Question

(I) Why do transition metals show variable oxidation states? (II) Out of Mn$^{2+}$ and Ti$^{2+}$ which will be more paramagnetic and why? Atomic No.: Ti = 22, Mn = 25 (III) Which ion is the strongest oxidising agent in the options given below: \[ Cr^{3+},\ V^{3+},\ Mn^{3+} \] Atomic No.: Cr = 24, V = 23, Mn = 25 (ii) Complete and balance the following equations: (I) \[ 2MnO_2+4KOH+O_2\rightarrow ? \] (II) \[ 5C_2O_4^{2-}+2MnO_4^-+16H^+\rightarrow ? \]

Hint:

Important points for transition metals: \[ \boxed{(n-1)d \text{ and } ns \text{ orbitals have similar energies}} \] \[ \boxed{\text{More unpaired electrons } \Rightarrow \text{ more paramagnetic}} \] Half-filled configuration: \[ \boxed{d^5 \text{ is highly stable}} \] Therefore: \[ Mn^{2+}=3d^5 \] is especially stable.

Answer 1

Step 1: Variable oxidation states and paramagnetism.
Transition metals have $(n-1)d$ and $ns$ orbitals of comparable energy; electrons from both can participate in bonding, giving variable oxidation states. $Mn^{2+}$ has configuration $3d^5$ (5 unpaired electrons) while $Ti^{2+}$ has $3d^2$ (2 unpaired electrons), so $Mn^{2+}$ is more paramagnetic.
Step 2: Strongest oxidising agent.
Among $Mn^{3+}$, $Cr^{3+}$, $V^{3+}$: $Mn^{3+}$ accepts an electron most readily because the product $Mn^{2+}$ has the exceptionally stable half-filled $3d^5$ configuration. Hence $Mn^{3+}$ is the strongest oxidising agent.
Step 3: Balance equation (I).
$MnO_2$ is oxidised to manganate in alkali: \[ 2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O \]
Step 4: Balance equation (II).
$MnO_4^-$ (reduced to $Mn^{2+}$) oxidises $C_2O_4^{2-}$ to $CO_2$ in acid: \[ \boxed{2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O} \]