So, two of those pairs form a double bond. Now, the original lone pair has become a bonding pair. There is 4 pairs of shared electrons around the central nitrogen and no lone pair remaining. The nitrate ion forms by the loss of the hydrogen ion and so its structure looks as below. If we observe the bonding around the nitrogen carefully, we will see that one of the bonds is entirely formed from the lone pair on the nitrogen. Since the nitrate ion comes from the nitric acid, so let us start from the structure of nitric acid. Thus, the order of bond length becomes NO 3 - > NO 2 - > NO 2 +

The order of the bond order can be given by, NO 2 + > NO 2 - > NO 3. Thus it has the bond order of (6–0)/2 = 3.įor NO 2 -, the bond order is given by, (6–3)/2=1.5, andįor NO 3 - the bond order is given by (6–4)/2=1 It has 6 electrons in 4th bonding orbitals and zero electrons in antibonding. Then, NO 2 + has 2 electrons, which is lesser than NO. Thus, it has the bond order of (6–1)/2 = 2.5. Now, NO molecule has 6 electrons in the bonding orbitals and 1 electron in the antibonding orbital. Now, to calculate these, we should consider the molecular order of a heteronuclear diatomic molecule NO first. So, Bond order= (Number of bonding electrons - Number of antibonding electrons)/2. The higher the bond order results, the shorter the bond length. Let us calculate the bond orders before going to find the order. īond Lengths Order For NO 2 +, NO 2 -, And NO 3. It means, for ‘number of bonds’ = 3 and the hybridization is sp 2. Referring to the table under ‘type of hybrid orbital’, this is where the hybridization should be. In the Lewis structure, we can see 3 bonds between the atoms (the dots show between atoms are the bonds). Now, we should count the number of bonds between the atoms on the Lewis structure and then check the table. Now, by using the following table, we need to find the hybridization. We should then position the bonds with the right number of electrons and fill in the lone pairs around the atoms to get the right number of total electron count (electrons, represented as dots). Now, without forgetting the '−,' we need to calculate the number of electrons present in the NO 3.

To know the central atom's hybridization in NO 3 -, let us take the approach of drawing the Lewis structure of it, which looks like follows. Hybridization of the Central Atom in NO 3. Thus, NO 3 - molecular geometry is trigonal planar and is slightly bent. In essence, nitrate has 3 electron domains with zero lone pairs. There is one central atom in nitrate which is surrounded by 3 identically-bonded oxygen atoms that lie at the triangle corners and a similar one-dimensional plane.

The oxygen atoms will also have two p orbitals that will accommodate a lone pair of electrons. The p orbital of nitrogen produces a double bond with three oxygen atoms where the three electron pairs are shared between the p orbital of the nitrogen and one p orbital of oxygen atom each. The three sp 2 orbitals of nitrogen overlap with one s orbital of the oxygen atom during bonding. Moreover, if we check the Lewis structure further, one of the nitrogen-oxygen bonds is a double bond besides the other two are of single bonds. If we see NO 3 -, the central atom is bonded with three oxygen atoms, and there exist no lone pairs. After drawing the structure, we need to count the number of electron pairs, and the bonds that exist in the central nitrogen atom. The easiest way to determine the hybridization of nitrate is by drawing the Lewis structure.

Nitrate ion bonding