Question

Which of the following elements has two electrons in Q shell in its ground state?

• A. Ba
• B. Ra
• C. La
• D. Pb

Hint:

Remember the shell sequence: \[ K(1),\,L(2),\,M(3),\,N(4),\,O(5),\,P(6),\,Q(7) \] The Q-shell represents the outermost \(n=7\) shell.

Answer 1

The Correct Option is B

Step 1: Map shell letters to n.
The shells K, L, M, N, O, P, Q correspond to $n=1,2,3,4,5,6,7$, so the Q shell is simply the seventh shell.

Step 2: Locate each element's period.
Ba ($Z=56$), La ($Z=57$) and Pb ($Z=82$) all sit in period 6, so none of these three places any electron in $n=7$.

Step 3: Check Ra.
Ra ($Z=88$) is the first element of period 7, with configuration $[Rn]\,7s^2$. Its outermost subshell $7s^2$ contributes exactly $2$ electrons to the $n=7$ shell.

Step 4: Conclude.
Only radium places electrons in the Q shell, and the count is
\[ \boxed{2} \]
corresponding to option (2), Ra.