Ammonium [NH4]+ ion Lewis dot structure, molecular geometry or shape, electron geometry, bond angle, hybridization, formal charges, polar vs non-polar
[NH4]+ is the chemical formula that represents the ammonium ion. The ammonium ion is an important nitrogen source for plants. It is used in the chemical industry for fertilizer manufacturing.
The ammonium [NH4]+ ion is the conjugate acid of a weak base i.e., NH3. A coordinate covalent bond between NH3 and H+ results in the formation of NH4+.
In this article, we will discuss how to draw the Lewis structure of [NH4]+, what is its molecular geometry or shape, electron geometry, bond angle, hybridization, formal charges, etc. So, let’s dive into it and start reading!
|Name of Molecular ion||Ammonium|
|Molecular geometry of [NH4]+||Tetrahedral|
|Electron geometry of [NH4]+||Tetrahedral|
|Total Valence electron in [NH4]+||8|
|Overall Formal charge in [NH4]+||+1|
How to draw lewis structure of NH4+?
The Lewis structure of ammonium [NH4]+ ion consists of a nitrogen (N) atom at the center. It is bonded to four atoms of hydrogen (H) at the sides. There are a total of 4 electron density regions around the central N-atom in [NH4]+ lewis structure.
All 4 electron density regions are constituted of N-H bond pairs which denote there is no lone pair of electrons on the central N-atom in [NH4]+ Lewis dot structure.
Drawing the Lewis dot structure of ammonium [NH4]+ ion is a super easy task if you follow the simple guidelines given below.
Steps for drawing the Lewis dot structure of [NH4]+
1. Count the total valence electrons in [NH4]+
The very first step while drawing the Lewis structure of [NH4]+ is to find the total valence electrons present in the concerned elemental atoms.
As NH4+ consists of atoms from two different elements i.e., nitrogen (N) and hydrogen (H) so you just need to look for these elements in the Periodic Table.
Nitrogen (N) is present in Group V-A of the Periodic Table so it has a total of 5 valence electrons. On the other hand, hydrogen (H) lies at the top of the Periodic Table containing a single valence electron only.
- Total number of valence electrons in Hydrogen = 1
- Total number of valence electrons in Nitrogen = 5
The [NH4] + ion consists of 1 N-atom, 4 H-atoms, and carries a positive (+1) charge which means 1 valence electron will be subtracted. Therefore, the total valence electrons available for drawing the Lewis structure of [NH4]+ = 1(5) + 4(1) – 1 = 8 valence electrons.
2. Choose the central atom
While drawing the Lewis structure of a molecule or a molecular ion, usually the least electronegative atom is chosen as the central atom.
This is because electronegativity refers to the ability of an atom to attract a shared pair of electrons from a covalent chemical bond.
So, the least electronegative atom is least likely to attract electrons and more likely to share its electrons with other atoms in its vicinity.
However, in the case of NH4+, there are only two types of atoms involved. Out of the two elemental atoms, hydrogen is undoubtedly less electronegative than nitrogen but it cannot be chosen as the central atom.
A hydrogen (H) atom can accommodate only 2 electrons so it can form a bond with a single adjacent atom only. This denotes that H is always placed as an outer atom in a Lewis structure.
Consequently, nitrogen (N) is chosen as the central atom in the [NH4]+ while the 4 H-atoms are placed in its surroundings, as shown below.
3. Connect outer atoms with the central atom
Now we need to connect the outer atoms with the central atom using single straight lines.
As hydrogen atoms are the outer atoms while the nitrogen atom is the central atom in the NH4+, so all four H-atoms are joined to the central N-atom via straight lines, as shown in the diagram below.
Each straight line represents a single covalent bond i.e., a bond pair containing 2 electrons. There are a total of 4 single bonds in the above diagram.
∴ As 4(2) = 8 valence electrons which means all 8 valence electrons available are now consumed.
4. Complete the duplet of outer atoms
Each H-atom needs a total of 2 valence electrons in order to achieve a stable duplet electronic configuration.
The Lewis structure obtained till this step already shows 2 electrons around each H-atom.
This means each H-atom already has a complete duplet and we do not need to make any changes regarding H-atoms in this structure. Also, there is no lone pair of electrons around any H-atom in the [NH4]+ Lewis dot structure.
5. Complete the octet of the central atom
- Total valence electrons available – electrons used till step 4 = 8 – 8 = 0 valence electrons.
All the valence electrons are already used up thus there is no lone pair on the central N-atom in [NH4]+ Lewis structure.
Now you may notice that in addition to a complete duplet of the outer H-atoms, the central N-atom also has a complete octet electronic configuration with 4 single bonds and no lone pair.
The final step is to check the stability of the NH4+ Lewis structure obtained. Let us do that using the formal charge concept.
6. Check the stability of the NH4+ Lewis structure using the formal charge concept
The less the formal charge on the atoms of a molecule, the better the stability of its Lewis structure.
The formal charge can be calculated using the formula given below.
- Formal charge = [ valence electrons – nonbonding electrons- ½ (bonding electrons)]
Now let us use this formula and the Lewis structure obtained in step 5 to determine the formal charges on an ammonium [NH4]+ ion.
For nitrogen atom
- Valence electrons of nitrogen = 5
- Bonding electrons = 4 single bonds = 4 (2) = 8 electrons
- Non-bonding electrons = no lone pair = 0 electrons
- Formal charge = 5-0-8/2 = 5-0-4 = 5-4 = +1
For hydrogen atom
- Valence electron of hydrogen = 1
- Bonding electrons = 1 single bond = 2 electrons
- Non-bonding electrons = no lone pairs = 0 electrons
- Formal charge = 1-0-2/2 = 1-0-1 = 1-1 = 0
The above calculation shows that a zero formal charge is present on each H-atom in the NH4+ Lewis structure. However, a +1 formal charge is present on the central N-atom which is equivalent to the charge present on the ammonium ion overall.
Therefore, the above Lewis structure is enclosed in square brackets and a +1 charge is placed at the top right corner, as shown below. The final Lewis structure obtained is the correct and most stable Lewis representation of the ammonium ion (NH4+).
Also check –
What are the electron and molecular geometry of NH4+?
The ammonium [NH4]+ ion has an identical electron pair geometry and molecular geometry or shape i.e., tetrahedral. There is no lone pair of electrons on the central N-atom in the ammonium ion thus there is no distortion witnessed in the shape and geometry of the ion.
Molecular geometry of [NH4]+
The ammonium [NH4]+ ion has a tetrahedral molecular geometry or shape.
The four hydrogen atoms lie at the four vertices of a regular tetrahedron while the nitrogen atom is present at the center. Refer to the figure below.
N-H bond pair-bond pair repulsions exist in the NH4+ molecule which keeps the terminal H-atoms as far apart from one another as possible.
However, there is no lone pair of electrons on the central N-atom therefore no lone pair-bond pair and lone pair-lone pair electronic repulsions are present in the molecule. The shape of the molecule (NH4+) thus stays intact, identical to its ideal electron pair geometry.
Electron geometry of [NH4]+
According to the valence shell electron pair repulsion (VSEPR) theory of chemical bonding, the ideal electron geometry of a molecule or a molecular ion containing a total of 4 electron density regions around the central atom is tetrahedral.
In the ammonium [NH4]+ ion, there are 4 single bonds around the central nitrogen atom which makes a total of 4 electron density regions. Thus, its electron geometry is also tetrahedral.
A shortcut to finding the electron and the molecular geometry of a molecule or a molecular ion is by using the AXN method.
AXN is a simple formula to represent the number of atoms bonded to the central atom in a molecule and the number of lone pairs present on it.
It is used to predict the geometry or shape of a molecule using the VSEPR concept.
AXN notation for [NH4]+ molecular ion
- A in the AXN formula represents the central atom. In the [NH4]+ ion, nitrogen is present at the center so A=N.
- X denotes the atoms bonded to the central atom. In [NH4]+, four hydrogens (H) atoms are bonded to the central N so X=4.
- N stands for the lone pairs present on the central atom. As per the Lewis structure of [NH4]+, there is no lone pair on central nitrogen so N=0.
So, the AXN generic formula for the [NH4]+ ion is AX4.
Now, you may have a look at the VSEPR chart below.
The VSEPR chart confirms that the ideal electron geometry and molecular geometry or shape of a molecule or molecular ion with AX4 generic formula are identical i.e., tetrahedral, as we already noted down for the [NH4]+ ion.
Hybridization of [NH4]+
The central nitrogen atom has sp3 hybridization in the [NH4]+ ion.
The electronic configuration of a nitrogen (N) atom is 1s2 2s2 2p3.
During chemical bonding, the 2s atomic orbital of nitrogen hybridizes with three half-filled 2p orbitals to yield four sp3 hybrid orbitals. Each sp3 hybrid orbital has a 25 % s-character and a 75% p-character.
Three sp3 hybrid orbitals contain a single electron only. These sp3 hybrid orbitals overlap with the s-orbital of hydrogen on each side of the ammonium ion to form three N-H sigma (σ) bonds by sp3-s overlap.
The fourth sp3 hybrid orbital contains paired electrons. The central nitrogen atom shares this electron pair with an H+ to form a coordinate covalent bond.
However, you may remember that once the coordinate covalent bond is formed it is treated the same as any other normal covalent chemical bond.
Another shortcut to find the hybridization present in a molecule or a molecular ion is by using its steric number against the table given below. The steric number of central N in [NH4]+ is 4 so it has sp3 hybridization.
The [NH4]+ bond angle
The bonded atoms in [NH4]+ ion form ideal bond angles as expected in a symmetrical tetrahedral molecule. The H-N-H bond angle is 109.5°. All the N-H bond lengths are also equal i.e., 103 pm.
Also check:- How to calculate bond angle?
Is NH4+ polar or nonpolar?
A specific electronegativity difference of 0.84 units exists between the bonded nitrogen (E.N = 3.04) and hydrogen (E.N = 2.20) atoms in each N-H bond in [NH4]+. Nitrogen being more electronegative attracts the shared electron cloud from each N-H bond.
Consequently, the central N-atom gains a partial negative (δ–) charge while each outer H-atom gains a partial positive (δ+) charge. Each N-H bond thus possesses a particular dipole moment value (symbol µ).
However, it is due to the symmetrical tetrahedral shape of [NH4]+ that the individual N-H dipole moments get canceled. The electron cloud stays uniformly distributed in the molecular ion overall thus [NH4]+ is non-polar (net µ =0).
Read in detail–
What is the NH4+ Lewis structure?
There is no lone pair on the central N-atom or on any outer H-atom in the NH4+ Lewis structure.
Why is nitrogen positively charged in NH4+ if it is more electronegative than hydrogen?
A lone pair of electrons is present on nitrogen (N) in ammonia (NH3). NH3 uses this lone pair to form a covalent bond with the fourth hydrogen in NH4+.
This covalent bond is a coordinate covalent bond where nitrogen acts as the donor while hydrogen ion is the acceptor.
As the central N-atom is acting as a donor, sharing a complete electron pair with an adjacent H-atom thus this N carries a positive charge in [NH4]+.
Why is the shape of NH3 pyramidal but the shape of NH4+ is tetrahedral?
The ammonia (NH3) molecule has a trigonal pyramidal shape due to the presence of a lone pair of electrons on the central N-atom. Lone pair-bond pair electronic repulsions exist in the molecule that pushes the N-H bonds away from the center.
Contrarily, there is no lone pair of electrons on the central N-atom in NH4+ so it has a tetrahedral shape and geometry.
Why is the bond angle of NH4+ greater than NH3?
The H-N-H bond angle in NH4+ is 109.5° because it is an ideal tetrahedral molecular ion. There is no lone pair on the central N-atom in NH4+ so no distortion is present in its shape and geometry.
On the other hand, the lone pair of electrons present on the central N-atom in NH3 leads to lone pair-bond pair repulsions, and the H-N-H bond angle decreases and becomes equal to 107.5°.
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- The total number of valence electrons available for drawing ammonium [NH4]+ ion Lewis structure is 8.
- The positive 1 charge present on the ion accounts for 1 valence electron removed in its Lewis structure.
- The [NH4]+ ion has an identical electron geometry and molecular geometry or shape i.e., tetrahedral.
- The NH4+ ion has sp3 hybridization.
- The NH4+ ion is overall non-polar (net µ= 0) due to its symmetrical shape and geometry.
- +1 formal charge is present on the central nitrogen atom while zero formal charges are present on each hydrogen atom in the NH4+ Lewis structure
- This accounts for an overall positive charge on the monovalent cation i.e., the ammonium ion.
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