Question

Considering that there are 111 elements in the present periodic table, how many more elements are to be filled if an element with an electron entering into 'g' subshell has to be placed?

• A. 10
• B. 12
• C. 8
• D. 14

Answer 1

The correct option is A 10

${}^{\prime }{g}^{\prime }$ subshell filling starts after the ${}^{\prime }7p^{\prime }$ subshell is completely filled. That means $s,p,d,f$ subshells of 4th and 5th shells are completely filled $s,p,d$ subshell. 's' subshell of $8^{th}$ shell is filled. The total number of electrons would be 120. The atomic number of the element with one electron in the 5g subshell is 121. If 111 is the last element, 10 more elements are to be discovered.${111}^{th}$ element belongs to' block. There is a vacancy for one more element in' block 6 elements go to ${}^{\prime }{p}^{\prime }$ block. 2 elements go to ${}^{\prime }{s}^{\prime }$ block. The next element contains differentiating electron in the $5g$ subshell. It belongs to $8^{th}$ period.