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How to draw NaI Lewis Structure? - Science Education and Tutorials

How to draw NaI Lewis Structure?

The sodium iodide chemical formula is NaI. Drawing NaI Lewis Structure is very easy to by using the following method. Here in this post, we described step by step method to construct NaI Lewis Structure. The iodine and sodium elements come as members of the halogen and alkaline metal family groups from the periodic table respectively. The valence electrons in iodine and sodium are seven and one respectively. Sodium iodide is used to make chemical reagents for organic chemical reactions as a brominating agent in organic chemistry.

Key Points To Consider When Drawing The NaI Electron Dot Structure

A three-step approach for drawing the NaI Lewis structure can be used. The first step is to sketch the Lewis structure of the NaI molecule, to add valence electrons around the iodine atom; the second step is to add valence electrons to the one sodium atom, and the final step is to combine the step1 and step2 to get the NaI Lewis Structure.

The NaI Lewis structure is a diagram that illustrates the number of valence electrons and bond electron pairs in the NaI molecule. The geometry of the NaI molecule can then be predicted using the Valence Shell Electron Pair Repulsion Theory (VSEPR Theory), which states that molecules will choose the NaI geometrical shape in which the electrons have from one another.

Finally, you must add their bond polarities to compute the strength of the one Na+—I- single bonds (dipole moment properties of the NaI molecule). The sodium-iodine ionic bonds in sodium iodide(NaI), for example, are polarised toward the more electronegative iodine in NaI molecule, and because both bonds have the same size and are located around one sodium terminal of the linear structure with three lone pairs (in total eight electrons) on the iodine atom as iodide ion(I-), their sum of dipole moment is nonzero due to the NaI molecule’s bond dipole moment and less electron polarity to the sodium atoms and become sodium ion. Because Na+—-I- bonds polarity is not canceled in the NaI molecule due to the presence of four lone pairs of electrons in the linear structure. The sodium iodide(NaI) molecule is classified as a polar ionic molecular crystal.

How to draw NaI Lewis Structure?

The molecule of sodium iodide(linear-shaped molecular geometry) is tilted, the bond angles between iodine (as iodide ion) and sodium (as sodium ion) are 180 degrees. It has a difference in electronegativity values between iodine and sodium atoms, with central iodine’s pull being higher than terminal sodium’s in the NaI ionic molecule. But they do not cancel each other due to the linear structure with four lone pairs in the molecular geometry of the NaI molecule.

As a result, it has the nonzero dipole moment. The NaI molecule has a nonzero dipole moment due to an unequal charge distribution of negative and positive charges. But both iodine and sodium atoms fall on the halogen and alkaline metal family groups in the periodic table respectively. The iodine atom is a more electronegative value than sodium in the NaI ionic molecule. The NaI molecule has the net dipole moment of non-zero value in the ground state energy level.

NaI molecule has one Na+—–I- single bond. Its dipole moment in the ground state is totally different as compared with the excited state. If it absorbs light may be from visible or UV light. It undergoes pi to pi star and n to pi star transition from ground state energy level to excited state energy level. In the excited state energy level, the NaI ionic molecule shows a definite dipole moment. But it is very dynamic in nature.

Molecules can be classified as polar or nonpolar. The molecule polar behaves in a different manner as compared to nonpolar.

Overview: NaI Lewis Structure

The central atom is iodine, which is bordered on two terminals with sodium atoms(linear structural geometry), and four lone pairs on the central iodine atom in the linear molecular geometry. Iodine has seven outermost valence electrons, indicating that it possesses seven electrons in its outermost shell, whereas sodium also has one valence electron in its outermost shell. To complete the octet of the iodine atom requires one valence electron on each of their outermost shell.

One sodium atom establishes covalent connections with the central iodine atom as a result, leaving the iodine atom with three lone pairs. There are three lone pairs of electrons on the iodine central atom that resists the bond pairs of the Na+—-I- bond. According to VSEPR theory, the Na+—-I- ionic bond pairs polarity lead the NaI molecule to take on the linear geometry structure.

The NaI molecule’s one Na+—-I- bonds are arranged in asymmetrical polarity order around the linear molecular geometry, giving rise to the NaI molecular shape. The NaI molecule has a linear molecular geometry because there is an electrical repulsion between the lone pairs of electrons in iodine and one single ionic bond pair(Na-I) of the NaI molecule.

Lewis structure of NaI has dot electron representative structure. Valence electrons of atoms undergo orbitals mixing in the chemical reactions, giving new types of molecular species of NaI. The molecule is nothing but a bundle of valence electrons from the atoms. But it is converted to ionic bond pairs and lone pairs in the molecular structure.

Electronegative value Difference Calculation of NaI Molecule:

Iodine and sodium Electronegative difference in NaI:

The iodine atom has an electronegativity of 2.66, while sodium has an electronegativity of 0.93 in the NaI ionic molecule. The difference in electronegativity of iodine and sodium can be estimated using the method below.

The electronegative value difference between iodine and sodium in NaI molecule

Electronegativity value of iodine= 2.66

Electronegativity value of sodium= 0.93

Difference of electronegativity value between iodine and sodium in NaI molecule = 2.66 – 0.93 = 1.73

Electronegativity difference between Na-I ionic bond calculation of NaI molecule

The electronegative difference between iodine and sodium is greater than 0.5. This indicated the bond polarity moves near to polar nature. Na-I ionic bond polarity in the NaI molecule is polar.

Because of this difference in electronegativity of iodine and sodium atoms, the NaI molecule’s Na+—I- bond becomes nonpolar. The total net dipole moment of the NaI molecule is nonzero due to the cancellation of the bond dipole moment in the linear geometry due to the presence of three lone pairs of electrons. The electronegativity of an atom is the strength with which it may attract bound electron pairs to its side.

As a result, the Na-I bond’s dipole moment is higher due to the polarization of the bonds and four lone pairs of electrons on iodine, and all Na-I bonds’ dipoles are arranged in the asymmetrical NaI molecular geometry. The NaI molecule has a nonzero net dipole moment.

The electron dot structure of the NaI molecule is also known as the NaI Lewis structure. It determines the number of outermost valence electrons as well as the electrons engaged in the NaI ionic molecule’s bond formation. The outermost valence electrons of the NaI molecule must be understood while constructing the Lewis structure of the molecule.

The iodine atom is the middle element in NaI molecular geometry, with seven electrons in its outermost valence electron shell, whereas the sodium atom has one electron in its outermost valence electron shell. The sodium atom has one valence electron.

The NaI has a total of 8 valence electrons as a result of the foregoing above-said reasoning. With the core central iodine atom, the one terminals with one sodium atom form ionic bonds, leaving the iodine atom with four lone pairs in the middle of linear geometry.

Because three lone pairs on the terminal iodine atoms create interaction with Na-I bond pairs(but it is negligible in the ground state of the NaI molecule). The bond angle of the Na-I bond in the linear molecular geometry is approximately 180 degrees. This angle is greater than the CH4 molecule bond angle. The Na-I bond length is 250 pm(picometer).

To sketch the NaI Lewis structure by following these instructions:

Step-1: NaI Lewis dot Structure by counting valence electrons on the iodine atom

To calculate the valence electron of each atom in NaI, look for its periodic group from the periodic table. The halogen and alkaline metal group families, which are the 17th and 1st groups in the periodic table, are both made up of iodine and sodium atoms respectively. In their outermost shells, sodium and iodine have one and seven valence electrons respectively.

Calculate the total number of valence electrons in the NaI molecule’s outermost valence shell. The first step is to determine how many electrons are in the NaI Lewis structure’s outermost valence shell. An electron in an atom’s outermost shell is known as a valence electron. It is represented by dots in the NaI Lewis diagram. The NaI molecule’s core iodine atom can be represented as follows:

Total outermost valence shell electron of iodine atom in NaI= 7

Total outermost valence shell electron of sodium atom in NaI= 1

The NaI molecule has one central iodine and one sodium atoms. Then the total outermost valence shell electrons can be calculated as follows

∴ Total outermost valence shell electrons available for NaI Lewis structure( dot structure) = 7+1*1= 8 valence electrons  in NaI.  

calculation of total valence electron of NaI molecule

Choose the atom with the least electronegative value atom and insert it in the center of the molecular geometry of NaI. We’ll choose the least electronegative value atom in the NaI molecule to place in the center of the NaI Lewis structure diagram in this phase.

But in this case, sodium is the least electronegative than iodine. Sodium loses one electron and forms sodium positive ions(Na+). So that iodine stays in the central molecular structure. The electronegativity value in periodic groups grows from left to right in the periodic table and drops from top to bottom.

How to draw NaI Lewis Structure?

The first step is to put seven valence electrons around the sodium atom as given in the figure.

Step-2: Lewis Structure of NaI for counting valence electrons around the terminal sodium atoms

As a result, iodine is the fourth atom in the periodic table’s halogen family group. Sodium is the first member of the alkaline metal family. It is the first element in the periodic table. The electronegative value of the iodine atom is higher than that of the sodium atom in the NaI molecule. Furthermore, sodium has a one-electron limit since it is the less electronegative element in the NaI molecule.

In the NaI Lewis structure diagram, the iodine atom can be the center atom of the molecule. As a result, central iodine in the NaI Lewis structure, with one sodium atom arranged in a linear geometry.

Iodine accepts one electron and forms an iodide ion(I-). The total lone pair of electrons in the iodide ion is eight. They are negatively charged.

How to draw NaI Lewis Structure?

Add valence electron around the iodine atom, as given in the figure.

Step-3: Lewis dot Structure for NaI generated from step-1 and step-2

Connect the exterior and core central atom of the NaI molecule with one single Na-I bond. In this stage, use one sodium atom on the outside of the NaI molecule to the central iodine atom in the middle.

Count how many electrons from the outermost valence shell have been used in the NaI structure so far. Na-I single bond carries two electrons because the iodine atom is connected to one sodium atom by Na-I single bonds. Bond pairings of Na-I are what they’re called.

So, out of the total of 8 valence electrons available for the NaI Lewis structure, we used four electrons for the NaI molecule’s one Na-I single bond. The NaI molecule has three lone pairs of electrons in the central iodine atom.

Place the valence electrons in the Na-I bond pair starting with the core iodine, one sodium atom in the NaI molecule. In the NaI Lewis structure diagram, we always begin by introducing valence electrons from the central iodine atom(in step 2). As a result, wrap around the central iodine atom’s bond pair valence electrons first (see figure for step2).

The iodine atom in the molecule gets only 8 electrons around its molecular structure. This central iodine atom is octet stable. But it has three lone pairs. Iodine gas(I2) is a brownish liquid gas. Iodine is very corrosive in nature. It is one of the very reactive chemical reagents.

Sodium metal is in a soft solid state at normal temperature and pressure. It is used as a reducing agent in the field of organic chemistry. It is highly flammable in nature. It is very reactive in water and alcohol. Water reacts with sodium metal and forms sodium hydroxide. Similarly, alcohol(such as methanol) reacted with sodium and forms sodium methoxide.

Iodine requires 8 electrons in its outermost valence shell to complete the molecular octet stability, two electrons bond pairs in one Na-I single bond, and three lone pairs in the central iodine atom. No lone pairs of electrons on the sodium atoms of the NaI molecule are placed in a linear geometry. Iodine already shares 8 electrons to the one Na-I single bonds. Then place the valence electron in the sodium atoms, it placed around one electron on each atom(step-1). There are no valence electrons placed around one sodium atom as lone pair of electrons.

We’ve positioned 8 electrons around the one-terminal sodium atoms(step-3), which is represented by a dot, in the NaI molecular structure above. The iodine atom completes its molecular octet stability in the NaI molecule because it possesses 2 electrons in its (one Na-I single ionic bond pairs) bond pairs with one sodium in the outermost valence shell.

How to draw NaI Lewis Structure?

Count how many outermost valence shell electrons have been used so far using the NaI Lewis structure. One electron bond pairs are shown as dots in the NaI chemical structure, whereas one single bond contains two electrons. The outermost valence shell electrons of the NaI molecule(bond pairs) are 2 as a result of the calculation. The total valence electron in an iodine atom is 8.

So far, we’ve used 8 of the NaI Lewis structure’s total 8 outermost valence shell electrons. Three lone pairs of electrons on the iodine atom in the linear or tetrahedral geometry of the NaI molecule.

Complete the middle NaI atom stability and, if necessary, apply a covalent bond. The central iodine atom undergoes octet stability(due to one single bond pair of electrons).

The core atom in the NaI Lewis structure is iodine, which is bonded to the one sodium atom by single bonds (one Na-I). With the help of one single bond, it already shares 8 electrons. As a result, the iodine follows the octet rule and has 8 electrons surrounding it on the one terminal of the NaI molecule’s linear or tetrahedral geometry.

How to calculate the formal charge on iodine and sodium atoms in NaI Lewis Structure?

Calculating formal charge on the iodine of NaI molecule:

The formal charge on the NaI molecule’s iodine central atom often corresponds to the actual charge on that iodine central atom. In the following computation, the formal charge will be calculated on the central iodine atom of the NaI Lewis dot structure.

To calculate the formal charge on the central iodine atom of the NaI molecule by using the following formula:

The formal charge on the iodine atom of NaI molecule= (V. E(I)– L.E(I) – 1/2(B.E))

V.E (I) = Valence electron in a iodine atom of NaI molecule

L.E(I) = Lone pairs of an electron in the iodine atom of the NaI molecule.

B.E = Bond pair electron in Br atom of NaI molecule

calculation of formal charge on iodine atom in NaI molecule

The iodine core atom (one single bond connected to one sodium atom) of the NaI molecule has seven valence electrons, three lone pairs of electrons(six electrons), and 2 bonding pairing valence electrons. Put these values for the iodine atom in the formula above.

Formal charge on iodine atom of NaI molecule = (7- 8-(0/2)) = -1

In the Lewis structure of NaI, the formal charge on the central iodine atom is -1 (negative charge).

Calculating formal charge on the sodium atom of NaI molecule:

The formal charge on the NaI molecule’s sodium terminal atoms often corresponds to the actual charge on that sodium terminal atoms. In the following computation, the formal charge will be calculated on the terminal sodium atom of the NaI Lewis dot structure.

To calculate the formal charge on the terminal sodium atom of the NaI molecule by using the following formula:

The formal charge on the sodium atom of NaI molecule= (V. E(Na)– L.E(Na) – 1/2(B.E))

V.E (Na) = Valence electron in a sodium atom of NaI molecule

L.E(Na) = Lone pairs of an electron in the sodium atom of the NaI molecule.

B.E = Bond pair electron in H atom of NaI molecule

calculation of formal charge on sodium atom in NaI molecule

The sodium terminal atoms of the NaI molecule have one valence electron, no lone pairs of electrons(zero electrons), and two bonding pairing valence electrons(single bond). Put these values for the sodium atom in the formula above.

Formal charge on sodium atom of NaI molecule = (1- 0-(0/2)) =+1

In the Lewis structure of NaI, the formal charge on the terminal sodium atom is +1 (positive charge).

Summary:

In this post, we discussed the method to construct the NaI Lewis structure. First, the valence electrons are placed around the sodium atom and lose one electron. Then it becomes sodium positive ions. Second, place the valence electron on the iodine atoms and the iodine atom accepts one electron in its valence shell. Finally, when we combined the first and second steps. It gives NaI Lewis structure. Need to remember that, if you follow the above-said method, you can construct molecular dot structure very easily.

What is the NaI Lewis structure?

NaI Lewis structure is dot representation

What is the formal charge on the NaI Lewis structure?

Zero charges on the NaI molecular structure

The polarity of the molecules

The polarity of the molecules are listed as follows

Lewis Structure and Molecular Geometry

Lewis structure and molecular geometry of molecules are listed below

External Reference:

Information on sodium iodide(NaI)

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