Chemistry, 18.12.2019 01:31 wypozworski

Challenge question: this question is worth 6 points. as you saw in problem 9 we can have species bound to a central metal ion. these 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.

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Answer from: solomander4

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

Answer from: Quest

the cold front replaces the warmer air at the ground level changing the wind to be breezy and denser.there is also an immediate drop in the temperature.

Answer from: Quest

remember that hf is only the energy of one photon. not two. not three. one. 1) 5.12 * 10^3 photons find the energy of one photon, then determine how many photons are needed to reach the sensitivity. e = hf = (6.626 * 10^-34 js)(5.95 * 10^19 hz) = 3.94247 * 10^-14 j per photon (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons round to three sig figs. 2) 5.77 * 10^-17 j energy of pulse = energy per photon * number of photons = hfn = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250) = 5.774559 * 10^-17 j round to three sig figs. 3) 4.07 * 10^8 kj/mol 10^12 pm = 1 m c = fλ f = c/λ e = hf = hc/λ = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol) = 40702571.4 kj/mol round to three sig figs. 4) 9.94 * 10^-1 m e = hc/λ λ = hc / e = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j) = 0.9939 m round to three sig figs. 5) 9.06 * 10^3 photons energy of pulse = energy per photon * number of photons number of photons = energy of pulse / energy per photon = energy of pulse / (hf) = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)] = 9055.23695 photons round to three sig figs. 6) 2.78 * 10^-15 j this is problem 1, except in reverse. sensitivity = minumum number of photons * energy per photon = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz) = 2.78070758 × 10^-15 j round to three sig figs.

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