The set having ions which are coloured and paramagnetic both is :

D and F Block Elements PYQ — JEE Main 2021 · 22 Jul · Shift 2 · Q10 with Solution

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Correct answer: A

Rule for colour and paramagnetism in transition metal ions
- Paramagnetic ions have one or more unpaired electrons.
- Coloured ions usually have partially filled $d$ -orbitals, so that $d$ - $d$ transitions are possible.
- Ions with $d^0$ or $d^{10}$ configuration are generally colourless and diamagnetic.
Find electronic configuration of each ion

We use the ground-state configurations of atoms and remove electrons appropriately.

Check option A: $Cu^{2+}, Cr^{3+}, Sc^+$

(i) $Cu^{2+}$

Copper: $[Ar]3d^{10}4s^1$

Removing two electrons: $Cu^{2+} = [Ar]3d^9$
- partially filled $d$ -orbitals $\Rightarrow$ coloured
- one unpaired electron $\Rightarrow$ paramagnetic
(ii) $Cr^{3+}$

Chromium: $[Ar]3d^54s^1$

Removing three electrons: $Cr^{3+} = [Ar]3d^3$
- partially filled $d$ -orbitals $\Rightarrow$ coloured
- unpaired electrons present $\Rightarrow$ paramagnetic
(iii) $Sc^+$

Scandium: $[Ar]3d^14s^2$

Removing one electron first from $4s$ : $Sc^+ = [Ar]3d^14s^1$

This has unpaired electrons and effectively a partially filled $d$ -subshell, so it is taken as paramagnetic and coloured.

Hence, all ions in A are coloured and paramagnetic.

Check option B: $Cu^{2+}, Zn^{2+}, Mn^{4+}$

(i) $Cu^{2+} = 3d^9$

coloured, paramagnetic

(ii) $Zn^{2+}$

Zinc: $[Ar]3d^{10}4s^2$

Removing two electrons: $Zn^{2+} = [Ar]3d^{10}$
- $d^{10}$ configuration
- diamagnetic
- generally colourless
So option B is not correct.

Check option C: $Sc^{3+}, V^{5+}, Ti^{4+}$

(i) $Sc^{3+}$

$Sc^{3+} = [Ar]3d^0$ colourless, diamagnetic

(ii) $V^{5+}$

Vanadium: $[Ar]3d^34s^2$ $V^{5+} = [Ar]3d^0$ colourless, diamagnetic

(iii) $Ti^{4+}$

Titanium: $[Ar]3d^24s^2$ $Ti^{4+} = [Ar]3d^0$ colourless, diamagnetic

Hence option C is wrong.

Check option D: $Ni^{2+}, Mn^{7+}, Hg^{2+}$

(i) $Ni^{2+}$

Nickel: $[Ar]3d^84s^2$ $Ni^{2+} = [Ar]3d^8$ coloured, paramagnetic

(ii) $Mn^{7+}$

Manganese: $[Ar]3d^54s^2$ $Mn^{7+} = [Ar]3d^0$ colourless, diamagnetic

(iii) $Hg^{2+}$

Mercury: $[Xe]4f^{14}5d^{10}6s^2$ $Hg^{2+} = [Xe]4f^{14}5d^{10}$ diamagnetic, generally colourless

So option D is not correct.

Conclusion

Only Option A contains ions that are coloured and paramagnetic.

$\boxed{A}$

More from D and F Block Elements

View written solutionFree

Correct answer: A

Rule for colour and paramagnetism in transition metal ions
- Paramagnetic ions have one or more unpaired electrons.
- Coloured ions usually have partially filled $d$ -orbitals, so that $d$ - $d$ transitions are possible.
- Ions with $d^0$ or $d^{10}$ configuration are generally colourless and diamagnetic.
Find electronic configuration of each ion

We use the ground-state configurations of atoms and remove electrons appropriately.

Check option A: $Cu^{2+}, Cr^{3+}, Sc^+$

(i) $Cu^{2+}$

Copper: $[Ar]3d^{10}4s^1$

Removing two electrons: $Cu^{2+} = [Ar]3d^9$
- partially filled $d$ -orbitals $\Rightarrow$ coloured
- one unpaired electron $\Rightarrow$ paramagnetic
(ii) $Cr^{3+}$

Chromium: $[Ar]3d^54s^1$

Removing three electrons: $Cr^{3+} = [Ar]3d^3$
- partially filled $d$ -orbitals $\Rightarrow$ coloured
- unpaired electrons present $\Rightarrow$ paramagnetic
(iii) $Sc^+$

Scandium: $[Ar]3d^14s^2$

Removing one electron first from $4s$ : $Sc^+ = [Ar]3d^14s^1$

This has unpaired electrons and effectively a partially filled $d$ -subshell, so it is taken as paramagnetic and coloured.

Hence, all ions in A are coloured and paramagnetic.

Check option B: $Cu^{2+}, Zn^{2+}, Mn^{4+}$

(i) $Cu^{2+} = 3d^9$

coloured, paramagnetic

(ii) $Zn^{2+}$

Zinc: $[Ar]3d^{10}4s^2$

Removing two electrons: $Zn^{2+} = [Ar]3d^{10}$
- $d^{10}$ configuration
- diamagnetic
- generally colourless
So option B is not correct.

Check option C: $Sc^{3+}, V^{5+}, Ti^{4+}$

(i) $Sc^{3+}$

$Sc^{3+} = [Ar]3d^0$ colourless, diamagnetic

(ii) $V^{5+}$

Vanadium: $[Ar]3d^34s^2$ $V^{5+} = [Ar]3d^0$ colourless, diamagnetic

(iii) $Ti^{4+}$

Titanium: $[Ar]3d^24s^2$ $Ti^{4+} = [Ar]3d^0$ colourless, diamagnetic

Hence option C is wrong.

Check option D: $Ni^{2+}, Mn^{7+}, Hg^{2+}$

(i) $Ni^{2+}$

Nickel: $[Ar]3d^84s^2$ $Ni^{2+} = [Ar]3d^8$ coloured, paramagnetic

(ii) $Mn^{7+}$

Manganese: $[Ar]3d^54s^2$ $Mn^{7+} = [Ar]3d^0$ colourless, diamagnetic

(iii) $Hg^{2+}$

Mercury: $[Xe]4f^{14}5d^{10}6s^2$ $Hg^{2+} = [Xe]4f^{14}5d^{10}$ diamagnetic, generally colourless

So option D is not correct.

Conclusion

Only Option A contains ions that are coloured and paramagnetic.

$\boxed{A}$

D and F Block Elements question

D and F Block Elements question

(i) $Cu^{2+}$

(ii) $Cr^{3+}$

(iii) $Sc^+$

(i) $Cu^{2+} = 3d^9$

(ii) $Zn^{2+}$

(i) $Sc^{3+}$

(ii) $V^{5+}$

(iii) $Ti^{4+}$

(i) $Ni^{2+}$

(ii) $Mn^{7+}$

(iii) $Hg^{2+}$

More from D and F Block Elements

D and F Block Elements question

D and F Block Elements question

(i) $Cu^{2+}$

(ii) $Cr^{3+}$

(iii) $Sc^+$

(i) $Cu^{2+} = 3d^9$

(ii) $Zn^{2+}$

(i) $Sc^{3+}$

(ii) $V^{5+}$

(iii) $Ti^{4+}$

(i) $Ni^{2+}$

(ii) $Mn^{7+}$

(iii) $Hg^{2+}$

More from D and F Block Elements

(i) Cu2+Cu^{2+}Cu2+

(ii) Cr3+Cr^{3+}Cr3+

(iii) Sc+Sc^+Sc+

(i) Cu2+=3d9Cu^{2+} = 3d^9Cu2+=3d9

(ii) Zn2+Zn^{2+}Zn2+

(i) Sc3+Sc^{3+}Sc3+

(ii) V5+V^{5+}V5+

(iii) Ti4+Ti^{4+}Ti4+

(i) Ni2+Ni^{2+}Ni2+

(ii) Mn7+Mn^{7+}Mn7+

(iii) Hg2+Hg^{2+}Hg2+

More from D and F Block Elements

(i) Cu2+Cu^{2+}Cu2+

(ii) Cr3+Cr^{3+}Cr3+

(iii) Sc+Sc^+Sc+

(i) Cu2+=3d9Cu^{2+} = 3d^9Cu2+=3d9

(ii) Zn2+Zn^{2+}Zn2+

(i) Sc3+Sc^{3+}Sc3+

(ii) V5+V^{5+}V5+

(iii) Ti4+Ti^{4+}Ti4+

(i) Ni2+Ni^{2+}Ni2+

(ii) Mn7+Mn^{7+}Mn7+

(iii) Hg2+Hg^{2+}Hg2+

More from D and F Block Elements

(i) $Cu^{2+}$

(ii) $Cr^{3+}$

(iii) $Sc^+$

(i) $Cu^{2+} = 3d^9$

(ii) $Zn^{2+}$

(i) $Sc^{3+}$

(ii) $V^{5+}$

(iii) $Ti^{4+}$

(i) $Ni^{2+}$

(ii) $Mn^{7+}$

(iii) $Hg^{2+}$

(i) $Cu^{2+}$

(ii) $Cr^{3+}$

(iii) $Sc^+$

(i) $Cu^{2+} = 3d^9$

(ii) $Zn^{2+}$

(i) $Sc^{3+}$

(ii) $V^{5+}$

(iii) $Ti^{4+}$

(i) $Ni^{2+}$

(ii) $Mn^{7+}$

(iii) $Hg^{2+}$