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: Recall shell notation.
The shells are designated as: \[ K,L,M,N,O,P,Q \] corresponding to \[ n=1,2,3,4,5,6,7 \] Hence, the Q-shell is the seventh shell \((n=7)\).

Step 2: Write the electronic configurations of the given elements.
For Barium (\(Ba\), \(Z=56\)): \[ [Xe]\,6s^2 \] For Radium (\(Ra\), \(Z=88\)): \[ [Rn]\,7s^2 \] For Lanthanum (\(La\), \(Z=57\)): \[ [Xe]\,5d^1 6s^2 \] For Lead (\(Pb\), \(Z=82\)): \[ [Xe]\,4f^{14}5d^{10}6s^2 6p^2 \]

Step 3: Identify electrons present in the Q-shell.
Since Q-shell corresponds to \[ n=7, \] we count only electrons having principal quantum number \(7\).
For \(Ba\): \[ 7^{th}\text{ shell electrons}=0 \] For \(La\): \[ 7^{th}\text{ shell electrons}=0 \] For \(Pb\): \[ 7^{th}\text{ shell electrons}=0 \] For \(Ra\): \[ 7s^2 \] Thus, \[ 7^{th}\text{ shell electrons}=2 \]

Step 4: Compare all options.
Only Radium contains exactly two electrons in the Q-shell.

Step 5: Final conclusion.
Therefore, the required element is \[ \boxed{\text{Ra}} \] which corresponds to option (2).