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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<u>Answer:</u> Nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.
<u>Explanation:</u>
To calculate the number of moles, we use the equation
....(1)
- <u>For ozone:</u>
Given mass of ozone = 0.827 g
Molar mass of ozone = 48 g/mol
Putting values in above equation, we get:
- <u>For nitric oxide:</u>
Given mass of nitric oxide = 0.635 g
Molar mass of nitric oxide = 30.01 g/mol
Putting values in above equation, we get:
For the given chemical equation:
By Stoichiometry of the reaction:
1 mole of ozone reacts with 1 mole of nitric oxide.
So, 0.0172 moles of ozone will react with = of nitric oxide
As, given amount of nitric oxide is more than the required amount. So, it is considered as an excess reagent.
Thus, ozone is considered as a limiting reagent because it limits the formation of product.
- Amount of excess reagent (nitric oxide) left = 0.0211 - 0.0172 = 0.0039 moles
By Stoichiometry of the reaction:
1 mole of ozone produces 1 mole of nitrogen dioxide.
So, 0.0172 moles of ozone will react with = of nitrogen dioxide
Now, calculating the mass of nitrogen dioxide from equation 1, we get:
Molar mass of nitrogen dioxide = 46 g/mol
Moles of nitrogen dioxide = 0.0172 moles
Putting values in equation 1, we get:
Hence, nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.