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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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The work function for metallic potassium is 2.3 eV. Calculate the velocity in km/s for electrons ejected from a metallic potassi
mote1985 [20]

Answer:

932.44 km/s

Explanation:

Given that:

The work function of the magnesium = 2.3 eV

Energy in eV can be converted to energy in J as:

1 eV = 1.6022 × 10⁻¹⁹ J

So, work function = 2.3\times 1.6022\times 10^{-19}\ J=3.68506\times 10^{-19}\ J

Using the equation for photoelectric effect as:

E=\psi _0+\frac {1}{2}\times m\times v^2

Also, E=\frac {h\times c}{\lambda}

Applying the equation as:

\frac {h\times c}{\lambda}=\psi _0+\frac {1}{2}\times m\times v^2

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

m is the mass of electron having value 9.11\times 10^{-31}\ kg

\lambda is the wavelength of the light being bombarded

\psi _0=Work\ function

v is the velocity of electron

Given, \lambda=260\ nm=260\times 10^{-9}\ m

Thus, applying values as:

\frac{6.626\times 10^{-34}\times 3\times 10^8}{260\times 10^{-9}}=3.68506\times 10^{-19}+\frac{1}{2}\times 9.11\times 10^{-31}\times v^2

3.68506\times \:10^{-19}+\frac{1}{2}\times \:9.11\times \:10^{-31}v^2=\frac{6.626\times \:10^{-34}\times \:3\times \:10^8}{260\times \:10^{-9}}

\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=7.64538\times 10^{-19}-3.68506\times 10^{-19}

\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=3.96032462\times 10^{-19}

v^2=0.869446\times 10^{12}

v = 9.3244 × 10⁵ m/s

Also, 1 m = 0.001 km

<u>So, v = 932.44 km/s</u>

5 0
3 years ago
The complete orbital notation diagram of an atom is shown.
Elan Coil [88]

values of the quantum numbers: -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6

location of the electron: In the 7th energy level away from the nucleus.

Explanation:

From the description of the problem, the magnetic number is given is as -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6 and the electron is located in the 7th energy level away from the nucleus. Basically, the problem is testing for the understanding of the principal quantum numbers which gives the location of electrons and the magnetic quantum number that shows the spatial orientation of the orbitals.

           The orbital designation of the describe electron is 7d

  1. Magnetic quantum number is limited by the azimuthal quantum number which is the quantum number describing the possible shapes. The azimuthal is given as L= n-1. "n" is the principal quantum number which is 7. Therefore L is 6 and the magnetic quantum numbers are -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6
  2. The position of the electron is given by the principal quantum number which represents the main energy level in which the orbital is located or the average distance from the nucleus. Here it is 7.

Learn more:

brainly.com/question/9288609

#learnwithBrainly

5 0
3 years ago
calculate how many Liters of 0.50 M silver nitrate solution you will need to provide the 2.4x10^-3 moles of silver nitrate
wlad13 [49]

Answer:

4.8x10⁻³ Liters are required

Explanation:

Molarity is an unit of concentration in chemistry defined as the ratio between moles of solute (In this case, silver nitrate) and liters of solution.

The 0.50M solution contains 0.50 moles of silver nitrate per liter of solution.

To provide 2.4x10⁻³ moles Silver nitrate are required:

2.4x10⁻³ moles * (1L / 0.50 moles) =

<h3>4.8x10⁻³ Liters are required</h3>
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What are two characteristics of acids
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Answer:it’s C

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