Cyanate [OCN]- Lewis structure, molecular geometry or shape, electron geometry, bond angle, hybridization, formal charges, polar vs non-polar
The cyanate ion is represented by the chemical formula [OCN]–, also usually written as OCN–. It is a Lewis base. It is made up of three different atoms i.e., nitrogen, oxygen, and carbon.
Both nitrogen and oxygen carry lone pairs of electrons, therefore OCN– can act as an electron donor in acid-base reactions. It is also a very useful ambidentate ligand in organometallic chemistry.
Considering the versatility of the cyanate ion, we will introduce you to OCN–, starting from the basics including how to draw its Lewis dot structure, what is its molecular geometry or shape, electron geometry, bond angle, hybridization, formal charges, polarity nature, etc.
|Name of Molecule||Cyanate|
|Molecular geometry of OCN–||Linear|
|Electron geometry of OCN–||Linear|
|Total Valence electron in OCN–||16|
|Overall Formal charge in OCN–||-1|
How to draw lewis structure of OCN-?
The Lewis structure of [OCN]– is composed of three different elemental atoms i.e., a carbon (C) atom at the center which is bonded to an oxygen (O) atom, and a nitrogen (N) atom, one on either side.
There are a total of two-electron density regions around the central C-atom in the OCN– Lewis structure. Both the electron density regions are bond pairs which denote there is no lone pair of electrons on the central C-atom in the OCN– Lewis dot structure.
Drawing the Lewis dot structure of OCN– is not a difficult task at all. So you may grab a paper and pencil and draw this Lewis structure along with us, using the following simple steps.
Steps for drawing the Lewis dot structure of [OCN]–
1. Count the total valence electrons in OCN–
The Lewis dot structure of a molecule is referred to as a simplified representation of all the valence electrons present in it. Therefore, the very first step while drawing the Lewis structure of OCN– is to count the total valence electrons present in the concerned elemental atoms.
As OCN– consists of three different elements i.e., oxygen (O), nitrogen (N), and carbon (C) so you just need to look for these elements in the Periodic Table.
Carbon (C) is present in Group IV A so it has a total of 4 valence electrons. Nitrogen (N) is present in Group V A so it has 5 valence electrons while oxygen (O) is situated in Group VI A of the Periodic Table which means it has a total of 6 valence electrons in each atom.
- Total number of valence electrons in Carbon = 4
- Total number of valence electrons in Nitrogen = 5
- Total number of valence electrons in Oxygen = 6
The [OCN]– ion consists of 1 C-atom, 1 N-atom, and 1 O-atom. Thus, the valence electrons in the Lewis dot structure of [OCN]– = 1(4) + 1(5) + 1(6) = 15 valence electrons.
However, the twist here is that the [OCN]– ion carries a negative (-1) charge which means 1 extra valence electron is added in this Lewis structure.
∴ Hence, the total valence electrons available for drawing the Lewis dot structure of [OCN]– = 15+1 = 16 valence electrons.
2. Choose the central atom
In the second step of drawing the Lewis structure of a molecule or a molecular ion, we need to place the least electronegative atom at the center.
As electronegativity refers to the ability of an elemental atom to attract a shared pair of electrons from a covalent chemical bond therefore the least electronegative atom is the one that is most likely to share its electrons with other atoms.
As carbon is less electronegative than both nitrogen and oxygen, so, in the Lewis structure of [OCN]–, we will place the C atom at the center while N and O atoms are placed in its surroundings, as shown in the figure below.
3. Connect outer atoms with the central atom
Now we need to connect the outer atoms with the central atom of a Lewis structure using single straight lines. As both nitrogen and oxygen atoms are the outer atoms in [OCN]– Lewis structure while the carbon atom is the central atom, so we will connect N and O atoms with the central C-atom using straight lines, as shown below.
Each straight line represents a single covalent bond i.e., a bond pair containing 2 electrons. There are a total of 2 single bonds in the above diagram which means a total of 2(2) = 4 valence electrons are used till this step, out of the 16 initially available.
- Total valence electrons available – electrons used till step 3 = 16-4 = 12 valence electrons.
- This means we still have 12 valence electrons to be accommodated in the Lewis dot structure of [OCN]–.
4. Complete the octet of outer atoms
As we already identified, the nitrogen and the oxygen atoms act as outer atoms in OCN– Lewis structure and both N and O atoms need a total of 8 valence electrons to achieve a stable octet electronic configuration. Single bonds with the central C-atom on each side of the ion show both N and O atoms already have 2 electrons. So, both N and O atoms require 6 more valence electrons to complete their octet.
Thus, these 6 electrons are placed as 3 lone pairs around each outer atom in OCN– Lewis structure, as shown in the figure below.
5. Complete the octet of the central atom and make a covalent bond if necessary
- Total valence electrons used till step 4 = 2 single bonds + electrons placed around N-atom + electrons placed around O-atom = 2(2) + 6 + 6 = 16 valence electrons.
- Total valence electrons available – electrons used till step 4 = 16-16= 0 valence electrons.
As all the 16 valence electrons initially available are already used up in drawing the Lewis dot structure of OCN– so there is no lone pair on the central C-atom.
But the problem here is that there are only 2 single bonds around the central C-atom which mean there are only 4 valence electrons around it. This denotes that this carbon has an incomplete octet and it still needs 4 more electrons to achieve a stable octet.
We can solve this problem by converting 2 lone pairs present on an outer atom into covalent bonds between the central C-atom and the concerned outer atom.
Now if we have to choose which of the two outer atoms (N or O) will agree to share its electrons with the central atom to fulfill its deficiency? Then, we can again take into account the electronegativity factor.
Both nitrogen and oxygen are highly electronegative atoms but in comparison, the electronegativity of N (E.N = 3.04) is less than that of O (E.N = 3.44). Hence, nitrogen is more likely to convert its non-bonded electrons into bond pairs in the OCN– Lewis structure. So, 2 lone pairs of the outer N-atom are converted into bond pairs between C and N atoms.
In this way, there is a triple covalent bond between the central C and the outer N atom. The central C-atom now has a complete octet (1 triple bond +1 single bond). The octets of the outer N and O atoms are also complete with 1 triple bond and a lone pair on nitrogen and 1 single bond and 3 lone pairs on oxygen respectively.
The final step is to check the stability of the Lewis structure obtained till step 5. Let us do that using the formal charge concept.
6. Check the stability of the OCN– 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 cyanate [OCN]– ion.
For carbon atom
- Valence electrons of carbon = 4
- Bonding electrons = 1 single bond + 1 triple bond = 2 + 3(2) = 8 electrons
- Non-bonding electrons = no lone pair = 0 electrons
- Formal charge = 4-0-8/2 = 4-0-4 = 4-4 = 0
For nitrogen atom
- Valence electrons of nitrogen = 5
- Bonding electrons = 1 triple bonds = 3 (2) = 6 electrons
- Non-bonding electrons = 1 lone pair = 2 electrons
- Formal charge = 5-2-6/2 = 5-2-3 = 5-5 = 0
For oxygen atom
- Valence electrons of oxygen = 6
- Bonding electrons = 1 single bond = 2 electrons
- Non-bonding electrons = 3 lone pairs = 3(2) = 6 electrons
- Formal charge = 6-6-2/2 = 6-6-1 = 6-7 = -1
This calculation shows that zero formal charges are present on the carbon and nitrogen atoms in OCN– Lewis structure. However, a -1 formal charge is present on the oxygen atom which is also the charge present on the cyanate ion, overall.
Henceforth, the OCN– Lewis structure is enclosed in square brackets and a negative 1 charge is placed at the top right corner, as shown below. This ensures that it is a correct and stable Lewis representation for the cyanate [OCN]– ion.
A fascinating fact to remember is that the actual structure of OCN– is a hybrid of the following resonance structures. Each resonance structure is a way of representing the Lewis structure of a molecule or a molecular ion. There are 3 most common resonance structures of OCN–.
These three resonance forms are non-equivalent as the negative charge keeps circulating from one position to another, along with the delocalization of the triple covalent bond. Consequently, the magnitude of charge present on an outer atom differs in different resonance forms. However, the overall formal charge on the cyanate ion stays the same i.e., -1.
Because of these resonance forms, the cyanate (OCN–) ion is also sometimes written as NCO– and is called isocyanate. But, don’t get confused if you find written somewhere CNO–, because OCN– and NCO– may be the same in connectivity but CNO– is a completely different chemical entity. Rather, CNO– is known as a fulminate ion, it is a much less stable isomer of the cyanate ion.
Now, that we have discussed everything about cyanate ion Lewis structure, we are good to proceed to the next section of this article where we will discuss the shape and geometry of [OCN]–.
Also check –
What are the electron and molecular geometry of OCN-?
The cyanate [OCN]– ion has an identical electron and molecular geometry or shape i.e., linear. There is no lone pair present on the central C-atom in the cyanate molecule therefore there is no distortion present in the shape and geometry of the molecule.
Molecular geometry of [OCN]–
The cyanate [OCN]– ion has a linear shape and molecular geometry. Both the nitrogen and oxygen atoms bonded to the central C-atom lie in a straight line, in a planar arrangement.
There is no lone pair of electrons on the central carbon atom therefore no bond pair-lone pair and lone pair-lone pair repulsions exist in the molecular ion. O-C and C-N bond pair-bond pair electronic repulsions exist which push the bond pairs as far apart from each other as possible and the O and N atoms occupy the terminals, as shown in the figure below.
Electron geometry of [OCN]–
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 2 electron density regions around the central atom is linear.
In the cyanate [OCN]– ion, there is 1 single bond and 1 triple bond around the central carbon atom which makes a total of 2 electron density regions. Thus, its electron geometry is also linear.
An easy way to find the shape and geometry of the molecule is to use 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 [OCN]– molecular ion
- A in the AXN formula represents the central atom. In the [OCN]– ion, carbon (C) is present at the center so A = Carbon.
- X denotes the atoms bonded to the central atom. In [OCN]–, 1 nitrogen (N) atom and 1 oxygen (O) atom are bonded to the central C so X = 1+1 =2.
- N stands for the lone pairs present on the central atom. As per the Lewis structure of [OCN]– there is no lone pair on central carbon so N=0.
Thus, the AXN generic formula for the [OCN]– ion is AX2.
Now, you may have a look at the VSEPR chart below.
The VSEPR chart reaffirms that the ideal electron geometry and molecular geometry or shape of a molecule with AX2 generic formula are identical i.e., linear, as we already noted down for the [OCN]– ion.
Hybridization of [OCN]–
The cyanate [OCN]– ion has sp hybridization.
The electronic configuration of a carbon (C) atom is 1s2 2s2 2p2.
During chemical bonding, the 2s electrons get unpaired and one of the two 2s electrons of carbon shifts to the empty 2p orbital. This results in half-filled 2s and 2p atomic orbitals.
Consequently, one 2s and one 2p orbital mix to yield two equivalent sp hybrid orbitals. Each sp hybrid orbital contains a single electron only and possesses a 50% s-character and a 50% p-character.
One sp hybrid orbital of carbon forms the C-O sigma (σ) bond while the other sp hybrid orbital forms the C-N sigma (σ) bond by sp-p and sp-sp overlap with oxygen and nitrogen atoms respectively.
The unhybridized p orbitals of carbon form the required pi (π) bonds in the C-N triple bond by p-p side-by-side overlap.
The outer nitrogen atom is also sp hybridized in OCN– which uses one of the two sp hybrid orbitals for the C-N sigma bond, as discussed above, while the other sp hybrid orbital contains a lone pair of electrons, while uses its unhybridized orbitals for C-N pi bonds, as shown in the figure below.
A shortcut to finding 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 C in [OCN]– is 2 so it has sp hybridization.
The [OCN]– bond angle
As all three bonded atoms (O, C, and N) lie on a straight line in the linear [OCN]– ion therefore they form a mutual bond angle of 180°. The C-O single bond is longer in length as compared to the stronger and shorter in length C-N triple covalent bond.
Also check:- How to find bond angle?
Is OCN- polar or nonpolar?
Both O-C and C-N bonds are individually polar in the cyanate [OCN]– ion. Pauling’s electronegativity scale states that a bond containing an electronegativity difference between 0.5 to 1.6 units between the bonded atoms is polar in nature.
An electronegativity difference of 0.89 units exists between an oxygen (E.N = 3.44) and a carbon (E.N = 2.55) atom while an electronegativity difference of 0.49 units exists between the nitrogen (E.N = 3.04) and carbon atoms.
Therefore, both O-C and C-N bonds are polar and possess a specific dipole moment value (symbol µ). However, the O-C bond is strongly polar while the C-N bond is only weakly polar.
Although the OCN– is linear in shape as the the C-N dipole moment is much smaller than an O-C dipole moment therefore the two dipole moments do not get canceled in the molecular ion overall. Consequently, the electron cloud stays non-uniformly distributed over the ion which makes it a polar ion (net µ = 1.622 D).
Read in detail–
What is the Lewis structure for OCN–?
There are 3 lone pairs of electrons on the single-bonded oxygen atom, 1 lone pair on the triple-bonded nitrogen atom, and no lone pair on the central C-atom.
What are the possible number of Lewis structures for OCN–?
There are 3 possible resonance structures for OCN–. Each resonance structure is a way of representing the Lewis structure of OCN–. The best possible Lewis structure of OCN– is a hybrid of these resonance structures.
What are the electron and molecular geometry of OCN–?
|The cyanate [OCN]– ion possesses an identical electron and molecular geometry or shape i.e., linear. The O-C-N bond angle is 180° in this linear shape.|
Why does the OCN– ion have a shape identical to its ideal electron pair geometry?
There is no lone pair of electrons present on the central C-atom in the OCN– molecular ion. Thus, the AXN generic formula for OCN– is AX2 so according to the VSEPR concept, it has an identical shape to its ideal electron pair geometry i.e., linear.
There is a single bond and a triple covalent bond around the central C-atom in OCN– and no lone pair of electrons is present on this central atom thus no lone pair-bond pair and/or lone pair-lone pair electronic repulsions are present in the ion that may distort its shape.
Why is the fulminate CNO– ion less stable than the cyanate OCN– ion?
The central nitrogen (N) atom carries a +1 formal charge in the fulminate CNO– ion while a -1 charge formal is present on each of the outer C and O atoms.
On the other hand, a -1 formal charge is present on the outer O-atom while the outer N and the central C-atoms carry no formal charge in the cyanate OCN– ion.
The fewer the formal charges on the bonded atoms, the greater the stability of that structure. Thus cyanate is more stable than its isomer fulminate.
Also, the second and very important factor is that the OCN– ion is resonance stabilized while CNO– is not.
- N2H4 lewis structure and its molecular geometry
- CH2Cl2 lewis structure and its molecular geometry
- CH3COOH lewis structure and its molecular geometry
- C2H2Cl2 lewis structure and its molecular geometry
- CHCl3 lewis structure and its molecular geometry
- CH3F lewis structure and its molecular geometry
- CF2Cl2 lewis structure and its molecular geometry
- CH3CN lewis structure and its molecular geometry
- CH2O lewis structure and its molecular geometry
- The total valence electrons available for drawing cyanate [OCN]– ion Lewis structure are 16.
- The negative 1 charge present on the ion accounts for 1 extra valence electron in its Lewis structure.
- The Lewis structure of [OCN]– is composed of three different elemental atoms i.e., Carbon, Oxygen, and Nitrogen.
- The [OCN]– ion has an identical electron and molecular geometry or shape i.e., linear.
- The O-C-N atoms lie in a planar arrangement, on a straight line therefore the O-C-N bond angle is 180° in OCN–
- The OCN– ion has sp hybridization.
- The OCN– ion is overall polar in nature. Both O-C single bond and the C-N triple bond are polar as an electronegativity difference exists between the bonded atoms. The molecule possesses a linear shape but the dipole moments of O-C and C-N bonds do not get canceled equally so net µ > 0.
- -1 formal charge is present on the oxygen atom while zero formal charges are present on the nitrogen and carbon atoms which makes the cyanate ion an anion possessing an overall negative charge.
About the author
Vishal Goyal is the founder of Topblogtenz, a comprehensive resource for students seeking guidance and support in their chemistry studies. He holds a degree in B.Tech (Chemical Engineering) and has four years of experience as a chemistry tutor. The team at Topblogtenz includes experts like experienced researchers, professors, and educators, with the goal of making complex subjects like chemistry accessible and understandable for all. A passion for sharing knowledge and a love for chemistry and science drives the team behind the website. Let's connect through LinkedIn: https://www.linkedin.com/in/vishal-goyal-2926a122b/