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Question 60

Given below are two statements :
Statement I : The number of pairs among [Ti$$^{4+}$$, V$$^{2+}$$], [V$$^{2+}$$, Mn$$^{2+}$$], [Mn$$^{2+}$$, Fe$$^{3+}$$] and [V$$^{2+}$$, Cr$$^{2+}$$] in which both ions are coloured is 3.
Statement II : The number of pairs among [La$$^{3+}$$, Yb$$^{2+}$$], [Lu$$^{3+}$$, Ce$$^{4+}$$] and [Ac$$^{3+}$$, Lr$$^{3+}$$] ions in which both are diamagnetic is 3.
In the light of the above statements, choose the correct from the options given below :

Statement I
Transition metal ions are typically coloured if they have partially filled $$d$$-orbitals ($$d^1$$ to $d^9$$ configuration), which allows for $$d\text{-}d$$ electronic transitions.

Ions with a $$d^0$$ or $$d^{10}$$ configuration are generally colourless.

  • $$\text{Ti}^{4+}$$: $$[\text{Ar}] 3d^0$$ $$\rightarrow$$ Colourless (no $$d$$-electrons)
    $$\text{}$$
  • $$\text{V}^{2+}$$: $$[\text{Ar}] 3d^3$$ $$\rightarrow$$ Coloured 
    $$\text{}$$
  • $$\text{Mn}^{2+}$$: $$[\text{Ar}] 3d^5$$ $$\rightarrow$$ Coloured
    $$\text{}$$
  • $$\text{Fe}^{3+}$$: $$[\text{Ar}] 3d^5$$ $$\rightarrow$$ Coloured
    $$\text{}$$
  • $$\text{Cr}^{2+}$$: $$[\text{Ar}] 3d^4$$ $$\rightarrow$$ Coloured
    $$\text{}$$
  • $$[\text{Ti}^{4+}, \text{V}^{2+}]$$: $$\text{Ti}^{4+}$$ is colourless $$\rightarrow$$ Not both coloured
    $$\text{}$$
  • $$[\text{V}^{2+}, \text{Mn}^{2+}]$$: Both have partially filled $$d$$-orbitals $$\rightarrow$$ Both coloured
    $$\text{}$$
  • $$[\text{Mn}^{2+}, \text{Fe}^{3+}]$$: Both have partially filled $$d$$-orbitals $$\rightarrow$$ Both coloured
    $$\text{}$$
  • $$[\text{V}^{2+}, \text{Cr}^{2+}]$$: Both have partially filled $$d$$-orbitals $$\rightarrow$$ Both coloured
  • There are exactly 3 pairs where both ions are coloured.

    Statement I is CORRECT.

      Statement II

      An ion is diamagnetic if it has no unpaired electrons (i.e., all its orbitals are completely empty or completely filled).

    • $$\text{La}^{3+}$$ ($$Z = 57$$): $$[\text{Xe}] 4f^0$$ $$\rightarrow$$ No unpaired electrons $$\rightarrow$$ Diamagnetic
    • $$\text{Yb}^{2+}$$ ($$Z = 70$$): $$[\text{Xe}] 4f^{14}$$ $$\rightarrow$$ Completely filled $$f$$-orbital $$\rightarrow$$ Diamagnetic
    • $$\text{Lu}^{3+}$$ ($$Z = 71$$): $$[\text{Xe}] 4f^{14}$$ $$\rightarrow$$ Completely filled $$f$$-orbital $$\rightarrow$$ Diamagnetic
    • $$\text{Ce}^{4+}$$ ($$Z = 58$$): $$[\text{Xe}] 4f^0$$ $$\rightarrow$$ No unpaired electrons $$\rightarrow$$ Diamagnetic
    • $$\text{Ac}^{3+}$$ ($$Z = 89$$): $$[\text{Rn}] 5f^0$$ $$\rightarrow$$ No unpaired electrons $$\rightarrow$$ Diamagnetic
    • $$\text{Lr}^{3+}$$ ($$Z = 103$$): $$[\text{Rn}] 5f^{14}$$ $$\rightarrow$$ Completely filled $$f$$-orbital $$\rightarrow$$ Diamagnetic
      1. $$[\text{La}^{3+}, \text{Yb}^{2+}]$$: Both are diamagnetic $$\rightarrow$$ True
      2. $$[\text{Lu}^{3+}, \text{Ce}^{4+}]$$: Both are diamagnetic $$\rightarrow$$ True
      3. $$[\text{Ac}^{3+}, \text{Lr}^{3+}]$$: Both are diamagnetic $$\rightarrow$$ True

      There are exactly 3 pairs where both ions are diamagnetic.

      • Statement II is CORRECT.

      Option A: Both Statement I and Statement II are correct

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