The reduced form of a metal ion M in a complex is NMR active. On oxidation, the complex gives an EPR signal with $g_{\parallel}=2.2$ and $g_{\perp}=2.0$. Mossbauer spectroscopy cannot characteristic the metal complex. The M is
(a)Zn
(b)Sn
(c)Cu
(d)Fe
Answer
Answer: C ✓ checked by 4AB · confidence high
Explanation
Cu(I) is d¹⁰ (diamagnetic, ⁶³/⁶⁵Cu NMR-active); Cu(II) is d⁹ with an axial EPR spectrum (g∥ > g⊥ > 2); copper has no Mössbauer isotope.