MCQ Questions for Class 11 Medical Chemistry Redox Reactions Quiz 6

Analyze the diagram. The left beaker represents the start of the reaction and the right beaker represents the end of the reaction.
Classify this reaction.

0%
Acid-Base Neutralization reaction.
0%
Redox reaction.
0%
Precipitation reaction.
0%
More information is needed.

The oxidation number of chromium in

$K_{2}CrO_{4}$ is:

0%
$+2$
0%
$+3$
0%
$+6$
0%
$+8$

Explanation

$K_2CrO_4$ ,

Oxidation no. of

$K=+1.$

Oxidation no. of

$O=-2.$

Oxidation no. of

$Cr$ be

$x$ .

$2(1)+x+4(-2)=0,\\ 2+x-8=0,\\ x-6=0,\\x=6.$

$\therefore$ Oxidation no. of

$Chromium$ in

$K_2CrO_4$ is

$6$ .

$KMnO_4$ and

$K_2Cr_2O_7$ are _____agents and

$Na_2S_2O_7$ is ______ agent.

0%
reducing , oxydising
0%
oxydising , reducing
0%
reducing , reducing
0%
oxydising , oxydising

Explanation

An oxidizing agent (oxidant, oxidizer) is a substance that has the ability to oxidize other substances (cause them to lose electrons).

${ KMnO }_{ 4 },\quad { K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }$ are oxidizing agent.

A reducing agent (also called a reductant or reducer) is an element (such as calcium) or compound that loses (or "donates") an electron to another chemical species in a redox chemical reaction. Since the reducing agent is losing electrons, it is said to have been oxidized.

${ Na }_{ 2 }{ S }_{ 2 }{ O }_{ 3 }$ is a reducing agent.

$2Pb{ O }_{ \left( s \right) }+{ C }_{ \left( s \right) }\rightarrow 2{ Pb }_{ \left( s \right) }+C{ O }_{ 2\left( g \right) }$
Which of the following statements are correct for the above chemical reaction?
a) Lead is reduced
b) Carbon dioxide is oxidized
c) Carbon is oxidized
d) Lead oxide is reduced

0%
(a) and (b)
0%
(a) and (c)
0%
(a), (b) and (c)
0%
All

Explanation

$2PbO_{(s)}+C_{(s)}\longrightarrow 2Pb_{(s)}+CO_{2(g)}$

$\therefore$

$C$ is oxidized from

$0$ to

$+4$ and

$Pb$ is reduced from

$+2$ to

$0.$

Correct answer

$(a)$ and

$(c).$

1048190_575090_ans_d18505db6e954016b9299976a578db10.png

A sample of oleum is labelled 109%. The % of free $SO_3$ in the sample is :

0%
40%
0%
80%
0%
60%
0%
9%

Explanation

Oleum is a combination of sulphuric acid

$+$ sulphuric trioxide and free

$SO_3$ in oleum on reaction with

$H_2O \longrightarrow H_2SO_4$ .

$SO_3+H_2O \longrightarrow H_2SO_4 \longrightarrow (1)$

Let us take

$109$ % sample has

$100gm$ of oleum and we added

$9gms$ of water.

By reaction

$(1)$

$18gms$ of water

$\longrightarrow 80gm$

$9gm$ of water

$\longrightarrow 40gm$

As per our assumption

$40gm=40$ % .

Which of the following compounds we use in our laboratory as a standard solution (titrant) ?

0%
$KMnO_4$
0%
$K_2Cr_2O_7$
0%
$Na_2S_2O_3$
0%
All of these

Explanation

A reagent, called the titrant or titrator is prepared as a standard solution. A known concentration and volume of titrant reacts with a solution of analyte or titrand to determine concentration.

${ KMnO }_{ 4 }$ , ${ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }$ , ${ Na }_{ 2 }{ S }_{ 2 }{ O }_{ 3 }$ etc. are compounds we use in our laboratory as a standard solution.

If the 1.58g of

$KMnO_4$ in acidic medium completely reacts with ferrous oxalate what weight (in gm) of ferrous oxalate is required?

0%
2.73 gm
0%
4.73 gm
0%
11.19 gm
0%
5.62 gm

Explanation

$FeC_2O_4\Rightarrow n_f=3$

$KMnO_4\Rightarrow n_f=5$

Moles of

$KMnO_4=\dfrac{1.58}{158}=0.01$

Equivalents of

$KMnO_4=moles\times n_f=0.01\times5=0.05$

Equivalents of

$FeC_2O_4=moles\times n_f=n\times3=0.05$

$\Rightarrow n=\dfrac{0.05}{3}$

Weight of

$FeC_2O_4=n\times142=2.73\ gm$

$H_3PO_4$ is a tribasic acid and one of its salts is

$NaH_2PO_4$ . What volume of 1 M

$NaOH$ solution should be added to 12 g of

$NaH_2PO_4$ to convert it into

$Na_3PO_4$ ?

0%
100 ml
0%
200 ml
0%
300 ml
0%
50 ml

Explanation

$\textbf{Step 1:}$ Calculation of the number of moles of

$NaOH$ .

$NaH_2PO_4 + 2NaOH \rightarrow Na_3PO_4 + H_2O$

Molar mass of

$NaH_2PO_4 = 23+(2\times1)+31+(4\times16)= 120\ g\ mol^{-1}$

Number of moles =

$\cfrac{\text{weight of substance}}{\text{Molecular weight}}$

Weight of

$NaH_2PO_4 = 12\ g$

Number of moles of

$NaH_2PO_4 (n) = \dfrac{w}{M}$ =

$\cfrac{12}{120} = 0.1$ mole

From the stoichiometric coefficient of above reaction we can say that,

One mole of

$NaH_2PO_4$ reacts with 2 moles of

$NaOH$ .

So, number of moles of

$NaOH$ required

$= 0.1 \times 2 = 0.2$ mole

$\textbf{Step 2:}$ Calculation of the volume of NaOH

Molarity of

$NaOH = 1M$

Molarity =

$\cfrac{n}{V (in\ L)} = \cfrac{0.2}{V}$

$\Rightarrow1 = \cfrac{0.2}{V}$

$\Rightarrow V = 0.2\ L=0.2 \times1000\ mL = 200\ mL$

Thus, volume of

$NaOH$ to be added is

$200\ mL$ .

Since, the titration is completely based on volume measurement, it is a volumetric analysis.

0%
True
0%
False

In a redox titration, the compound of known concentration is called as analyte or titrand while the standard solution is called as titrant or titrator

0%
True
0%
False

What is the change in the oxidation state of Mn, in the reaction of

$MnO_4^-$ with

$H_2O_2$ in acidic medium?

0%
$7\rightarrow 4$
0%
$6 \rightarrow 4$
0%
$7 \rightarrow 2$
0%
$6 \rightarrow 2$

Explanation

In acidic medium:

$3H_2O_2 + 2KMnO_4 \longrightarrow 2MnO_2 + 2KOH + 3O_2 + 2H_2O$

$\overset{7+}{Mn}O_4^- \rightarrow Mn^{2+}$

The oxidation state of

$Mn$ changes from +7 to +2.

Hence, option C is correct.

Which titrant is used in the Iodometric titration which involves

$I_2$ ?

0%
$KMnO_4$
0%
$K_2Cr_2O_7$
0%
$Na_2S_2O_3$
0%
All of them

What is the oxidation state of central atom in

$[Co(NH_3)_4(OH_2)I]SO_4$ ?

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

$[CO{ (N{ H }_{ 3 }) }_{ 4 }({ H }_{ 2 }O)I]S{ O }_{ 4 }$

$\Downarrow$

${ [CO{ (N{ H }_{ 3 }) }_{ 4 }({ H }_{ 2 }O)I] }^{ 2+ }{S{ O }_{ 4 }}^{2-}$

Central atom

$=CO$

Oxidation state of

$NH_3$ in general

$=0,H2O=0$ and

$I=-1$

Applying formula,

$\text{Sum of total oxidation state of all atoms } = \text{ Overall charge on the compound}.$

Let oxidation state of

$CO=x\\ x+4(0)+1\times (0)+1\times (-1)=+2,\\x=3$

$\therefore$ Oxidation state of

$CO=+3.$

The oxidation state of the underlined element in the given compound is:

$Na_2\underline{Mo}O_4$

0%
+4
0%
+5
0%
+6
0%
none of these

Explanation

$Na_2MnO_4$

Let oxidation state of Mo be

$x$

$2 (+1) + x + 4 (-2) = 0$

$x = +6$

What is the oxidation state of central atom in

$K[Cr(NH_3)_2Cl_4]$ ?

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

$K[Cr{ ({ NH }_{ 3 }) }_{ 2 }C{ l }_{ 4 }]$

$\Downarrow$

${ { K }^{ + }[Cr{ ({ NH }_{ 3 }) }_{ 2 }C{ l }_{ 4 }]} ^{-1}$

Central atom

$=Cr$

Oxidation state of

$NH_3$ in general

$=0,Cl=-1$

Applying formula,

$\text{Sum of total oxidation state of all atoms } = \text{ Overall charge on the compound}.$

Let oxidation state of

$Cr=x\\ x+2\times(0)+4\times (-1)=-1,\\x=+3$

$\therefore$ Oxidation state of

$Cr=+3.$

What is the oxidation state of

$Mn$ in

$[Mn(CN)_5]^{2-}$ ?

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

${ [Mn{ (CN) }_{ 5 }] }^{ 2- }$

Central atom

$=Mn$

Oxidation state of

$CN$ in general

$=-1$

Applying formula,

$\text{Sum of total oxidation state of all atoms } = \text{ Overall charge on the compound}.$

Let oxidation state of

$Mn=x\\ x+5(-1)=-2,\\x=+3$

$\therefore$ Oxidation state of

$Mn=+3.$

What is the oxidation state of central atom in

$Ca[PtCl_4]$ ?

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

$Ca[PtC{ l }_{ 4 }]\Leftrightarrow C{ a }^{ +2 }{ [PtC{ l }_{ 4 }] }^{ 2- }$

Take

${ [PtC{ l }_{ 4 }] }^{ 2- }.$ Central atom

$=Pt.$

Let

$x$ be the oxidation no. of

$Pt$ ,

$Cl$ oxidation no. is

$-1$ .

$x+4(-1)=-2$

$x-4=-2$

$x=2$

In alkaline medium, the reaction of hydrogen peroxide with potassium permanganate produces a compound in which the oxidation state of

$Mn$ is:

0%
$0$
0%
$+2$
0%
$+3$
0%
$+4$

Explanation

$2Mn{ O }_{4}^{-}(aq) + 3H_{2}O_{2}(aq) \rightarrow 2MnO_{2}(s) + 2H_{2}O(l) + 3O_{2}(g) + 2O{ H }^{-}_{(aq)}$

Oxidation state of

$Mn$ in

$MnO_{2}$ is

$+4$ .

Hence, option

$D$ is correct.

Oxygen gas reacts with hydrogen to produce water. The reaction is represented by the equation:

${ O }_{ 2 }\left( g \right) +{ H }_{ 2 }\left( g \right) \longrightarrow { H }_{ 2 }O\left( g \right)$
The above reaction is an example of
a) Oxidation of hydrogen
b) Reduction of oxygen
c) Reduction of hydrogen
d) Redox reaction

0%
(a), (b) and (c)
0%
(b), (c) and (d)
0%
(a), (c) and (d)
0%
(a), (b) and (d)

Oxidation number of nitrogen in which among the oxides of nitrogen is the lowest?

0%
Nitric oxide
0%
Nitrous oxide
0%
Nitrogen dioxide
0%
Dinitrogen trioxide

Explanation

Nitrogen can exhibit +1 to +5 oxidation state due to its half-filled orbital.

The least oxidation state will be

$+1$ which is for

$N_2O$ ( Nitrous oxide. )

So, the correct option is

$B$

The oxidation number of iron in

${Fe}_{3}{O}_{4}$ is:

0%
$+2$
0%
$+3$
0%
$\cfrac{8}{3}$
0%
$\cfrac{2}{3}$

Explanation

$Fe_3O_4$ is basically a mixture of

$Fe_2O_3$ and

$FeO$ .

$Fe_2O_3$ , the two

$Fe$ atom has

$3$ oxidation state.

$2x - 6 = 0$

$x = +3$

$FeO$ , Fe has

$2$ oxidation state.

$x-2 = 0$

$x = 2$

Average oxidation state of

$Fe = \dfrac{3 + 3+ 2}{3}$

$= \dfrac 83$

Hence, option

$C$ is correct.

What is the number of coulombs required for the conversion of one mole of

$MnO_{4}^{-}$ to one mole of

$Mn^{2+}$ ?

0%
$5\times 96500$
0%
$3\times 96500$
0%
$96500$
0%
$9650$

Explanation

$MnO^-_4$ is converted to

$Mn^{+2}$ .

$MnO^-_4$ , oxidation state of

$Mn$ be

$x.$

$x+4(-2)=-1,\\ x-8=-1,\\x=+7.$

Oxidation state changes from

$+7$ to

$+2$ .

Change in oxidation state

$= 5$ .

$1$ Faraday (

$96500$ coulombs) to change one oxidation state.

To change

$5$ oxidation state, no. of coulombs required,

$=5\times 1 \text{ Faraday }\\=5\times 96500 \text{ coulombs},$

The half-cell reaction is the one that:

0%
takes place at one electrode
0%
consumes half a unit of electricity
0%
involves half a mole of electrolyte
0%
goes half way to completion

Oxidation state of

$S$ in

${SO_4}^{2-}$ is:

0%
$+6$
0%
$+3$
0%
$+2$
0%
$-2$

Explanation

Let oxidation state of

$S$ is x.

and oxidation state of

$O$ in

$SO^{2-}_4$ is -2.

and the net charge on the compound is -2.

$\therefore x+4(-2)=-2$

$x=+6$ .

The oxidation number of chromium in

$\displaystyle Cr{ O }_{ 5 }$ is:

0%
+6
0%
+5
0%
+10
0%
0

Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half-cell reactions and their standard potentials are given below:

$Mn{O}_{4}^{-} (aq.) + 8H^{+}(aq.) + 5e^{-} \rightarrow Mn^{2+}(aq.) + 4H_{2}O(l); E^{\circ} = 1.51\ volt$

$Cr_{2}O_{7}^{2-}(aq.) + 14H^{+}(aq.) + 6e^{-} \rightarrow 2Cr^{3+}(aq.) + 7H_{2}O(l)' E^{\circ} = 1.38\ volt$

$Fe^{3+}(aq.) + e^{-}\rightarrow Fe^{2+}(aq.); E^{\circ} = 0.77\ volt$

$Cl_{2}(g) + 2e^{-} \rightarrow 2Cl^{-}(aq.); E^{\circ} = 1.40\ volt$
Identity the only incorrect statement regarding the quantitative estimation of aqueous

$Fe(NO_{3})_{2}$ .

0%
$Mn{O}_{4}^{-}$ can be used in aqueous $HCl$
0%
$Cr_{2}O_{7}^{2-}$ can be used in aqueous $HCl$
0%
$Mn{O}_{4}^{-}$ can be used in aqueous $H_{2}SO_{4}$
0%
$Cr_{2}O_{7}^{2-}$ can be used in aqueous $H_{2}SO_{4}$

Explanation

According to the given half cell reactions for the quantitative estimation of aqueous

$Fe(NO_3)_2$

That

$Mn{O}_4^-$ cannot be used in aqueous

$HCl$ because

$Mn{O}_4^-$ wii not oxidise

$Fe^{+2}$ to

$Fe^{+3}$ instead of that it also oxidise

$Cl^-$ to

$Cl_2$ .

Whereas all other reactions are feasible.

$Ni(CO)_4$ , the oxidation state of Ni is?

0%
$4$
0%
Zero
0%
$2$
0%
$8$

Explanation

In nickel tetracarbonyl, the oxidation state for nickel is assigned as zero. The formula conforms to the 18-electron rule. The molecule is tetrahedral, with four carbonyl (carbon monoxide) ligands attached to nickel.

Hence, the correct option is

$B$

In which of the following compounds, iron has an oxidation state of

$+3$ ?

0%
$Fe(NO_3)_2$
0%
$FeC_2O_4$
0%
$[Fe(H_2O)_6]Cl_3$
0%
$(NH_4)_2SO_4\cdot FeSO_4\cdot 6H_2O$

Explanation

Let

$Fe$ has

$x$ oxidation state

$Fe{ (N{ O }_{ 3 }) }_{ 2 }$ :

$x-2(+1)=0,x=+2$

$Fe{ C }_{ 2 }{ O }_{ 4 }$ :

$x-2=0,x=+2$

$[Fe{ ({ H }_{ 2 }O) }_{ 6 }{ Cl }_{ 3 }]$ :

$x+6(0)+3(-1)=0, x=+3$

$({ { NH }_{ 4 } })_{ 2 }{ SO }_{ 4 }.{ FeSO }_{ 4 }.6{ H }_{ 2 }O$ :

$x-2=0, x=+2$ (In

$({ { NH }_{ 4 } })_{ 2 }{ SO }_{ 4 }.{ FeSO }_{ 4 }.6{ H }_{ 2 }O$ for calculation of oxidation state we would only consider

${ FeSO }_{ 4 }$ )

So in option C iron has an oxidation state of +3.

$One\ mole$ of

$N_2H_4$ loses

$10\ moles$ of electrons to form a new compound

$Y$ . Assuming that all nitrogen appear in the new compound, what is the oxidation state of nitrogen in compound

$Y$ ?

[There is no change in the oxidation state of hydrogen.]

0%
$-1$
0%
$-3$
0%
$+3$
0%
$+5$

Explanation

$\text{One}$

$\text{Mole}$ of

$N_2H_4$ loses

$10$

$\text{moles}$ of

$e^-\longrightarrow Y$

The oxidation state of

$H$ remains unchanged and all the

$\text{Nitrogen"}$ appears in

$Y.$

$\therefore$

$e^-$ must be lost from

$\text{nitrogen}$ ,

$1$

$\text{mole}$ of

$N_2H_4$ contains

$2$

$\text{moles}$ of

$\text{nitrogen}$ .

$\Longrightarrow 2$

$\text{moles}$ of

$N$ loses

$10$

$\text{moles}$ of

$e^-$ .

$\therefore 1$

$\text{mole}$ loses

$5$

$\text{moles}$ of

$e^-$

Initial oxidation state of

$N$ in

$N_2H_4=-2$

when it loses

$5$

$\text{moles}$ of

$e^-$

Final oxidation state of each

$N=-2+5=+3.$

(since it loses

$e^-$ , therefore it must acquire positive charge).

Hence, the correct option is

$C$

When

$SO_2$ is passed through an acidified solution of

$K_2Cr_2O_7$ , then chromium sulphate is formed. Change in oxidation state of Cr is from.

0%
$+4$ to $+2$
0%
$+6$ to $+3$
0%
$+7$ to $+2$
0%
$+5$ to $+3$

Explanation

In acidic medium

$Cr$ is reduced from +6 to +3 as:

${ { Cr }_{ 2 }{ O }_{ 7 } }^{ -2 }+14{ H }^{ + }+6{ e }^{ - }\rightarrow 2{ Cr }^{ +3 }+7{ H }_{ 2 }O$

and

${ SO }_{ 2 }$ is oxidised to

${ SO }_{ 3 }$ as:

${ SO }_{ 2 }+{ H }_{ 2 }O\rightarrow S{ O }_{ 3 }+2{ H }^{ + }+{ 2e }^{ - }$

Arrange the following in the increasing order of oxidation state of

$Mn.$
(i)

$Mn^{2+}$
(ii)

$MnO_2$
(iii)

$KMnO_4$
(iv)

$K_2MnO_4$ .

0%
(i) $>$ (ii) $>$ (iii) $>$ (iv)
0%
(i) $<$ (ii) $<$ (iv) $<$ (iii)
0%
(ii) $<$ (iii) $<$ (i) $<$ (iv)
0%
(iii) $<$ (i) $<$ (iv) $<$ (ii)

Explanation

Oxidation state of given compounds is as follow:

$\underset{(+2)}{Mn^{2+}} < \underset{(+4)}{MnO_2} < \underset{(+6)}{K_2MnO_4} < \underset{(+7)}{KMnO_4}$ .

When

$KMnO_4$ is reduced with oxalic acid in acidic solution, the oxidation number of Mn changes from.

0%
$7$ to $2$
0%
$7$ to $4$
0%
$7$ to $6$
0%
$6$ to $2$

Explanation

In acidic medium it reacts as follows:

$Mn{ { O }_{ 4 } }^{ - }+8{ H }^{ + }+5{ e }^{ - }\rightarrow { Mn }^{ +2 }+4{ H }_{ 2 }O$

So oxidation state of

$Mn$ changes from +7 to +2.

When P reacts with caustic soda, the products are

$PH_3$ and

$NaH_2PO_2$ . The reaction is an example of.

0%
Oxidation
0%
Reduction
0%
Both oxidation and reduction
0%
Neutralisation

Explanation

${ P }_{ 4 }+3NaOH+3{ H }_{ 2 }O\rightarrow P{ H }_{ 3 }+3Na{ H }_{ 2 }P{ O }_{ 2 }$

In reactant P is present in (0) oxidation state and in

$P{ H }_{ 3 }$ , it is present in (-3) oxidation state and in

$Na{ H }_{ 2 }P{ O }_{ 2 }$ it is present in (+1) oxidation state.

$P$ undergoes both oxidation and reduction.

Oxidation number of fluorine in

$F_2O$ is:

0%
$+1$
0%
$+2$
0%
$-1$
0%
$-2$

Oxidation number of Fe in

$Fe_{0.94}O$ is?

0%
$200$
0%
$200/94$
0%
$94/200$
0%
None of these

Explanation

Oxidation number of oxygen

$=-2$

Let oxidation number of

$Fe$ is

$x$ so,

$0.94(x)+(-2)=0$

$x=\dfrac { 2 }{ 0.94 }$

$x=\dfrac { 200 }{ 94 }$

Hence, the correct option is

$B$

In the compounds

$KMnO_4$ and

$K_2Cr_2O_7$ , the highest oxidation state is of the element.

0%
Potassium
0%
Chromium
0%
Oxygen
0%
Manganese

Explanation

$KMn{ O }_{ 4 }$ ,

$K$ is in +1 oxidation state,

$O$ is in -2 oxidation state

$Mn$ is present in +7 oxidation state.

${ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }$ ,

$K$ is in +1 oxidation state,

$O$ is in -2 oxidation state

$Cr$ is present in +6 oxidation state.

$Mn$ is present in highest oxidation state.

The oxidation numbers of C in

$CH_4, CH_3Cl, CH_2Cl_2, CHCl_3$ and

$CCl_4$ are respectively.

0%
$+4, +2, 0, -2, -4$
0%
$+2, +4, 0, -4, -2$
0%
$-4, -2, 0, +2, +4$
0%
$-2, -4, 0, +4, +2$

Explanation

Let oxidation state of carbon is

$x$ .

$C{ H }_{ 4 }$ ,

$x+4(+1)=0$

$x=-4$

$C{ H }_{ 3 }Cl$ ,

$x+3(+1)+(-1)=0$

$x=-2$

$C{ Cl }_{ 4 }$ ,

$x+4(-1)=0$

$x=+4$

$CH{ Cl }_{ 3 }$ ,

$x+1+3(-1)=0$

$x=+2$

$C{ H }_{ 2 }{ Cl }_{ 2 }$ ,

$x+2(+1)+2(-1)=0$

$x=+0$

So option C) is correct answer.

The oxidation state of the most electronegative element in the products of the reaction between

$BaO_2$ and

$H_2SO_4$ are:

0%
$0$ and $-1$
0%
$-1$ and $-2$
0%
$-2$ and $0$
0%
$-2$ and $+1$

Explanation

$Ba{ O }_{ 2 }+{ H }_{ 2 }S{ O }_{ 4 }\rightarrow { BaSO }_{ 4 }+{ H }_{ 2 }{ O }_{ 2 }$

The most electronegative element in the product is oxygen.

The oxidation state of oxygen in

$BaS{ O }_{ 4 }$ is -2 and in

${ H }_{ 2 }{ O }_{ 2 }$ is -1.

$Note$ : Oxidation state of hydrogen is always +1 so in

${ H }_{ 2 }{ O }_{ 2 }$ oxidation of oxygen is -1 instead of -2.

Hence, the correct option is

$\text{B}$

It is found that V forms a double salt isomorphous with Mohr's salt. The oxidation number of V in this compound is_________.

0%
$+3$
0%
$+2$
0%
$+4$
0%
$-4$

Explanation

Double salt of

$V$ isomorphous with Mohr's salt is

${ (N{ H }_{ 4 }) }_{ 2 }V{ { (SO }_{ 4 } })_{ 2 }.6{ H }_{ 2 }O$ . When double salt dissolves in water it dissociates to give

$NH_4^+$ ,

$V^{+2}$ and

$SO_4^{-2}$ .So oxidation state of vanadium is +2.

Oxygen has an oxidation state of

$+2$ in.

0%
$H_2O_2$
0%
$OF_2$
0%
$SO_2$
0%
$H_2O$

Explanation

Oxidation state of oxygen is always -2 except in peroxides,superoxides and when it reacts with fluorine.

$H_2O_2$ ,oxidation state of

$H$ is +1, so oxidation state of oxygen is -1.

$OF_2$ ,oxidation state of

$F$ is -1, so oxidation state of oxygen is +2.

$SO_2$ and

$H_2O$ ,oxidation state of oxygen is -2.

So B) is correct answer.

Bromine reacts with the hot aqueous alkali to give bromide and bromate. What is the change that is brought about in the oxidation state of bromine to bromate?

0%
$-1$ to $+5$
0%
$0$ to $+5$
0%
$0$ to $+7$
0%
None of these

Explanation

$3{ Br }_{ 2 }+6NaOH\rightarrow 5NaBr+NaBr{ O }_{ 3 }+3{ H }_{ 2 }O$

Oxidation number of bromine changes from 0 in

$Br_2$ to +5 in

$NaBr{ O }_{ 3 }$ .

Oxidation number of bromine in bromate can be calculated with the help of known oxidation state of oxygen (-2) and

$Na$ (+1).

Hence, the correct option is

$B$ .

Oxidation number of C in HNC is_________.

0%
$+2$
0%
$-3$
0%
$+3$
0%
Zero

Explanation

Oxidation number of hydrogen is +1. As nitrogen is more electronegative than carbon so its oxidation number is -3. Net charge on compound is zero.

Now we can find oxidation number of carbon:

Let oxidation number of carbon is

$x$ ,

$1+(-3)+x=0$

$x=+2$

Hence, the correct option is

$A$

Oxidation number of iodine in

$IO^-_3, IO^-_4, KI$ and

$I_2$ respectively are____________.

0%
$-1, -1, 0, +1$
0%
$+3, +5, +7, 0$
0%
$+5, +7, -1, 0$
0%
$-1, -5, -1, 0$
0%
$-2, -5, -1, 0$

Explanation

Oxidation number of oxygen is

$-2$ and

$K$ is

$+1$ .

$IO_3^{-}$ ,oxidation number of iodine is

$(x+3(-2)=-1,x=+5) +5$ .

$IO_4^{-}$ ,oxidation number of iodine is

$(x+4(-2)=-1,x=+7) +7$ .

$KI$ ,oxidation number of iodine is

$(1+x=0,x=-1) -1$ .

$I_{2}$ ,oxidation number of iodine is

$0$ .

Nitrogen forms a variety of compounds in all oxidation states ranging from.

0%
$-3$ to $+5$
0%
$-3$ to $+3$
0%
$-3$ to $+4$
0%
$-3$ to $+6$

Explanation

Hint:

The nitrogen atom has five electrons in its valence shell.

Explanation:

A nitrogen atom can donate all five electrons to form bonds with other elements.
As it requires three electrons to satisfy the octet rule, it can receive three electrons from other elements to form a molecule.
It loses five electrons to show $+5$ oxidation state. And it accepts three electrons to show $-3$ oxidation state.
Some compounds that exhibits these oxidation state are

$NH_3$ have

$-3$ state,

$N_2H_4$ have

$-2$ state,

$NH_2OH$ have

$-1$ state,

$N_2O_5$ have

$+5$ state,

$NO_2$ have

$+4$ state,

$N_2O_3$ have

$+3$ state,

$NO$ have

$+2$ state and

$N_2O$ have

$+1$ state.

Hence, nitrogen exhibits oxidation states that ranges from $-3$ to $+5$ .

Final Answer:

Therefore, the correct answer is option $A.$

The oxidation number of sulphur in

$S_8, S_2F_2$ and

$H_2S$ respectively are__________.

0%
$0, +1$ and $-2$
0%
$+2, +1$ and $-2$
0%
$0, +1$ and $+2$
0%
$-2, +1$ and $-2$

Explanation

Oxidation number of sulphur in

$S_8$ is 0.

Oxidation number of

$F$ is -1 so oxidation number of sulphur in

$S_2F_2$ is +1.

Oxidation number of

$H$ is +1 so oxidation number of sulphur in

$H_2S$ is -2.

On reduction of

$KMnO_4$ by oxalic acid in acidic medium, the oxidation number of Mn chnages. What is the magnitude of this change?

0%
$7$ to $2$
0%
$6$ to $2$
0%
$5$ to $2$
0%
$7$ to $4$

Explanation

In acidic medium reduction of

$KMnO_4$ takes place as follows:

$Mn{ O }_{ 4 }^{ - }+5{ e^{ - }+8{ H }^{ + } }\rightarrow { Mn }^{ +2 }+{ 4H }_{ 2 }O+5{ e^{ - } }$

So oxidation state of

$Mn$ changes from +7 to +2.

Which of the following shows nitrogen with its increasing order of oxidation number?

0%
$NO < N_2O < NO_2 < NO^-_3, NH^+_4$
0%
$NH^+_4 < N_2O < NO_2 < NO^-_3 < NO$
0%
$NH^+_4 < N_2O < NO < NO_2 < NO^-_3$
0%
$NH^+_4 < NO < N_2O < NO_2 < NO^-_3$

Explanation

Oxidation state of

$N$ in

$NO$ is +2,in

$N_2O$ is +1,in

$NO_2$ is +4,in

$NO_3^{-}$ is +5 and in

$NH_4^{+}$ is -3.

Nitrogen with increasing number of oxidation number is:

$NH_4^{+}$ <

$N_2O$ <

$NO$ <

$NO_2$ <

$NO_3^{-}$

The oxidation numbers of the sulphur atoms in peroxy monosulphuric acid

$(H_2SO_5)$ and peroxydisulphuric and

$(H_2S_2O_8)$ are respectively.

0%
$+8$ and $+7$
0%
$+3$ and $+3$
0%
$+6$ and $+6$
0%
$+4$ and $+6$

Explanation

By looking the structure of

$H_2SO_5$ , we can observe that there are two oxygen atoms which are linked by peroxide linkage so their oxidation numbers are -1.Rest oxygen atoms attached normally so their oxidation state is -2. The oxidation number of hydrogen is +1.So oxidation number of sulphur is

$2(+1)+x+2(-1)+3(-2)=0,x=+6$

By looking the structure of

$H_2S_2O_8$ , we can observe that there are two oxygen atoms which are linked by peroxide linkage so their oxidation numbers are -1.Rest oxygen atoms attached normally so their oxidation state is -2. Oxidation number of hydrogen is +1.So oxidation number of sulphur is

$2(+1)+2(x)+2(-1)+6(-2)=0,x=+6$

$CrO_5$ has structure as shown, The oxidation number of chromium in the compound is?

0%
$+4$
0%
$+5$
0%
$+6$
0%
$+10$
0%
$0$

Explanation

By looking the above structure we can observe that 4 oxygen atoms are linked by peroxide linkage. So there oxidation state is -1 as in peroxide.

One oxygen atom is attached normally so its oxidation state is -2.

So oxidation state of

$Cr$ is

$x+4(-1)+(-2)=0,x=+6$

The oxidation state of the underlined element in the given compound is:

$\underline{C_{12}}H_{22}O_{11}$

0%
+5
0%
+4
0%
0
0%
none of these

Explanation

$C_{12}H_{22}O_{11}$ = Sucrose

$\Rightarrow 12x+22 \times 1- 11 \times 2=0$

$\Rightarrow 12x+22-22=0$

$\Rightarrow x=0$ .

CBSE Questions for Class 11 Medical Chemistry Redox Reactions Quiz 6 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Redox Reactions Quiz 6 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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