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

Solution

Chemistry ma’am · 3 steps

Turn the configuration into an atomic number first, then spell that number digit by digit.

1. Count the electrons. The core [Rn] holds 86 electrons. Add the ones written after it: 14 + 1 + 2 = 17. So Z = 86 + 17 = 103.
2. Digit roots. IUPAC gives each digit a root: 0 = nil, 1 = un, 2 = bi, 3 = tri, 4 = quad. Write one root per digit, then add the ending -ium.

   Option	Digits	Atomic number
   (a) Unnilbium	un + nil + bi	102
   (b) Unnilunium	un + nil + un	101
   (c) Unnilquadium	un + nil + quad	104
   (d) Unniltrium	un + nil + tri	103

3. So. Element 103, which is lawrencium, has the IUPAC name Unniltrium, option (d).

The idea to keep. For any heavy element, add the noble-gas core to the electrons written after it to get Z. Then read Z digit by digit: nil, un, bi, tri, quad, pent, hex, sept, oct, enn, and add -ium.

Answer from the NTA final answer key. Solution by PrepWiser.