The element with the electronic configuration 4\(f^{14} 5d^{10} 6s^2 6p^4\) is the last element of the p-block in the 6th period, which corresponds to the element Lead (Pb). This configuration represents the completion of the 4f, 5d, and 6s subshells, with the last electron entering the 6p orbital, completing the p-block.
So, the correct answer is (D): \(f^{14} 5d^{10} 6s^2 6p^4\)
The last element in the p-block of the \(6^{th}\) period is Polonium (Po), which is in Group 16 (the chalcogen group). The atomic number of Polonium is 84, and its electronic configuration can be written as: \[ \text{Po:} \quad [Xe] 4f^1 5d^1 6s^2 6p^4 \] Thus, the outermost electronic configuration of Polonium (Po) is 4f\(^1\) 5d\(^1\) 6s\(^2\) 6p\(^4\), making option (D) the correct answer.
The Group 18 elements, also known as noble gases except for Helium, are inert in nature because they have completely filled ns2 np6 electronic configuration in their valence shells. These gases also have high ionization enthalpy and more positive electron gain enthalpy. All these elements are chemically unreactive i.e. they don’t form many compounds.
Group 18 Elements consist of six elements. They are as mentioned below:
The electronic configuration of Group 18 Elements along with their symbol and atomic number is given in the tabulated form below: