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vfiekz [6]
2 years ago
14

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)64-: here we also have Fe2+, but in this case all the electrons are paired, yielding a diamagnetic species. How can you explain this?
A. There are 3 low lying d orbitals, which will be filled with 6 electrons before filling the 2, assumed to be degenerate, higher energy orbitals.

B. There are 4 low lying d orbitals, which will be filled with 8 electrons before filling the 1 higher energy orbital.

C. There is 1 low lying d orbital, which will be filled with two electrons before filling the 4, assumed to be degenerate, higher energy orbitals.

D. All the d orbitals are degenerate.

E. There are 2 low lying d orbitals, which will be filled with 4 electrons before filling the 3, assumed to be degenerate, higher energy orbitals.
Chemistry
1 answer:
Llana [10]2 years ago
3 0

Answer:

A

Explanation:

Iron has the ground state electronic configuration [Ar]3d64s2

Fe2+ has the electronic configuration [Ar]3d6.

In an octahedral crystal field, there are two sets of degenerate orbitals; the lower lying three t2g orbitals, and the higher level two degenerate eg orbitals. Strong field ligands cause high octahedral crystal field splitting, there by separating the two sets of degenerate orbitals by a tremendous amount of energy. This energy is much greater than the pairing energy required to pair the six electrons in three degenerate orbitals. Since CN- is a strong field ligand, it leads to pairing of six electrons in three degenerate orbitals

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Which statement is true of a reversible reaction at equilibrium?
Vedmedyk [2.9K]

Answer:

D.

The concentration of reactants and the concentration of products are constant.

Explanation:

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7 0
2 years ago
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One of the most important chemical reactions is the Haber process, in which N2 and H2 are converted to ammonia which is used in
Lera25 [3.4K]

Answer:

c) 22

Explanation:

Let's consider the following balanced equation.

N₂(g) + 3 H₂(g) ----> 2 NH₃(l)

According to the balanced equation, 34.0 g of NH₃ are produced by 1 mol of N₂. For 170 g of NH₃:

170gNH_{3}.\frac{1molN_{2}}{34.0gNH_{3}} =5.00molN_{2}

According to the balanced equation, 34.0 g of NH₃ are produced by 3 moles of H₂. For 170 g of NH₃:

170gNH_{3}.\frac{3molH_{2}}{34.0gNH_{3}} =15.0molH_{2}

The total gaseous moles before the reaction were 5.00 mol + 15.0 mol = 20.0 mol.

We can calculate the pressure (P) using the ideal gas equation.

P.V = n.R.T

where

V is the volume (50.0 L)

n is the number of moles (20.0 mol)

R is the ideal gas constant (0.08206atm.L/mol.K)

T is the absolute temperature (400.0 + 273.15 = 673.2K)

P=\frac{n.R.T}{V} =\frac{20.0mol\times (0.08206atm.L/mol.K)\times 673.2K ) }{50.0L} =22.0atm

7 0
2 years ago
Consider the hypothetical atom x. if the molecular formula for lithium carbonate is li2co3 and the formula of x chloride is xcl3
eduard
First step is to determine the valency of each of x and CaCO3 from the given compounds:

1- As for Li2CO3: we can deduce that the valency of lithium is one while that of CO3 is two

2- As for XCl3: we can deduce that the valency of chlorine is one while that of X is three

Second step is to write the required compound: 
X : CO3  (elements involved)
3 : 2        (write the valency of each)
Then write the positive ion (X) followed by the valency of the negative ion (2) and then the negative ion (CO3) followed by the valency of the positive ion (3).
The final x carbonate is written as: X2(CO3)3
7 0
2 years ago
(a) calculate %ic of the interatomic bonds for the intermetallic compound al6mn. (b) on the basis of this result what type of in
Tatiana [17]

The answer is:

a)0.25%

b) metallic bond

The explanation:

A) The percentage of ionic character in a compound having some covalent character can be calculated by the following equation. The percent ionic character = Observed dipole moment/Calculated dipole moment assuming 100% ionic bond × 100.

-The percent ionic character is a function of the electron negativities of the ions XA and XB .  The electronegativities for Al and Mn  are 1.5 and 1.6, respectively :

when %IC = [1-exp(-1/4) (XB-XA)^2].100

so, %IC =  [1  -  exp(- 0.25)(1.6- 1.5)^2] .  100 = 0.25%

 

(b) Because the percent ionic character is so small (0.25%) and this intermetallic compound is composed of two metals Al and Mn, the bonding is completely metallic.

Metallic bond:

• Metallic bonding can be either weak (68 kJ/mole or 0.7 eV/atom for Hg) or strong (850 kJ/mole or 8.8 eV/atom for W)

• Metallic bonding gives rise to high electrical and thermal conductivity.

-The electrons are loosely held since each atom has several unoccupied valence orbitals; it is relatively easy for the electrons to move about. In this manner the electrons allow atoms to slide past each other.

6 0
2 years ago
When you tapped the stick on one end of the tape, did you see a wave that
rusak2 [61]

Answer:

Yes.

Explanation:

Yes, I did see a wave that formed and rolled down the line because of tapping the stick on one end of the tape. The wave that had been created goes from one side of the tape to the other end of the tape. This movement occur in the tape due to the disturbance at one end of the tape so we can say that wave formed in the tape and rolled down the line.

7 0
2 years ago
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