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If the electronic configuration of \(M^{3+}\) is [Xe] \(4 f^3\), then \(M^{3+}\) is
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The correct answer is:
\(\mathrm{Nd}^{3+}\)
Given, the electronic configuration of \(M^{3+}\) is [Xe] \(4 \int^3\).
The electronic configuration of metals given in options in \(\mathrm{M}^{3+}\) are as follows:
(a) \(\mathrm{Nd}^{3+}=[\mathrm{Xe}] 4 f^3\)
(b) \(\operatorname{Pr}^{3+}=[\mathrm{Xe}] 4 f^2\)
(c) \(\mathrm{Sm}^{3+}=[\mathrm{Xe}] 4 f^5\)
(d) \(\mathrm{Dy}^{3+}=[\mathrm{Xe}] 4 f^9\)
Thus, \(M^{3+}\) among the given options will be \(\mathrm{Nd}^{3+}\).
The electronic configuration of metals given in options in \(\mathrm{M}^{3+}\) are as follows:
(a) \(\mathrm{Nd}^{3+}=[\mathrm{Xe}] 4 f^3\)
(b) \(\operatorname{Pr}^{3+}=[\mathrm{Xe}] 4 f^2\)
(c) \(\mathrm{Sm}^{3+}=[\mathrm{Xe}] 4 f^5\)
(d) \(\mathrm{Dy}^{3+}=[\mathrm{Xe}] 4 f^9\)
Thus, \(M^{3+}\) among the given options will be \(\mathrm{Nd}^{3+}\).
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