The IUPAC nomenclature of an element with electronic configuration [Rn] 5f14 6d1 7s2 is:

🧠 Count the electrons to find Z, then translate Z digit-by-digit using IUPAC roots The configuration tells you the atomic number. The IUPAC system for elements above Z=100 spells the digits of the atomic number using fixed Latin/Greek roots.

🗺️ Find Z $[\ce{Rn}]\,5f^{14}\,6d^{1}\,7s^{2}$. Rn has $Z = 86$. Add the extra electrons: $14 + 1 + 2 = 17$. So $Z = 86 + 17 = 103$.

Decode 103 with IUPAC roots The roots are: 0 = nil, 1 = un, 2 = bi, 3 = tri, 4 = quad, 5 = pent, 6 = hex, 7 = sept, 8 = oct, 9 = enn. Digits of 103: 1, 0, 3 → un + nil + tri + ium → Unniltrium.

⚠️ The trap Option (a) Unnilunium is for $Z = 101$ (1-0-1). Option (b) Unnilbium is for $Z = 102$. Option (d) Unnilquadium is for $Z = 104$. Students who miscount one electron land on a neighbouring name. Always recount the electrons in the configuration twice before applying the roots.

$\boxed{\text{Answer: (c)}}$