NaI Lewis Structure & Characteristics (13 Complete Facts)

NaI or sodium iodide is an inorganic salt of an alkali earth metal having a molecular weight of 149.89 g/mol. We will learn more about NaI in this article.

NaI is an ionic molecule and the bond between Na and I is formed by the donation of one electron from the iodide as it contains one negative charge and that electron will be accepted by the Na atom to form the molecule. It is a 1:1 compound having a fixed stoichiometric proportion.

In the crystal form, it consists of a halite lattice structure where each Na atom is surrounded by four iodides, and each iodine is surrounded by four Na atoms. Now we discuss the structure, hybridization, polarity, ionic nature, and solubility of NaI in the following part with proper explanation.

1.     How to draw NaI lewis structure

Lewis structure for a covalent molecule can give us an idea of the bonding nature and the electrons involved in bonding. Let us try to draw the lewis structure of NaI.

Counting the total valence electrons

To draw a lewis structure of a molecule we should count the total valence electrons of the molecule by counting the valence electrons of each atom. The total number of the valence electrons of NaI will be 8, which is the summation of the valence electrons of Na and I separately added together.

Choosing the central atom

The 2^nd step for drawing the lewis structure is to choose the central atom among all the atoms. The central atom is chosen based on larger size and electronegativity. Iodine is chosen as the central atom here because it has larger in size and has higher electronegativity and Na is the terminal atom here.

Satisfying the octet rule

Each atom should follow the octet while bonding formation by completing their valence electrons with suitable electrons. Na and I both are completing their valence shell by sharing electrons in the bond formation. The electrons required for the octet are 2+8 = 10, which can be shared by the bond formation.

Satisfying the valency

The electrons required for completion of an octet is 10 and the available valence electrons for the NaI is 8, so there will be 10-8 = 2 electrons left and these electrons need 2/2 =1 bond. So, there will be a minimum of one bond is required in the NaI to satisfy the valency. Na and I both are monovalent.

Assign the lone pairs

Na is lack lone pairs because it has only one valence electron which is satisfied by its valency. Iodine has seven electrons, after satisfying the monovalency of iodine remaining six electrons exist as three pairs of lone pairs over the iodine atom and are not involved in the bond formation.

2.     NaI valence electrons

Ionic or covalent, every molecule has valence electrons which are the contribution from their constituent atom. let us count the total valence electrons for NaI.

The total number of valence electrons for the NaI is 8, those electrons exist as the valence shell of each atom and participate in the bond formation. We just count the valence electrons of each atom and then added them together by multiplying their stoichiometric proportion in the molecule.

• The valence electrons present over the Na atom is 1 (as it has 1 electron in its valence s orbital)
• The valence electrons present over the Iodine atom are 7 (as it has 7 electrons in its valence 5s and 5p orbital)
• So, the total number of valence electrons present over the NaI molecule is, 1+7 = 8

3.     NaI lewis structure lone pairs

The lone pairs over a molecule are those valence electrons that are not involved in the bond formation but involve in the reaction. Let us count the total lone pairs of NaI.

The total lone pairs present over the NaI molecule are 6 electrons, which means there are three pairs of lone pairs present over the molecule. All the lone pairs are contributed from the iodine site because Na has zero lone pairs and I have excess electrons than its stable valence to show lone pairs.

• The formula to be calculated for the lone pairs is, lone pairs = electrons present in the valence orbital – electrons involved in the bond formation
• The lone pairs present over the Na atom are, 1-1 = 0
• The lone pairs present over the I atom are, 7-1 = 6
• So, the total lone pairs present over the NaI molecule are 6 electrons or three pairs of lone pairs.

4.     NaI lewis structure octet rule

The octet is for the formation of the bond by completing the valence electrons of each tom by a suitable number of electrons. Let us see the octet of NaI.

NaI follows the octet because it has lesser electrons in the valence orbital than the electrons required for the octet. It needs 10 electrons in the octet formation (two for the s block element and 8 for the p block element). The reaming electrons are satisfied by the suitable number of bonds.

The reaming electrons will be 10 – 8 = 2 which can be satisfied by the 2/2 = 1 bond. In the NaI, there will be a minimum of 1 bond present between Na and Iodine, and if needed then an extra bond will be formed as per valency. But via 1 bond Na and Iodine both complete their valence orbital and octet also.

5.     NaI lewis structure shape

The molecular shape of the ionic molecule by its lattice structure and for covalent molecule by the VSEPR theory. Lets us predict the shape of the NaI.

The molecular shape of the NaI is linear according to the VSEPR theory which can be discussed in the below table.

According to the VSEPR (Valence Shell Electrons Pair Repulsion) theory, the molecular geometry will be tetrahedral as it is an AX₄ type of molecule which means tetra coordinated but there are 3 lone pairs present so it changes its best geometry to linear and for this reason, the hybridization will be changed.

6.     NaI lewis structure angle

The bond angle of an ionic molecule depends on the arrangement of lattice crystal and covalent on hybridization. Let us calculate the bond angle for NaI.

The bond angle for the NaI molecule is 180⁰, which clear indication that the molecule has linear geometry, and only for linear geometry the angle should be 180⁰. There is only one bond present between Na and Iodine and no other elements are present so the possible bond angle makes by two atoms is 180⁰.

• Now we merge the theoretical bond angle with the calculated bond angle value by the hybridization value.
• The bond angle formula according to Bent’s rule is COSθ = s/(s-1).
• The Iodine is sp³ hybridized but due to linear geometry, it adopts sp hybridization.
• The central atom Iodine is sp hybridized, so the s character here is 1/2^th
• So, the bond angle is, COSθ = {(1/2)} / {(1/2)-1} =-( 1)
• Θ = COS^-1(-1/2) = 180⁰
• So, the value of the bond angle is calculated value and the theoretical value is equal.

7.     NaI lewis structure formal charge

The formal charge present over each atom is calculated by assuming the same electronegativity of all atoms. Let us calculate the formal charge of NaI.

The formal charge over NaI is zero because the charge present over each atom can be neutralized by the same magnitude value but opposite in sign. In the NaI, Na atoms released one electron to form Na⁺ and iodine takes that electron to form I^–. But the charge is nullified because the molecule is neutral.

• The molecule is neutral on the calculation of formal charge by the formula, Formal charge = N_v – N_l.p. -1/2 N_b.p
• The formal charge present over the Na⁺ ion is 1-0-(0/2) = +1
• The formal charge present over the Iodide ion is, 7-8-(0/2) = -1
• So, the formal charge of Na⁺ and I^– +1 and -1 respectively, so the value is the same but opposite in sign, so they neutralized each other and make the molecule neutral.

8.     NaI hybridization

Hybridization for the covalent molecule for the proper bonding of different energy-containing atomic orbitals. Let us see the hybridization of the iodine in NaI.

The central iodine is sp³ hybridized here and it is can be confirmed from the following table.

• We can calculate the hybridization by the convention formula, H = 0.5(V+M-C+A),
• So, the hybridization of central Iodine is, ½(7+1+0+0) = 4 (sp³)
• One s orbital and three p orbitals of N are involved in the hybridization.
• The lone pairs over the Iodine are involved in the hybridization.

9.     NaI solubility

The solubility for an ionic compound is breaking bond between two toms and getting soluble in the particular solvent. Let us see whether NaI is soluble in water or not.

NaI is soluble in water because it is an ionic molecule and ionic compounds are polar so soluble in polar solvents. Also, the hydration of energy of NaI is higher than its bond enthalpy, so when it breaks into ions then the water molecule is attracted to the Na+ ions as it has a higher ionic potential.

Apart from water, NaI can be soluble in other polar solvents like,

• CCl₄
• CHCl₃
• DMSO
• Methanol
• Ethanol
• Toluene

10. Is NaI solid or gas?

Most of the ionic compounds are solid in nature due to strong internuclear interaction between the constituent atoms. Let us see whether NaI is solid or not.

NaI is a solid crystalline molecule. It exists as a white crystalline powder form because it consists of a halite lattice and for this reason, the structure of lattice and ionic interaction in the molecule is very high and all constituent atoms are present very close to each other and exist as a solid form.

The nature of the crystal is very hard and required very large energy to break the crystal of NaI.

11. Is NaI polar or nonpolar?

All the ionic compounds are polar because the nature of the bond present between atoms is highly polar due to ionic interaction. Let us see whether NaI is polar or not.

NaI is a polar molecule as it is an ionic compound, so the nature of the bond between Na-I is polar character. Due to linear structure, the dipole-moment flows from electropositive Na to electronegative Iodine atom and there is no other dipole-moment will work. So, there is a resultant dipole-moment value.

Also, there is a huge electronegativity difference observed between Na and Iodine because one is most electropositive and the other is halogen so highly electronegative.

12. Is NaI acidic or basic?

Most of the ionic compounds are salt in nature because they don’t have any properties like acid or base. Let us see whether NaI is acidic, basic or neutral.

NaI is neither acidic nor base, rather it is an ionic inorganic salt. NaI is formed by the reaction of strong base sodium hydroxide with strong acid hydrogen iodide. So, they form salt along with water, and NaI is salt and its nature is very strong and can be ionized easily to form a strong charged particle.

13. Is NaI electrolyte?

Ionic compounds are electrolytes due to ionization in the aqueous solution and most of the salt is ionic in nature. Let us discuss whether NaI is an electrolyte or not.

NaI is a strong electrolyte because it can be ionized into Na⁺ and I^–, those ions have higher mobility and higher charge density so they can easily carry electricity through the solution when they are dissolved in a solution. Also, it is an ionic salt so therefore it can carry electricity.

14. Is NaI ionic or covalent?

Most salt is ionic in nature and forms bonds by the complete ionization and donation of electrons and the bonds are polar. Let us see whether NaI is ionic or not.

NaI is a pure ionic molecule because it is formed by the donation of one electron by the iodide ion to the Na⁺ atom. So, the bond formed between them by the ionic interaction and the ionic potential of Na⁺ is very high so it can easily polarize larger polarizable anion iodide easily and show ionic character.

Conclusion

NaI is a strong ionic inorganic salt. It can be used as strong electrolytes. Also, it can be used in organic chemistry for the transformation of alkyl chloride to alkyl iodide, because it is a good source of iodine.

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Biswarup Chandra Dey

Hi……I am Biswarup Chandra Dey, I have completed my Master’s in Chemistry from the Central University of Punjab. My area of specialization is Inorganic Chemistry. Chemistry is not all about reading line by line and memorizing, it is a concept to understand in an easy way and here I am sharing with you the concept about chemistry which I learn because knowledge is worth to share it.