[CN]– is the chemical formula for the cyanide ion. A carbon (C) atom triple bonded to a nitrogen (N) atom is known as the cyano or a nitrile group. Organic compounds containing the cyanide ion smell like bitter almonds and are extremely toxic to living organisms.
In this article, we will discuss how to draw the Lewis dot structure and the molecular orbital diagram of the cyanide [CN]– ion. You will also learn other interesting facts about the molecular geometry or shape, electron geometry, bond order, formal charges, hybridization, polarity, etc. of the CN– ion.
So without any further delay, let’s start reading!
| Name of Molecule | Cyanide ion |
| Chemical formula | [CN]– |
| Molecular geometry of [CN]– | Linear |
| Electron geometry of [CN]– | Linear |
| Hybridization | sp |
| Nature | Polar molecule |
| Bond angle (C-N) | 180° |
| Total Valence electron in [CN]– | 10 |
| Overall Formal charge in [CN]– | -1 |
| Bond order of [CN]– | 3 |
The Lewis structure of a cyanide [CN]– ion consists of a carbon (C) atom and a nitrogen (N) atom. The two atoms are connected via a triple covalent bond. There are a total of 3 bond pairs and 1 lone pair around both C and N atoms respectively in CN– lewis structure.
The 3 bond pairs are considered a single electron domain while determining the shape and/or geometry of the molecular ion.
In this article, we will discuss the Lewis structure of CN and CN+ as well, in addition to drawing the Lewis structure of CN–.
But let us first start with the simple steps given below to draw the Lewis dot structure of [CN]–.
Steps to draw the lewis dot structure for CN–
1. Count the total valence electrons in [CN]–
The Lewis dot structure of a molecule or a molecular ion is a simplified representation of all the valence electrons present in it. Therefore, the very first step while drawing the Lewis structure of the cyanide [CN]– ionis tocount the total valence electronspresent in its concerned elemental atoms.
There are 2 different elements present in [CN]– i.e., carbon (C) and nitrogen (N). When you search through the Periodic Table of elements, you will find that carbon is present in Group IV A so it has a total of 4 valence electrons in each atom. In contrast to that, nitrogen belongs to Group V A of the Periodic Table so it has a total of 5 valence electrons in each atom.
- Total number of valence electrons in carbon = 4
- Total number ofvalence electrons in nitrogen= 5
∴[CN]– consists of 1 C-atom, 1 N-atom, and -1 charge which accounts for 1 extra valence electron. Thus, the valence electrons present in [CN]– Lewis structure = 4 + 5 + 1 = 10 valence electrons.
2. Select the central atom
In the cyanide [CN]– Lewis structure, we do not need to find the least electronegative atom to choose the central atom as we do so in other molecules or molecular ions. As there are only two atoms present in [CN]– so we place them next to each other, as shown below.
3. Connect the two atoms using a single straight line
In this step, we need to connect the carbon and nitrogen atom using a single straight line, as shown below.
The straight line represents a single covalent bond i.e., a bond pair containing 2 electrons. 2 electrons consumed out of the 10 initially available leaves behind 10-2 = 8 valence electrons. So, we still need to accommodate these 8 electrons in the Lewis dot structure of CN–. Let’s see how we can do so in the upcoming steps.
4. Complete the octet of the more electronegative atom
Nitrogen is more electronegative than carbon. A nitrogen (N) atom needs a total of 8 valence electrons in order to achieve a stable octet electronic configuration. A C-N single bond represents that this N-atom already has 2 electrons.
It is thus short of 6 more electrons to complete its octet. Thus, these 6 electrons are placed as 3 lone pairs around the N-atom, as shown below.
5. Complete the octet of the less electronegative atom
- Total valence electrons used till step 4 = 1 single bond + electrons placed around the N-atom (shown as dots) = 2 + 6 = 8 valence electrons.
- Total valence electrons available – electrons used till step 4 = 10 – 8 = 2.
So these two valence electrons are now placed as a lone pair on the less electronegative carbon atom.
But the problem here is that in the above structure, the C-atom has 1 single bond and 1 lone pair which makes a total of 2 +2 = 4 valence electrons. This means it is still short of 4 valence electrons in order to achieve a complete octet.
This problem can be solved if we convert 2 lone pairs present on the N-atom into two covalent chemical bonds between the bonded C and N-atoms, as shown below.
The figure above shows that both the C and N-atoms now have a complete octet having 1 triple bond + 1 lone pair each.
As a final step, we need to check the stability of this Lewis structure. Let’s do that using the formal charge concept.
6. Check the stability of [CN]–Lewis structure using the formal charge concept
The less the formal charge on the atoms of a molecule or molecular ion, 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)]
Let us demonstrate how we can use this formula and the Lewis structure obtained instep5to calculate the required formal charges.
For carbon atom
- Valence electrons of carbon = 4
- Bonding electrons = 1 triple bond = 3(2) = 6 electrons
- Non-bonding electrons = 1 lone pair = 2 electrons
- Formal charge = 4-2-6/2 = 4-2-3 = 4-5 = -1
For nitrogen atom
- Valence electrons of nitrogen = 5
- Bonding electrons = 1 triple bond = 3(2) = 6 electrons
- Non-bonding electrons = 1 lone pair = 2 electrons
- Formal charge = 5-2-6/2 = 5-2-3 = 5-5 = 0
The above calculation shows that a zero formal charge is present on the nitrogen atom. Contrarily, a -1 formal charge is present on the carbon atom which is also the charge present on the ion overall. Consequently, the Lewis structure obtained above is correct. It is finally enclosed in square brackets and a -1 charge is placed at the top right corner.
By following the same guidelines as given for drawing [CN]– anion Lewis structure, you can also draw the Lewis structures of the CN molecule and CN+ cation, using a total of 9 and 8 valence electrons respectively.
Now that we have successfully drawn the Lewis dot structure of CN–, let us proceed forward and find out how we can draw its molecular orbital diagram.
Also check –
- How to draw a lewis structure?
- Formal charge calculator
- Lewis structure generator
The molecular orbital theory (MOT) of chemical bonding states that after bond formation, the individual atomic orbitals combine to form molecular orbitals. In this way, the electrons are situated in molecular orbitals that belong to the entire molecule rather than their placement in an individual orbital belonging to a single atom only.
The linear combination of atomic orbitals yields two types of molecular orbitals (MOs):
- Bonding molecular orbitals
- Anti-bonding molecular orbitals
A bonding molecular orbital lies at lower energy. It is formed by the combination of two atomic orbitals in the same phase.
Contrarily, an antibonding molecular orbital is formed by the combination of two atomic orbitals in different phases. It is thus less stable and possesses higher energy than the bonding molecular orbital.
Electrons first occupy the bonding molecular orbitals lying at lower energy followed by their placement in high-energy anti-bonding molecular orbitals.
The different numbers of electrons of a molecule present in bonding and/or antibonding molecular orbitals can be represented in the form of a molecular orbital diagram.
The cyanide [CN]– ion is a heteronuclear diatomic molecular ion as it consists of 2 atoms from two different elements, chemically bonded to each other. The molecular orbital diagram of CN– can be used to determine valuable information about the chemical bonding present in CN– in terms of bond order which is in turn used to determine its bond length, bond stability, etc.
You can use the simple steps given below to draw the molecular orbital diagram of the CN– ion.
Steps for drawing the molecular orbital diagram of [CN]–
1. Find the valence electrons of each atom in [CN]–
As already determined (in the previous section), [CN]– is made up of a carbon atom and a nitrogen atom only. The C-atom consists of 4 valence electrons while a total of 5 valence electrons are present in the N-atom.
- The valence shell electronic configuration of carbon = 2s22p2
- The valence shell electronic configuration of nitrogen = 2s22p3
2. Determine whether the molecule or molecular ion is hom*onuclear or heteronuclear
The molecule (CN–) consists of two different elemental atoms. So surely, it forms a heteronuclear diatomic molecular orbital. The linear combination of two s atomic orbitals (2s orbital of carbon and 2s orbital of nitrogen) yields two molecular orbitals i.e., a bonding molecular orbital (σ2s) and an anti-bonding molecular orbital (σ*2s).
Similarly, the six p atomic orbitals, three of each atom (2p orbitals of carbon and 2p orbitals of nitrogen) combine to form six molecular orbitals. Out of these six molecular orbitals, there are three bonding molecular orbitals (σ2p and π2p) and three anti-bonding molecular orbitals (π*2p and σ*2p).
3. Fill the molecular orbitals using the energy and bonding properties of the overlapping atomic orbitals.
A total of 4 valence electrons are present in the 2s atomic orbitals of carbon and nitrogen combined. So 2 of these 4 electrons are placed in the bonding MO (σ2s) while the remaining 2 electrons are placed in anti-bonding MO (σ*2s).
One extra electron gained by carbon makes a total of 6 valence electrons in the 2p atomic orbitals (3 in carbon and 3 in nitrogen). Out of these 6 electrons, 2 electrons are first placed in π 2px and π 2py, these electrons are then paired. 6- 4 = 2 so the remaining 2 electrons are consequently placed in σ2p bonding MO, as shown below.
The highest energy anti-bonding MOs (π*2p and σ*2p) stay empty in the molecular orbital diagram of cyanide [CN]–.
Now let’s see how we can use this molecular orbital diagram to find the bond order of CN–.
Bond order of [CN]–
The bond order formula is:
∴Bond order = (Nb –Na)/2
- Nb = Electrons present in the bonding MOs (Bonding electrons).
- Na= Electrons present in the anti-bonding MOs (Anti-bonding electrons).
According to [CN]– molecular orbital diagram, Nb = 10, Na = 4.
∴ So bond order of [CN]– = (10 – 4)/2 = 6/2 = 3.
Thus, there is a triple covalent bond present between the bonded C and N atoms in the CN– ion.
The molecular orbital configuration of [CN]– is (σ1s2)(σ*1s2)(σ2s2)(σ*2s2)(π2px2)(π2py2)(σ 2p2).
The bond order of [CN]+ is 2 because it contains 12 valence electrons.
The molecular orbital diagram of CN– also reveals the diamagnetic character of a cyanide ion as it does not possess any unpaired electrons in its molecular orbitals.
The cyanide [CN]– ion has an identical electron pair geometry and molecular geometry or shape i.e., linear. There are only two atoms bonded to each other with one lone pair of electrons present on each atom.
Molecular geometry of [CN]–
The cyanide [CN]– molecular ion adopts a linear shape that allows a symmetrical distribution of all the valence electrons present in it. There is the only type of bond present in the molecule so no bond pair-bond pair electronic repulsions exist.
An equal number of lone pairs is present on both carbon and nitrogen. So these lone pairs are kept as far apart from each other as possible in order to minimize the lone pair-lone pair repulsive effect.
Electron geometry of [CN]–
There are a total of 2 electron density regions around each C and N atom in [CN]–. So, with respect to each atom, CN– has a linear electron pair geometry.
The two bonded atoms lie on a straight line and form an ideal bond angle of 180°.
The electron and molecular geometry or shape of CN– can also be determined using the AXN method.
The AXN generic formula for CN– isAXN1 or simply AXN.
The VSEPR chart given below indicates that a molecule or a molecular ion having an AXN1 generic formula has an identical electron and molecular geometry i.e., linear.
Hybridization of [CN]–
The hybridization present in [CN]– can be determined using its steric number against the table given below. The steric number of both C and N atoms in [CN]– is 2 so it has sp hybridization.
Steric number | Hybridization |
2 | sp |
3 | sp2 |
4 | sp3 |
5 | sp3d |
6 | sp3d2 |
Two sp hybrid orbitals are formed by the mixing of an s and a p atomic orbitals. Each sp hybrid orbital exhibits 50% s-character and 50% p-character.
The carbon and nitrogen atoms present in CN– use one of their two sp hybrid orbitals for C-N sigma bond formation by sp-sp overlap. While a lone pair of electrons is situated in the other sp hybrid orbital of each atom.
