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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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A 0.050 M solution of AlCl3 had an observed osmotic pressure of 3.85 atmatm at 20°C.Calculate the van't Hoff factor iii for AlCl
Alja [10]

Answer:

The actual Van't Hoff factor for AlCl3 is 3.20

Explanation:

Step 1: Data given

Molarity of AlCl3 = 0.050 M

osmotic pressure = 3.85 atm

Temperature = 20 °C

Step 2: Calculate the Van't Hoff factor

AlCl3(aq) → Al^3+(aq) + 3Cl^-(aq)

The theoretical value is 4 ( because 1 Al^3+ ion + 3 Cl- ions) BUT due to the interionic atractions the actual value will be less

Osmotic pressure depends on the molar concentration of the solute but not on its identity., and is calculated by:

π = i.M.R.T

 ⇒ with π = the osmotic pressure = 3.85 atm

⇒ with i = the van't Hoff factor

⇒ with M = the molar concentration of the solution = 0.050 M

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 20 °C = 293.15 Kelvin

i = π /(M*R*T )

i = (3.85) / (0.050*0.08206*293.15)

i = 3.20

The actual Van't Hoff factor is 3.20

6 0
2 years ago
Which solution contains the largest number of moles of chloride ions?
elixir [45]
Molarity = number of moles of solute/liters of solution
number of moles of solute = molarity x liters of solution

Part (a): <span>30.00 ml of 0.100m Cacl2
number of moles of CaCl2 =  0.1 x 0.03 = 3x10^-3 moles
1 mole of CaCl2 contains 2 moles of chlorine, therefore 3x10^-3 moles of CaCl2 contains 6x10^-3 moles of chlorine

Part (b): </span><span>10.0 ml of 0.500m bacl2
number of moles of BaCl2 = 0.5 x 0.01 = 5x10^-3 moles
1 mole of BaCl2 contains 2 moles of chlorine, therefore 5x10^-3 moles of BaCl2 contains 10x10^-3 moles of chlorine

Part (c): </span><span>4.00 ml of 1.000m nacl
number of moles of NaCl = 1 x 0.004 = 0.004 moles
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Part (d): </span><span>7.50 ml of 0.500m fecl3
number of moles of FeCl3 = 0.5 x 0.0075 = 3.75x10^-3 moles
1 mole of FeCl3 contains 3 moles of chlorine, therefore 3.75x10^-3 moles of FeCl3 contains 0.01125 moles of chlorine

Based on the above calculations, the correct answer is (d)</span>
5 0
2 years ago
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trans-2-Butene does not exhibit a signal in the double-bond region of the spectrum (1600–1850 cm-1); however, IR spectroscopy is
spayn [35]

Answer:

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4 0
2 years ago
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weeeeeb [17]
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I hope this helps.  Let me know if anything is unclear.
4 0
2 years ago
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ahrayia [7]

Answer:

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