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

Step 1: Understand the Mossbauer setup for iron.
Mossbauer spectroscopy of \(^{57}\mathrm{Fe}\) needs a gamma-ray source that decays into the correct excited nuclear state of iron. The parent isotope \(^{57}\mathrm{Co}\) decays by electron capture to an excited state of \(^{57}\mathrm{Fe}\), which then drops to the ground state and gives out a 14.4 keV gamma-ray. This gamma-ray is resonantly absorbed by \(^{57}\mathrm{Fe}\) nuclei in the sample, so \(^{57}\mathrm{Co}\) is the standard source. Statement (A) is correct.

Step 2: Check the resonance energy value.
The nuclear transition used in \(^{57}\mathrm{Fe}\) Mossbauer spectroscopy is the \(3/2 \rightarrow 1/2\) transition of the \(^{57}\mathrm{Fe}\) nucleus, and its energy is fixed at 14.4 keV. This is the standard, well known value quoted for all iron Mossbauer work. Statement (B) is correct.

Step 3: Look at the quadrupole splitting of \(\mathrm{K_2[Fe(CN)_5NO]}\).
Quadrupole splitting comes from an electric field gradient at the iron nucleus, which grows when the electron density around iron is not symmetric. In \(\mathrm{K_2[Fe(CN)_5NO]}\) (the nitroprusside type ion), the strong pi-acceptor NO ligand is very different from the five CN ligands, so the field around iron is heavily distorted along the Fe-NO direction. This asymmetry gives a large electric field gradient, and hence a large quadrupole splitting, one of the biggest seen among iron nitrosyl-cyanide complexes. Statement (C) is correct.

Step 4: Compare isomer shifts of \(\mathrm{FeSO_4 \cdot 7H_2O}\) and \(\mathrm{K_3[Fe(CN)_6]}\).
Isomer shift depends on the s-electron density at the iron nucleus: more s-density gives a smaller (more negative) isomer shift. \(\mathrm{FeSO_4 \cdot 7H_2O}\) has high spin \(\mathrm{Fe^{2+}}\) (\(d^6\)), and its extra d-electrons shield the s-electrons from the nucleus, lowering the s-density there and pushing the isomer shift to a larger, positive value (about 1.2 to 1.4 mm/s). \(\mathrm{K_3[Fe(CN)_6]}\) has low spin \(\mathrm{Fe^{3+}}\) (\(d^5\)), where the strong field CN ligands pair up the electrons and the smaller d-shielding lets s-density at the nucleus stay higher, giving a small isomer shift close to zero (about -0.1 mm/s). So the isomer shift of \(\mathrm{FeSO_4 \cdot 7H_2O}\) is actually larger, not smaller, than that of \(\mathrm{K_3[Fe(CN)_6]}\). Statement (D) is wrong.

Final Answer:
Statements (A), (B) and (C) are correct; (D) is incorrect. \[ \boxed{\text{A, B, C}} \]