Match the metal/ion given in List-I with properties given in List-II
|
List-I Metal |
List-II Properties |
|
(A) \( Sc^{3+} \) |
(I) \( 3d^3 \) |
|
(B) \( Cr^{3+} \) |
(II) Colorless |
|
(C) \( Cu^{2+} \) |
(III) \( 3d^8 \) |
|
(D) \( Ni^{2+} \) |
(IV) \( 3d^9 \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
To find the 3d electrons of an ion, write the configuration of the neutral atom from its atomic number and remove the 4s electrons first, then 3d electrons. An ion with no unpaired d electron (3d0 or 3d10) has no d–d transition and is colourless.
(A) Sc3+ — This matches (II) Colorless. Sc (Z = 21) is \( \mathrm{[Ar]\,3d^{1}\,4s^{2}} \); losing three electrons gives \( \mathrm{3d^{0}} \). With no d electron, no d–d transition is possible.
(B) Cr3+ — This matches (I) 3d3. Cr (Z = 24) is \( \mathrm{[Ar]\,3d^{5}\,4s^{1}} \); losing the 4s electron and two 3d electrons gives \( \mathrm{3d^{3}} \).
(C) Cu2+ — This matches (IV) 3d9. Cu (Z = 29) is \( \mathrm{[Ar]\,3d^{10}\,4s^{1}} \); losing the 4s electron and one 3d electron gives \( \mathrm{3d^{9}} \).
(D) Ni2+ — This matches (III) 3d8. Ni (Z = 28) is \( \mathrm{[Ar]\,3d^{8}\,4s^{2}} \); losing the two 4s electrons gives \( \mathrm{3d^{8}} \).
Shortcut: \( \ce{Sc^{3+}} \) has no d electron, so (A)-(II); this pair appears only in Option (2).