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

Part (i):

(I) Variable oxidation states of transition metals Transition elements have incompletely filled $d$ orbitals. The general electronic configuration of transition metals is: \[ (n-1)d^{1-10}ns^{0-2} \] The energy difference between $(n-1)d$ and $ns$ orbitals is very small. Therefore, electrons from both orbitals can be removed or shared during chemical reactions. For example: \[ Fe^{2+}\text{ and }Fe^{3+} \] show different oxidation states. Hence, transition metals exhibit variable oxidation states.

(II) Magnetic behaviour of Mn$^{2+}$ and Ti$^{2+}$

Titanium: Atomic number: \[ Ti=22 \] Electronic configuration: \[ Ti=[Ar]3d^24s^2 \] After losing two electrons: \[ Ti^{2+}=[Ar]3d^2 \] It contains: \[ \boxed{2\text{ unpaired electrons}} \]

Manganese: Atomic number: \[ Mn=25 \] Electronic configuration: \[ Mn=[Ar]3d^54s^2 \] After losing two electrons: \[ Mn^{2+}=[Ar]3d^5 \] It contains: \[ \boxed{5\text{ unpaired electrons}} \] Since paramagnetism depends on the number of unpaired electrons: \[ \boxed{Mn^{2+}\text{ is more paramagnetic than }Ti^{2+}} \]

(III) Oxidising strength of ions An oxidising agent is a species that accepts electrons and gets reduced. The reduction reactions are: \[ Mn^{3+}+e^-\rightarrow Mn^{2+} \] \[ Cr^{3+}+e^-\rightarrow Cr^{2+} \] \[ V^{3+}+e^-\rightarrow V^{2+} \] $Mn^{3+}$ has the greatest tendency to accept electrons because the resulting $Mn^{2+}$ has a stable half-filled $3d^5$ configuration. Therefore: \[ \boxed{Mn^{3+}\text{ is the strongest oxidising agent}} \]

Part (ii): Balancing reactions

(I) Formation of potassium manganate In alkaline medium: \[ MnO_2 \] is oxidised to manganate ion: \[ MnO_4^{2-} \] The balanced equation is: \[ \boxed{ 2MnO_2+4KOH+O_2 \rightarrow 2K_2MnO_4+2H_2O } \]

(II) Reaction of permanganate with oxalate ion In acidic medium: \[ MnO_4^- \] is reduced to: \[ Mn^{2+} \] and oxalate ion is oxidised to carbon dioxide. The balanced ionic equation is: \[ \boxed{ 2MnO_4^-+5C_2O_4^{2-}+16H^+ \rightarrow 2Mn^{2+}+10CO_2+8H_2O } \]

Final Answer:

(i) \[ \boxed{\text{Transition metals show variable oxidation states due to similar energies of }(n-1)d\text{ and }ns\text{ orbitals}} \] \[ \boxed{Mn^{2+}\text{ is more paramagnetic because it has five unpaired electrons}} \] \[ \boxed{Mn^{3+}\text{ is the strongest oxidising agent}} \]

(ii) \[ \boxed{ 2MnO_2+4KOH+O_2 \rightarrow 2K_2MnO_4+2H_2O } \] \[ \boxed{ 2MnO_4^-+5C_2O_4^{2-}+16H^+ \rightarrow 2Mn^{2+}+10CO_2+8H_2O } \]