While coming to NO2(-), that is, nitrite ion, the N-atom has sp2 hybridization; thus, it adopts the bent geometry, for NO2(-), and the actual O-N-O bond angle is 115 ° (slightly deviated from the expected 120° due to the repulsion between lone pair of electrons and interacting bond pairs). N2O3: sp^2 hybridization. sp Hybridization. We call it as a radical - fairly unstable, prone to stealing a single electron from the other molecule (creating another radical and continuing - this kind of chain reaction is how we destroyed the ozone layer almost with a small number of chemicals.) For example there is a question.. Since NO2 has an extra electron in an orbital on the nitrogen atom it will result in a higher degree of repulsions. Hydrolysis. If there happens to be a surplus electron around, NO2 is stabilized by adding it on. So, the repulsions are unidentical. In nitrogen dioxide, there are 2 sigma bonds and 1 lone electron pair. Therefore, the bond order of both N and O bonds is 2. There are 17 valence electrons to account for. After drawing the diagram, we need to count the number of electron pairs and the bonds present in the central nitrogen atom. That’s why nitrite exists and why its charge is -1. What is the hybridization of nitrogen in each of the following: (a) \mathrm{NO} ;(\mathrm{b}) \mathrm{NO}_{2} ;(\mathrm{c}) \mathrm{NO}_{2}^{-} ? Dear Student, In NO 2, Nitrogen atom needs three hybridised orbitals to accomodate two sigma bonds and a single electron, so it has sp 2 hybridisation. As that set of atoms are more stable in that configuration while an additional electron is available, the “octet rule” clarifies that an atom is more stable if it has a filled outermost orbital of s and p. There are one s, and 3 p’s, and each orbital carries 2 electrons, so, as a result, eight is the good number. For sp2 hybridization, there must be either 3 sigma bonds or two sigma bonds and one lone pair of electrons in the molecules or ions.In BF3 molecule, a number of sigma bond is 3 ie, sp2 hybridization. If we apply the hybridization rule now, then it states that if the sum of the number of sigma bonds, the electrons’ lone pair, and odd electrons is equal to three, then the hybridization is sp. The first step in determining hybridization is to determine how many "charge centres" surrounds the atoms in question, by looking at the Lewis structure. This molecule is tetrahedral in structure as well as in shape, since there are no lone pairs and the number of σ-bonds is equal to the steric number. Check this link to learn drawing of NO 2 molecule. However, this atom does not have an octet as it is short on electrons. There are two regions of valence electron density in the BeCl 2 molecule that correspond to the two covalent Be–Cl bonds. has 17 valence electrons. This results in sp2 hybridization. Here we will notice that the nitrogen atom is the centre atom and has only one lone electron. 3) On dividing by 8 you will get a remainder and quotient, add quotient + (remainder / 2) 4) Now get the hybridization corresponding to the number what you got. STEP-5: Assign hybridization and shape of molecule . Check this link to learn drawing of NO 2 molecule. We will also find that in nitrogen dioxide, there are two sigma bonds and one lone electron pair. Since we consider odd electron a lone pair like in $\ce{NO2}$ therefore hybridisation is coming to be $\ce{sp^3}$. A few basic key points that are to remember are listed below. The hybridization of the central atom was developed to explain the geometry of simple molecules and ions. It means it will only have 7 valence electrons. As suggested by the weakness of the N–O bond, NO 2 is a good oxidizer. Therefore NO 2 F is polar. 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