Q132 / 200 · MHT CET - Medical 2026 (22 April Shift 2) · Chemistry
Which of the following is NOT the electronic configuration of Lanthanoid?
- A$[\text{Xe}]\ 4f^{10}\ 6s^2$
- B$[\text{Xe}]\ 4f^1\ 5d^1\ 6s^2$
- Correct answerC$[\text{Xe}]\ 4f^{14}\ 5d^{10}\ 6s^1$
- D$[\text{Xe}]\ 4f^7\ 5d^1\ 6s^2$
Solution
The general electronic configuration of lanthanoids is $[\text{Xe}]\ 4f^{1-14}\ 5d^{0-1}\ 6s^2$.
$[\text{Xe}]\ 4f^{10}\ 6s^2$ corresponds to Dysprosium (Dy, $Z=66$), which is a lanthanoid.
$[\text{Xe}]\ 4f^1\ 5d^1\ 6s^2$ corresponds to Cerium (Ce, $Z=58$), which is a lanthanoid.
$[\text{Xe}]\ 4f^{14}\ 5d^{10}\ 6s^1$ corresponds to Gold (Au, $Z=79$), which is a d-block transition element, not a lanthanoid.
$[\text{Xe}]\ 4f^7\ 5d^1\ 6s^2$ corresponds to Gadolinium (Gd, $Z=64$), which is a lanthanoid.
Therefore, $[\text{Xe}]\ 4f^{14}\ 5d^{10}\ 6s^1$ is not the electronic configuration of a lanthanoid.
Answer: $[\text{Xe}]\ 4f^{14}\ 5d^{10}\ 6s^1$
$[\text{Xe}]\ 4f^{10}\ 6s^2$ corresponds to Dysprosium (Dy, $Z=66$), which is a lanthanoid.
$[\text{Xe}]\ 4f^1\ 5d^1\ 6s^2$ corresponds to Cerium (Ce, $Z=58$), which is a lanthanoid.
$[\text{Xe}]\ 4f^{14}\ 5d^{10}\ 6s^1$ corresponds to Gold (Au, $Z=79$), which is a d-block transition element, not a lanthanoid.
$[\text{Xe}]\ 4f^7\ 5d^1\ 6s^2$ corresponds to Gadolinium (Gd, $Z=64$), which is a lanthanoid.
Therefore, $[\text{Xe}]\ 4f^{14}\ 5d^{10}\ 6s^1$ is not the electronic configuration of a lanthanoid.
Answer: $[\text{Xe}]\ 4f^{14}\ 5d^{10}\ 6s^1$
Hence, the correct answer is option: 'C'.