Q30 / 160 · TS EAMCET 2024 (10 May Shift 2) · Chemistry

Which of the following orders is correct for the property given?

  • A
    $\mathrm{Cr} \lt \mathrm{Mn} \lt \mathrm{Fe}$ - standard electrode potential value of $\mathrm{M}^{3+} / \mathrm{M}^{2+}$
  • B
    $\mathrm{Cr}^{2+} \lt \mathrm{Mn}^{2+} \lt \mathrm{Fe}^{2+}-$ magnetic moments
  • Correct answer
    C
    $\mathrm{VO}_2^{+} \lt \mathrm{Cr}_2 \mathrm{O}_7^{2-} \lt \mathrm{MnO}_4^{-}$- oxidizing power
  • D
    $\mathrm{Ti} \lt \mathrm{V} \lt \mathrm{Cr}$ - first ionization enthalpy.

Solution

Standard electrode potential value for $\mathrm{M}^{3+} / \mathrm{M}^{2+}$ is $\mathrm{Mn}\gt\mathrm{Fe}\gt\mathrm{Cr}$. Magnetic moment $(\mu)=\sqrt{\mathrm{n}(\mathrm{n}+2}$ $\mathrm{n}=$ no. of unpaired electron. So $\mathrm{Mn}^{2+}=3 \mathrm{~d}^5=5$ unpaired election $\mathrm{Cr}^{2+}=3 \mathrm{~d}^4=4$ unpaired electron $\mathrm{Fe}^{2+}=3 \mathrm{~d}^6=4$ unpaired electron So, $\mathrm{Mn}^{2+}\gt\mathrm{Cr}^{2+} \approx \mathrm{Fe}^{2+}$ $(+5) \quad(+6) \quad(+7)$ $\mathrm{VO}_2^{+} \lt \mathrm{Cr}_2 \mathrm{O}_7^{2-} \lt \mathrm{MnO}_4^{-}$ $\mathrm{Mn}^{+7}$ have highest oxidation state then $\mathrm{Cr}^{+6}$ and $\mathrm{V}^{+5}$ so it is act as strongest oxidising agent. First ionisation enthalpy $\mathrm{Ti}\gt\mathrm{Cr}\gt\mathrm{V}$

Hence, the correct answer is option: 'C'.

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