If 10 gm of V₂O₅ is dissolved in acid and is reduced to V²⁺ by zinc metal, how many mole of I₂ could be reduced by the resulting solution if it is further oxidised to VO²⁺ ions ? [Assume no change in state of Zn²+ions] (V = 51, O =16, I = 127) :

(1) 0.11 mole of I₂

(2) 0.22 mole of I₂

(3) 0.055 mole of I₂

(4) 0.44 mole of I₂

Oxidation numbers of P in ${\mathrm{PO}}_4^{3-}$, of S in ${\mathrm{SO}}_4^{2-}$ and that of Cr in ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$ are respectively,

(a) +5, +6 and +6                      (b) +3, +6 and +5

(c) +5, +3 and +6                      (d) -3, +6 and +6

The number of mole of KMnO₄ that will be needed to react with one mole of sulphite ion in acidic solution is:

(a) $\frac{3}{5}$

(b) $\frac{4}{5}$

(c) $\frac{2}{5}$

(d) 1

For the redox reaction $Mn{O}_4^{-}+C_2{O}_4^{-2}+H^{+}\to M{n}^{2+}+C{O}_2+H_{2}O$ the correct coefficients of the reactants for the balanced reaction are [IIT 1988, 92; BHU 1995; CPMT 1997; RPMT 1999; DCE 2000; MP PET 2003]

The number of moles of KMnO₄ reduced by one mole of KI in alkaline medium is: [CBSE PMT 2005]

(1) One fifth

(2) five

(3) One

(4) Two

The compound ${\mathrm{YBa}}_2{\mathrm{Cu}}_3{\mathrm{O}}_7$ which shows superconductivity has copper in oxidation state ........ Assume that the rare earth element Yttrium is in its usual +3 oxidation state [IIT 1994]

(1) 3/7

(2) 7/3

(3) 3

(4) 7

In a balanced equation $H_{2}S{O}_4+xHI\to H_{2}S+y{I}_2+z{H}_{2}O$, the values of x, y, z are [EAMCET 2003]

(1) x = 3, y = 5, z = 2

(2) x = 4, y = 8, z = 5

(3) x = 8, y = 4, z = 4

(4) x = 5, y = 3, z = 4