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

The problem asks for the group number in the periodic table for an element given its IUPAC systematic name, "Unununnium".

Concept Used:

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

The numerical roots are:

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

The name is formed by concatenating the roots and adding the suffix "-ium".

Step-by-Step Solution:

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

The name "Unununnium" is broken down into its constituent roots:

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

Combining these digits gives the atomic number:

\[ Z = 111 \]

The element is Roentgenium (Rg).

Step 2: Determine the block and period of the element.

To find the position of the element, we can write its electronic configuration. The noble gas preceding the element with Z = 111 is Radon (Rn, Z = 86).

The electronic configuration of Rn is \([Xe] 4f^{14} 5d^{10} 6s^2 6p^6\). After Rn, we start filling the 7th period.

The configuration for Z = 111 is:

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

(Note: Due to stability, the actual configuration is likely \([Rn] \, 7s^1 \, 5f^{14} \, 6d^{10}\), similar to Gold in the same group, but either configuration will lead to the same group number).

Since the last electron enters the d-orbital (6d), the element belongs to the d-block. The highest principal quantum number is 7, so it is in the 7th period.

Step 3: Calculate the group number.

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

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

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

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

Alternatively, using 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 place the element in Group 11.

Final Computation & Result:

The element with the IUPAC name "Unununnium" has an atomic number of 111 and belongs to Group 11 of the periodic table.

The problem states that the element belongs to the nth group, so \(n = 11\).

The value of n is 11.