# NEET Questions Solved

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 12

(2) 0.22 mole of 12

(3) 0.055 mole of12

(4) 0.44 mole of 12

(1) 6e- + 10H+ + V2O5 → 2V2+ + 5 H2 O

Zn → Zn2+ + 2e- 3 × 3

V2O5 + 3 Zn + 10 H+ → 3Zn2+ + 2V2+ + 5H2O    … (1)

Now H2O + V2+ → VO2+ + 2H+ + 2e-

2e- + I2 → 2r-

V2+ + I2 + H2O → 2I- + VO2+ + 2H+

so we have 1 moles of V2O5 will reduce 2 moles of iodine

so $\left(\frac{10}{102\text{\hspace{0.17em}}+\text{\hspace{0.17em}}80}\right)\text{×\hspace{0.17em}2}$ moles of will be reduced by given amount of V2O5 = 0.11moles of 12

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NEET - 2009

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

Key Idea (i) Sum of oxidation states of all atoms = charge of ion.

(ii) Oxidation number of oxygen = -2

Let the oxidation state of P in ${\mathrm{PO}}_{4}^{3-}$ be x.

${\mathrm{PO}}_{4}^{3-}$

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NEET - 2007

The number of messages 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

5 moles of sulphite ion react with = 2 moles of ${\mathrm{MnO}}_{4}^{-}$

So, 1 mole of sulphite ion react with =$\frac{2}{5}\mathrm{moles}$ of ${\mathrm{MnO}}_{4}^{-}$

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For the redox reaction $Mn{O}_{4}^{-}+{C}_{2}{O}_{4}^{-2}+{H}^{+}\to \text{\hspace{0.17em}}M{n}^{2+}+C{O}_{2}+{H}_{2}O$ 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]

 $Mn{O}_{4}^{-}$ ${C}_{2}{O}_{4}^{2-}$ ${H}^{+}$ (1) 2 5 16 (2) 16 5 2 (3) 5 16 2 (4) 2 16 5

(1) $Mn{O}_{4}^{-}+8{H}^{+}+5{e}^{-}\to M{n}^{2+}+4{H}_{2}O×2$

${C}_{2}{O}_{4}^{2-}\to 2C{O}_{2}+2{e}^{-}×5$

$\overline{2Mn{O}_{4}^{-}+5{C}_{2}{O}_{4}^{2-}+16{H}^{+}\to 2M{n}^{2+}+10C{O}_{2}+8{H}_{2}O}$

Thus the coefficient of $Mn{O}_{4}^{-}$, ${C}_{2}{O}_{4}^{2-}$ and H+ in the above balanced equation respectively are 2, 5, 16.

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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

(4) In alkaline medium

$2KMn{O}_{4}+KI+{H}_{2}O\to 2Mn{O}_{2}+2KOH+KI{O}_{3}$.

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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

(2) $B{a}_{2}\stackrel{\ast \text{\hspace{0.17em}\hspace{0.17em}}}{C{u}_{3}}{O}_{7}$

$3+2×2+3x-\left(2×7\right)=0$

$3+4+3x-14=0$

$3x=7$

$x=\frac{7}{3}$.

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In a balanced equation ${H}_{2}S{O}_{4}+x\text{\hspace{0.17em}}HI\to {H}_{2}S+y\text{\hspace{0.17em}}{I}_{2}+z\text{\hspace{0.17em}}{H}_{2}O$, 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

(3) The values of x, y, z are 8, 4, 4 respectively hence the reaction is

${H}_{2}S{O}_{4}+8HI\to {H}_{2}S+4{I}_{2}+4{H}_{2}O$

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