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 10102+80× 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 PO43-, of S in SO42- and that of Cr in Cr2O72- 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 PO43- be x.

                        PO43-

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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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) 35

(b) 45

(c) 25

(d) 1

 

2KMnO4 +3H2SO4           K2SO4 + 2MnSO4 + 2MnSO4 +3H2O +5OMnO4- + 8H+ + 5e-           Mn2++4H2O × 2    SO32- + H2O           SO42- + 2H++2e- × 5_______________________________________2MnO4-+6H+ + 5SO32-           2Mn2+ + SO42-+3H2O

5 moles of sulphite ion react with = 2 moles of MnO4-

So, 1 mole of sulphite ion react with =25moles of MnO4-

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For the redox reaction MnO4+C2O42+H+Mn2++CO2+H2O 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]

 

MnO4  

C2O42  

H+

(1)  

 2

5

16

(2)

16

5

2

(3)

5

16

2

(4)

2

16

5

(1) MnO4+8H++5eMn2++4H2O×2

C2O422CO2+2e×5

2MnO4+5C2O42+16H+2Mn2++10CO2+8H2O¯

Thus the coefficient of MnO4, C2O42 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 

2KMnO4+KI+H2O2MnO2+2KOH+KIO3.  

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The compound YBa2Cu3O7 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) Ba2Cu3  O7 

3+2×2+3x(2×7)=0

3+4+3x14=0

3x=7

x=73.

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In a balanced equation H2SO4+xHIH2S+yI2+zH2O, 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

H2SO4+8HIH2S+4I2+4H2O 

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