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3 years ago

Consider the three ligand field spectra corresponding to octahedral complexes A, B, and C, all formed from the same metal ion.

Chemistry

1 answer:

scoundrel [369]3 years ago

8 0

Answer:

Ni^2+ is most likely

Ti^3+ is very unlikely

Explanation:

The Crystal Field Stabilization Energy almost always favors octahedral over tetrahedral in very many cases, but the degree of this favorability varies with the electronic configuration. In other words, for d1 there is only a small gap between the octahedral and tetrahedral lines, whereas at d3 and d8 is a very big gap. However, for d0, d5 high spin and d10, there is no crystal field stabilization energy difference between octahedral and tetrahedral. The ordering of favorability of octahedral over tetrahedral is:

d3, d8 > d4, d9> d2, d7 > d1, d6 > d0, d5, d10. This explains the answer choices above.

Ti^3+ being a d1 specie is least likely to exist in octahedral shape while Ni2+ a d8 specie is more likely to exist in octahedral shape.

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The Balmer series, named after Johann Balmer, is a portion of the hydrogen emission spectrum produced from the transitions betwe

horrorfan [7]

Explanation:

The wavelength of the balmer series is calculated using the following steps;

- Find the Principle Quantum Number for the Transition

- Calculate the Term in Brackets

- Multiply by the Rydberg Constant

- Find the Wavelength

The Balmer series in a hydrogen atom relates the possible electron transitions down to the n = 2 position to the wavelength of the emission that scientists observe.

The λ symbol represents the wavelength, and RH is the Rydberg constant for hydrogen, with RH = 1.0968 × 107 m−1

n=7 to n=2

- The principal quantum numbers are 2 and 7.

- (1/2²) − (1 / n²₂)

For n₂ = 7, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 7²)

= (1/4) − (1/49)

= 0.2230

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.2230

= 2445864 m−1

- λ = 1 / 2445864 m−1

= 4.08 × 10−7 m

= 408 nanometers

≈ 410nm

n=6 to n=2

- The principal quantum numbers are 2 and 6.

- (1/2²) − (1 / n²₂)

For n₂ = 6, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 6²)

= (1/4) − (1/36)

= 0.2222

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 3/16

= 2437090 m−1

- λ = 1 / 2437090 m−1

= 4.10 × 10−7 m

= 410 nanometers

n=5 to n=2

- The principal quantum numbers are 2 and 5.

- (1/2²) − (1 / n²₂)

For n₂ = 5, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 5²)

= (1/4) − (1/25)

= 0.21

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.21

= 2303280 m−1

- λ = 1 / 2303280 m−1

= 4.34 × 10−7 m

= 434 nanometers

n=4 to n=2

- The principal quantum numbers are 2 and 4.

- (1/2²) − (1 / n²₂)

For n₂ = 4, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 4²)

= (1/4) − (1/16)

= 0.1875

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.1875

= 2056500 m−1

- λ = 1 / 2056500 m−1

= 4.86 × 10−7 m

= 486 nanometers

n=3 to n=2

- The principal quantum numbers are 2 and 3.

- (1/2²) − (1 / n²₂)

For n₂ = 3, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 3²)

= (1/4) − (1/9)

= 0.13889

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.13889

= 1523345 m−1

- λ = 1 / 1523345 m−1

= 6.56 × 10−7 m

= 656 nanometers

7 0

2 years ago

Which half-reaction correctly shows the oxidation of iron

pishuonlain [190]

Answer:

What is the oxidation half reaction for iron?

The two elements involved, iron and chlorine, each change oxidation state; iron from +2 to +3, chlorine from 0 to -1. There are then effectively two half-reactions occurring. These changes can be represented in formulas by inserting appropriate electrons into each half-reaction: Fe2+ → Fe3+ + e.

Hope this helps..

8 0

2 years ago

What is the specific heat of a substance if 25 grams rises in temperature from 10 degrees Celsius to 25 degrees with the additio

Marina86 [1]

Answer:

0.75 cal/g°c

Explanation:

for specific heat we have formula:

Amount of heat absorbed or released = mass x specific heat of a substance x change in temperature.

ΔQ=m x c x ΔT

where c= specific heat

m= mass of a substance

ΔT = total temperature

ΔQ = Amount of heat

so for specific heat,

c= ΔQ/mxΔT

c= 280/25x (25-10)

c= 280/375

c= 0.75 cal/g°c

4 0

3 years ago

A teacher wrote the following part of a balanced chemical equation: Cu+2AgNO 3 To show the conservation of matter in the chemica

Llana [10]

2, because there are 2 silver atoms on the left

7 0

2 years ago

Read 2 more answers

Which one of the following statements is not true concerning 2.00 L of a 0.100 M solution of Ca3(PO4)2?

Len [333]

Answer: The correct answer is Option B.

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .......(1)

For A:

Molarity of calcium phosphate solution = 0.100 M

Volume of solution = 2.00 L

Putting values in equation 1, we get:

$0.100M=\frac{\text{Moles of }Ca_3(PO_4)_2}{2.00}\\\\\text{Moles of }Ca_3(PO_4)_2=(0.100mol/L\times 2.00L)=0.200mol$

Moles of calcium phosphate = 0.200 moles

For B:

1 mole of calcium phosphate contains 3 moles of calcium atoms, 2 moles of phosphate atoms and 8 moles of oxygen atoms.

So, 0.200 moles of calcium phosphate will contain = $(8\times 0.200)=1.6$ moles of oxygen atoms.

Moles of oxygen atoms = 1.6 moles

For C:

Molarity of calcium phosphate solution = 0.100 M

Volume of solution = 1.00 L

Putting values in equation 1, we get:

$0.100M=\frac{\text{Moles of }Ca_3(PO_4)_2}{1.00}\\\\\text{Moles of }Ca_3(PO_4)_2=(0.100mol/L\times 1.00L)=0.100mol$

Moles of calcium ions = $(0.100\times 3)=0.300$ moles

For D:

Molarity of calcium phosphate solution = 0.100 M

Volume of solution = 5.00 L

Putting values in equation 1, we get:

$0.100M=\frac{\text{Moles of }Ca_3(PO_4)_2}{5.00}\\\\\text{Moles of }Ca_3(PO_4)_2=(0.100mol/L\times 5.00L)=0.500mol$

Moles of phosphorus atoms = $(0.500\times 2)=1.00$ moles

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of atoms

Number of phosphorus atoms in 0.500 moles of calcium phosphate will be = $(1.00\times 6.022\times 10^{23})=6.022\times 10^{23}$

For E:

1 mole of calcium phosphate contains 3 moles of calcium ions and 2 moles of phosphate ions.

So, 0.200 moles of calcium phosphate will contain = $(3\times 0.200)=0.600$ moles of calcium ions

Moles of calcium ions = 0.600 moles

Hence, the correct answer is Option B.

6 0

2 years ago