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Novay_Z [31]
3 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]3 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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3. At a pressure of 405 kPa, the volume of a gas is 6.00 cm
monitta

The pressure gets increased to 486 kPa from 405 kPa, when the volume is decreased from 6 cm³ to 4 cm³.

Explanation:

In the present problem, the temperature is said to remain at constant and there is change in the pressure. So according to Boyle's law, the relationship between pressure and volume of any gaseous objects are inversely related to each other. In other words, the pressure attained by gas molecules in a container will be inversely proportional to the volume of the gas molecules occupied in the container, at constant temperature.

V=\frac{1}{P}

So, if two volumes V₁ and V₂ are considered, then their respective pressure will be represented as P₁ and P₂. Then, as per Boyle's law,

V_{1}P_{1}=V_{2}P_{2}

So let us consider, V₁ = 6 cm³ and V₂ = 4 cm³ and pressure P₁ = 405 kPa and we have to determine P₂.

Then,  6*405=5*P_{2}\\ \\P_{2}=\frac{2430}{5} =486 kPa

So, the pressure at new volume of 4 cm³ is 486 kPa. It can be seen that as there is decrease in the volume, there is an increase in the pressure. So it satisfied the Boyle's law.

Thus, the pressure gets increased to 486 kPa from 405 kPa, when the volume is decreased from 6 cm³ to 4 cm³.

6 0
3 years ago
The website of a popular stargazers club notes that on Wednesday the moon will rise at 6:23 PM and set at 6:12 PM on Thursday. W
astraxan [27]

It's C that's correct because it's very true...

6 0
3 years ago
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Which layer of rock contains more humus
sleet_krkn [62]
It is the horizon layer
~Hope this helps~
4 0
3 years ago
Uranium has an atomic weight of 238 amu.
shepuryov [24]

Answer:

U= 238g/mol

U2O5= 556g/mol

Explanation:

Since U= 238

O=16

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4 0
3 years ago
If a sample contains 21.2g N how many moles of N does it contain
tigry1 [53]

Hey there!

The molar mass of nitrogen is 14.007.

Convert grams to moles:

21.2 ÷ 14.007 = 1.51

The sample containing 21.2g of nitrogen contains 1.51 moles of nitrogen.

Hope this helps!

8 0
3 years ago
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