Question

The IUPAC nomenclature of an element with electronic configuration $[ Rn ] 5 f ^{14} 6 d ^1 7 s ^2$ is :

• A. Unnilbium
• B. Unnilunium
• C. Unnilquadium
• D. Unniltrium

Answer 1

The Correct Option is D

The given electronic configuration is \([Rn] \, 5f^{14} \, 6d^1 \, 7s^2\). This configuration can be deciphered as follows:

1. \([Rn]\) refers to the noble gas Radon, which is used here as a core reference for electronic configuration.
2. \(5f^{14}\) indicates the f-block is completely filled in the 5th energy level.
3. \(6d^1\) shows that there is one electron in the d sub-shell in the 6th energy level.
4. \(7s^2\) means there are two electrons in the s sub-shell in the 7th energy level.

To find the element's identity, we calculate its atomic number. The atomic number of Radon is 86, and the additions from the configuration are:

• \(5f^{14}\) adds 14 electrons.
• \(6d^1\) adds 1 electron.
• \(7s^2\) adds 2 electrons.

Thus, the total atomic number is:

\(86 + 14 + 1 + 2 = 103\)

The element with atomic number 103 is Lawrencium (Lr). In the IUPAC naming system for elements beyond 100, each digit of the atomic number is translated into a syllable:

• '1' is translated to "Un".
• '0' is translated to "nil".
• '3' is translated to "trium".

Combining these syllables, the systematic IUPAC name for element 103 is "Unniltrium". Therefore, the correct answer is:

Unniltrium