What type of orbitals are used to form the N–N bond in hydrazine, N2H4?
N is Nitrogen, H is Hydrogen
I think you are looking for N-H and N-N bonds as sigma bonds.
To determine the type of orbitals used to form the N-N bond in hydrazine (N2H4), we first need to understand the electronic configuration of nitrogen (N) and how it bonds with other atoms.
Nitrogen has the atomic number 7, which means it has seven electrons. Its electron configuration can be written as 1s^2, 2s^2, 2px^1, 2py^1, 2pz^1. Nitrogen can form three covalent bonds by hybridizing its atomic orbitals.
In the case of hydrazine (N2H4), there are two nitrogen atoms. For each nitrogen atom, three hybrid orbitals are formed by combining one s orbital and two p orbitals. This process is called sp2 hybridization, which results in the formation of three sp2 hybrid orbitals.
In hydrazine, each nitrogen atom uses two of its sp2 hybrid orbitals to form sigma (σ) bonds with the two hydrogen atoms. The third sp2 hybrid orbital of each nitrogen atom overlaps with the third sp2 hybrid orbital of the other nitrogen atom to form a sigma (σ) bond between them. This is the N-N bond in hydrazine.
Therefore, the N-N bond in hydrazine is formed by the overlap of sp2 hybrid orbitals.