The correct answer is Option (2) → (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
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List-I Metal
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List-II Properties
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(A) $Sc^{3+}$
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(II) Colorless
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(B) $Cr^{3+}$
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(I) $3d^3$
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(C) $Cu^{2+}$
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(IV) $3d^9$
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(D) $Ni^{2+}$
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(III) $3d^8$
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(A) $Sc^{3+}$ — (II) Colorless
- Neutral Scandium ($Sc$): $[Ar] 3d^1 4s^2$
- $Sc^{3+}$ Ion: By removing 3 electrons (two from $4s$ and one from $3d$), we get $[Ar] 3d^0$.
- Property: Because there are no electrons in the $d$-orbital, $d-d$ transitions are impossible. This makes $Sc^{3+}$ compounds colorless.
(B) $Cr^{3+}$ — (I) $3d^3$
- Neutral Chromium ($Cr$): $[Ar] 3d^5 4s^1$ (Exception due to half-filled stability)
- $Cr^{3+}$ Ion: Removing 3 electrons (one from $4s$ and two from $3d$) leaves $[Ar] 3d^3$.
(C) $Cu^{2+}$ — (IV) $3d^9$
- Neutral Copper ($Cu$): $[Ar] 3d^{10} 4s^1$ (Exception due to fully-filled stability)
- $Cu^{2+}$ Ion: Removing 2 electrons (one from $4s$ and one from $3d$) leaves $[Ar] 3d^9$.
(D) $Ni^{2+}$ — (III) $3d^8$
- Neutral Nickel ($Ni$): $[Ar] 3d^8 4s^2$
- $Ni^{2+}$ Ion: Removing the 2 electrons from the $4s$ orbital leaves $[Ar] 3d^8$.
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