d and f Block Elements | JEE Main 2024 Solution

JEE MAIN 2024

27-01-2024 SHIFT-1

Question

The electronic configuration for Neodymium is:
[Atomic Number for Neodymium 60]

Select the correct option:

$[Xe]4{f^4}6{s^2}$

$[{\rm{Xe}}]5{{\rm{f}}^4}7\;{{\rm{s}}^2}$

$[Xe]4{f^6}6{s^2}$

$[{\rm{Xe}}]4{{\rm{f}}^1}5\;{{\rm{d}}^1}6\;{{\rm{s}}^2}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Electronic configuration of $Nd(Z=60)$ is; $[Xe]4f^46s^2$

Question Tags

JEE Main

Chemistry

Easy

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