JEE MAIN - Chemistry (2024 - 27th January Morning Shift - No. 20)

The electronic configuration for Neodymium is:

[Atomic Number for Neodymium 60]

$$[\mathrm{Xe}] 4 \mathrm{f}^4 6 \mathrm{~s}^2$$
$$[\mathrm{Xe}] 5 \mathrm{f}^4 7 \mathrm{~s}^2$$
$$[\mathrm{Xe}] 4 \mathrm{f}^6 6 \mathrm{~s}^2$$
$$[\mathrm{Xe}] 4 \mathrm{f}^1 5 \mathrm{~d}^1 6 \mathrm{~s}^2$$

Explanation

The electronic configuration of an element is determined by adding electrons to atomic orbitals following the Aufbau principle, the Pauli exclusion principle, and Hund's rule. For atoms with atomic numbers higher than that of xenon (Xe, atomic number 54), the electrons start filling the outer orbitals following the xenon core. In this case, we're looking at neodymium (Nd), which has an atomic number of 60.

To find the electronic configuration of neodymium, we first note the configuration of xenon, which serves as the core, and then add the additional electrons beyond xenon to the relevant orbitals.

Neodymium has 60 - 54 = 6 electrons more than xenon. These additional electrons will fill the 4f and 6s orbitals. Based on its position in the periodic table, neodymium’s electrons start filling the 4f orbitals before the 5d. Since the 6s orbital is at a lower energy level than the 4f orbital, it gets filled before the 4f. Neodymium will have 2 electrons in the 6s orbital, and the remaining 4 electrons will go into the 4f orbital.

Therefore, the correct electronic configuration for neodymium is:

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

So, Option A is the correct answer.

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