Question

The correct statement(s) about Mossbauer spectroscopy of iron compounds is(are)

• A. \(^{57}\mathrm{Co}\) is used as a source
• B. Their Mossbauer spectra are obtained using \(\gamma\)-ray with resonance energy of 14.4 keV
• C. \(\mathrm{K_2[Fe(CN)_5NO]}\) shows large quadrupole splitting
• D. Isomer shift of \(\mathrm{FeSO_4 \cdot 7H_2O}\) is smaller than that of \(\mathrm{K_3[Fe(CN)_6]}\)

Hint:

Recall the source isotope and transition energy for \(^{57}\mathrm{Fe}\) Mossbauer spectroscopy, and remember that isomer shift falls as s-electron density at the nucleus rises while quadrupole splitting grows with an asymmetric ligand field.

Answer 1

The Correct Option is A, B, C

Mossbauer spectroscopy is a nuclear technique. To judge each statement, remember three ideas: which isotope pair produces the right gamma energy, what causes quadrupole splitting, and what controls isomer shift.

1. Option A ($^{57}$Co as source): Iron Mossbauer spectroscopy runs on the $^{57}$Fe nucleus. $^{57}$Co is the parent isotope: it captures an electron and decays into excited $^{57}$Fe, which emits the gamma-ray used to probe the sample. This makes $^{57}$Co the standard, correct source. True.
2. Option B (14.4 keV resonance energy): The nuclear excited state of $^{57}$Fe reached from $^{57}$Co decay sits 14.4 keV above the ground state. The gamma-ray released when this state relaxes, and the one resonantly absorbed by $^{57}$Fe nuclei in the sample, has exactly this 14.4 keV energy. True.
3. Option C (large quadrupole splitting in $\mathrm{K_2[Fe(CN)_5NO]}$): Quadrupole splitting needs an uneven charge distribution around the nucleus. The nitrosyl ligand NO is electronically very different from the five cyanide ligands around it, so the ligand field is strongly distorted along the Fe-NO direction. This large asymmetry produces one of the biggest quadrupole splittings seen among iron complexes. True.
4. Option D (isomer shift comparison): Isomer shift falls as s-electron density at the nucleus rises. High spin $\mathrm{Fe^{2+}}$ in $\mathrm{FeSO_4 \cdot 7H_2O}$ carries more d-electron shielding, so s-density at the nucleus is lower and the isomer shift is larger (roughly 1.2 to 1.4 mm/s). Low spin $\mathrm{Fe^{3+}}$ in $\mathrm{K_3[Fe(CN)_6]}$ has less shielding, so its isomer shift is small, near zero. So the isomer shift of $\mathrm{FeSO_4 \cdot 7H_2O}$ is bigger, not smaller. False.

Putting the four checks together, A, B and C hold and D fails.

Let's summarize:

• $^{57}$Co is the source isotope and 14.4 keV is the fixed resonance energy for $^{57}$Fe Mossbauer work.
• Asymmetric ligand fields, like NO next to five CN groups, give large quadrupole splitting.
• More d-electron shielding means lower s-density at the nucleus and a larger isomer shift, so high spin Fe(II) beats low spin Fe(III) in isomer shift, not the other way round.

The correct statements are A, B and C.