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In ${{N}_{2}}{{H}_{4}}$ molecule type of overlapping present between N-H bond:-
(A) $s-p$
(B) $s{{p}^{2}}-p$
(C) $s{{p}^{3}}-s$
(D) $s{{p}^{2}}-s$

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Last updated date: 13th Jun 2024
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Answer
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Hint: The molecule ${{N}_{2}}{{H}_{4}}$ is called hydrazine. From drawing the Lewis structure of ${{N}_{2}}{{H}_{4}}$, we could identify that each nitrogen atom is $s{{p}^{3}}$ hybridized and uses two $s{{p}^{3}}$ orbitals to form N–H bonds and hydrogen has only one electron in its 1s orbital.

Complete step by step solution:
- Hydrazine is a colourless liquid with an ammoniacal odour and the hydrazine is miscible with water in all proportions. Also, its aqueous solutions are highly alkaline in nature.
- In order to get an idea of overlapping present between N-H bonds in ${{N}_{2}}{{H}_{4}}$ molecules, we need to look at the concept of hybridization. It is the process in which the overlap of bonding orbitals takes place and as a result, the formation of stronger bonds occur. Using the model of hybridization we would be able to predict the shapes of certain molecules.
- In hybridization, the atomic orbitals which have similar energy but not equivalent are combined mathematically in such a way to produce sets of equivalent orbitals which are properly oriented to form bonds. Since they are produced by hybridizing two or more atomic orbitals from the same atom, these new combinations are called hybrid atomic orbitals.
- The Lewis structure of ${{N}_{2}}{{H}_{4}}$ molecule is given below
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- From the Lewis structure it’s clear that every nitrogen atom has one lone pair of electrons and is bonded to three other atoms including two hydrogens and one nitrogen atom.
- As we know the Steric Number is the sum of the number of lone electron pairs on the central atom and number of atoms bonded to the central atom. So the steric number for nitrogen is $4(3+1)$. This implies that four hybrid orbitals are needed by every nitrogen atom and this is possible only through $s{{p}^{3}}$ hybridization.
- Since hydrogen has only one s orbital, it has an s overlapping. Thus the overlapping present between N-H bonds in hydrazine will be $s{{p}^{3}}-s$.

Thus the answer is option (C). $s{{p}^{3}}-s$.

Note: It should be noted that the bond between nitrogen atoms in hydrazine is single, not a triple bond. It’s because we only have one p-orbital per nitrogen atom that is available to form a pi-bond, and in order to form two pi-bonds, we need at least two p-orbitals per each nitrogen atom to be available as in the case of acetylene.