LiBr Lewis Structure & Characteristics (13 Helpful Facts)

LiBr or lithium bromide is the halogenated inorganic salt of alkali metal having a molecular weight of 86.84 g/mol. Let us discuss about LiBr in detail.

LiBr is an ionic salt that adopts a cubic crystal structure where each Lithium-ion is surrounded by four Br atoms and also each Br atom is surrounded by four Li atoms. It can be prepared by the reaction of hydrobromic acid and lithium hydroxide, so the nature of the salt is very strong.

LiBr is hygroscopic and magnetic in nature having magnetic susceptibility of -34.3 *10^-6 cm³/mol. In this article, we will learn about lewis structure, polarity, bond angle, hybridization, ionic nature, acidity, and molecular shape of the LiBr with proper explanation in the following part.

1.     How to draw LiBr 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 LiBr.

Counting the total valence electrons

The total 8 electrons are the valence electrons for LiBr, which are involved in the bond formation and other chemical properties of the molecule. Basically, Li and Br contributed a total of 8 electrons from their valence orbital and added them together. Those electrons are considered valence electrons.

Choosing the central atom

After counting the total valence electrons, in the 2^nd step, there is a required for the central atom. to choose the central atom some conditions should be verified, one is electronegativity and the other is size. Based on all the considerations we have select Bromine as the central atom in the LiBr molecule.

Satisfying the octet rule

Each atom should follow the octet while bonding formation by completing their valence electrons with suitable electrons. Li and Br 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 LiBr 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 LiBr to satisfy the valency. Li and Br both are monovalent.

Assign the lone pairs

Li is lack lone pairs because it has only one valence electron which is satisfied by its valency. Br is satisfied here by its stable monovalency where it forms only one single bond with li by sharing one electron. The remaining six valence electrons are exist as lone pairs by the paired up form.

2.     LiBr 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 LiBr.

Total valence electrons for the LiBr are 8, which electrons are involved in the bond formation and present in the outer shell. We just count the valence electrons of each atom and then added them together by multiplying their stoichiometric proportion in the molecule.

• The valence electron present over the Li atom is 1 (because it contains only 1 electron in its valence s orbital)
• The valence electrons present over the Bromine atom are 7 (because it contains 7 electrons in its valence 5s and 5p orbital together)
• So, the total number of valence electrons present over the LiBr molecule is, 1+7 = 8

3.     LiBr 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 LiBr.

There are 3 pairs of lone pairs present over the LiBr molecule which are the remaining valence electrons after bond formation. All the lone pairs are contributed from the Bromine site because Li has 0 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 Li atom are, 1-1 = 0
• The lone pairs present over the Br atom are, 7-1 = 6
• So, the total lone pairs present over the LiBr molecule are 6 electrons or three pairs of lone pairs.

4.     LiBr 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 LiBr.

In LiBr each atom has not filled the valence orbital so it follows the octet to complete its valence orbital by electrons. 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 LiBr, there will be a minimum of 1 bond present between Li and Bromine, and if needed then an extra bond will be formed as per valency. But via 1 bond Li and Bromine both complete their valence orbital and octet also.

5.     LiBr lewis structure shape

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

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

Molecular Formula	No. of bond pairs	No. of lone pairs	Shape	Geometry
AX	1	0	Linear	Linear
AX2	2	0	Linear	Linear
AXE	1	1	Linear	Linear
AX3	3	0	Trigonal planar	Trigonal Planar
AX2E	2	1	Bent	Trigonal Planar
AXE2	1	2	Linear	Trigonal Planar
AX4	4	0	Tetrahedral	Tetrahedral
AX3E	3	1	Trigonal pyramidal	Tetrahedral
AX2E2	2	2	Bent	Tetrahedral
AXE3	1	3	Linear	Tetrahedral
AX5	5	0	trigonal bipyramidal	trigonal bipyramidal
AX4E	4	1	seesaw	trigonal bipyramidal
AX3E2	3	2	t-shaped	trigonal bipyramidal
AX2E3	2	3	linear	trigonal bipyramidal
AX6	6	0	octahedral	octahedral
AX5E	5	1	square pyramidal	octahedral
AX4E2	4	2	square pyramidal	octahedral

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.     LiBr 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 LiBr.

The bond angle for the LiBr molecule is 180⁰, which is a 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 Li and Bromine and no other elements are present so the possible bond angle makes by two atoms is 180⁰.

• By the help of hybridization value we can merge the theoretical and calculated bond angle value for LiBr.
• The bond angle formula according to Bent’s rule is COSθ = s/(s-1).
• The Bromine is sp³ hybridized but due to linear geometry, it adopts sp hybridization.
• The central atom Bromine 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.     LiBr 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 LiBr.

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

• Let us see how 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 Li⁺ ion is 1-0-(0/2) = +1
• The formal charge present over the Bromide ion is, 7-8-(0/2) = -1
• So, the formal charge of Li⁺ 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.     LiBr hybridization

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

The central Br is sp³ hybridized here which is conformed from the following table,

Structure	Hybridization value	State of hybridization of central atom	Bond angle
1.Linear	2	sp /sd / pd	1800
2.Planner trigonal	3	sp2	1200
3.Tetrahedral	4	sd3/ sp3	109.50
4.Trigonal bipyramidal	5	sp3d/dsp3	900 (axial), 1200(equatorial)
5.Octahedral	6	sp3d2/ d2sp3	900
6.Pentagonal bipyramidal	7	sp3d3/d3sp3	900,720

• We can calculate the hybridization by the convention formula, H = 0.5(V+M-C+A),
• So, the hybridization of central Bromine 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 Bromine are involved in the hybridization.

9.     LiBr solubility

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

LiBr 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 LiBr is higher than its bond enthalpy, so when it breaks into ions then the water molecule is attracted to the Li⁺ ions as it has a higher ionic potential.

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

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

10. Is LiBr 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 LiBr is solid or not.

LiBr 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 LiBr.

11. Is LiBr 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 LiBr is polar or not.

LiBr is a polar molecule as it is an ionic compound, so the nature of the bond between Li-Br is polar character. Due to linear structure, the dipole-moment flows from electropositive Li to electronegative Bromine 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 Li and Bromine because one is most electropositive and the other is halogen so highly electronegative.

12. Is LiBr 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 LiBr is acidic, basic or neutral.

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

13. Is LiBr 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 LiBr is an electrolyte or not.

LiBr is a strong electrolyte because it can be ionized into Li⁺ and Br^–, 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 LiBr 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 LiBr is ionic or not.

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

Conclusion

LiBr is an ionic inorganic salt that can be used as a desiccant in an air controlling system. In different organic syntheses, LiBr is used as a reagent, it is also used in catalysts for the oxidation and hydroformylation reactions.

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