the bond will break
The bond will dissolve (break) if the electron absorbs a photon and is moved from a bonding molecular orbital to an antibonding orbital since there is no longer an overall stabilizing interaction.
<h3>What is an antibonding orbital?</h3>
An antibonding molecular orbital is the molecular orbital created by the destructive overlapping of atomic orbitals.
<h3>Why is it called antibonding orbital?</h3>
- Every atom will add one electron to the bond that makes up the lower energy bond.
- To prevent interacting with the other two electrons, the additional electron will occupy a higher energy state.
- The antibonding orbital is the name of this higher energy orbital.
<h3>What orbitals form an antibond?</h3>
- The bonding orbitals are home to electrons that spend the majority of their time between the nuclei of two atoms, whereas the antibonding orbitals are home to electrons that spend the majority of their time outside the nuclei of two atoms.
<h3>When an electron was elevated to the antibonding orbital, what happened?</h3>
- In contrast, putting electrons in antibonding orbitals will make the molecule less stable.
- The energy levels of the orbitals will determine how many electrons are filled.
- The lower energy orbitals will be filled first, and then the higher energy orbitals.
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Answer:
10 liters of the 1% solution and 8 liters of the 10% solution she should mix to make the 5% solution.
Explanation:
Let the volume of 1% acid solution used to make the mixture = x L
So, the volume of 10% acid solution used to make the mixture = y L
Total volume of the mixture = <u>x + y = 18 L .................. (1)</u>
For 1% acid solution:
C₁ = 1% , V₁ = x L
For 10% acid solution :
C₂ = 10% , V₂ = y L
For the resultant solution of sulfuric acid:
C₃ = 5% , V₃ = 18 L
Using
C₁V₁ + C₂V₂ = C₃V₃
1×x + 10×y = 5×18
So,
<u>x + 10y = 90 .................. (2)</u>
Solving 1 and 2 we get,
<u>x = 10 L</u>
<u>y = 8 L</u>
Thus,
<u>10 liters of the 1% solution and 8 liters of the 10% solution she should mix to make the 5% solution.</u>
<h3><u>Answer;</u></h3>
pOH = 3.08
<h3><u>Explanation;</u></h3>
NX3 + H2O <----> NHX3+ + OH-
Kb = 4.0 x 10^-6
Kb = c(NH₄⁺) · c(OH⁻) ÷ c(NH₃).
c(NH₄⁺) = c(OH⁻) = x.
x² = Kb · c(NH₃)
x² = 4.0 × 10⁻⁶ × 0.175 = 7.0 × 10⁻⁷.
x = c(OH⁻) = √(7.0 × 10⁻⁷)
= 8.367 × 10⁻⁴
pOH = -log(c(OH⁻))
=- log ( 8.367 × 10⁻⁴)
<u>= 3.08</u>
