If IUPAC name of an element is "Unununnium" then the element belongs to nᵗʰ group of Periodic table. The value of n is ________.

Solution

Chemistry ma’am · 5 steps

Turn the name into digits first: un, un, un gives 111, and 111 sits right below gold.

1. Decode the name. IUPAC roots stand for digits: nil = 0, un = 1, bi = 2, tri = 3, and so on, with "-ium" at the end. "Un-un-un-ium" reads as 1, 1, 1, so Z = 111. This is roentgenium, Rg.
2. Find the period. Radon closes period 6 at Z = 86, and oganesson closes period 7 at Z = 118. So Z = 111 is in period 7.
3. Find the group. In period 7, the d-block runs from Z = 104 (Rf, group 4) to Z = 112 (Cn, group 12), one group per element. From 104 onwards, group number = Z - 100, so 111 - 100 = 11.
4. Check. Group 11 is the coinage-metal group: Cu, Ag, Au. Element 111 sits directly below gold, with the outer configuration (n-1)d⁹ns², expected like gold to be (n-1)d¹⁰ns¹.
5. So. Unununnium is in group 11, so n = 11.

The common slip: 10. Reading the name as un-un-nil (110, the old ununnilium) gives group 10, not 11. Every root in "unununnium" is "un", so Z = 111.

The idea to keep. Decode the IUPAC name into Z, then place it. For elements 104 to 118 in period 7, the group number is just Z - 100.

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