I think:
H2CCH2
H H
| |
C_C
| |
H H
H2NNH2: H-H-:N-N:-H-H
Two Nitrogen atoms connected by single bond with a lone pair and 2 Hydrogen atoms at each ends
The answer to the quest is not zero
the answer is HNNH the reason for this is the double bond if you draw out the lewis structure
Yes, ammonia is an electron pair donor and a Lewis base.
Bases, with partial negative charges and overshielded nuclei, grab protòns from watter. In azo molecules, lone elèctronic pairs on the nitrogen are these negative ends. ammonia: H3N + H2O → H4N+ + HO− hydrazine hydrate: H2NNH2·nH2O + H2O → H2NHNH2+ + n H2O + HO−
The bond in N2 dinitrogen is a triple bond, with a length of 109.76 pm. An indicator of the length of a single N-N bond is given by hydrazine. H2NNH2, where the N-N is 145pm, this indicates the shortening in N2 due to the extra pi bonds.
yes, I think
Hydrazine. Keep up the hard work!
HNNH, it has a double bond while H2NNH2 has a single, and doubles are stronger than single bonds, just as triple is stronger than double.
The answer to the quest is not zero
HNNH (100% sure of it) :)
the answer is HNNH the reason for this is the double bond if you draw out the lewis structure
HNNHThe first compound(H2NNH2) has a single bong between the nitrogen atoms and follows the octet rule allowing nitrogen's orbital to hold 8 electrons. The second compound (HNNH) requires nitrogen to be double bonded. Double bonds are stronger than single bonds.Source:Yahoo Answers
Yes, ammonia is an electron pair donor and a Lewis base.
Bases, with partial negative charges and overshielded nuclei, grab protòns from watter. In azo molecules, lone elèctronic pairs on the nitrogen are these negative ends. ammonia: H3N + H2O → H4N+ + HO− hydrazine hydrate: H2NNH2·nH2O + H2O → H2NHNH2+ + n H2O + HO−
oxidation value of each N atom in hydrazine (H2NNH2) is -2 and in NO3- it is +5(standard in compounds: H is +1, O is -2, but these stay unchanged)The difference in oxidation value is (-2) -(+5) = -7 = 7 electrons (e-) per N atom (or 14 e- per 2N's in one N2H4).So 7 electrons are to be donated to - (and taken or accepted by) one NO3- ion.oxidator: 2NO3- + 16 H+ + 14 e- --> N2H4 + 6 H2Oreductor (donator): XXX --> YYY + 14 e-
The bond in N2 dinitrogen is a triple bond, with a length of 109.76 pm. An indicator of the length of a single N-N bond is given by hydrazine. H2NNH2, where the N-N is 145pm, this indicates the shortening in N2 due to the extra pi bonds.