Question

If IUPAC name of an element is “Unununnium” then the element belongs to nth group of periodic table. The value of n is______

Answer 1

Correct Answer: 11

This task requires identifying the periodic table group for an element based on its IUPAC systematic name, "Unununnium".

Underlying Principle:

An element's IUPAC systematic name is derived from its atomic number (Z). Each digit in the atomic number corresponds to a specific numerical root. The group number can be determined from the atomic number either by analyzing its electronic configuration or by locating its position within its period on the periodic table.

The numerical roots are as follows:

• 0 = nil
• 1 = un
• 2 = bi
• 3 = tri
• 4 = quad
• 5 = pent
• 6 = hex
• 7 = sept
• 8 = oct
• 9 = enn

The systematic name is constructed by concatenating these roots and appending the suffix "-ium".

Sequential Breakdown:

Step 1: Ascertain the atomic number (Z) from the IUPAC name.

The name "Unununnium" is decomposed into its root components:

• Un \(\rightarrow\) 1
• un \(\rightarrow\) 1
• un \(\rightarrow\) 1

The combination of these digits yields the atomic number:

\[ Z = 111 \]

This element is known as Roentgenium (Rg).

Step 2: Establish the element's block and period.

The element's location can be found by examining its electronic configuration. The noble gas preceding the element with Z = 111 is Radon (Rn, Z = 86).

Radon's electronic configuration is \([Xe] 4f^{14} 5d^{10} 6s^2 6p^6\). Following Rn, the filling of the 7th period commences.

The configuration for Z = 111 is:

\[ [Rn] \, 7s^2 \, 5f^{14} \, 6d^9 \]

(Note: Due to stability considerations, the probable configuration is \([Rn] \, 7s^1 \, 5f^{14} \, 6d^{10}\), mirroring Gold's configuration in the same group. Either configuration will result in the same group number determination.)

Since the final electron occupies a d-orbital (6d), the element is categorized within the d-block. The highest principal quantum number is 7, indicating its placement in the 7th period.

Step 3: Compute the group number.

For elements in the d-block, the group number is the sum of the electrons in the outermost s-orbital and the penultimate d-orbital.

\[ \text{Group Number} = (\text{electrons in } ns) + (\text{electrons in } (n-1)d) \]

Utilizing the configuration \([Rn] \, 7s^2 \, 5f^{14} \, 6d^9\):

\[ \text{Group Number} = 2 (\text{from } 7s^2) + 9 (\text{from } 6d^9) = 11 \]

Alternatively, employing the more stable configuration \([Rn] \, 7s^1 \, 5f^{14} \, 6d^{10}\):

\[ \text{Group Number} = 1 (\text{from } 7s^1) + 10 (\text{from } 6d^{10}) = 11 \]

Both configurations assign the element to Group 11.

Concluding Calculation and Outcome:

The element designated by the IUPAC name "Unununnium" possesses an atomic number of 111 and resides in Group 11 of the periodic table.

As stated in the problem, the element is located in the nth group, therefore \(n = 11\).

The determined value of n is 11.