JEE MAIN - Chemistry (2023 - 31st January Morning Shift - No. 4)

$$\mathrm{Nd^{2+}}$$ = __________
$$\mathrm{4f^4 6s^2}$$
$$\mathrm{4f^3}$$
$$\mathrm{4f^4}$$
$$\mathrm{4f^2 6s^2}$$

Explanation

Nd(60) = [Xe] 4f4 5d0 6s2

Nd2+ = [Xe] 4f4 5d0 5s0

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