Nitrogen has no d-orbital in its valence shell and therefore it cannot1) Exhibits the oxidation state of +52) Have covalency greater than 33) Exhibit orbital Hybridization4) Form oxides with oxidation states greater than +5Select the correct answer using codes given below:
Nitrogen, being a p-block element, has the electronic configuration \(1s^2 2s^2 2p^3\). Its valence shell does not possess any d-orbitals, limiting its ability for certain chemical behavior and states. Let us evaluate the statements provided in the question:
Exhibits the oxidation state of +5: Nitrogen can have a maximum oxidation state of +5 due to the complete utilization of its five valence electrons. For example, in \(NF_3\), nitrogen exhibits this oxidation state.
Have covalency greater than 3: Covalency relates to the number of shared electron pairs. Given the lack of d-orbitals, which could be involved in bonding, nitrogen typically cannot exceed a covalency of 4. However, realistically in stable compounds, it is usually not seen sharing more than three bonds, e.g., in \(NH_3\), where nitrogen has a covalency of 3. Therefore, nitrogen can have a maximum covalency of 4 but does not have more than 3 in stable common compounds.
Exhibit orbital Hybridization: Despite having no d-orbitals, nitrogen can undergo hybridization such as \(sp^3\) in molecules like ammonia (\(NH_3\)). Thus, it does exhibit orbital hybridization.
Form oxides with oxidation states greater than +5: The maximum oxidation state nitrogen can reach is +5, as seen in compounds like \(N_2O_5\) (dinitrogen pentoxide). Therefore, nitrogen does not form oxides with oxidation states greater than +5.
Considering these analyses, we can conclude that out of the provided statements, only statement 2 is incorrect. Thus, the correct option is:
