In the transformation of oxyhaemoglobin to deoxyhaemoglobin
(a)$\mathrm{Fe}^{2+}$ in the low spin state changes to $\mathrm{Fe}^{2+}$ in the high spin state
(b)$\mathrm{Fe}^{2+}$ in the low spin state changes to $\mathrm{Fe}^{3+}$ in the low spin state.
(c)$\mathrm{Fe}^{2+}$ in the high spin state changes to $\mathrm{Fe}^{2+}$ in the low spin state
(d)$\mathrm{Fe}^{2+}$ in the high spin state changes to $\mathrm{Fe}^{3+}$ in the high spin state.
Answer
Answer: A ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
Oxyhaemoglobin has low-spin Fe(II) (in-plane, six-coordinate); on losing O₂ it becomes high-spin Fe(II), which is too large for the porphyrin hole and moves out of plane — the trigger of the cooperative T-state change.