The nitrogen atoms in N2 participate in multiple bonding whereas those in hydrazine, N2H4, do not. a.) Draw Lewis structures for
both molecules. b.) What is the hybridization of the nitrogen atoms in each molecule? c.) Which molecule has a stronger N-N bond?
1 answer:
Answer:
a) Attached images.
b) N₂: sp; N₂H₄: sp³
c) N₂
Explanation:
a) In the attached images are the Lewis structures.
<em>N₂:</em> there is a triple covalent bond between the N atoms.
<em>N₂H₄:</em> there is a single covalent bond between the N atoms.
b)
<em>N₂:</em> N has 2 electron domains. The corresponding hybridization is sp. 1 sp orbital form 1 sigma bonds whereas 2 p orbitals from 2 pi bonds.
<em>N₂H₄:</em> N has 4 electron domains. The corresponding hybridization is sp³. The 3 sp³ orbitals form 3 sigma bonds.
c) The triple bond is stronger than the single bond. Then, the bond in N₂ is stronger than the bond in N₂H₄.
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Answer : It takes less amount of heat to metal 1.0 Kg of ice.
Solution :
The process involved in this problem are :
Now we have to calculate the amount of heat released or absorbed in both processes.
<u>For process 1 :</u>
where,
= amount of heat absorbed = ?
m = mass of water or ice = 1.0 Kg
= enthalpy change for fusion =
Now put all the given values in , we get:
<u>For process 2 :</u>
where,
= amount of heat absorbed = ?
m = mass of water = 1.0 Kg
= specific heat of liquid water =
= initial temperature =
= final temperature =
Now put all the given values in , we get:
From this we conclude that, that means it takes less amount of heat to metal 1.0 Kg of ice.
Hence, the it takes less amount of heat to metal 1.0 Kg of ice.