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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 years ago
How many amps are required to produce 75. 8 g of iron metal from a solution of aqueous iron(iii)chloride in 6. 75 hours?
Shalnov [3]

The amount of current required to produce 75. 8 g of iron metal from a solution of aqueous iron (iii)chloride in 6. 75 hours is 168.4A.

The amount of Current required to deposit a metal can be find out by using The Law of Equivalence. It states that the number of gram equivalents of each reactant and product is equal in a given reaction.

It can be found using the formula,

m = Z I t

where, m = mass of metal deposited = 75.8g

            Z = Equivalent mass / 96500 = 18.6 / 96500 = 0.0001

             I is the current passed

              t is the time taken = 75hour = 75 × 60 = 4500s

On subsituting in above formula,

75.8 = E I t / F

⇒ 75.8 = 0.0001 × I × 4500

⇒ I = 168.4 Ampere (A)

Hence, amount of current required to deposit a metal is 168.4A.

Learn more about Law of Equivalence here, brainly.com/question/13104984

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7 0
1 year ago
How many molecules are in 2.38g of SO2
Murrr4er [49]

2.63 x 10^22 molecules of SO2.

To find this, start with what you know.

2.38g of SO2.

You need to first convert this into Moles since you cannot directly convert grams into molecules. In order to convert grams to moles, you need to find the molecular mass of SO2 - 64.066.

This is because Sulfur has the mass of 32.066 and Oxygen has a mass of 16, but since there are two Oxygen atoms, it's going to be 32.

Your equation should currently appear as so:

2.38g of SO2 = 1 Mole of SO2 / 64.066

Now, you need to convert this to molecules.

Whenever you are searching for molecules without a given amount, you always use Avogadro's number: 6.02 x 10^23

Now, your equation should appear as so:

2.38g SO2 = 1 Mole of SO2 / 64.066 next to a new fraction which is 6.02 x 10^23 / 1 Mole of SO2

Now, multiply across (2.38 x 1 x 6.02 x 10^23). When using Avogadro's number, don't forget to place parenthesis around it.

Then, divide that number by the bottom: 64.066.

Thus, your final answer is that there is 2.63 x 10^22 molecules of SO2.

Don't forget your units!

Hope this helps!

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