Give the Lewis structure for N2H4 (Hydrazine) which is used in rocket fuels?

To draw the Lewis structure for hydrazine (N₂H₄), we first need to account for the total number of valence electrons.

Nitrogen (N) contributes 5 valence electrons and hydrogen (H) contributes 1 valence electron. Therefore, for N₂H₄:
2(N) + 4(H) = 2×5 + 4×1 = 10 + 4 = 14 valence electrons.

Next, we’ll place the atoms in a suitable arrangement. A common structure is to put the two nitrogen atoms in the center, bonded to each other, and then attach four hydrogen atoms around them:

N - N
 |  |
H  H
 |  |
H  H

Now, we begin allocating the 14 valence electrons. Each single bond (N-H or N-N) accounts for 2 electrons:

There are 6 electrons used for the three N-H bonds and 2 electrons for the N-N bond, totaling 8 electrons:

2(N-H bonds) + 1(N-N bond) = 8 electrons

This leaves us with 6 electrons remaining. To satisfy the octet rule for each nitrogen, we can form a bond between the two nitrogens:

H2N-NH2

Since each N now forms three bonds (one with each H and one with the other N), they have sufficient electrons to satisfy the octet rule. Each N atom shares a pair of electrons with the other N atom, completing the structure.

In summary, the Lewis structure for N₂H₄ is:

H2N-NH2

This structure indicates that hydrazine consists of a single bond between the two nitrogen atoms, with each nitrogen bonded to two hydrogen atoms.