1
MCQ (Single Correct Answer)

AIPMT 2009

Oxidation numbers of P in PO$$_4^{3 - }$$, of S in SO$$_4^{2 - }$$ and that of Cr in Cr2O$$_7^{2 - }$$ are respectively
A
+3, +6 and +5
B
+5, +3 and +6
C
$$-$$3, +6 and + 6
D
+5, +6 and +6

Explanation

Let oxidation number of P in PO4 3– be x.

$$ \therefore $$ x + 4(–2) = –3

$$ \Rightarrow $$ x = +5

Let oxidation number of S in SO4 2– be y.

$$ \therefore $$ y + 4(–2) = –2

$$ \Rightarrow $$ y = +6

Let oxidation number of Cr in Cr2O7 2– be z.

$$ \Rightarrow $$ 2z + 7(–2) = –2

$$ \Rightarrow $$ z = +6
2
MCQ (Single Correct Answer)

AIPMT 2008

Number of moles of MnO$$_4^{ - }$$ required to oxidize one mole of ferrous oxalate completely in acidic medium will be
A
7.5 moles
B
0.2 moles
C
0.6 moles
D
0.4 moles

Explanation

The balanced ionic equation for oxidation of ferrous oxalate by MnO4 in acidic medium is as follows :

3MnO4- + 5FeC2O4 + 24H+ $$ \to $$

3Mn2+ + 10CO2 + 12H2O + 5Fe3+

Thus, 5 moles of FeC2O4 require 3 moles of MnO4-.

So, 1 mole of FeC2O4 requires

= $${3 \over 5}$$ = 0.6 moles of MnO4-
3
MCQ (Single Correct Answer)

AIPMT 2004

Which is the best description of the behaviour of bromine in the reaction given below?
H2O + Br2 $$ \to $$ HOBr + HBr
A
Proton acceptor only
B
Both oxidised and reduced
C
Oxidised only
D
Reduced only

Explanation

H2O + $${\mathop {Br}\limits^0}$$2 $$ \to $$ HO$$\mathop {Br}\limits^{ + 1} $$ + H$$\mathop {Br}\limits^{ - 1} $$

In the above reaction the oxidation number of Br2 increases from zero (in Br2 ) to +1 (in HOBr) and decreases from zero (in Br2 ) to –1 (in HBr). Thus Br2 is oxidised as well as reduced and hence it is a redox reaction.
4
MCQ (Single Correct Answer)

AIPMT 2003

The oxidation states of sulphur in the anions

SO3$${_3^{2 - }}$$, S2O$${_4^{2 - }}$$ and S2O$${_6^{2 - }}$$ follow the order
A
S2O$${_4^{2 - }}$$ < SO$${_3^{2 - }}$$ < S2O$${_6^{2 - }}$$
B
SO$${_3^{2 - }}$$ < S2O$${_4^{2 - }}$$ < S2O$${_6^{2 - }}$$
C
S2O$${_4^{2 - }}$$ < S2O$${_6^{2 - }}$$ < SO$${_3^{2 - }}$$
D
S2O$${_6^{2 - }}$$ < S2O$${_4^{2 - }}$$ < SO$${_3^{2 - }}$$

Explanation

SO32– : oxidation state of ‘S’ is +4

S2O42– : oxidation state of ‘S’ is +3.

S2O62– : oxidation state of ‘S’ is +5.

So, the order is S2O$${_4^{2 - }}$$ < SO$${_3^{2 - }}$$ < S2O$${_6^{2 - }}$$.

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