard water, preventing the formation of insoluble soap scum. This property makes phosphate compounds valuable in detergents, as they enhance the cleaning efficiency and maintain the quality of the washed fabrics.

Additionally, the Lewis structure of PO4 3- is also relevant in the field of battery technology. Phosphate-based materials, such as lithium iron phosphate (LiFePO4), are used as cathodes in lithium-ion batteries due to their stability and high energy density. The understanding of the chemical bonding and molecular geometry of phosphate ions aids in the development of efficient battery materials, contributing to the advancement of renewable energy storage systems.

The Lewis structure of PO4 3- has diverse real-world applications in biological systems, environmental chemistry, and industrial processes. Understanding the chemical bonding and molecular geometry of phosphate ions enables us to comprehend their role in DNA, RNA, nutrient cycling, water quality, fertilizers, detergents, and battery technology. This knowledge facilitates the development of innovative solutions and technologies that benefit various sectors and contribute to sustainable development.

Frequently Asked Questions

Q1: What is the Lewis dot structure of a PO4 3- molecule?

The Lewis dot structure of a PO4 3- molecule, also known as a phosphate ion, involves one phosphorus atom at the center surrounded by four oxygen atoms. The phosphorus atom shares covalent bonds with each oxygen atom, and there are double bonds with one oxygen atom. The remaining three oxygen atoms carry a negative charge each, accounting for the 3- charge of the ion.

Q2: What is the significance of valence electrons in drawing Lewis structures?

Valence electrons play a crucial role in drawing Lewis structures. They are the electrons in the outermost shell of an atom and are involved in forming covalent bonds. The number of valence electrons determines how many bonds an atom can form and its placement in the Lewis structure.

Q3: What is the molecular geometry of the PO4 3- molecule?

The molecular geometry of the PO4 3- molecule is tetrahedral. This is because the phosphorus atom is surrounded by four electron pairs, three of which are bonded to oxygen atoms and one is a lone pair.

Q4: How does the octet rule apply to the PO4 3- molecule?

The octet rule states that atoms tend to form bonds in such a way that each atom has eight electrons in its valence shell. In the case of the PO4 3- molecule, the phosphorus atom shares its five valence electrons with four oxygen atoms to achieve a stable electron configuration.

Q5: What are resonance structures and how do they apply to the PO4 3- molecule?

Resonance structures are different forms of a molecule where the chemical connectivity is the same but the electrons are distributed differently around the structure. In the case of the PO4 3- molecule, there are four possible resonance structures, each with the double bond located between the phosphorus atom and a different oxygen atom.

Q6: What is the role of covalent bonds in the PO4 3- molecule?

Covalent bonds are formed when atoms share electrons. In the PO4 3- molecule, the phosphorus atom forms covalent bonds with the oxygen atoms, sharing electrons to achieve a stable electron configuration.

Q7: How does the negative charge in the PO4 3- molecule affect its structural representation?

The negative charge in the PO4 3- molecule is due to the three extra electrons from the oxygen atoms. This charge is often represented in Lewis structures as three lone pairs, one on each of the three oxygen atoms.

Q8: How does chemical notation help in understanding the PO4 3- molecule?

Chemical notation, like the Lewis dot structure, provides a visual representation of the molecule. It shows how the atoms are connected, the type of bonds between them, and the presence of any charge or lone pairs. This helps in understanding the molecule’s structure, bonding, and properties.

Q9: Why are polyatomic ions like PO4 3- important in chemistry education?

Polyatomic ions like PO4 3- are important in chemistry education because they introduce concepts like covalent bonding, molecular geometry, and resonance structures. They also demonstrate how atoms achieve stability through the sharing of electrons and the formation of ions.

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OH- ion is a diatomic anion known as hydroxide ion. It is basic in nature. It is also known as hydroxyl group while determining functional group of any structure.

Hydroxide (OH-) ion is occurs naturally in water. It is basically electrically neutral molecule. It can ionizes itself to get H+ ions (cations) which is a major constituent in acid formation. This ion can produce salts some of which can dissociates in aqueous solutions. Here, in this editorial we are learning about OH- lewis structure and its detailed facts.

How to draw lewis structure for OH-?

Following are the points mentioned to draw the lewis structure of any molecule:

1. Counting of total valence electrons available on structure.
2. Central position is occupied by the element with lowest electronegativity.
3. Make connection of bonding within all elements present in structure.

Chemical formula for hydroxide ion is OH-.

Molecular weight of OH- is 17.007 g mol^-1.

Molecular geometry of OH- is linear in shape.

OH- has sp hybridization.

OH- is polar in nature.

In OH- ion, only two elements involved i.e. it is diatomic containing one oxygen and one hydrogen atoms. There is a single covalent bond between the O and H atoms. Oxygen atom carrying -1 charge on it which shows the generally as whole charge on OH- ion. In OH- lewis structure there are lone electron pairs present on O atom.

• Counting of total valence electrons available on OH- ion structure.

Now we will calculate the total valence electrons present on hydroxide (OH-) ion. OH- ion has only two elements hydrogen atom and oxygen atom. Hydrogen atom belongs to the 1^st group of periodic table and oxygen atom belongs to the 16^th group of periodic table. Therefore, hydrogen atom contains one valence electron and oxygen atom contains six valence electrons.

Also during valence electrons calculation of OH- ion we have to add extra one electron for the minus (-) charge on OH- group.

Therefore, Oxygen atom valence electrons = 6 x 1 (O) = 6

                  Hydrogen atom valence electrons = 1 x 1 (H) = 1

      Add one more electron for negative charge present on OH-

Hence, OH- ion has total valence electrons = 6 (O) + 1 (H) + 1 (-) = 8

So, OH- lewis structure has eight total valence electrons.

Now, as we have total 8 valence electrons on OH- lewis structure, so we have to calculate the total electron pairs. For that we have to divide it by 2 we get.

Total electron pairs = Total valence electrons on OH- ion / 2 = 8 / 2 = 4

Therefore, we have to otal four electron pairs.

• Central position is occupied by the element with lowest electronegativity.

This point is not applied in case of OH- lewis structure because it has only two atoms present and they cannot take central position of structure. Even the central atom should attached with at least two more atoms and there is only single bond is present within oxygen and hydrogen atom.

So, we can place oxygen and hydrogen atom beside each other in OH- lewis structure.

• Make connection of bonding within all elements present in structure.

Now we have to make a link between the two elements present in OH- lewis structure by making a bond between them as shown in the following image.

OH- lewis structure octet rule

In the above OH- lewis structure we already make a single covalent bond between O and H atom. So, two valence electrons get involved in bonding between O and H atom out of total eight valence electrons. Now, we are left with only six valence electrons for electron distribution to complete the octet of O atom.

As we already know that hydrogen atom has the capacity to have only two electrons in its valence shell. So, hydrogen atom already filled its valence orbital with two bonding electrons of O-H single covalent bond, as single bond consists of two valence electrons.

Now, we have to distribute all the remaining six valence electrons i.e. three electron pairs with oxygen atom only. So we get the following structure.

From the above structure we can see that the hydrogen atom has two bonding electrons means its valency is satisfied and the oxygen atom has total eight electrons means it has a complete octet with three lone pairs and one bond pair electrons.

OH- lewis structure formal charges

Any lewis structure is more stable if it has a small formal charge on it. To calculate the formal charge on any molecule or structure there is a formula as follows:

Formal charge = (valence electrons – non-bonding electrons – ½ bonding electrons)

Now we have to calculate the formal charge for hydroxide ion i.e. OH- lewis structure. So first let us calculate formal charge on hydrogen atom.

Hydrogen atom: Valence electrons on Hydrogen atom = 01

                           Non-bonding electrons on Hydrogen atom = 00

                           Bonding electrons with Hydrogen atom =02

Formal charge on Hydrogen atom = (1 – 20– 2/2) = 00

Thus, hydrogen atom has zero formal charge on it.

Oxygen atom:Valence electron on oxygen atom = 06

                        Non-bonding electron pair on oxygen atom = 06

                        Bonding electrons on oxygen atom =2

Formal charge on oxygen = (06 – 06 – 2/2) = -1

Hence, the oxygen atom of OH- ion has -1 formal charge

So, in OH- lewis structure hydrogen has zero formal charge and oxygen has -1 formal charge.

OH- lewis structure lone pairs

As we see in the above structure, there is a one bond between hydrogen and oxygen atoms containing two valence electrons i.e. they become bond pair. The remaining electron pairs which then goes on oxygen atoms to complete its octet. As the hydrogen atom has two electrons in its bond pair and its valence shell is gets filled as it cannot accept more than two electrons.

So, all the remaining three electrons pairs will goes to oxygen atom. Thus the oxygen atom of OH- lewis structure has total three lone electron pairs.

OH- lewis structure shape

Any lewis structure can be determined by the help of representations of VSEPR theory if any molecular diagram has more than two atoms in it. Here, in OH- lewis structure involvement of two elements are there i.e. oxygen and hydrogen. OH- ions follow the VSEPR theory generic formula AXE.

So, according to VSEPR theory geometry of ions the OH- shows linear shape. In OH- lewis structure, there is both the valence shell of O and H atoms get filled with sufficient electrons as per its capacity. Hydrogen has two electrons and oxygen atom has six electrons but if we count bond pairs with O atom so it has complete octet with eight electrons. So, after bonding there is three lone pair electrons remain on O atom.

So, there is no repulsion between any bond pair and lone pair of electrons in OH- lewis structure due to the presence of single covalent bond between O and H atoms. Therefore, OH- lewis structure has linear shape.

OH- Hybridization

Until now we have discussed about OH- lewis structure bonding, lone pairs, molecular geometry, shape and formal charges. Now, let us discuss about its hybridization.

If we look upon the OH- lewis structure we have discussed yet, we can see that there is no double or triple bond present so there is no pi-bond in OH- ion. It has only one single covalent bond within H and O atoms i.e. presence of only sigma (σ) bond within it and there are three lone electron pairs are present on one oxygen atom.

Any structure or molecules hybridization is depend upon its steric number of central atom. The OH- hybridization is depend on both atoms O and H atoms steric number as there is no central atom present.

Steric number is the sum of total no. of bonded atoms associated with central element and its lone electron pair present on it.

Steric number of OH- = (no. of bonded elements or atoms attached central element + central atom having lone pair of electrons)

As there is no middle element in OH- lewis structure, there is only one bonded atom and three lone electron pairs are present.

So, the Steric number for OH- ion = 1 (bonded H atom) + 3 (lone pair electron of O)

Therefore, the oxygen atom shows sp3 hybridization in OH- lewis structure. But if we see the shape of OH- ion which has linear shape than it should have sp hybridization according to its shape. Basically hybridization of OH- ion is depends upon the molecules attached to it for ex. CH₃-OH, H-OH, etc.

Means if the geometry of structure or molecule is linear then OH- is sp hybridised, if trigonal planar then it is sp2 hybridised, if it is tetrahedral then it is sp3 hybridised, and so on depends upon the molecule containing OH- ion.

OH- lewis structure resonance

The resonance structure of any molecule shows the variation of distribution of electrons in the same molecule. There are some rules to draw the resonance structure of any molecule is that, the molecule should contain multiple bonds in its structure like double or triple bonds and also the connected atoms have at least one electron lone pair. If the molecule accomplished all these conditions then it has a possible resonance structure.

In case of OH- lewis structure, the oxygen atom contain three lone pairs with one negative charge on it and linked with hydrogen with a single covalent bond, which shows the more stable form of OH- lewis structure. But generally while writing the chemical formula for hydroxide ion the negative charge gets shown on whole molecule.

The negative charge on oxygen atom is due the completion of octet (presence of eight) electrons. So, the OH- ion does not show any typical resonance structure as there is no multiple bond and do not obey the resonance structure rules. But we can draw the comparative structure to show the charge transformation.

OH- polar or nonpolar

Now, let we learn about the non-polar / polar nature of OH- ion. The electronegativity of oxygen atom is 3.44 and the electronegativity of hydrogen atom is 2.2. there is a huge difference in electronegativity of oxygen and hydrogen molecules.

As we talk about the polarity of OH- ion, the OH- ion shows polar nature mostly if present in any molecule. If we see the example of methanol and water, in which the OH- ions are present in its structure. In both methanol and water it shows polar nature as the oxygen atom is more electronegative than hydrogen atom.

Therefore, due to more electronegativity of O atom, the shared electrons makes crowd on the O atom, which results in the creation of partial negative charge on O atom and partial positive charge on H atom. Means the whole electron density attracts by O atom towards itself and the polarity creates on O-H bond.

Thus the OH- ion or lewis structure shows the polar nature.

OH- lewis structure bond angle

As we already discussed earlier about the shape and hybridization of OH- lewis structure. The hybridization and shape of OH- ion depends upon the nature of compound or molecule in which the OH- ion is present.

As per the shape of OH- ion discussed above is basically linear. So, its bond angle should have to be nearly 180 degree. But the bond angle of Oh- ion is actually depends upon the structure and orientation of that particular molecule in which the OH- ion is present.

OH- lewis structure electron geometry

The electron geometry of OH- ion shows the presence of eight electrons in it. All the eight electrons is surrounded to oxygen atom means all the electron density cover to oxygen atom. Eight valence electrons it means four electron pairs. Thus, in OH- ion there is four electron densities surrounding O atom.

So, we consider the tetrahedral shape for electron geometry of OH- ion. But still it depends on electron geometry of the molecule in which the OH- is present.

OH- Uses

• Caustic soda which is the form of sodium hydroxide are used in paper and pulp industry, textile industry, potable drinking water, soap bars and detergents and as a septic and wash basin drain cleaner.
• In fuel cells hydroxides are used.
• Hydroxide ions are used in disinfectants.
• In agriculture potassium hydroxide is used.
• Limonite and goethite which are low grade brown iron ore is produced by the use of iron hydroxides.
• Hydroxides are used as food preservatives to protect the food from various bacteria’s and moulds.
• In extractation of alumina hydroxide is used.
• Aluminium hydroxides are used in aluminium ore bauxite.

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In this article Mg²⁺ ion , its lewis structure, shape, angle, and other facts are discussed.

Magnesium ion or Mg²⁺ is a magnesium cation, divalent metal cation and a monoatomic dication. It find various application in several fields especially in medical field.

how to draw MG2+ lewis structure

The Lewis structure of a molecule helps to understand the electronic distribution in a bond formation of a  molecule. To draw the  lewis structure of a molecule, valence electrons are considered. All other electrons are well protected from chemical bond formation. The electrons denoted as dots  in  lewis structure  called lewis symbols.

Mg²⁺ or Magnesium ion is formed by the loss of two electrons from the Mg atom. The electronic configuration of  Mg  can be shown as 1s² 2s² 2p⁶ 3s² and that of  Mg²⁺ is  found to be 1s² 2s² 2p⁶. The electron distribution among various shells of Mg and Mg²⁺ is given below.

	K	L	M	N
Mg	2	8	2
Mg2+	2	8

In Mg²⁺ only 10 electrons are present in K and L shell respectively. So the lewis structure of Mg2+ ion will be the following one.

MG2+ lewis structure resonance

Resonance structure is two or more way of representation of  lewis structure of the given molecule. Since Mg²⁺ is an ion it has only one resonance structure and it is same as its lewis structure.

MG2+ lewis structure shape

Mg2+ has no shape eventhough its structure is same as that of its lewis structure.

MG2+ lewis structure formal charge

Formal charge is the imaginary charge allotted to an atom  when all its valence electrons are perfectly shared in a chemical bond.

Formal charge of a molecule can be found out by an equation

Formal charge of an atom = No. of valence electrons – No. of dots or electrons –  No.of  bonds  formed.

Mg²⁺  is not a molecule and its an ion. It doesn’t take part in any chemical bond. So the Mg²⁺ ion doesn’t have any formal charge but there is  a net charge present to this ion.

MG2+ lewis structure angle

Bond angle is the angle formed between the 3 bonds when different or same atoms combine together to form a compound. Here in the case of Mg²⁺there is no bond formation with other atom so the concept of bond angle cannot seen here. Therefore Mg²⁺ ion doesn’t have any bond angle.

MG2+ lewis structure octet rule

According to Octet Rule, the number of valence electrons or the electrons present on the outermost shell must be 8. If there are 8 electrons are present among the outermost shell then its octet is fulfilled.

In the case of Mg²⁺ ion its octet is fulfilled. Because the when two electrons from Mg is lost then the outer most shell of Mg²⁺ is L shell and it has 8 electrons present. So Mg²⁺ ion obeys the octet rule.

MG2+ lewis structure lone pairs

Lone pair of electrons are the electrons that doesn’t took part in any chemical reaction. The lone pair of electron in any molecule is given as .

Lone pair of electron in any atom =  (Valence electrons – No. of electrons shared by the atom) /2

 There is no lone pair of electron in Magnesium ion.

MG2+ valence electrons

Valence electrons are the electrons present in the outermost shell of any atom or ion that can take part in any chemical reaction. A molecule or an atom attain valency by losing or gaining  electrons from its outer most shell to attain noble gas configuration.

The electronic configuration of Mg²⁺  ( 2, 8). Since Mg²⁺ is formed by the loss of 2 electrons from Mg. The number of valence electrons present to Mg²⁺ is zero and the total number of electrons present to Mg²⁺ ion will be 10.

MG2+ hybridization

Hybridisation is a concept that actually we apply among covalent bonded compounds. Since Mg²⁺ is an ionic compound. So the concept of hybridisation doesn’t work here.

MG2+ uses

Magnesium ions are required by several enzymes to perform their function.Magnesium ions  are vital  for the transfer, storage, and utilisation of energy in human metabolic activities. Mg²⁺ ion regulates and catalyses most of the enzymatic functions in mammals.

The energy storage of our cell ATP need Mg2+ ion to combine with it to perform its biological function. Magnesium containing compounds are also found importance in laxatives, antacids. Magnesium compounds are also used for certain health treatment like stabilising abnormal nerve excitation.

Is MG2+ ionic or covalent

Mg²⁺ is an ionic compound. Magnesium ion or Mg²⁺ is a positively charged cation.

Is mg2 stable?

Mg²⁺ is a stable compound. Magnesium ion is formed by the loss of two electrons from the Mg. Then its electronic configuration is ( 2, 8). Since both the shells are completely filled and obeying octet rule Mg²⁺ ion is a stable ion.

Read more about Americium Electron Configuration.

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In this article, “icl2- lewis structure”, different facts like hybridization, shape, formal charge calculation, stability of ICl₂^– with some detailed explanations are discussed thoroughly.

ICl₂^– is an interhalogen compound in which iodine is connected with two chlorine atoms by two covalent bonds. In it, iodine is sp³d hybridized with a linear structure. It has two bond pair and three lone pair and angle between two I-Cl is 180⁰. Due to its linear shape, it is a nonpolar molecule.

Let’s focus on the following relevant topics on ICl₂^–.

How to draw ICl₂^– lewis structure?

Lewis structure is one of the structural representation of any molecule in which nonbonding electrons are shown around each of the atom and the bonds between the atoms.

The process of drawing the lewis structure is described below-

1. Finding out the valance electrons: The first step of lewis structure drawing is to determine the valance shell electron. In ICl₂^–, both the halogen atom, iodine and chlorine have seven electrons in their respective outer most shell. Iodine is in -1 oxidation state. Thus, it has eight electrons in its valance shell.
2. Determine the bonding electrons: ICl₂^– has total two covalent bonds in it. Thus, (2×2=4) electrons are involved in two covalent bond formation
3. Determine the nonbonding electrons: In ICl₂^–, total three lone pairs and two bond pairs are present between iodine and two chlorine atoms. Thus, iodine has (3×2=6) nonbonding electrons and each of the chlorine atom has six electrons remain as nonbonding.

Total nonbonding electrons in ICl₂^– is { 6 + (6×2) } = 18

ICl₂^– Lewis Structure Shape

The factor “hybridization” plays the most significant role in structure determining. Molecular shape is changed with the change of central atom hybridization (shown below).

Hybridization of central atom	Structure
sp	Linear
sp2	Trigonal planar
sp3	Tetrahedral
sp3d	Trigonal bipyramidal
sp3d2	Octahedral

 Not only hybridization, some times lone pair bond pair repulsion also determines the structure of any molecule. Three types of repulsion are responsible in the structure determination-

• Lone pair- lone pair repulsion
• Lone pair-bond pair repulsion
• Bond pair-bond pair repulsion

The increasing order of the above repulsion is-

Lone pair -lone pair repulsion > Lone pair – bond pair repulsion > Bond pair- bond pair repulsion.

In ICl₂^–, iodine has three lone pairs. These lone pairs face repulsion from each other and from the bonding electrons also. Iodine is sp³d hybridized in ICl₂^–. Thus, its actual geometrical structure should be trigonal bipyramidal or TBP. But due to presence of these lone pair, the actual structure is distorted and it becomes linear.

In TBP, there is two types of position for the substituent atoms. One is axial position and another one is equatorial position. Three lone pairs are placed in the three equatorial position and two Cl atoms are placed in the two axial position of TBP structure according to the VSEPR theory due to minimize the lone pair bond pair repulsion. Placing the two Cl atoms in the axial position make the molecule, ICl₂^– a linear shaped and bond angle between two I-Cl bonds becomes 180⁰.

ICl₂^– Lewis Structure Formal Charge

Formal charge of any molecule is calculated to check if the electrons in all the chemical bonds are shared equally to all over the atoms or not. Calculation of formal charge also helps to find out the most stable lewis structure of any 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 iodine = 7 – 6 – (4/2) = -1
• Formal charge of each of the chlorine atom = 7 – 6 – (2/2) = 0

Iodine is in -1 oxidation state in ICl₃^–. Thus, it contains eight electrons in its valance shell and each of the two chlorine atoms has seven electrons in their valance shell. Iodine is connected by two covalent bonds with two chlorine atoms. So, 4 electrons (2×2) and 2 electrons (2×1) are involved in bond formation for iodine and chlorine atom respectively.

ICl₂^– Lewis Structure Angle

Bond angle is nothing but the angle between two covalent bonds. Hybridization also helps to determine the bond angle in a molecule. Hybridization is responsible to change the bond angle in any molecule.

 Three lone pair and two lone pair are present in ICl₂^– molecule. Due to presence of two lone pair of iodine, two chlorine atoms are placed in the two axial position of TBP structure. The three atom, iodine and two chlorine are oriented in a straight line keeping the three lone pair in three equatorial position.

Due to having linear structure the angle between two I-Cl bond is 180⁰ and the angle between one lone pair with a bond pair is 90⁰ (right angle).

ICl₂^– Lewis Structure Octet Rule

Octet rule is one of the most significant rule in chemistry as it says that any atom should have eight electrons in its valance shell to match the electron configuration with its nearest noble gas. This special electron configuration gives any atom an extra stability.

Octet rule is not satisfied in ICl₂^– molecule. Three lone pairs and two bond pairs are involved in this molecular species. Thus, it has ten electrons in its valance shell (including nonbonding and bonding electrons). This number of electrons do not resemble with the nearest noble gas Xenon or Xe (5s² 5p⁶).

But each of the chlorine atom  obeys octet rule in ICl₂^– molecule. The chlorine atom has three pair of nonbonding electrons and one of the valance electron of it is involved in covalent bond formation with iodine atom. So, it gains total eight electrons in its valance shell that matches with the electron configuration of nearest noble gas Argon or Ar (3s² 3p⁶).

ICl₂^– Lewis Structure Lone Pairs

Lone pairs or nonbonding electrons are two similar words. Both of these words are carrying same meaning in chemistry. Lone pairs are those valance electrons who are not responsible for bond formation. Basically, nonbonding electrons or lone pairs are those valance electrons left after bond formation.

• Nonbonded electron = Total number of valance electron – number of bonded electrons.
• Nonbonding electrons on Iodine (I) = 8 – 2 = 6 or 3 lone pair
• Nonbonding electrons on each of the chlorine atoms (Cl) = 7 – 1 = 6 or three lone pairs.

Being halogen atom, both the iodine and chlorine have same number of electrons in their valance shell (ns² np⁵). But iodine is negatively charged in ICl₂^– and I^– has eight electrons in its valance shell.

Thus, total number of nonbonding electrons in ICl₂^– is { 6 + (2×6)} = 18 or 9 lone pairs.

ICl₂^– Valence Electrons

Valance electrons are those who are revolving around the nucleus from the outer most shell of any atom. Atoms take part in any reaction just because of these valance electrons due to the lesser attraction of nucleus on them. Inner shell electrons are very strongly attracted by the nucleus. Thus, they can’t participate in any reaction.

As iodine and halogen are two halogen compounds, they have same electrons in their valance shell. Iodine and chlorine have electron configuration in their valance shell are 5s2 5p5 and 3s2 3p5. Due to achieving -1 oxidation state, I^– has eight electrons in its outer most shell.

Thus, total number of valance electrons in ICl₂^– are {8 + (7×2)} = 22.

ICl₂^– Hybridization

Hybridization is introduced in chemistry as the mixing of two atomic orbitals having similar energies, size and symmetry. There are five basic types of hybridization. Thy are-

1.  Planar (sp)
2. Trigonal Planar (sp²)
3. Tetrahedral (sp³)
4. Trigonal bipyramidal (sp³d²)
5. Octahedral (sp³d²)

In ICl₂^–, central atom, iodine is sp³d hybridized having three lone pair and two bond pair. One s orbital, three p orbital and one d orbital of iodine are involved in this sp³d hybridization. Each of the chlorine atom shares their only one valance electron with iodine in this hybridization.

Sp³d hybridization directs a molecule to be TBP structured but due to presence of three lone pair, its actual geometrical structure is violated and becomes linear.

Is ICL₂^– ionic or covalent?

The main difference between a covalent and ionic compound is that electrons are shared in covalent compound but they are completely donated in ionic compound from electropositive element to electronegative element.

In ICl₂^–, electrons are shared between iodine and two chlorine atoms. Not only that it has one of the most important features of covalent compound that iodine and chlorine both are nonmetal compounds. The electronegativity difference between iodine and chlorine is also not high (The electronegativity of iodine and chlorine are 2.66 and 3.16 respectively). But for being an ionic compound this difference must be high and the participating atoms must be metal and nonmetal combination.

Thus, ICl₂^– is a covalent compound.

Is ICL₂^– stable?

ICl₂^– is moderately stable compound. It is less stable than any di halogen compound like I₂ or Cl₂ due to comparatively poor overlap of atomic orbitals.

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This article contains the NH4+ lewis structure and its hybridization, shape, bond angle, and detailed explanations. In NH4+ lewis structure, the molecule is sp³ hybridized and has a bond angle of 109.5⁰, so the molecule’s shape is tetrahedral. This structure has no double bond, so the molecule could not show a different resonance structure. There is a shortage of electrons over N in this molecule. So, the positive charge is on N only.

Some facts about NH₄⁺

NH₄⁺ is known as ammonium ion. It is a cationic form of ammonia. The molar mass of ammonium ion is 18.039 g/mol. The ammonium ion is a white and soluble salt. Ammonium ion behaves as a weak acid. The pka value is 9.25. The conjugate base is ammonia.

 The protonation of ammonia synthesizes ammonium ions.

During the metabolism of animals, ammonium ions produce as a by-product. Ammonium ions behave similarly to alkali metals.

Method of drawing the Lewis structure for NH₄⁺

We always count the total valence electron for every molecule to draw its lewis structure.

In NH4+ lewis structure, there are one N, and four H atoms are present. The central atom is decided by electronegativity, but four H atoms surround one N atom. So, the central atom is N here. All the H atoms make a single bond with the N atom. There is a shortage of electrons over N, so the positive charge of the molecule resides over N only.

The octet of H is completed by forming a single bond with N. Now. We added the valence electrons of N and four H atoms. The ammonium ion is cationic, so the equal number of the electron that is one electron should be removed from this molecule. After making the single bond, there is no lone pair in this molecule. A positive charge is put on N to complete its octet, so, no need to put a double or triple bond here.

N has five electrons in its outermost orbital. But here, only four electrons make a four sigma bond with four each H atom. So a positive charge resides on the N atom. Each H atom has one electron, and they are for a single bond with a single electron of the N atom.

NH₄⁺ lewis structure shape

In NH4+ lewis structure ,four H atoms surround the central N atom. The shape of this molecule is tetrahedral. Four H atoms are present in four corners of the tetrahedral site. 

There are four H atoms present so that the structure can be a square planner, but in square planner geometry, the bond angle of the molecule will be 90⁰. In tetrahedral geometry, the bond angle is 109.5⁰. Increasing the bond angle will decrease repulsion as the N atom is small. The shape of the molecule also indicates the hybridization.

Again, from VSEPR theory, the total electron count for this molecule is (5+4-1) =8, so the molecule’s structure having an electron count of 8 is tetrahedral.

NH₄⁺ valence electron

NH4+ lewis structure the distribution of electrons in N is 1s22s22p3, and  H atom is 1s1. So, there are five valence electrons for N and one for H atom. But there is a positive charge in this ammonium ion, which means an electron shortage here. N donates its one electron. The valence electrons of N are four and for four H atoms, the total valence electrons are 4. So total valence electron of ammonium ion is 4+ (1*4) =8.

NH₄⁺ lewis structure formal charge

The following formula can calculate the formal charge of any molecule, 

F.C. = N_v – N_l.p. -1/2 N_b.p.

N_v = number of electrons in the valence shell of the free atom

N_l.p = number of electrons in lone pair

N_b.p = number of electrons involved in the bond formation.

F.C. of H atom = 1-0-2/2 =0

F.C. of N atom = 5-0-8/2 =1

NH4+ lewis structure as the molecule is ionic species, there will be a formal charge on the molecule equal to the number of positive charges, which is 1.

Generally, the formal charge is equal to its charge for ionic species.

Number of lone pairs in an NH₄⁺ lewis structure

NH4+ lewis structure after the bond formation, if there are pairs of electrons in a single shell, it is considered a lone pair for an individual atom.

Among five valence electrons of N, four of them form a bond with four H atoms in this molecule. H atom contains only one electron. So, there is no chance that H contains lone pair. One electron of N in the valence shell is donated. So, the molecule is lack of lone pairs.

Hybridization of NH₄⁺  

Hybridization is mixing atomic orbitals of similar and different energy to give new hybrid orbitals.

NH4+ lewis structure for hybridization of NH4+, we should count the valence electron of the central atom that is N. In the ground state, the distribution of electron of N is 1s²2s²2p³. 

In an excited state, it donates one electron from its p orbital, and one electron fromthe s orbital will shift to this vacant p orbital. This way it has four single unpaired electrons in its valence shell. These four unpaired electrons, one from s and one from p orbital under hybridization with four single electrons of four H atoms. 

After mixing these orbitals, a new hybrid orbital sp³ is formed. So, the hybridization of this molecule is sp³.

The molecule is sp3 hybridized, so the molecule’s structure is tetrahedral. Here only the sigma bond is considered in hybridization.

The central N atom is single-bonded with surrounded by four H atoms only.

Structure	state of hybridization of central atom	Bond angle
Linear	sp /sd / pd	1800
Planner trigonal	sp2	1200
tetrahedral	sd3/ sp3	109.50
Trigonal bipyramidal	sp3d/dsp3	900 (axial), 1200(equatorial)
Octahedral	sp3d2/ d2sp3	900
pentagonal bipyramidal	sp3d3/d3sp3	900,720

Again, from the table, we can see that the tetrahedral shape molecule will be sp³ hybridized. So, ammonium ion is sp³ hybridized.

NH₄⁺ bond angle

As the central N atom is sp3 hybridized in ammonium ion, the H-N-H bond angle is expected to be 109.5⁰.

In NH4 lewis structure there is no deviation from this original data cause there are no lone pairs present in this molecule, so there is no chance of lone pair-bond pair repulsion. Besides, N is small in size, and four H atoms are attached with a single bond. Due to the larger bond angle, there is no repulsion.

The VSEPR theory and hybridization table confirm that the bond angle in ammonium ion is 109.5⁰.

NH₄⁺ octet rule

The octet rule is to complete the valence shell by donating or accepting electrons and gaining the nearest noble gas configuration.

NH4+ lewis structure H has one electron. It forms a covalent bond with N by sharing one electron and completing its octet and gaining the electronic configuration as He. N has five electrons in its valence orbital. In ammonium, ion N forms four sigma bonds with four H atoms by sharing one electron. But its octet is not fulfilled. To satisfy its octet, one electron is required. So, a positive charge resides on the N atom only. This way, N and H complete their octet.

NH₄⁺ resonance structure

NH4+ lewis structure contains no π bond. So, we cannot draw the resonance structure its cause there is no chance of shifting the electronic clouds. Even if there is no lone pair of electron density present, we can shift them to draw a resonating structure.

Uses of NH₄⁺ 

• Ammonium ion maintains the acid-base balance in the human body. The kidney produces ammonium ions to combine with many anions in the body to maintain the acid-base balance.
• Using ammonium ion, we can synthesize the quaternary ammonium salt which can use in phase transfer catalysts in many organic reactions.
• Ammonium salt is highly soluble in water, using this fact we can use them as a cleaning agent, surfactant, etc.
• It is a good Nitrogen source for many plants, so we can use it in soil, but in limiting conditions, the excess will be poisonous.

Compare the N-H length in NH₃ and NH₄⁺.

NH₄⁺ , 5+4-1 = 8 (tetrahedral) , NH₃ 5+3 = 8 (tetrahedral)

Using VSEPR, both NH₃ and NH₄⁺ should have a tetrahedral structure. In both cases, the N atoms should be in sp³ hybridized. In NH₃, since the hybrid orbital win in which the lone pair is present is of greater “s” character (Bent’s rule), the hybrid orbital of N directed towards the 3 H atoms in NH₃ will have lesser “s” character w.r.t. sp². This is unlike the case of ammonium, where all the hybrid orbitals of N are sp³ hybridized. Thu the N-H length in NH₃ will be longer.

Compare the acidic character of NH₃ and NH₄⁺.

Ammonia is basic, it has lone pair to be donated in any vacant site of other species, So it behaves like a lewis base.

But in ammonium ion, there will be no lone pair, and it has a positive charge and has an extra proton, so it can easily release this proton and behave like an acid.

Compare the bond angle of NH₃, NH₄⁺, and NH₂^–.

NH₄⁺ > NH₃ > NH₂^–

For ammonium ion there is no lone pair, in ammonia molecule, there is one lone pair and for an for amino ion there are two pairs of lone pair. As the number of lone pairs increases, then the repulsion also increases, and to minimize the repulsion the bond angle decreases.

Frequently Asked Questions

What is the Lewis structure of NH4+ ion?

The Lewis structure of NH4+ (Ammonium ion) consists of a central Nitrogen atom surrounded by four Hydrogen atoms, all connected by single covalent bonds. There are no lone pairs on the Nitrogen atom in this structure. The positive charge comes from the loss of one valence electron by Nitrogen to achieve a stable octet structure.

How does the VSEPR theory apply to the NH4+ Lewis structure?

According to the VSEPR (Valence Shell Electron Pair Repulsion) theory, the shape of a molecule is determined by the repulsion between the electron pairs in its valence shell. For NH4+, all four valence electron pairs are bonded, resulting in a tetrahedral molecular geometry.

What is the formal charge on the NH4+ Lewis structure?

The formal charge on the NH4+ Lewis structure is +1. This is calculated by subtracting the total number of valence electrons in the isolated atoms from the total number of valence electrons in the molecule, considering both bonding and non-bonding electrons.

How does the octet rule apply to the NH4+ Lewis structure?

The octet rule states that atoms tend to form bonds in such a way that each atom has eight electrons in its valence shell, similar to the electron configuration of a noble gas. In the NH4+ Lewis structure, the central Nitrogen atom shares its five valence electrons with four Hydrogen atoms, achieving a stable octet structure.

What is the molecular geometry of the NH4+ Lewis structure?

The molecular geometry of the NH4+ Lewis structure is tetrahedral. This is due to the four Hydrogen atoms symmetrically surrounding the central Nitrogen atom, with bond angles of approximately 109.5 degrees.

What is the electron geometry of the NH4+ Lewis structure?

The electron geometry of the NH4+ Lewis structure is also tetrahedral. This is because electron geometry considers both bonding and non-bonding electron pairs, and in the case of NH4+, all electron pairs are bonding.

How are resonance structures used in the NH4+ Lewis structure?

Resonance structures are not applicable to the NH4+ Lewis structure. Resonance occurs when there are multiple valid ways to place the pi bonds and non-bonding lone pairs of electrons. In NH4+, all bonds are single and there are no lone pairs on the central atom.

What is the bond angle in the NH4+ Lewis structure?

The bond angle in the NH4+ Lewis structure is approximately 109.5 degrees, which is characteristic of a tetrahedral molecular geometry.

Is the NH4+ Lewis structure polar or nonpolar?

The NH4+ Lewis structure is nonpolar. Despite the individual N-H bonds being polar, the molecule as a whole is nonpolar due to its symmetric tetrahedral geometry which causes the dipole moments of the individual bonds to cancel out.

How do you draw the NH4+ Lewis structure?

To draw the NH4+ Lewis structure, start with a central Nitrogen atom and surround it with four Hydrogen atoms. Connect each Hydrogen to the Nitrogen with a single line representing a single covalent bond. Indicate the positive charge of the ion by writing a + sign next to the structure.

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In this article we want to discuss about the if2- lewis structure including its drawing, hybridization, shape, charges, pairs and some FAQS.

IF₂^– is a poly halide in which iodine act as a central atom and 2 fluorine atoms act as terminal atom.

IF2– Lewis Structure Drawing

In the if2- lewis structure we see that as iodine becomes larger in size and less electronegative than F it goes in the center of the Lewis structure.

Iodine has 7 valance electrons out of which 2 electrons take part in bonding with F atom and form 2 covalent sigma bond and still 3 electron pair there on I atom which does not take part in bonding F atom and exist as lone pair of electron. F has also 7 electron in their valance shell out of which only 1 electron make covalent bond with central I atom and remaining 6 electrons present as lone pair on I atom.

IF2– Lewis Structure Resonance

Resonance means shifting of electron pair from one atom to another atom and the structure obtained by this process is called resonating structure.

IF₂^– has 3 resonating structure in which each I-F bond gets partial double bond character by the process in which F atom donate its lone pair of electron into vacant d orbital of I atom to form p(pi)-d(pi) back bonding.

IF2– Lewis Structure Shape

According to VSEPR theory the shape of IF₂^–is linear in which iodine is the central atom around which 2 F atoms surround it. As 3 lone pair present in the central I atom the ideal TBP geometry becomes distorted. If 3 lone pair is not present in IF₂^–  the shape becomes TBP but as 3 lone pair present on central I atom the shape becomes distorted and the actual shape is linear.

IF2– Lewis Structure Formal Charge

The formal charge on any atom can be calculated by the formula given below:

Formal Charge (f) =V-B/2-N

Where,

V= No of valance electrons, B= No of bonding electrons, N= No of nonbonding electrons.

Hence formal charge on I atom in IF₂^–=7-4/2-6= -1

Formal charge on each F atom in IF₂^– =7 -2/2-6= 0

Hence formal charge on I atom is -1 and on each F atom is 0 making the whole compound is negatively charged.

IF2– Lewis Structure Angle

if2- lewis structure has distorted TBP geometry hence normal TBP bond angle 1200 and 900 does not arises here. Here in this compound lone pair will occupy the equatorial position and F atoms will occupy the axial position.

This is because according to Bent’s rule more electronegative atom will occupy the hybrid orbital having less s character and less electronegative atom will occupy the hybrid orbital having more s character.

Hence we know that lone pair has 0 electronegativity for that it will occupy equatorial position where % of s character is 33.3 and more electronegative F atom will occupy the axial position where % of s character is 0 making the F-I-F bond angle is 180⁰.

IF2– Lewis Structure Octet Rule

In if2- lewis structure we see that each F atom has 8 valance electrons and completes their octet. In I forms 2 I-F bonds and each bond contain 2 electrons. There is also 3 lone pairs that present on I atom making a total of 10 electrons around I atom. As I is a member of 3^rd period and we also know that 3^rd period element can increase their octet more than 8 electrons. Hence according to octet rule IF₂^– is a stable compound.

IF2– Lewis Structure Lone Pairs

The valance electron that does not take place in bonding is defined as lone pair of electron or nonbonding electrons.

The formula through which we can calculate the lone pair of electron is given below:

For central atom,

No of lone pairs=Total no of valance electron of the atom-no of bonding electron formed by that atom

In if2- lewis structure , lone pair present on I atom=7-4=3

For terminal atom,

No of lone pairs=Total no of valance electron-no of bonds formed by that atom

 Lone pair present on each F atom=7-1=6 i.e. 3 lone pair

These lone pairs are shown In the Lewis structure of IF₂^– on the given atoms as dots.

IF2– Valance Electrons

At first, to calculate the total valance electron in IF₂^–, it is essential to know the electronic configuration of I and F atom.

The electronic of I is [Kr₃₆]4d¹⁰5s²5p⁵ and from electronic configuration we see that there are 7 electrons in the outermost shell of I atom. The electronic configuration of F atom is [He₂]2s²2p⁵ and has 7 electrons. As both I and F both belong to same group i.e. 17, there are 7 valance electrons in both I and F atom. There is also 1 negative charge.

The total valance electron in the compound will be equal to the sum of the valance electron of I and F atom + 1 negative charge i.e. (7*1) + (7*2) + 1=22. There are 22 valance electrons in this species.

IF2– Hybridization

Hybridization is the process of mixing of same energy atomic orbitals to form an equal number of hybrid orbitals.

The ground state valance shell electronic configuration of I is 5s²5p⁵. In the ground state of I we see that there is only 1 unpaired electron and to make IF₂^– 2 unpaired electron is required. In the excited state I send its 1 p electron into d orbital making a total of 3 unpaired electrons. In the next step 2 F atoms give their 1 unpaired electron to form 2 I-F sigma bonds and sp³d hybridization takes place according to VSEPR theory.

In this compound I uses sp³d hybrid orbital to make I-F bonds. According to sp³d hybridization the geometry should be TBP but the actual structure is linear due to the presence of 3 lone pair of electron in the equatorial position of if2- lewis structure.

IF2– Uses

if2- mainly used in making eye drops. It is used as a fluorinating agent. It is used in explosive material.

FAQS about IF2–

Is IF2– Ionic or Covalent?

IF₂^– is a covalent compound. This is because it is formed by covalent sigma bonds. In the formation of ionic compound there occurs shifting of electron from electropositive atom to electronegative atom. In if2- lewis structure it is not possible because there occurs mutual sharing of electron between I and F atom to form sigma bonds so that no ions is formed. Hence it is a covalent compound.

Is IF2– Stable?

IF₂^– is an unstable compound. This is due to the presence of 3 lone pair electron around I atom and there occurs severe LP-LP repulsion. Also I-F bonds are not too strong due to large electronegativity difference and poor orbital overlap between I and F atoms. Due to these two reason this compound is not stable.

Is IF2– polar in nature?

IF₂^– is a nonpolar compound. A compound is said to be polar when its dipole moment is not equal to 0. The electronegativity of I and F is 2.66 and 3.98 respectively that’s why I-F bond moment lies towards F atom. But as the shape is linear, 2 I-F bond moments lie in opposite direction and cancel each other making the molecule nonpolar.

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The azide ion (N₃⁻) has a linear structure with a total of 16 valence electrons. It consists of three nitrogen (N) atoms, with the middle N atom bonded to the other two N atoms through double bonds. The Lewis structure depicts these double bonds and a lone pair of electrons on each terminal N atom. The central N atom carries a negative charge. Azide ion is resonance stabilized, with three equivalent resonance structures, each showing a different N atom bearing the negative charge. The electronegativity of nitrogen (3.04) and the high electron density contribute to N₃⁻’s reactivity and its role as a nucleophile in chemical reactions.

Only three nitrogen atoms make up the azoide ion (N₃^–). Two N=N bonds are present in the Lewis structure of the N₃^– ion. Outside nitrogen atoms contain two lone pairs, whereas the core nitrogen atom has none.

1. N₃^– Lewis Structure
2. N₃^– lewis structure octet rule
3. N₃^– lewis structure resonance
4. N₃^– Molecular Geometry
5. N₃^– lewis structure formal charge
6. N₃^– lewis structure angle
7. N₃^– Hybridization

The left and right nitrogen atoms have a negative (-1) charge, whereas the central nitrogen atom has a positive (+1) charge.

1. N3– Lewis Structure:

Here’s a step-by-step guide on drawing the N3– Lewis structure.

Step 1: draw sketch

• To begin, count the total amount of valence electrons.

Nitrogen is in group 15 of the periodic table. As a result of this, nitrogen has five valence electrons.

Because N₃^– contains three nitrogen atoms,

Three nitrogen atoms’ valence electrons = 5 × 3 = 15

Because the N₃^– now has a negative (-1) charge, we must add another electron.

As a result, there are 15 + 1 = 16 valence electrons altogether

• Next, calculate the total number of electron pairs.

In total, there are 16 valence electrons. Divide this value by two to get the total number of electron pairs.

Total electron pairs = total valence electrons divided by 2

As a result, there are 16 ÷ 2 = 8 total electron pairs.

• Decide on the centre atom third.

We may consider any of the three atoms as the centre atom because they are all nitrogen.

Let’s pretend that the centre atom is nitrogen.

• Finally, do a rough drawing.

Step 2: Identify lone pairings

We have a total of eight electron pairs here. Two N-N bonds have previously been identified. As a result, we just need to label the remaining six electron pairs on the drawing as lone pairs.

Remember that because nitrogen is a period 2 element, it can only have 8 electrons in its final shell.

Always begin by identifying lone pairs from exterior atoms. The outer atoms are nitrogen, both left and right.

So there are three lone pairs for left and right nitrogen, and zero lone pairs for canter nitrogen because all six electron pairs have been used up.

On the drawing, draw the following lone pairs:

Step 3: Charges for marking

Using the following formula, calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

Formal charge = 5 – 6 –½ (2) = -2 for left and right nitrogen atoms.

Formal charge for core nitrogen atom = 5 – 0 – ½ (4) = +3

Because all nitrogen atoms have charges in this situation, draw them as follows:

Because all nitrogen atoms have charges, the structure below is not a stable Lewis structure. As a result, convert lone pairs to bonds to lower the costs.

Step 4: reduce charges

Make a new N-N bond with the centre nitrogen atom by converting a lone pair of the left nitrogen atom.

Step 5: Charges should be reduced once again

Because nitrogen atoms have charges, build a new N-N bond with the centre nitrogen atom by converting a lone pair of the right nitrogen atom as follows:

The core atom (centre nitrogen) forms an octet in the structure shown above. As a result, the octet rule is met.

On the atoms, there are still charges.

This is acceptable since the ideal Lewis structure has a negative charge on the most electronegative element. Nitrogen is the most electronegative element in this circumstance.

As a result, this structure is N3-‘s most stable Lewis structure.

Because the N3- has a negative (-1) charge, add brackets to the Lewis structure to indicate that charge:

2. N3– lewis structure octet rule:

A N₃^– ion has eight valence electrons in total.

N has 5 valence electrons, and its configuration is 1s²2s²2p³, therefore N^-3 has 3 additional electrons, and its octet is complete.

3. N3– lewis structure resonance:

The azide ion has resonance structures that may be calculated. These are displayed below.

The presence of triple bonds and the lack of an octet in the aforementioned resonance configurations make the ion unstable. A double charge on a tiny atom-like Nitrogen is unusual among the triple bound resonance structures.

The lack of an octet in the later two causes relative instability. As a result, we’ll look at the Lewis structure with double bonds and a full octet illustrated above.

4. N3– Molecular Geometry:

The Lewis structure of a chemical provides information on the molecular geometry and electronic shape of that molecule.

Three nitrogen atoms make up the Azide Lewis structure. With the nitrogen atoms around it, the central nitrogen atom creates two double bonds. The atoms will reject each other, resulting in a Linear Geometry, according to the VSEPR hypothesis.

The A-X-N approach can be used to validate this.

The core atom Nitrogen is represented by the letter ‘A’. As a result, ‘A’ equals 1.

The number of atoms bound to the centre atom is denoted by the letter ‘X.’ In this example, two additional Nitrogen atoms are connected to the core nitrogen atom.

As a result, X = 2.

The number of lone pairs connecting to the centre atom is represented by the symbol “N.”

There are no lone pairings in this example, thus N = 0.

As a result, the Azide ion would be designated as AX₂ (N₃^–)

An AX₂ arrangement may be shown to match to a Linear Molecular geometry.

5. N3– lewis structure formal charge:

Formal charge = (Number of valence e⁻ in free neutral atom) −1/2(Number of bonding e⁻) − (Number of non-bonding e⁻)

We can determine formal charge on using the preceding equation.

N₁: 5−2−4 = −1

N₂: 5−4−0 = +1

N₃: 5−2−4 = −1

6. N3– lewis structure angle:

According to the VSEPR theory, the nitrogen atoms present will reject each other and arrange themselves in a linear pattern. This results in 180° bond angles.

7. N3– Hybridization:

To identify the hybridization of the central atom, the Lewis structure of the Azide ion must be investigated.

Through double bonds, the core Nitrogen atom is chemically connected to two nearby Nitrogen atoms. We can rapidly establish the hybridization from this data since we already explored the idea of electron regions.

The centre Nitrogen atom is surrounded by two areas. As a result, the Azide ion’s hybridization is determined to be sp.

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• Ph3 lewis structure

In this article, we are going to study PH3 lewis structure and various facts about it.

PH3 or phosphine is a compound of phosphorus that is classified under pnictogen hydride. Phosphine or phosphane forms bonds by covalent bonding. We can study the bonding in the molecule of PH3 by taking into consideration lewis method. We will study the PH3 lewis structure and understand the concept.

Some facts about Phosphane

PH3 has a molar mass equal to 33.99 g/mol. It exists as a gas that is colorless and has an odor like that of a rotten fish.

The density of PH3 is 1.37 g/L. The melting point of the compound is said to be-132 degrees Celsius and the observed boiling point is somewhere around-87 degrees Celsius. Talking about its solubility, it is soluble in water at a temperature of 17 degrees Celsius (31.2 mg/100mL).

It is also soluble in organic compounds like alcohol, chloroform, benzene, etc. Coming to its preparation, it can be prepared by the treatment of phosphorus (white) and sodium hydroxide. This method is usually used at the industrial level.

How to draw lewis structure for PH3?

To write the PH3 lewis structure one should know the total of all the valence electrons that could be present in the molecule of PH3.

In the structure of phosphane, we can see that there are 3 atoms of hydrogen element one phosphorus element atom is present. Coming to the contribution of phosphorus element in bond formation, we can see there is only one P atom. Meaning the contribution of valence electrons will be only 5 from the P atom. Now counting the contribution of hydrogen element atoms, there are 3 H atoms present.

Coming to the H atom as there are also H atom present in the structure. There are total 3 H atoms, hence contribution by H valence electrons will be 3 times more. Meaning 3×1, which comes up to 3. Now the total valence electrons contribution from both elements it will be 8. As we know the total valence electrons, now we need to know or understand which elements atom will come in the middle or be the central one.

We can predict by taking into account electronegativity of atoms under consideration. The concept that is used to predict that which atom will come in the middle is electronegativity. Meaning the atoms of element participating in bonding must having some electronegativity. What we have to see is among them the atom with less electronegativity will be the middle one. Meaning it will be placed as the central atom.

The other atoms are placed in a manner to satisfy their valence. In PH3 lewis structure the P atom will be placed as the central atom the reason we have already understood. And the hydrogen (H) atom is seen to be the surrounding atom that is placed to satisfy the valency. In this molecule we can see there is sharing of one electron pair between each hydrogen and Phosphorus element atoms. Since one pair of electrons is shared or used for forming bond the resultant bond will be a single bond.

Phosphane lewis structure

We can determine the shape of PH3 by taking into account to concepts by the lewis structure concept and valence shell electron pair repulsion theory concept. 

We will analyze the structure of phosphane by taking into consideration the concept of valence shell electron pair repulsion theory. According to this concept electron pair will prefer to maintain a considerable distance between each other, as this is said to reduce the repulsion in between the valence electrons. And this makes the structure of the molecule stable.

So we cay lone pair tend to affect the shape of the molecule to quite an extent. The number of lone pairs in phosphane molecule are one and this will prefer to maintain a distance from the other bonds. This causes the shape of the molecule to be trigonal pyramid.

PH3 lewis structure formal charges

What is the meaning of the term formal charge or what we understand by the term formal charge?

It means there is some type of difference in between the valence electrons and all the (total) electrons (of the atom that is under consideration). Also in this concept it is presumed that in the bonding process when electrons are being shared (in a molecule) they are done in a equal manner. Meaning shared equally. Formula is given below:

V is said to be valence electrons of the atom of the molecule.

N is said to be the figure indicating the unbound (valence) electrons.

B is said to be or includes all the electrons that are shared with other atom.

Number of lone pairs in PH3 lewis structure

The number of lone pairs present in the molecule have a lot of role in the structure prediction.

In the lewis structure of phosphane we can see there are 5 electrons with P as valence electrons and during the process of bonding P will be surrounded by 3 H atoms forming single bonds. We will see that one pair of electrons will remain unbonded, meaning will not participate in bonding. Hence the number of valence electrons in phosphane molecule will be one.

Hybridization in phosphane

We know that hybridization is the concept where in atomic orbitals combine to form hybrid atomic orbitals.

Coming to hybridization in the molecule of phosphane, there is no hybridization observed in this molecule. The reason for no hybridization is shape of orbitals of this molecule is quite distinct. And also according to Drago’s rule concept if there is only one lone pair (minimum) at the central atom, then the chances of hybridization are very less.

PH3 lewis structure resonance

By following resonance concept we can explain the delocalised electrons that are present in the molecule.

PH3 Lewis structure octet rule

In the concept of Octet rule an atom must have a complete octet, means the outermost shell should be filled.

So in the molecule of phosphane, the valency of phosphorus is 3, so it needs more 3 electrons in order to complete its octet. Hence it takes electrons from three hydrogen and forms three single bonds.

PH3 polar or nonpolar

Phosphane is a polar molecule, because there is one lone pair present which has electron-electron repulsion which leads to the bent structure of the molecule.

PH3 lewis structure bond angle

As the phosphane structure has a trigonal pyramid shape, the bond angle present in the molecule is equal to 93 degrees.

PH3 uses

• It is used as a dopant in the industries of semiconductors.
• Phosphane is also used in plastic industries.

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