$KMnO_{4}$ in Acidic Medium (5$e^{-}$ transfer):

$\text{MnO}_4^- + 8\text{H}^+ + 5e^- \longrightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$

Overall with $FeSO_{4}$: $\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \longrightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}$

$KMnO_{4}$ in Neutral Medium (3$e^{-}$ transfer):

$\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \longrightarrow \text{MnO}_2 + 4\text{OH}^-$

$KMnO_{4}$ in Basic Medium (1$e^{-}$ transfer):

$\text{MnO}_4^- + e^- \longrightarrow \text{MnO}_4^{2-}$

$K_{2}Cr_{2}O_{7}$ in Acidic Medium:

$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \longrightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$

Overall with $FeSO_{4}$: $\text{Cr}_2\text{O}_7^{2-} + 6\text{Fe}^{2+} + 14\text{H}^+ \longrightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}$

Chromate-Dichromate Equilibrium:

$2\text{CrO}_4^{2-} + 2\text{H}^+ \rightleftharpoons \text{Cr}_2\text{O}_7^{2-} + \text{H}_2\text{O}$

Cisplatin Hydrolysis (activation step inside cell):

$[\text{Pt(NH}_3)_2\text{Cl}_2] + \text{H}_2\text{O} \longrightarrow [\text{Pt(NH}_3)_2\text{Cl(H}_2\text{O})]^+ + \text{Cl}^-$

Determination of Oxidation State in Complex:

For $[Co(NH_3)_4Cl_2]^+$:

$x + 4(0) + 2(-1) = +1 \implies x = +3$

Co is in +3 oxidation state.

Magnetic Moment Formula:

$\mu_s = \sqrt{n(n+2)}\text{ BM}$

For $Fe^{3+}$ with $CN^{-}$ (low spin, $d^{5}$ → 1 unpaired): $\mu = \sqrt{1 \times 3} = \sqrt{3} \approx 1.73\text{ BM}$

For $Fe^{3+}$ with $F^{-}$ (high spin, $d^{5}$ → 5 unpaired): $\mu = \sqrt{5 \times 7} = \sqrt{35} \approx 5.92\text{ BM}$

Crystal Field Splitting Relationship:

$\Delta_t = \frac{4}{9}\Delta_o$

CFSE for $d^{6}$ low spin octahedral:

$\text{CFSE} = 6 \times (-0.4\Delta_o) + 0 \times (0.6\Delta_o) = -2.4\Delta_o$