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In which of the following processes is the change in oxidation state equal to five?
Calculate the change in oxidation state for each:
- (A) C in C$_2$O$_4^{2-}$: $+3$; C in CO$_2$: $+4$. Change $= +1$ per C atom (2 C, so total = 2, but per atom = 1).
- (B) Mn in MnO$_4^-$: $+7$; Mn in Mn$^{2+}$: $+2$. Change $= 7 - 2 = \mathbf{5}$ β
- (C) Cr in Cr$_2$O$_7^{2-}$: $+6$; Cr in Cr$^{3+}$: $+3$. Change $= 3$ per Cr atom.
- (D) Cr in CrO$_4^{2-}$: $+6$; Cr in Cr$^{3+}$: $+3$. Change $= 3$.
The change of 5 occurs in the conversion $\text{MnO}_4^- \to \text{Mn}^{2+}$.
Suppose $x, y, z$ are in arithmetic progression (A.P.) and $\tan^{-1}x,\ \tan^{-1}y,\ \tan^{-1}z$ are also in A.P. Then:
If $\tan^{-1}x, \tan^{-1}y, \tan^{-1}z$ are in A.P., then $2\tan^{-1}y=\tan^{-1}x+\tan^{-1}z$.
Also $x,y,z$ in A.P. means $2y=x+z$.
From $2\tan^{-1}y=\tan^{-1}x+\tan^{-1}z$: using the addition formula, $\tan(2\tan^{-1}y)=\tan(\tan^{-1}x+\tan^{-1}z)$, i.e., $\dfrac{2y}{1-y^2}=\dfrac{x+z}{1-xz}$.
Substituting $x+z=2y$: $\dfrac{2y}{1-y^2}=\dfrac{2y}{1-xz}$. So $1-y^2=1-xz$, giving $y^2=xz$.
Combined with $x+z=2y$ (A.P.) and $y^2=xz$ (G.P.) $\Rightarrow x=y=z$.
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