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Alecsey [184]
4 years ago
8

C) when two electrons occupy the bonding molecular orbital above, what type of bond results? explain.

Chemistry
2 answers:
WARRIOR [948]4 years ago
7 0
There are only two possible bonds when molecular orbitals combine with each other. It can either be sigma bond, or a pi bond. You can differentiate between them by looking at the way they bond together. Sigma bond are much stronger because the orbitals are superimposed on each other. On the other hand, pi bond are overlapped only sideways.
Brrunno [24]4 years ago
3 0

Two atomic orbitals can overlap in two ways depending on their phase relationship. Orbital phases are a direct consequence of the nature of waves like electrons. In graphical representations of orbitals, the orbital phases are represented by plus or minus signs (which have no connection with electric charges) or by incising a lobe. The phase sign itself has no physical meaning except when mixing orbitals to form molecular orbitals.

<h2>Further Explanation </h2>

Molecular orbital diagrams, or OM diagrams, are a qualitative descriptive tool that explains chemical bonds in molecules in terms of general molecular orbital theory and linear combinations of atomic orbitals (LCAO) in particular. The basic principle of these theories is that when atoms are bound to form molecules, a number of atomic orbitals combine to form the same number of molecular orbitals, even though electrons can be redistributed between orbitals. This tool is very suitable for simple diatomic molecules such as dihydrogen, dioxygen, and carbon monoxide but becomes more complex when discussing relatively simple polyatomic molecules, such as methane. OM diagrams can explain why some molecules exist and others don't. They can also predict the strength of bonds, as well as electronic transitions that can occur.

In MO theory molecular orbitals are formed by overlapping atomic orbitals. Because σ bonds have greater overlap than π bonds, σ, and σ * bonds and antibonding orbitals have greater energy separation (separation) from pemisahan and π * orbitals. The energy of atomic orbitals correlates with electronegativity because more electronegative atoms hold their electrons more tightly, reducing their energy. MO modeling only applies if atomic orbitals have comparable energy; When the energy is very different, the bonding mode becomes ionic. The second condition for overlapping atomic orbitals is that they have the same symmetry.

Learn more

molecular orbital brainly.com/question/4816397

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Grade:  College

Subject:  Chemistry

keywords: molecular, orbital.

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What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb
Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

(\text{CH}_3)_3\text{N} in this question acts as a weak base. As seen in the equation in the question, (\text{CH}_3)_3\text{N} produces \text{OH}^{-} rather than \text{H}^{+} when it dissolves in water. The concentration of \text{OH}^{-} will likely be more useful than that of \text{H}^{+} for the calculations here.

Finding the value of [\text{OH}^{-}] from pH:

Assume that \text{pK}_w = 14,

\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}.

[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}.

Solve for [(\text{CH}_3)_3\text{N}]_\text{initial}:

\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}

Note that water isn't part of this expression.

The value of Kb is quite small. The change in (\text{CH}_3)_3\text{N} is nearly negligible once it dissolves. In other words,

[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}.

Also, for each mole of \text{OH}^{-} produced, one mole of (\text{CH}_3)_3\text{NH}^{+} was also produced. The solution started with a small amount of either species. As a result,

[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}.

\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3},

[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b},

[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}.

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Explanation

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When electromagnetic radiation of wavelength 300nm falls on the surface of sodium electrons are emitted with a KE of 1.68 * 10 5
gtnhenbr [62]

Answer:

3.83 × 10⁻¹⁹ J;  518 nm  

Step-by-step explanation:

The equation for the <em>photoelectric effect</em> is

hf = Φ + KE  

<em>Data: </em>

λ = 300 nm = 300 × 10⁻⁹ m

KE = 1.68 × 10⁵ J/mol

Calculations:

Part 1. Minimum energy to remove an electron

(a) Calculate the <em>energy of the photon</em>

fλ = c  

 f = c/λ     Divide each side by λ

E = hf

E = hc/λ

E = (6.626× 10⁻³⁴ × 2.998 × 10⁸)/(300 × 10⁻⁹)

E = 6.622 × 10⁻¹⁹ J

(b) Calculate the <em>KE of one electron</em>

KE = 1.68 × 10⁵ × 1/(6.022 × 10²³)

KE = 2.790 × 10⁻¹⁹ J

(c) Calculate the work function

hf = Φ + KE     Subtract KE from each side

Φ = 6.622 × 10⁻¹⁹  - 2.790 × 10⁻¹⁹

Φ = 3.83 × 10⁻¹⁹ J

The minimum energy to remove an electron from a sodium atom

is 3.83 × 10⁻¹⁹ J.

Part 2. Maximum wavelength to remove an electron

The photon must have just enough energy to overcome the work function and leave the electron with zero kinetic energy.

    E = Φ

hc/λ = Φ                      Multiply each side by λ

  hc = Φ λ                   Divide each side by Φ

   λ = hc/ Φ

   λ = (6.626 × 10⁻³⁴ × 2.998 × 10⁸)/(3.83 × 10⁻¹⁹)

   λ = 5.18 × 10⁻⁷ m     Convert to nanometres

   λ = 518 nm

The maximum wavelength that will cause an electron to move is 518 nm.

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When balancing a chemical equation can you adjust the number that is subscripted to a substance formula?
Ugo [173]
No, you can't do that
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