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Novay_Z [31]
2 years ago
7

Challenge question: This question is worth 6 points. As you saw in problem 9 we can have species bound to a central metal ion. T

hese species are called ligands. In the past we have assumed all the d orbitals in some species are degenerate; however, they often are not. Sometimes the ligands bound to a central metal cation can split the d orbitals. That is, some of the d orbitals will be at a lower energy state than others. Ligands that have the ability to cause this splitting are called strong field ligands, CN− is an example of these. If this splitting in the d orbitals is great enough electrons will fill low lying orbitals, pairing with other electrons in a given orbital, before filling higher energy orbitals. In question 7 we had Fe2+, furthermore we found that there were a certain number (non-zero) of unpaired electrons. Consider now Fe(CN)6 4−: here we also have Fe2+, but in this case all the electrons are paired, yielding a diamagnetic species. How can you explain this?
Chemistry
1 answer:
GenaCL600 [577]2 years ago
5 0

Answer:

CN^- is a strong field ligand

Explanation:

The complex, hexacyanoferrate II is an Fe^2+ specie. Fe^2+ is a d^6 specie. It may exist as high spin (paramagnetic) or low spin (diamagnetic) depending on the ligand. The energy of the d-orbitals become nondegenerate upon approach of a ligand. The extent of separation of the two orbitals and the energy between them is defined as the magnitude of crystal field splitting (∆o).

Ligands that cause a large crystal field splitting such as CN^- are called strong field ligands. They lead to the formation of diamagnetic species. Strong field ligands occur towards the end of the spectrochemical series of ligands.

Hence the complex, Fe(CN)6 4− is diamagnetic because the cyanide ion is a strong field ligand that causes the six d-electrons present to pair up in a low spin arrangement.

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Be sure to answer all parts. one of the most important industrial sources of ethanol is the reaction of steam with ethene derive
lions [1.4K]

Answer: 2.17x10⁻³ atm

Explanation:

First, we must write the balanced chemical equation for the process:

C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g)

The chemical reactions that occur in a closed container can reach a state of <u>chemical equilibrium</u> that is characterized because the concentrations of the reactants and products remain constant over time. The <u>equilibrium constant</u> of a chemical reaction is the value of its reaction quotient in chemical equilibrium.

The equilibrium constant (K) is expressed as <u>the ratio between the molar concentrations (mol/L) of reactants and products.</u> Its value in a chemical reaction depends on the temperature, so it must always be specified.

<u>We will use the the equilibrium constant Kc of the reaction to calculate partial pressure of ethene.</u> The constant Kc for the above reaction is,

Kc = \frac{[C_{2} H_{5}OH]}{[H_{2}O][C_{2} H_{4}]}

According to the law of ideal gases,  

PV = nRT  

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the gas constant (0.082057 atm L / mol K) .

We can use the ideal gas law to determine the molar concentrations ([x] = n / V) from the gas pressures of ethanol and water, assuming that all gases involved behave as ideal gases. In this way,

PV = nRT → P = (n/V) RT → P = [x] RT → [x] = P / RT

So,  

[C_{2} H_{5}OH] = \frac{200 atm}{0.082057 \frac{atm L}{mol K} x 600 K } = 4.06 \frac{mol}{L}

[H_{2}O] = \frac{400 atm}{0.082057 \frac{atm L}{mol K} x 600 K } = 8.12 \frac{mol}{L}

So, the molar concentration of ethene (C₂H₄) will be,

[C_{2} H_{4}] = \frac{[C_{2} H_{5}OH]}{[H_{2}O] x Kc} = \frac{4.06 \frac{mol}{L} }{8.12 \frac{mol}{L}x9.00 x 10^{3} \frac{L}{mol} } = 5.56 x 10^{-5}\frac{mol}{L}

Then, according to the law of ideal gases,

P_{C_{2} H_{4}} = [C_{2} H_{4}]RT = 5.56 x 10^{-5} \frac{mol}{L}  x 0.082057 \frac{atm L}{mol K} x 600 K = 2.17x10^{-3} atm

So, when the partial pressure of ethanol is 200 atm and the partial pressure of water is 400 atm, the partial pressure of ethene at 600 K is 2.17x10⁻³ atm.

7 0
2 years ago
Substitution of an amino group on the para position of acetophenone shifts the cjo frequency from about 1685 to 1652 cm−1 , wher
cluponka [151]

Answer:

Here's what I get.

Explanation:

The frequency of a vibration depends on the strength of the bond (the force constant).

The stronger the bond, the more energy is needed for the vibration, so the frequency (f) and the wavenumber increase.

Acetophenone

Resonance interactions with the aromatic ring give the C=O bond in acetophenone a mix of single- and double-bond character, and the bond frequency = 1685 cm⁻¹.

p-Aminoacetophenone

The +R effect of the amino group increases the single-bond character of the C=O bond. The bond lengthens, so it becomes weaker.

The vibrational energy decreases, so wavenumber decreases to 1652 cm⁻¹.

p-Nitroacetophenone

The nitro group puts a partial positive charge on C-1. The -I effect withdraws electrons from the acetyl group.

As electron density moves toward C-1, the double bond character of the C=O group increases.

The bond length decreases, so the bond becomes stronger, and wavenumber  increases to 1693 cm¹.

6 0
2 years ago
5. A Dumas bulb is filled with chlorine gas at the ambient pressure and is found to contain 7.1 g of chlorine when the temperatu
kati45 [8]

Answer:

a. The original temperature of the gas is 2743K.

b. 20atm.

Explanation:

a. As a result of the gas laws, you can know that the temperature is inversely proportional to moles of a gas when pressure and volume remains constant. The equation could be:

T₁n₁ = T₂n₂

<em>Where T is absolute temperature and n amount of gas at 1, initial state and 2, final states.</em>

<em />

<em>Replacing with values of the problem:</em>

T₁n₁ = T₂n₂

X*7.1g = (X+300)*6.4g

7.1X = 6.4X + 1920

0.7X = 1920

X = 2743K

<h3>The original temperature of the gas is 2743K</h3><h3 />

b. Using general gas law:

PV = nRT

<em>Where P is pressure (Our unknown)</em>

<em>V is volume = 2.24L</em>

<em>n are moles of gas (7.1g / 35.45g/mol = 0.20 moles)</em>

R is gas constant = 0.082atmL/molK

And T is absolute temperature (2743K)

P*2.24L = 0.20mol*0.082atmL/molK*2743K

<h3>P = 20atm</h3>

<em />

7 0
1 year ago
What is the advantage of having large vertebrae at the base of the vertebral column?
WINSTONCH [101]
The advantage of having large vertebrae at the base of the vertebral column is having stability in terms of the center of gravity of the animal. If the animal has a large vertebrae, then it has an excellent balance and strength. 
3 0
1 year ago
A 1.0-gram sample of solid iodine is placed in a tube and the tube is sealed after all of the air is removed. The tube and the s
yKpoI14uk [10]

Answer:

27.0

Explanation:

Because Mass can neither be created nor be destroyed hence total mass of sample of iodine and tube remain equal as it is sealed.

7 0
2 years ago
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