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

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

Table of Contents

Key Points To Consider When Drawing The NaBr Electron Dot Structure

A three-step approach for drawing the NaBr Lewis structure can be used. The first step is to sketch the Lewis structure of the NaBr molecule, to add valence electrons around the bromine 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 NaBr Lewis Structure.

The NaBr Lewis structure is a diagram that illustrates the number of valence electrons and bond electron pairs in the NaBr molecule. The geometry of the NaBr molecule can then be predicted using the Valence Shell Electron Pair Repulsion Theory (VSEPR Theory), which states that molecules will choose the NaBr 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+—Br- single bonds (dipole moment properties of the NaBr molecule). The sodium-bromine ionic bonds in sodium bromide(NaBr), for example, are polarised toward the more electronegative bromine in NaBr 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 bromine atom as bromide ion(Br-), their sum of dipole moment is nonzero due to the NaBr molecule’s bond dipole moment and less electron polarity to the sodium atoms and become sodium ion. Because Na+—-Br- bonds polarity is not canceled in the NaBr molecule due to the presence of four lone pairs of electrons in the linear structure. The sodium bromide(NaBr) molecule is classified as a polar ionic molecular crystal.

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

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

NaBr molecule has one Na+—–Br- 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 NaBr 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: NaBr Lewis Structure

The central atom is bromine, which is bordered on two terminals with sodium atoms(linear structural geometry), and four lone pairs on the central bromine atom in the linear molecular geometry. Bromine 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 bromine atom requires one valence electron on each of their outermost shell.

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

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

Lewis structure of NaBr has dot electron representative structure. Valence electrons of atoms undergo orbitals mixing in the chemical reactions, giving new types of molecular species of NaBr. 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 NaBr Molecule:

Bromine and sodium Electronegative difference in NaBr:

The bromine atom has an electronegativity of 2.96, while sodium has an electronegativity of 0.93 in the NaBr ionic molecule. The difference in electronegativity of bromine and sodium can be estimated using the method below.

The electronegative value difference between bromine and sodium in NaBr molecule
Electronegativity value of bromine= 2.96
Electronegativity value of sodium= 0.93
Difference of electronegativity value between bromine and sodium in NaBr molecule = 2.96 – 0.93 = 2.03
Electronegativity difference between Na-Br ionic bond calculation of NaBr molecule

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

Because of this difference in electronegativity of bromine and sodium atoms, the NaBr molecule’s Na+—Br- bond becomes nonpolar. The total net dipole moment of the NaBr 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-Br bond’s dipole moment is higher due to the polarization of the bonds and four lone pairs of electrons on bromine, and all Na-Br bonds’ dipoles are arranged in the asymmetrical NaBr molecular geometry. The NaBr molecule has a nonzero net dipole moment.

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

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

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

To sketch the NaBr Lewis structure by following these instructions:

Step-1: NaBr Lewis dot Structure by counting valence electrons on the bromine atom

To calculate the valence electron of each atom in NaBr, 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 bromine and sodium atoms respectively. In their outermost shells, sodium and bromine have one and seven valence electrons respectively.

Calculate the total number of valence electrons in the NaBr molecule’s outermost valence shell. The first step is to determine how many electrons are in the NaBr 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 NaBr Lewis diagram. The NaBr molecule’s core bromine atom can be represented as follows:

Total outermost valence shell electron of bromine atom in NaBr= 7
Total outermost valence shell electron of sodium atom in NaBr= 1
The NaBr molecule has one central bromine and one sodium atoms. Then the total outermost valence shell electrons can be calculated as follows
∴ Total outermost valence shell electrons available for NaBr Lewis structure( dot structure) = 7+1*1= 8 valence electrons in NaBr.
calculation of total valence electron of NaBr molecule

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

But in this case, sodium is the least electronegative than bromine. Sodium loses one electron and forms sodium positive ions(Na+). So that bromine 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.

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

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

As a result, bromine is the third 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 bromine atom is higher than that of the sodium atom in the NaBr molecule. Furthermore, sodium has a one-electron limit since it is the less electronegative element in the NaBr molecule.

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

Bromine accepts one electron and forms a bromide ion(Br-). The total lone pair of electrons in the bromide ion is eight. They are negatively charged.

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

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

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

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

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

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

The bromine atom in the molecule gets only 8 electrons around its molecular structure. This central bromine atom is octet stable. But it has three lone pairs. Bromine gas(Br2) is a brownish liquid gas. Bromine is very corrosive in nature. It is one of the very reactive chemical reagent.

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.

Bromine requires 8 electrons in its outermost valence shell to complete the molecular octet stability, two electrons bond pairs in one Na-Br single bond, and three lone pairs in the central bromine atom. No lone pairs of electrons on the sodium atoms of the NaBr molecule are placed in a linear geometry. Bromine already shares 8 electrons to the one Na-Br 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 NaBr molecular structure above. The bromine atom completes its molecular octet stability in the NaBr molecule because it possesses 2 electrons in its (one Na-Br single ionic bond pairs) bond pairs with one sodium in the outermost valence shell.

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

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

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

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

How to calculate the formal charge on bromine and sodium atoms in NaBr Lewis Structure?

Calculating formal charge on the bromine of NaBr molecule:

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

To calculate the formal charge on the central bromine atom of the NaBr molecule by using the following formula:
The formal charge on the bromine atom of NaBr molecule= (V. E(Br)– L.E(Br) – 1/2(B.E))
V.E (Br) = Valence electron in a bromine atom of NaBr molecule
L.E(Br) = Lone pairs of an electron in the bromine atom of the NaBr molecule.
B.E = Bond pair electron in Br atom of NaBr molecule
calculation of formal charge on bromine atom in NaBr molecule

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

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

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

Calculating formal charge on the sodium atom of NaBr molecule:

The formal charge on the NaBr 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 NaBr Lewis dot structure.

To calculate the formal charge on the terminal sodium atom of the NaBr molecule by using the following formula:
The formal charge on the sodium atom of NaBr molecule= (V. E(Na)– L.E(Na) – 1/2(B.E))
V.E (Na) = Valence electron in a sodium atom of NaBr molecule
L.E(Na) = Lone pairs of an electron in the sodium atom of the NaBr molecule.
B.E = Bond pair electron in H atom of NaBr molecule
calculation of formal charge on sodium atom in NaBr molecule

The sodium terminal atoms of the NaBr 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 NaBr molecule = (1- 0-(0/2)) =+1

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

Summary:

In this post, we discussed the method to construct the NaBr 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 bromine atoms and the bromine atom accepts one electron in its valence shell. Finally, when we combined the first and second steps. It gives NaBr Lewis structure. Need to remember that, if you follow the above-said method, you can construct molecular dot structure very easily.