The Fe3+ ion has electronic configuration of d5. if all five electrons are unpaired. what is the spin state?

(a) singlet
(b) sextet
(c)septet
(d)Octet
(e) none of the above the answer is____

Also Which of the following 1H are most deshielded relative to the TMS standard at 0 ppm?

(a) CH3F
(b) Ch3Cl
(c) CH3Br
(d) CH3I
(e) none of the above the answer is________

M = 2s + 1

s = (1/2)*5
Looks like M = 6 to me.

Could you help me with the second one as well please?

The answer to the second one is (a) because its the most electronegative.

No, I can't help with part B.

To determine the spin state of the Fe3+ ion with an electronic configuration of d5, we need to consider Hund's rule. Hund's rule states that electrons will occupy separate orbitals with parallel spins before pairing up.

In the case of Fe3+ with d5 configuration, we have 5 unpaired electrons. This indicates that all 5 electrons are occupying separate orbitals with parallel spin. The maximum number of unpaired electrons for d-electrons is 5. Therefore, the spin state for Fe3+ with d5 configuration is a quintet.

So, the answer to the first question is (e) none of the above.

Moving on to the second question regarding the deshielding of protons:

The deshielding of protons refers to the shift of their NMR signal towards higher chemical shifts (upfield). The chemical shift is measured relative to a standard such as tetramethylsilane (TMS), which is assigned a chemical shift of 0 ppm.

When comparing the provided options: CH3F, CH3Cl, CH3Br, and CH3I, we need to consider electronegativity and the presence of electronegative atoms (in this case, F, Cl, Br, and I).

Electronegative atoms withdraw electron density from the neighboring carbon-hydrogen bonds, resulting in a higher deshielding effect. The more electronegative the atom, the stronger its deshielding effect.

In this case, the order of increasing electronegativity is F < Cl < Br < I. Therefore, the order of increasing deshielding effect will correspond to the opposite order: I < Br < Cl < F.

Thus, the most deshielded proton relative to the TMS standard at 0 ppm is found in the compound CH3F.

So, the answer to the second question is (a) CH3F.