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0.5 g of fuming H₂SO₄ (oleum) is diluted  with water. This solution is completely neutralised by 26.7 mL of 0.4 N NaOH. The percentage of free SO₃ in the sample is:
1. 30.6% 
2. 40.6%
3. 20.6%
4. 50%

MnO₂ + 4HCl $\to$MnCl₂+2H₂O+Cl₂, the equivalent wt. of HCl will be (M$\equiv$Mol wt of HCl)
1. M
2. M/2
3. 2M
4. M/4

The number of moles of MnO^-₄ required to oxidise one mole of ferrous oxalate completely in an acidic medium is-
1. 0.6 mole
2. 0.4 mole
3. 7.5 moles
4. 0.2 mole

Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below: 
 
Then the species undergoing disproportionation is:-

 ${\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{4-}\to {\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{3-}+{\mathrm{e}}^{-};$ ${\mathrm{E}}^{\mathrm{o}}=-0.35$ $\mathrm{V}$
${\mathrm{Fe}}^{2+}\to {\mathrm{Fe}}^{3+}+{\mathrm{e}}^{-};$ ${\mathrm{E}}^{\mathrm{o}}=-0.77$ $\mathrm{V}$
The strongest oxidizing agent in the above equation is:
1. ${\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{4-}$
2. ${\mathrm{Fe}}^{2+}$
3. ${\mathrm{Fe}}^{3+}$
4. ${\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{3-}$

In the conversion, H${}_2$SO${}_4$ $\to$ H${}_2$S${}_2$O${}_8$which process occurs?
1. Oxidation
2. Reduction
3. Oxidation as well as reduction
4. Neither oxidation nor reduction

In the reaction, 2 Cu${}^{+}$ $\to$ Cu + Cu${}^{2+}$, the equivalent weight of Cu${}^{+}$ is (M is the mol. wt. of Cu${}^{+}$)
1. $\mathrm{M}$
2.$\frac{\mathrm{M}}{2}$
3. $\frac{\mathrm{M}}{4}$
4. $2\mathrm{M}$

Valency factor $\left({\mathrm{V}}_t\right)$ for following redox reaction is respectively.
a. ${\mathrm{As}}_2{\mathrm{S}}_3 \to {\mathrm{AsO}}_4^{-3}+{\mathrm{SO}}_4^{2-}$
b. ${\mathrm{I}}_2 \to {\mathrm{I}}^{-}+{\mathrm{IO}}_3^{-}$
c. ${\mathrm{H}}_3{\mathrm{PO}}_2 \to {\mathrm{PH}}_3+{\mathrm{H}}_3{\mathrm{PO}}_3$
1. $28, \frac{5}{3},\frac{4}{3}$
2. $28,\frac{4}{3},\frac{5}{3}$
3. $6, 4, 7$
4. 4, 3, 9

The oxidation number of phosphorus in Ba(H₂PO₂)₂ is:
1. +3
2. +2
3. +1
4. -1

A solution (5 mL) of an acid X is completely neutralised by y mL of 1 M NaOH. The same volume (y mL) of 1 M NaOH is required to neutralise 10 mL of 0.6 M of $H_{2}S{O}_4$ completely. The normality (N) of the acid X is
1. 1.2
2. 2.4
3. 4.8
4. 0.6

The reaction,
$K_{2}C{r}_2{O}_7+mFeS{O}_4+n{H}_{2}S{O}_4\to C{r}_2{\left(S{O}_4\right)}_3+pF{e}_2{\left(S{O}_4\right)}_3$
$+K_{2}S{O}_4+q{H}_{2}O$
when balanced,m, n, p and q are, respectively
1. 6, 14, 3, 14
2. 6, 7, 3, 7
3. 3, 7, 2, 7
4. 4, 14, 2, 14

The species which is unable to show disproportionation reaction is-
1. $Cl{O}_3^{-}$
2. $Cl{O}_4^{-}$
3. $Cl{O}_2^{-}$
4. $Cl{O}^{-}$

Which of the following cannot act  as an oxidising agent?
1. $S^{2-}$
2. $B{r}_2$
3. $HS{O}_4^{-}$
4. $S{O}_3^{2-}$

$E^{⊝}$ values of some redox couples are given below. On the basis of these values choose the correct option.
E^⊝ values :   Br₂ / Br^- = + 1.90 ;   Ag⁺ / Ag = + 0.80   Cu⁺ / Cu = + 0.34 ;   I₂/ I^- = + 0.54
1. Cu will reduce Br^-
2. Cu will reduce Ag
3. Cu will reduce I^-
4. Cu will reduce Br₂

The oxidation state of Cr in $Cr{O}_6$ is -
1. -6
2. +12
3. +6
4. +4