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Chemistry M.C.Q (1 Marks)

Redox Reactions · UP Board (UPMSP) STD 11 Science (UPMSP - English Medium). 171 questions.

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M.C.Q (1 Marks)

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171 questions · auto-graded multiple-choice test.

MCQ 11 Mark
$\mathrm{H}_2 \mathrm{O}_2+\mathrm{O}_3 \rightarrow \mathrm{H}_2 \mathrm{O}+2 \mathrm{O}_2$, in this reaction :
  • A
    $\mathrm{H}_2 \mathrm{O}_2$ is bleached.
  • ✓
    $\mathrm{H}_2 \mathrm{O}_2$ is oxidised.
  • C
    $\mathrm{H}_2 \mathrm{O}_2$ is dehydrated.
  • D
    $\mathrm{H}_2 \mathrm{O}_2$ is neither oxidised nor reduced.
Answer
Correct option: B.
$\mathrm{H}_2 \mathrm{O}_2$ is oxidised.
Oxidation half reactions is,
$\mathrm{H}_2 \mathrm{O}_2 \rightarrow 2 \mathrm{H}^{+}+\mathrm{O}_2+2 \mathrm{e}^{-}$
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MCQ 21 Mark
What is the oxidation number of lithium in $\ce{LiCl}\ ?$
  • A
    $+3$
  • B
    $−1$
  • ✓
    $+1$
  • D
    $0$
Answer
Correct option: C.
$+1$
Oxidation number of $\ce{Li}$ in $\ce{LiCl } : x − 1 = 0$
$\Rightarrow x = +1$
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MCQ 31 Mark
In $\mathrm{Ni}(\mathrm{CO})_4,$ the oxidation state of $\ce{Ni}$ is?
  • ✓
    $4$
  • B
    Zero
  • C
    $2$
  • D
    $8$
Answer
Correct option: A.
$4$
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.
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MCQ 41 Mark
Which of the following elements does not show disproportionation tendency?
  • A
    $\ce{Cl}$
  • B
    $\ce{Br}$
  • ✓
    $\ce{F}$
  • D
    $\ce{I}$
Answer
Correct option: C.
$\ce{F}$
Being the most electronegative element, $F$ can only be reduced and hence it always shows an oxidation number of $-1$. Further, due to the absence of $d-$ orbitals, it cannot be oxidized and hence it does not show $+ve$ oxidation numbers. In other words, $F$ cannot be simultaneously oxidized as well as reduced and hence does not show disproportionation reactions. Thus, option $(c)$ is correct.
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MCQ 51 Mark
Which of the following is not an example of redox reaction?
  • A
    $\text{CuO}+\text{H}_2\xrightarrow{ \ \ \ \ \ \ \ \ \ }\text{Cu}+\text{H}_2\text{O}$
  • B
    $\text{Fe}_2\text{O}_3+3\text{CO}\xrightarrow{ \ \ \ \ \ \ \ \ \ }2\text{Fe}+3\text{CO}_2$
  • C
    $2\text{K}+\text{F}_2\xrightarrow{ \ \ \ \ \ \ \ }2\text{KF}$
  • ✓
    $\text{BaCl}_2+\text{H}_2\text{SO}_4\xrightarrow{ \ \ \ \ \ \ \ \ }\text{BaSO}_4+2\text{HCl}$
Answer
Correct option: D.
$\text{BaCl}_2+\text{H}_2\text{SO}_4\xrightarrow{ \ \ \ \ \ \ \ \ }\text{BaSO}_4+2\text{HCl}$
$\text{BaCl}_2+\text{H}_2\text{SO}_4\xrightarrow{ \ \ \ \ \ \ \ \ }\text{BaSO}_4+2\text{HCl}$ is not a redox reaction. It is an example of double displacement reactions.
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MCQ 61 Mark
Which of the following metal displacement reaction will not take place and why?
  • ✓
    $\text{Cu}+\text{Mg}^{2+}\xrightarrow{ \ \ \ \ \ \ \ }$
  • B
    $\text{Mg}+\text{Cu}^{2+}\xrightarrow{ \ \ \ \ \ \ \ }$
  • C
    $\text{Pb}+\text{Ag}^+\xrightarrow{ \ \ \ \ \ \ \ }$
  • D
    $\text{Zn}+\text{Cu}^{2+}\xrightarrow{ \ \ \ \ \ \ \ }$
Answer
Correct option: A.
$\text{Cu}+\text{Mg}^{2+}\xrightarrow{ \ \ \ \ \ \ \ }$
Will not take place because $'\text{Cu}\ '$ is less reactive than $\text{Mg}.$
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MCQ 71 Mark
In which of the following groups of iodine compounds shows increasing order of oxidation states :
  • ✓
    $\ce{HlO_4, ICl, I_2, Hl}$
  • B
    $\ce{Hl, I_2, IC, HIO_4}$
  • C
    $\ce{I_2, HI, HIO_4 , HI}$
  • D
    $\ce{ICl HIO_4,HI, I_2}$
Answer
Correct option: A.
$\ce{HlO_4, ICl, I_2, Hl}$
$\mathrm{HI}(-1), \mathrm{I}_2(\mathrm{O}), \mathrm{ICl}(+1), \mathrm{HIO}_4(+7)$
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MCQ 81 Mark
The formation of nitrous oxide from nitrogen and oxygen is the example for
  • A
    Decomposition reaction.
  • B
    Chemical combination of one element and one compound.
  • C
    Chemical combination of two compounds.
  • ✓
    Chemical combination of two elements.
Answer
Correct option: D.
Chemical combination of two elements.
The formation of nitrous oxide from nitrogen and oxygen is the example for chemical combination of two elements.
$\text{N}_2\text{O}\rightleftarrows\text{N}_2+\text{O}.$
$\text{N}_2\text{O}\rightleftarrows\text{N}_2+\text{O}_2$
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MCQ 91 Mark
Oxidation number of $C$ in $\text{HNC}$ is $ ..........$
  • ✓
    $+2$
  • B
    $−3$
  • C
    $+3$
  • D
    Zero
Answer
Correct option: A.
$+2$
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 $x,$
$1 + (−3) + x = 0$
$x = +2$
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MCQ 101 Mark
Which of the following is a redox reaction $($disproportionation reaction$).$
  • A
    $\text{PCl}_3+3\text{H}_2\text{O}\xrightarrow{ \ \ \ \ \ \ \ \ \ \ \ }\text{H}_3\text{PO}_3+3\text{HCL}$
  • B
    $\text{CO}^{3+}+6\text{NO}_2^-\xrightarrow{\ \ \ \ \ \ \ \ \ }[\text{CO(NO}_2)_6]^{3-}$
  • C
    $\text{Hg}_2\text{CrO}_4+2\text{OH}^-\xrightarrow{ \ \ \ \ \ \ \ \ }\text{Hg}_2\text{O}+\text{CrO}_4^{2-}+\text{H}_2\text{O}$
  • ✓
    $3\text{Br}_2+6\text{OH}^-\xrightarrow{ \ \ \ \ \ \ \ }5\text{Br}^-+\text{BrO}_3^-+3\text{H}_2\text{O}$
Answer
Correct option: D.
$3\text{Br}_2+6\text{OH}^-\xrightarrow{ \ \ \ \ \ \ \ }5\text{Br}^-+\text{BrO}_3^-+3\text{H}_2\text{O}$
$\because$ Oxidation state of $\ce{Br_2}$ is increasing from $0$ to $+5,$ decreasing from $0$ to $-1$.
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MCQ 111 Mark
Plumbous ion is represented as :
  • ✓
    $ \mathrm{Pb}^{+2}$
  • B
    $ \mathrm{~Pb}^{+4}$
  • C
    $ \mathrm{~Pb}^{+3} $
  • D
    $\mathrm{~Pb}^{+1}$
Answer
Correct option: A.
$ \mathrm{Pb}^{+2}$
$\mathrm{~Pb}^{2+}=$ Plumbous ion
$\mathrm{~Pb}^{4+}=$ Plumbic ion
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MCQ 121 Mark
What is the oxidation state of $\text{Mn}$ in the compound $\ce{K_2​MnO_4}\ ?$​
  • A
    $3$
  • B
    $4$
  • C
    $5$
  • ✓
    $6$
Answer
Correct option: D.
$6$
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MCQ 131 Mark
In which of the following, the highest oxidation state is not possible?
  • A
    $\left[\mathrm{XeO}_6\right]^{4-}$
  • ✓
    $\mathrm{XeF}_8$
  • C
    $\mathrm{OsO}_4$
  • D
    $\mathrm{RuO}_4$
Answer
Correct option: B.
$\mathrm{XeF}_8$
$\ce{Xe}$ shows $+8$ oxidation state in $\mathrm{XeF}_8$ but it does not exist because of steric hindrance of $8F$ atoms.
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MCQ 141 Mark
What is the oxidation number of $O$ in a diatomic molecule $\ce{(O_2)}\ ?$
  • A
    $+2$
  • B
    $±2$
  • C
    $+8$
  • ✓
    $0$
Answer
Correct option: D.
$0$
The oxidation state of any element in its native state is zero.
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MCQ 151 Mark
The reaction, $\text{2H}_2\text{O(I)}\xrightarrow{\ \ \ \ \Delta\ \ \ }\text{2H}_2\text{(g)}+\text{O}_2\text{(g)}$ is an example of :
  • A
    Addition reaction.
  • ✓
    Decomposition reaction.
  • C
    Displacement reaction.
  • D
    None of these.
Answer
Correct option: B.
Decomposition reaction.
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MCQ 161 Mark
The given reactions such as :
$i. \mathrm{Zn}+2 \mathrm{HCl} \rightarrow \mathrm{ZnCl}_2+\mathrm{H}_2$
$ii. \mathrm{Fe}+2 \mathrm{HCl} \rightarrow \mathrm{FeCl}_2+\mathrm{H}_2$
Are represented as :
  • A
    Displacement of zinc and iron metals.
  • B
    Displacement of only zinc metals.
  • C
    Displacement of only iron metals.
  • ✓
    Displacement of hydrogen.
Answer
Correct option: D.
Displacement of hydrogen.
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MCQ 171 Mark
Identify disproportionation reaction :
  • A
    $\text{CH}_4+2\text{O}_2\xrightarrow{ \ \ \ \ \ \ \ \ }\text{CO}_2+2\text{H}_2\text{O}$
  • B
    $\text{CH}_4+4\text{Cl}_2\xrightarrow{ \ \ \ \ \ \ \ \ }\text{CCl}_4+4\text{HCl}$
  • C
    $2\text{F}_2+2\text{OH}^-\xrightarrow{ \ \ \ \ \ \ \ \ }2\text{F}^-+\text{OF}_2+\text{H}_2\text{O}$
  • ✓
    $2\text{NO}_2+2\text{OH}^-\xrightarrow{ \ \ \ \ \ \ \ \ }\text{NO}^-_2+\text{NO}^-_3+\text{H}_2\text{O}$
Answer
Correct option: D.
$2\text{NO}_2+2\text{OH}^-\xrightarrow{ \ \ \ \ \ \ \ \ }\text{NO}^-_2+\text{NO}^-_3+\text{H}_2\text{O}$
Reactions in which the same substance is oxidized as well as reduced are called disproportionation reactions.
Writing the $O.N$. of each element above its symbol in the given reactions,
  1. $\stackrel{-4+1}{\text{CH}}_4+\stackrel{0}{\text{2O}}_2\xrightarrow{ \ \ \ \ \ \ \ \ }\stackrel{+4-2}{\text{CO}}_2+\stackrel{+1-2}{\text{2H}}_2\text{O}$
  2. $\stackrel{-4+1}{\text{CH}}_4+\stackrel{0}{\text{4Cl}}_2\xrightarrow{ \ \ \ \ \ \ \ \ }\stackrel{+4-1}{\text{CCl}}_4+\stackrel{+1-1}{\text{4HCl}}\text{}$
  3. $2\stackrel{0}{\text{F}}_2+\stackrel{-2+1}{\text{2OH}^-}\xrightarrow{ \ \ \ \ \ \ \ \ }\stackrel{-1}{\text{2F}^-+}\stackrel{+2-1}{\text{OF}_2}+\stackrel{+1-2}{\text{H}_2\text{O}}$
  4. $\stackrel{+4-2}{\text{2NO}_2}+\stackrel{-2+1}{\text{2OH}^-}\xrightarrow{ \ \ \ \ \ \ \ \ }\stackrel{+3-2}{\text{NO}^-_2}+\stackrel{+5-2}{\text{NO}^-_3}+\stackrel{+1-2}{\text{H}_2\text{O}}$
Thus, in reaction $(d), N$ is both oxidized as well as reduced since the $O . N $. of $N$ increases from $+4$ in $\mathrm{NO}_2$ to $+5$ in $ \mathrm{NO}^{-}3$ and decreases from $+4$ in $\mathrm{NO}_2$ to $+3$ in $\mathrm{NO}^{-} 2$.
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MCQ 181 Mark
Which among the following shows maximum oxidation state?
  • A
    $V$
  • B
    $Fe$
  • ✓
    $Mn$
  • D
    $Cr$
Answer
Correct option: C.
$Mn$
Metal Maximum Oxidation state
$V$ $+3$
$Cr$ $+6$
$Fe$ $+3$
$Mn$ $+7$
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MCQ 191 Mark
What is the oxidation state of central atom in $\ce{Ca[PtCl_4​]}\ ?$
  • A
    $1$
  • ✓
    $2$
  • C
    $3$
  • D
    $4$
Answer
Correct option: B.
$2$
$\mathrm{Ca}\left[\mathrm{PtCl}_4\right] \Leftrightarrow \mathrm{Ca}^{+2}[\mathrm{PtCl} 4]^{2-}$
$\text { Take }[\mathrm{PtCl} 4]^{2-} \text {. Central atom }=\mathrm{Pt} \text {. }$
Let $x$ be the oxidation no. of $\ce{Pt},$
$\ce{Cl}$ oxidation no. is $−1.$
$x + 4(−1) = −2$
$x − 4 = −2$
$x = 2$
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MCQ 201 Mark
The oxidation state of the underlined element in the given compound is : $\ce{BaCl_2}​$
  • ✓
    $+2$
  • B
    $-2$
  • C
    $0$
  • D
    None of these
Answer
Correct option: A.
$+2$
$\ce{BaCl_2}$
$\Rightarrow x + (−2) = 0$
$\Rightarrow x = 2$
As chlorine needs only one electron to get octet.
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MCQ 211 Mark
The oxidation numbers of the sulphur atoms in peroxy monosulphuric acid $\ce{(H_2 SO_5​)}$ and peroxydisulphuric and $\ce{(H_2S_2O_8)}$​ are respectively.
  • A
    $+8$ and $+7$
  • B
    $+3$ and $+3$
  • ✓
    $+6$ and $+6$
  • D
    $+4$ and $+6$
Answer
Correct option: C.
$+6$ and $+6$
By looking the structure of $\ce{H_2 SO_5},$ 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
$2(+1) + x + 2(−1) + 3(−2) = 0, x = +6$
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MCQ 221 Mark
In which of the following compounds, is the oxidation number of sulphur is the least?
  • A
    $ \mathrm{SO}_2 $
  • B
    $ \mathrm{SO}_3 $
  • ✓
    $ \mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_8 $
  • D
    $ \mathrm{H}_2 \mathrm{SO}_4 $
Answer
Correct option: C.
$ \mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_8 $
The oxidation number of sulphur in $\ce{SO_2​,SO_3,Na_2 SO_4}$​ and $\ce{H_2​SO_4}$​ are $+4,+6,+2.5$ and $+6$ respectively.
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MCQ 231 Mark
In the electrochemical series, metals are arranged in order of their tendency to :
  • A
    Release halogens from their salts.
  • B
    Become reduced.
  • C
    React with water.
  • ✓
    Lose electrons.
Answer
Correct option: D.
Lose electrons.
In the electrochemical series, metals are arrange in order of their tendency to lose electrons.
Oxidation potential increases down the group. $($tendency to lose electron increases$).$
Reduction potentials decreases down the group. $($tendency to gain $e−$ decreases$).$
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MCQ 241 Mark
The oxidation numbers of sulphur in $\ce{S_8, S_2 , F_2}$ and $\ce{F_2S}$ respectively, are :
  • A
    $0, +1$ and $−2$
  • B
    $+2, +1$ and $−2$
  • ✓
    $0, +1$ and $+2$
  • D
    $−2, +1$ and $−2$
Answer
Correct option: C.
$0, +1$ and $+2$
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MCQ 251 Mark
Oxygen has an oxidation state of $+2$ in.
  • A
    $\ce{H_2O_2}$
  • ✓
    $\ce{OF_2}$
  • C
    $\ce{SO_2}$
  • D
    $\ce{H_2O}$
Answer
Correct option: B.
$\ce{OF_2}$
Oxidation state of oxygen is always $-2$ except in peroxides,superoxides and when it reacts with fluorine.
In $\ce{H_2​O_2},$ oxidation state of $H$ is $+1,$ so oxidation state of oxygen is $-1.$
In $\ce{OF_2}$​, oxidation state of $F$ is $-1,$ so oxidation state of oxygen is $+2.$
In $\ce{SO_2}$ and $\ce{H_2​O}$, oxidation state of oxygen is $-2.$
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MCQ 261 Mark
Metals exhibit $........$ oxidation states in their compounds.
  • ✓
    Positive.
  • B
    Negative.
  • C
    Neutral.
  • D
    None of these.
Answer
Correct option: A.
Positive.
As metals have great tendency in donating electrons it will have positive oxidation states.
For example $\ce{Na}$ is metal has atomic number $11$
So electronic configuration is $2, 8, 1$
So after donating one electron its octet get completed and become stable and becomes $\ce{Na}^+$
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MCQ 271 Mark
Oxidation state of nitrogen in $\ce{NH_2OH}$ is :
  • A
    $−3$
  • ✓
    $−1$
  • C
    $+2$
  • D
    $3$
Answer
Correct option: B.
$−1$
Let $x$ be the oxidation state of $N$ in $\ce{NH_2​OH}$.
Since the overall charge on the complex is $0,$ the sum of oxidation states of all elements in it should be equal to $0.$
Therefore, $x + 2 − 1 = 0$ or, $x = −1$.
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MCQ 281 Mark
The oxidation number of chromium in $\ce{CrO_5}$​ is :
  • ✓
    $+6$
  • B
    $+5$
  • C
    $+10$
  • D
    $0$
Answer
Correct option: A.
$+6$
The oxidation number of chromium in chromium pentaoxide is $6.$
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MCQ 301 Mark
The sum of oxidation number of all the atoms in a neutral molecule must be zero.
  • ✓
    True.
  • B
    False.
  • C
    Ambiguous.
  • D
    None of these.
Answer
Correct option: A.
True.
The sum of oxidation number of all the atoms in a neutral molecule must be zero.
For example, neutral molecules such as $\mathrm{O}_2, \mathrm{P}_4, \mathrm{O}_3, \mathrm{~S}_8$ and $\mathrm{KMnO}_4$ have the sum of oxidation number of all the atoms equal to zero.
For an ion, the sum of oxidation number of all the atoms is equal to the charge on the ion.
For example, in cyanide ion $\ce{(CN}^-$), the sum of oxidation number of all the atoms is equal to $−1.$
In ammonium ion $\ce{(NH_4}​)+,$ the sum of oxidation number of all the atoms is equal to $+1.$
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MCQ 311 Mark
Which titrant is used in the Iodometric titration which involves $\ce{I_2} ?$
  • A
    $\mathrm{KMnO}_4 $
  • B
    $ \mathrm{~K}_2 \mathrm{Cr}_2 \mathrm{O}_7 $
  • ✓
    $ \mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3$
  • D
    All of them
Answer
Correct option: C.
$ \mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3$
Iodometry is one of the most important redox titration methods. Iodine reacts directly, fast and quantitively with many organic and inorganic substances.
Thanks to its relatively low, $\ce{pH}$ independent redox potential, and reversibility of the iodine/ iodide reaction, iodometry can be used both to determine amount of reducing agents $($by direct titration with iodine$)$ and of oxidizing agents $($by titration of iodine with thiosulfate$).$
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MCQ 321 Mark
In $\ce{MgCl_2}​,$ the oxidation number of chlorine is :
  • A
    $+1$
  • B
    $+2$
  • ✓
    $−1$
  • D
    $0$
Answer
Correct option: C.
$−1$
In $\ce{MgCl_2},$​ oxidation number of $\ce{Cl}$ is :
$\Rightarrow 2 + 2x = 0$
$\Rightarrow x = −1.$
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MCQ 331 Mark
The oxidation state of $\ce{Cr}$ in $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is :
  • A
    $+4$
  • B
    $+3$
  • ✓
    $+6$
  • D
    $+5$
Answer
Correct option: C.
$+6$
$\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$​
Let the oxidation state of $\ce{Cr}$ is $x.$
$2(+1) + 2x + 7(−2) = 0$
$+2 + 2x − 14 = 0$
$2x − 12 = 0$
$2x = 12$
$x = +6$
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MCQ 341 Mark
In the given reaction, $\mathrm{CH}_2=\mathrm{CH}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{CH}_3-\mathrm{CH}_3(\mathrm{~g})$ ethene undergoes :
  • ✓
    Reduction process.
  • B
    Oxidation process.
  • C
    Addition process.
  • D
    All of these.
Answer
Correct option: A.
Reduction process.
$\mathrm{CH}_2=\mathrm{CH}_2+\mathrm{H}-\mathrm{H} \rightarrow \mathrm{H}_3 \mathrm{C}-\mathrm{CH}_3$
$($Addition of hydrogen$)$
Because of the addition of hydrogen, there occurs reduction of ethylene.
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MCQ 351 Mark
Tailing of mercury is $ .......... $ redox change.
  • A
    Intramolecular.
  • ✓
    Intermolecular.
  • C
    Disproportion.
  • D
    None.
Answer
Correct option: B.
Intermolecular.
When ozone is passed through mercury, mercurous oxide $\ce{(Hg_2​O)}$ is formed.
Due to this, mercury loses its meniscus and starts sticking to the glass.
This phenomenon is known as Tailing of mercury.
$2 \mathrm{Hg}+\mathrm{O}_3 \rightarrow \mathrm{Hg}_2 \mathrm{O}+\mathrm{O}_2$.
In this reaction, the oxidation number of mercury changes from $0$ to $+1$. Thus, it is oxidized.
The oxidation number of oxygen changes from 0 to $-2$. Thus, it is reduced.
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MCQ 361 Mark
The half $-$ cell reaction is the one that :
  • ✓
    Takes place at one electrode.
  • B
    Consumes half a unit of electricity.
  • C
    Involves half a mole of electrolyte.
  • D
    Goes half way to completion.
Answer
Correct option: A.
Takes place at one electrode.
A half reaction is either the oxidation or reduction reaction component of a redox reaction. A half reaction is obtained by considering the change in oxidation states of individual substances involved in the redox reaction. Half $-$ reactions are often used as a method of balancing redox reactions.
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MCQ 371 Mark
The oxidation state of the most electronegative element in the products of the reaction between $\mathrm{BaO}_2$ and $\mathrm{H}_2 \mathrm{SO}_4$ are :
  • A
    $0$ and $−1$
  • ✓
    $−1$ and $−2$
  • C
    $−2$ and $0$
  • D
    $−2$ and $+1$
Answer
Correct option: B.
$−1$ and $−2$
$\mathrm{BaO}_2+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{BaSO}_4+\mathrm{H}_2 \mathrm{O}_2$
The most electronegative element in the product is oxygen.
The oxidation state of oxygen in $\mathrm{BaSO}_4$ is $-2$ and in $\mathrm{H}_2 \mathrm{O}_2$ is $-1 .$
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MCQ 381 Mark
Heat is released in :
  • ✓
    Exothermic reactions.
  • B
    Endothermic reactions.
  • C
    Both $A$ and $B.$
  • D
    None of these.
Answer
Correct option: A.
Exothermic reactions.
Heat is released in exothermic reactions.
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MCQ 391 Mark
When $P$ reacts with caustic soda, the products are $\ce{PH_3}$​ and $\ce{NaH_2​PO_2}$. The reaction is an example of.
  • A
    Oxidation.
  • B
    Reduction.
  • ✓
    Both oxidation and reduction.
  • D
    Neutralisation.
Answer
Correct option: C.
Both oxidation and reduction.
$\mathrm{P}_4+3 \mathrm{NaOH}+3 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{PH}_3+3\mathrm{NaH}_2 \mathrm{PO}_2$
In reactant $P$ is present in $(0)$ oxidation state and in $\ce{PH_3}$ , it is present in $(-3)$ oxidation state and in $\ce{NaH_2​PO_2}$​ it is present in $(+1)$ oxidation state.
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MCQ 401 Mark
The rods of transition metals such as copper and zinc where potential difference is generated, are termed as :
  • ✓
    Electrodes.
  • B
    Cathodes.
  • C
    Anodes.
  • D
    None of these.
Answer
Correct option: A.
Electrodes.
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MCQ 411 Mark
Standard reduction frotential of $\text{X, Y, Z}$ are ${-1.2v, +0.5v, -3.0v}$ respectively, the reducing power of the metals will be :
  • ✓
    $\text{Y > Z > X}$
  • B
    $\text{Y > X > Z}$
  • C
    $\text{Z > X > Y}$
  • D
    $\text{X > Y > Z}$
Answer
Correct option: A.
$\text{Y > Z > X}$
$'Z\ '$ is best because it has lowest standard reduction potential whereas $'Y\ '$ is weakest due to highest standard reduction potential.
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MCQ 421 Mark
A redox reaction is one in which :
  • A
    Both the substance are reduced.
  • B
    Both the substance are oxidised.
  • ✓
    One substance is reduced and other is oxidised.
  • D
    All of the above.
Answer
Correct option: C.
One substance is reduced and other is oxidised.
In a redox reaction both oxidation and reduction is happening together.
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MCQ 431 Mark
What is the oxidation number of $\ce{Si}$ in the compound $\ce{CaSiO_3} \ ?$
  • A
    $−4$
  • B
    $+2$
  • C
    $−2$
  • ✓
    $+4$
Answer
Correct option: D.
$+4$
$\ce{CaSiO_3}$
Total charge present $= 0.$
Oxidation no. of Oxygen is $−2.$
Oxidation no. od Calcium is $+2.$
Let, oxidation no. of Slilicon be $X.$
So, $[+2] + x + 3[−2] = 0$
$2 + x − 6 = 0$
$x = 4$
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MCQ 441 Mark
The difference in the oxidation numbers of the two types of sulphur atoms in $\mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6$ is :
  • ✓
    $5$
  • B
    $4$
  • C
    $3$
  • D
    $6$
Answer
Correct option: A.
$5$
In $\mathrm{Na}_2\mathrm{~S}_4 \mathrm{O}_6,$​ the oxidation number of end sulphur atoms is $+5$ each and the oxidation number of middle sulphur atoms is $0$ each.
The difference in the oxidation numbers of the two types of sulphur atoms is $5 − 0 = 0.$
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MCQ 451 Mark
Consider the given standard electrode potentials :
$\ce{i}.\frac{\text{K}^+}{\text{K}}=-3.02\text{V}$
$\ce{ii}.\frac{\text{Cu}^{2+}}{\text{Cu}}=+0.34\text{V}$
$\ce{iii}.\frac{\text{Hg}^{2+}}{\text{Hg}}=0.92\text{V}$
$\ce{iv}.\frac{\text{Cr}^{3+}}{\text{Cr}}=-0.74\text{V}$
Decreasing order of reducing power of these elements is :
  • A
    $\text{I > II > III > IV}$
  • B
    $\text{I > IV > III > II}$
  • ✓
    $\text{I > IV > II > III}$
  • D
    $\text{III > II > IV > I}$
Answer
Correct option: C.
$\text{I > IV > II > III}$
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MCQ 461 Mark
Which of the following pairs of ions cannot coexist in aqueous solution?
  • ✓
    $\text{Cr}^{2+}$ and $ \ \text{MnO}_4^-$
  • B
    $\text{Fe}^{3+}$ and $\ \text{Cr}_2\text{O}_7^{2-}$
  • C
    $\text{Cr}^{2+}$ and $\ \text{I}_3^-$
  • D
    $\text{Mn}^{2+}$ and $\ \text{Cl}^-$
Answer
Correct option: A.
$\text{Cr}^{2+}$ and $ \ \text{MnO}_4^-$
It is because $\text{Cr}^{2+}$ is strongly reducing, it will get oxidised by $\text{MnO}_4^-$ to $\text{Cr}^{3+}$
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MCQ 471 Mark
In the chemical reaction, $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+\mathrm{xH}_2 \mathrm{SO}_4+\mathrm{ySO}_2 \rightarrow \mathrm{K}_2 \mathrm{SO}_4+\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3+\mathrm{zH}_2 \mathrm{O}$
the value of $x + y + z$
  • A
    $6$
  • ✓
    $5$
  • C
    $7$
  • D
    $3$
Answer
Correct option: B.
$5$
The balanced redox reaction is : $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+\mathrm{xH}_2 \mathrm{SO}_4+\mathrm{3SO}_2 \rightarrow \mathrm{K}_2 \mathrm{SO}_4+\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3+\mathrm{zH}_2 \mathrm{O}$
$x = 1, y = 3, z = 1$
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MCQ 481 Mark
Which of the following compounds we use in our laboratory as a standard solution $($titrant$) \ ?$
  • A
    $\mathrm{KMnO}_4 $
  • B
    $ \mathrm{~K}_2 \mathrm{Cr}_2 \mathrm{O}_7 $
  • C
    $ \mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3 $​
  • ✓
    All of these
Answer
Correct option: D.
All of these
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.
$\mathrm{KMnO}_4, \mathrm{~K}_2 \mathrm{Cr}_2 \mathrm{O}_7, \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$​ etc. are compounds we use in our laboratory as a standard solution.
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MCQ 491 Mark
Which is the best description of behaviour of bromine in the given equation? $\mathrm{H}_2 \mathrm{O}+\mathrm{Br}_2 \rightarrow \mathrm{HBr}+\mathrm{HOBr}$
  • A
    Proton acceptor.
  • ✓
    Both oxidised and reduced.
  • C
    Oxidised.
  • D
    Reduced.
Answer
Correct option: B.
Both oxidised and reduced.
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MCQ 501 Mark
When tin $(IV)$ chloride is treated with excess of conc. hydrochloric acid, the complex ion $\ce{(SnCl_6)}^{2−}$ is formed. The oxidation state of tin in this complex ion is?
  • ✓
    $+4$
  • B
    zero
  • C
    $−2$
  • D
    $−4$
Answer
Correct option: A.
$+4$
Let oxidation state of $\text{Sn}$ is $x$ and we know oxidation of $\text{Cl}$ is $−1,$ so $x + 6(−1) = −2, x = +4.$
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MCQ 511 Mark
Which of the following reactions represent $(s)$ redox process?
  • A
    Electrochemical process for extraction of highly reactive metals and non $-$ metals.
  • B
    Manufacturing of caustic soda.
  • C
    Corrosion of metals.
  • ✓
    All of the above.
Answer
Correct option: D.
All of the above.
Electrochemical processes for the extraction of highly reactive metals and non $-$ metals, manufacturing of chemical compounds like caustic soda, operation of dry and wet batteries and corrosion of metals fall within the range of redox processes.
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MCQ 521 Mark
The correct statement for the molecule $\ce{CsI_3}$​ is that :
  • A
    It contains $\mathrm{Cs}^{3+}$ and $\mathrm{I}^{-}$ions.
  • B
    It contains $\mathrm{Cs}^{+}, \mathrm{I}^{-}$and lattice $\mathrm{I}_2$ molecule.
  • C
    It is a covalent molecule.
  • ✓
    It contains $\mathrm{Cs}^{+}$and $\mathrm{I}_3^{-}$ ions.
Answer
Correct option: D.
It contains $\mathrm{Cs}^{+}$and $\mathrm{I}_3^{-}$ ions.
$\mathrm{CsI}_3 \rightarrow \mathrm{Cs}^{+}+\mathrm{I}_3^{-}$
$\Rightarrow \ce{Cs}$ cannot show $+3$ oxidation state.
$\Rightarrow \ce{I_2}$​ molecules are too large to be accommodated in the lattice.
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MCQ 531 Mark
The given reaction, $\mathrm{CuSO}_4+\mathrm{Zn} \rightarrow \mathrm{Cu}+\mathrm{ZnSO}_4$ is an example of :
  • ✓
    Metal displacement reaction.
  • B
    Non $-$ metal displacement reaction.
  • C
    Metal addition reaction.
  • D
    Non $-$ metal addition reaction.
Answer
Correct option: A.
Metal displacement reaction.
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MCQ 541 Mark
Which is the best description of behaviour of bromine in the given equation? $\mathrm{H}_2 \mathrm{O}+\mathrm{Br}_2 \rightarrow \mathrm{HBr}+\mathrm{HOBr}$
  • A
    Proton acceptor.
  • ✓
    Both oxidised and reduced.
  • C
    Oxidised.
  • D
    Reduced.
Answer
Correct option: B.
Both oxidised and reduced.
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MCQ 551 Mark
What is the oxidation number of $\text{Br}$ in the compound $\ce{RbBrO_4}​\ ?$
  • A
    $−1$
  • ✓
    $+7$
  • C
    $+1$
  • D
    $+4$
Answer
Correct option: B.
$+7$
$\text{RbBrO}_4$
Total charge $= 0.$
$\text{Rb}$ is a group $I$ element.
So, its oxidation no. $= +1.$
Oxidation no. of oxygen $= −2.$
Let, Oxidation no. of $\text{Br}$ be $x.$
So, $1 + x + 4(−2) = 0$
$1 + x − 8 = 0$
$x − 7 = 0$
$x =7$
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MCQ 561 Mark
In redox reaction $\text{Cu}_\text{(s)}+2\text{Ag}^+\text{(aq)}\xrightarrow{\ \ \ \ \ \ \ \ \ }\text{Cu}^{2+}\text{(aq)}+2\text{Ag}$ the weakest oxidising agent and weakest reducing agents are :
  • ✓
    $\mathrm{Cu}^{2+}$ and $\ce{Ag}$ respectively.
  • B
    $\ce{Ag}$ and $\mathrm{Cu}^{2+}$ respectively.
  • C
    $\mathrm{Ag}^{+}$and $\mathrm{Cu}^{2+}$ respectively.
  • D
    $\mathrm{Ag}^{+}$and $\mathrm{Cult}^{2+}$ respectively.
Answer
Correct option: A.
$\mathrm{Cu}^{2+}$ and $\ce{Ag}$ respectively.
$\ce{Cu^{2+}}$ and $\ce{Ag}$ are weakest oxidising and reducing agents respectively because $\ce{Cu^{2+}}$ cannot gain electron easily and $\ce{Ag}$ cannot lose electron easily in above reaction.
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MCQ 571 Mark
In the reaction $,2 \mathrm{KClO}_3 \rightarrow 2 \mathrm{KCl}+3 \mathrm{O}_2,$ the elements which have been oxidised and reduced respectively are :
  • A
    Chlorine and oxygen.
  • ✓
    Oxygen and chlorine.
  • C
    Potassium and oxygen.
  • D
    Oxygen and potassium.
Answer
Correct option: B.
Oxygen and chlorine.
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MCQ 581 Mark
$\ce{Mg + CuO \rightarrow MgO + Cu}$ Which of the following is wrong relating to the above reaction?
  • A
    $\text{CuO}$ gets reduced.
  • B
    $\text{Mg}$ gets oxidised.
  • ✓
    $\text{CuO}$ gets oxidised.
  • D
    It is a redox reaction.
Answer
Correct option: C.
$\text{CuO}$ gets oxidised.
The reaction $\ce{Mg + CuO \rightarrow MgO + Cu}$ is a redox reaction.
$\text{CuO}$ loses an $O$ atom to form $\text{Cu}.$ Thus, it is reduced.
$\text{Mg}$ gains an $O$ atom. Thus, it is oxidized.
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MCQ 591 Mark
In which of the following does sulphur has the lowest oxidation state?
  • A
    $ \mathrm{H}_2 \mathrm{SO}_4 $
  • B
    $ \mathrm{SO}_2 $
  • C
    $\mathrm{H}_3 \mathrm{SO}_3$
  • ✓
    $ \mathrm{H}_2 \mathrm{~S}$
Answer
Correct option: D.
$ \mathrm{H}_2 \mathrm{~S}$
Sulphur has the lowest oxidation state in $ \mathrm{H}_2 \mathrm{~S}$ that is $−2.$
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MCQ 601 Mark
In the given reaction $,2 \mathrm{Na}+\mathrm{S} \rightarrow \mathrm{Na}_2 \mathrm{S},$ sulphur is :
  • A
    Oxidised.
  • ✓
    Reduced.
  • C
    Reducing agent.
  • D
    None of these.
Answer
Correct option: B.
Reduced.
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MCQ 611 Mark
Solution of potassium chloride or ammonium nitrate in salt $-$ bridge usually solidified by boiling with :
  • ✓
    Agar $-$ agar.
  • B
    Starch.
  • C
    Cellulose.
  • D
    Glycogen.
Answer
Correct option: A.
Agar $-$ agar.
A solution of potassium chloride or ammonium nitrate is solidified by boiling with agar $-$ agar and later cooling to a jelly like substance.
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MCQ 621 Mark
What is the oxidation number of chlorine in $\text{ClO}^-_3$​?
  • ✓
    $+5$
  • B
    $+3$
  • C
    $+4$
  • D
    $+2$
Answer
Correct option: A.
$+5$
Let $x$ be the oxidation number of chlorine in $\text{ClO}^-_3$
The oxidation number of oxygen is $−2.$
The sum of the oxidation numbers of chlorine and oxygen is $−1,$ which is equal to the charge on ion.
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MCQ 631 Mark
The more positive the value of $\text{E}^\ominus,$ the greater is the tendency of the species to get reduced. Using the standard electrode potential of redox couples given below find out which of the following is the strongest oxidising agent : $\text{E}^{\ominus}\text{values:} \text{ Fe}^{3+}/ \text{ Fe}^{2+}=+0.77;\text{I}_2(\text{S})/\text{ I}^-=+0.54; \text{Cu}^{2+}/\text{ Cu}=+0.34;\text{Ag}^+/\text{ Ag}=+0.80\text{V}$
  • A
    $ \mathrm{Fe}^{3+} $
  • B
    $\mathrm{I}_2(\mathrm{~S}) $
  • C
    $ \mathrm{Cu}^{2+} $
  • ✓
    $ \mathrm{Ag}^{+} $
Answer
Correct option: D.
$ \mathrm{Ag}^{+} $
Since $ \mathrm{Ag}^{+}\text{Ag}$ has highest positive value of $\text{E}^\ominus$, therefore, ​​​​​​​$ \mathrm{Ag}^{+} $ is the strongest oxidizing agent with highest tendency to get reduced.
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MCQ 641 Mark
On reduction of $\text{KMnO}_4$​ by oxalic acid in acidic medium, the oxidation number of $\text{Mn}$ chnages. What is the magnitude of this change?
  • ✓
    $7$ to $2$
  • B
    $6$ to $2$
  • C
    $5$ to $2$
  • D
    $7$ to $4$
Answer
Correct option: A.
$7$ to $2$
In acidic medium reduction of $\text{KMnO}_4$ takes place as follows :
So oxidation state of $\text{Mn}$ changes from $+7$ to $+2.$
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MCQ 651 Mark
The oxidation number of $\text{Cr}$ in $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is :
  • A
    $+3$
  • B
    $−3$
  • ✓
    $+6$
  • D
    $−6$
Answer
Correct option: C.
$+6$
The sum of the oxidation numbers in a polyatomic ion is equal to the charge on the ion.
The oxidation number of simple ions is equal to the charge on the ion.
The metals in Group $\text{IA (K)}$ atom has an oxidation number of $+1$.
Oxygen usually has an oxidation number of $−2.$
Oxidation no of $\text{Cr}$ in $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$​
$=2\times 1+2\times \text{x}+7\times (−2)=0$
$\text{x}=\frac{14-2}{2}=+6$
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MCQ 661 Mark
Which of the following solution is strongest oxidizing agent?
  • A
    $\text{MnO}_4^- $ in acidic medium.
  • ✓
    $\text{MnO}_4^-$ in basic medium.
  • C
    $\text{MnO}_2$ in basic medium.
  • D
    $\text{CrO2}_4^-$ in basic medium.
Answer
Correct option: B.
$\text{MnO}_4^-$ in basic medium.
$\text{MnO}_4^-+3\text{e}^-+2\text{H}_2\text{O}\xrightarrow{ \ \ \ \ \ \ \ \ }\text{MnO}_2+4\text{OH}^- \ 
\text{E}^\circ = 1.69V,$ which is highest.
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MCQ 671 Mark
The most common oxidation states of the most electronegative elements in their compounds are :
  • A
    $0$ and $-1$
  • ✓
    $-1$ and $-2$
  • C
    $-2$ and $0$
  • D
    $-2$ and $+1$
Answer
Correct option: B.
$-1$ and $-2$
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MCQ 681 Mark
The oxidation number of $\text{Mn}$ in potassium permanganate is :
  • A
    $+6$
  • ✓
    $+7$
  • C
    $+5$
  • D
    $+8$
Answer
Correct option: B.
$+7$
Let $x $ be the oxidation number of manganese in potassium permanganate. $\ce{(KMnO_4​)}$
The oxidation numbers of potassium and oxygen are $+1$ and $−2$ respectively.
The sum of the oxidation numbers in a neutral molecule is zero.
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MCQ 691 Mark
In a solution of $\text{KMnO}_4, \text{MnO}_4^-$ acts as a:
  • A
    Induced indicator.
  • ✓
    Self indicator.
  • C
    Spontaneous indicator.
  • D
    None of the above.
Answer
Correct option: B.
Self indicator.
In redox titrations, the reagent which itself is intense in colour,
e.g. permanganate ion, $[\text{MnO}]_4^-$ acts as self indicator.
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MCQ 701 Mark
One mole of $\mathrm{N}_2 \mathrm{H}_4$ loses $10$ moles of electrons to form a new compound $A.$ Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in $A\ ?$
$[$There is no change in the oxidation state of hydrogen$]$
  • A
    $+1$
  • B
    $−3$
  • ✓
    $+3$
  • D
    $+5$
Answer
Correct option: C.
$+3$
The oxidation state of $N$ in hydrazine is $-2.$
$1$ mole of hydrazine contains $2$ moles of $N$ and loses $10$ moles of electrons.
Hence, $1 N$ atom will lose $5$ electrons.
Hence, its oxidation number will increase by $5.$
Hence, the oxidation number of $N$ in compound $A$ will be $−2 + 5 = +3.$
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MCQ 721 Mark
$\text{E}^\ominus$ values of some redox couples are given below. On the basis of these values choose the correct option : $\text{E}^\ominus\text{values} : \text{Br}_2/ \text{ Br}^-=+1.90;\text{ Ag}^+/\text{ Ag(S)}=+0.80 \text{Cu}^{2+}/\text{ Cu(s)}=+0.34;\text{I}_2\text{(s)}/\text{ I}^-=0.54$
  • A
    $\text{Cu}$ will reduce $\text{Br} ^{-}$
  • B
    $\text{Cu}$ will reduce $\text{Ag}$
  • C
    $\text{Cu}$ will reduce $\text{I}^-$
  • ✓
    $\text{Cu}$ will reduce $\text{Br}_2$
Answer
Correct option: D.
$\text{Cu}$ will reduce $\text{Br}_2$
Copper will reduce $\text{Br}_2$, if the $E^\circ $ of the redox reaction, $2 \mathrm{Cu}+\mathrm{Br}_2 \mathrm{~CuBr}_2$ is $+\text{ve}.$
Now $\text{Cu}\xrightarrow{ \ \ \ \ \ \ }\text{Cu}^{2+}+2\text{e}^{-};\text{E}^{0}=-0.34\text{V}$
$\text{Br}_2+2\text{e}^{-}\xrightarrow{ \ \ \ \ \ \ \ \ \ }2 \text{Br}^{-};\text{E}^0=+1.90\text{V}$
$\underline{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }$
$\text{Cu + Br}_2\xrightarrow{ \ \ \ \ \ \ \ \ }\text{CuBr}_2;\text{E}^{0}=+1.56\text{V}$
Since $E^\circ$ of this reaction is $+\text{ve},$ therefore, $\text{Cu}$ can reduce $\text{Br}_2$ and hence option $(d)$ is correct.
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MCQ 731 Mark
Oxidation number of sulphur in marshall's acid $(\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_8)$ is :
  • A
    $+5$
  • B
    $+8$
  • ✓
    $+6$
  • D
    $+7$
Answer
Correct option: C.
$+6$
Let $x$ be the oxidation number of sulphur in marshall's acid $(\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_8)$​. It contains a peroxide linkage.
Thus, the oxidation number of two $O$ atoms is $−1$ and that of the remaining $O$ atoms is $−2$.
The oxidation number of $H$ is $+1.$
For the neutral molecule, the sum of the oxidation numbers is zero.
Therefore, $2(+1) + 2x + 2(−1) + 6(−2) = 0$ or, $x = +6$
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MCQ 741 Mark
The average oxidation number of iodine in $\text{l}_3^{-}$ ion is :
  • A
    $-1$
  • ✓
    $\frac{-1}{3}$
  • C
    $+1$
  • D
    $\frac{+1}{3}$
Answer
Correct option: B.
$\frac{-1}{3}$
$\text{I}_3^-$
$\because3\text{x}=1$
$\Rightarrow\text{x}=\frac{-1}{3}$
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MCQ 751 Mark
The oxidation number of $P$ in $\mathrm{Na}_4 \mathrm{P}_2 \mathrm{O}_7$​ is :
  • A
    $+3$
  • B
    $+2$
  • ✓
    $+5$
  • D
    $−3$
Answer
Correct option: C.
$+5$
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MCQ 761 Mark
In which of the following reactions, the oxidation state of nitrogen is not changed?
  • A
    Action of very dilute $\mathrm{HNO}_3$ on magnesium.
  • B
    Action of conc.$\mathrm{HNO}_3$​ on zinc.
  • ✓
    Action of conc.$\mathrm{HNO}_3$ on copper.
  • D
    Action of very dilute $\mathrm{HNO}_3$ on zinc.
Answer
Correct option: C.
Action of conc.$\mathrm{HNO}_3$ on copper.
(C)  Action of conc.$\mathrm{HNO}_3$ on copper. 
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MCQ 771 Mark
The state oxidation state of aluminium is $ .........$
  • A
    $+3$
  • ✓
    $+1$ and $+3$
  • C
    $+2$ and $+3$
  • D
    $+1$
Answer
Correct option: B.
$+1$ and $+3$
The vast majority of compounds with $\text{Al}$ minerals feature aluminium in the oxidation state $+3. \text{Al(III)}$ is six coordinate or tetra coordinate. Most $\text{Al}$ compounds exist in $+3$ oxidation state but a few exist in $+1$ oxidation state like $\text{AlF, AlCl, AlBr}$ when the trihalide is heated with $\text{Al}.$
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MCQ 781 Mark
Oxidation number of chlorine in $\ce{O}_3$​ is ?
  • ✓
    zero
  • B
    $+4$
  • C
    $+2$
  • D
    $-6$
Answer
Correct option: A.
zero
$\ce{O}_3$ Ozone is a molecule only consist of three oxygen atom.
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MCQ 791 Mark
An element if present in the free or the uncombined state, its each atom bears an oxidation number :
  • A
    More than $1$
  • B
    Less than $1$
  • C
    More than $2$
  • ✓
    Zero.
Answer
Correct option: D.
Zero.
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MCQ 801 Mark
When ammonium nitrate is gently heated, an oxide of nitrogen is formed. What is the oxidation state of nitrogen in this oxide?
  • A
    $+4$
  • B
    $+2$
  • C
    $+3$
  • ✓
    $+1$
Answer
Correct option: D.
$+1$
When ammonium nitrate is gently heated, dinitrogen monoxide $\ce{N_2O}$ is obtained.
Let $x$ be the oxidation state of $N$ in $\ce{N_2​O}.$
The oxidation state of oxygen is $−2.$
The sum of the oxidation states for neutral molecule is $0.$
Hence, $2x+(−2) = 0$ or, $x = +1.$
Therefore, the oxidation state of $N$ is $+1.$
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MCQ 811 Mark
In the reaction between copper nitrate solution and zinc, copper ions are reduced by gaining electrons from :
  • A
    Copper.
  • B
    Nitrogen.
  • ✓
    Zinc.
  • D
    Oxygen.
Answer
Correct option: C.
Zinc.
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MCQ 821 Mark
The value of $n$ in the molecular formula $\mathrm{Be_n} \mathrm{Al}_2 \mathrm{Si}_6 \mathrm{O}_{18}$ is :
  • A
    $1$
  • B
    $2$
  • ✓
    $3$
  • D
    $4$
Answer
Correct option: C.
$3$
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MCQ 831 Mark
Which of the following processes does not involve oxidation of iron?
  • A
    Rusting of iron sheets.
  • B
    Decolourisation of blue $\ce{CuSO}_4$ solution by $\ce{Fe}.$
  • ✓
    Formation of $\ce{Fe(CO)}_5$ from $\ce{Fe}.$
  • D
    Liberation of $H_2$ from steam by iron at high temperature.
Answer
Correct option: C.
Formation of $\ce{Fe(CO)}_5$ from $\ce{Fe}.$
$\because$ There is no change in oxidation state.
$\text{Fe}+5\text{CO}\xrightarrow{ \ \ \ \ \ \ }\text{Fe(CO)}_5$
$0\ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ \ 0$
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MCQ 841 Mark
The oxidation state of phosphorous is $...........$ in phosphoric acid.
  • A
    $+1$
  • B
    $+3$
  • ✓
    $+5$
  • D
    $-1$
Answer
Correct option: C.
$+5$
The oxidation state of $P$ is $+5$ in phosphoric acid.
The structure is given below
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MCQ 851 Mark
In which of the following substances, sulphur has the lowest oxidation number?
  • A
    $c\mathrm{H}_2 \mathrm{SO}_4$
  • B
    $\mathrm{SO}_2$
  • C
    $\mathrm{H}_2 \mathrm{SO}_3$
  • ✓
    $ \mathrm{H}_2 \mathrm{S} $
Answer
Correct option: D.
$ \mathrm{H}_2 \mathrm{S} $
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MCQ 861 Mark
The oxidation state of central atom in the anion of compound $\mathrm{Na} \mathrm{H}_2 \mathrm{PO}_2$ will be :
  • A
    $−3$
  • ✓
    $+1$
  • C
    $+3$
  • D
    $+5$
Answer
Correct option: B.
$+1$
$\mathrm{Na}\left(\mathrm{H}_2\right) \mathrm{PO}_2$
$= +1 + (2x + 1) + x + 2(−2) = 0$
$x − 1 = 0$
or $x = 1$
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MCQ 871 Mark
Oxygen shows oxidation state of $-1$ in the compound :
  • A
    $\mathrm{NO}_2$
  • B
    $\mathrm{MnO}_2$
  • C
    $\mathrm{PbO}_2$
  • ✓
    $\mathrm{Na}_2 \mathrm{O}_2$
Answer
Correct option: D.
$\mathrm{Na}_2 \mathrm{O}_2$
$\mathrm{Na}^2 \mathrm{O}^2$:
$2 + 2x = 0$
$\Rightarrow 2x = −2$
$\Rightarrow x = −1.$
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MCQ 881 Mark
Oxidation state of iodine in $\mathrm{ICl}_2{ }^{-}$​ is :
  • ✓
    $+1$
  • B
    $-1$
  • C
    $+2$
  • D
    $-2$
Answer
Correct option: A.
$+1$
$\text{Cl}$ is in $-1$ oxidation state since it has lower atomic number than iodine.
Let the oxidation state of iodine be $x.$
$x + (−1) \times 2 = −1$
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MCQ 891 Mark
Identify the strongest acids out of the following list.
  • ✓
    $\mathrm{HBrO}_4$
  • B
    $\mathrm{HOCl}$
  • C
    $\mathrm{HNO}_2$
  • D
    $\mathrm{H}_3 \mathrm{PO}_3$
Answer
Correct option: A.
$\mathrm{HBrO}_4$
As the oxidation state of central atom increases, acidity increases.
So, the strongest acid is the one which has maximum value of oxidation number of the central atom.
$\mathrm{HBrO}_4$
$\text{O.N}$ of $\text{Br} = +7$
$\text{HOCl; O.N}$ of $\text{Cl} = +1$
$\mathrm{HNO}_2​; \text{O.N}$ of $N = +3$
$\mathrm{H}_3 \mathrm{PO}_3​;\text{O.N}$ of $P = +3$
As, oxidation state of bromine is $+7,$ which is highest in all.
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MCQ 901 Mark
Carbon is in lowest oxidation state in :
  • A
    $\text{HCOOH}$
  • ✓
    $\text{HCHO}$
  • C
    $\text{CF}_4$
  • D
    $\text{CO}_2$
Answer
Correct option: B.
$\text{HCHO}$
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MCQ 911 Mark
Oxidation number of $\ce{Cr}$ in $\mathrm{CrO}_2 \mathrm{Cl}_2$ is $+6$. Oxidation number $($per atom$)$ of oxygen is :
  • ✓
    $−2$
  • B
    $+2$
  • C
    $−1$
  • D
    $+1$
Answer
Correct option: A.
$−2$
Let the oxidation state of $O$ be $x.$
Since, the compound $\mathrm{CrO}_2 \mathrm{Cl}_2$​ is neutral, sum of oxidaton states of all elements must be zero.
Hence, $6 + 2x − 2 = 0\ ($neutral compound$) =$
So, $x = −2$
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MCQ 921 Mark
The oxidation number of $\text{Ni}$ in $\mathrm{Ni}(\mathrm{CO})_4$ is $+2.$
  • A
    True
  • ✓
    False
  • C
    Anomalous
  • D
    None of these
Answer
Correct option: B.
False
Let $x$ be the oxidation number of $\text{Ni}$ in $\mathrm{Ni}(\mathrm{CO})_4$​.
Since the overall charge on the complex is $0,$ the sum of oxidation states of all elements in it should be equal to $0.$
Therefore, $x + 4 (0) = 0$
Hence, $x = 0.$
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MCQ 931 Mark
Which of the following processes takes place in oxidation?
  • ✓
    Addition of oxygen.
  • B
    Addition of hydrogen.
  • C
    Removal of oxygen.
  • D
    Removal of chlorine.
Answer
Correct option: A.
Addition of oxygen.
Addition of oxygen takes place in oxidation.
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MCQ 941 Mark
What electrolyte is used to electroplate silver onto a spoon?
  • A
    Pure water.
  • ✓
    Silver nitrate.
  • C
    Copper sulfate.
  • D
    Acidified water.
Answer
Correct option: B.
Silver nitrate.
In silver plating, the object $($eg.spood$)$ to be plated is made the cathode of an electrolytic cell. The anode is bar of silver metal and the electrodyte must contain $\text{Ag} +$ ions $($i.e. $\text{AgNO}_3),$ so that when the direct current is passed through the cell, positive $\text{(Ag}^+)$ ions move towards the negative anode $($spoon$)$ where they are neutralized by $e−$ and stick to spoon as silver metal.
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MCQ 951 Mark
The oxidation state of nitrogen in $\text{NaNO}_3$ is :
  • A
    $+3$
  • ✓
    $+5$
  • C
    $−4$
  • D
    $−5$
Answer
Correct option: B.
$+5$
The overall charge is zero $($because it is not an ion$).$
Since sodium is a $1A$ family member, the charge is $+1.$
The charge of oxygen is almost always $-2.$
Hence, $1 + \text{OS}$ of $N + (3 \times −2) = 0$
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MCQ 961 Mark
The oxidation number of cobalt in $\mathrm{K}\left[\mathrm{Co}(\mathrm{CO})_4\right]$ is :
  • ✓
    $−1$
  • B
    $−3$
  • C
    $+1$
  • D
    $+3$
Answer
Correct option: A.
$−1$
The oxidation number of colbalt in $\mathrm{K}\left[\mathrm{Co}(\mathrm{CO})_4\right]$ is :
$+1 + x + 0 \times 4 = 0$
$\therefore x = −1.$
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MCQ 971 Mark
To balance the charges which of the following is added to one side of the half $-$ reaction?
  • A
    Proton.
  • B
    Hydrogen.
  • C
    Oxygen.
  • ✓
    Electrons.
Answer
Correct option: D.
Electrons.
Electrons are added to one side of the half $-$ reaction to balance the charge.
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MCQ 981 Mark
In the compounds $\mathrm{KMnO}_4$ and $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7,$ the highest oxidation state is of the element.
  • A
    Potassium
  • B
    Chromium
  • C
    Oxygen
  • ✓
    Manganese
Answer
Correct option: D.
Manganese
In $\mathrm{KMnO}_4​, K$ is in $+1$ oxidation state, $O$ is in $-2$ oxidation state $\text{Mn}$ is present in $+7$ oxidation state.
In $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7​, K$ is in $+1$ oxidation state, $O$ is in $-2$ oxidation state $\text{Cr}$ is present in $+6$ oxidation state.
So $\text{Mn}$ is present in highest oxidation state.
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MCQ 991 Mark
Oxidation number of fluorine in $\ce{F_2O}$​ is :
  • A
    $+1$
  • B
    $+2$
  • ✓
    $−1$
  • D
    $2$
Answer
Correct option: C.
$−1$
Oxidation number of fluorine is always $-1$ as it is most electronegative element.
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MCQ 1001 Mark
The oxidation state of oxygen is maximum in $ .........$
  • A
    Bleaching powder $\left(\mathrm{CaOCl}_2\right)$
  • ✓
    Oxygen difluoride $\left(\mathrm{OF}_2\right)$
  • C
    Dioxygen difluoride $\left(\mathrm{O}_2 \mathrm{~F}_2\right)$
  • D
    Hydrogen peroxide $\left(\mathrm{H}_2 \mathrm{O}_2\right)$
Answer
Correct option: B.
Oxygen difluoride $\left(\mathrm{OF}_2\right)$
The oxidation states of oxygen in bleaching powder, oxygen difluoride, dioxygen difluoride and hydrogen peroxide are $−2, +2, +1$ and $−1$ respectively.
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MCQ 1011 Mark
The brown ring complex compound is formulated as $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5(\mathrm{NO})\right] \mathrm{SO}_4$​. The oxidation state of iron in this complex is :
  • A
    $0$
  • ✓
    $+1$
  • C
    $+2$
  • D
    $+3$
Answer
Correct option: B.
$+1$
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MCQ 1021 Mark
The oxidation state of iodine in $\mathrm{H}_4 \mathrm{IO}_6{ }^{\ominus}$ is :
  • ✓
    $+7$
  • B
    $−1$
  • C
    $+5$
  • D
    $+1$
Answer
Correct option: A.
$+7$
Let $x$ be the oxidation state of $I$ in $\mathrm{H}_4 \mathrm{IO}_6{ }^{\ominus}$​.
Since, the overall charge on the complex is $−1,$ the sum of oxidation states of all elements in it should be equal to $−1.$
Therefore,
$+4 +x + 6(−2) = −1$
or, $x = +7$
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MCQ 1031 Mark
When $\mathrm{SO}_2$ is passed through an acidified solution of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7,$ then chromium sulphate is formed. Change in oxidation state of $\text{Cr}$ is from.
  • ✓
    $+4$ to $+2$
  • B
    $+6$ to $+3$
  • C
    $+7$ to $+2$
  • D
    $+5$ to $+3$
Answer
Correct option: A.
$+4$ to $+2$
In acidic medium $\text{Cr}$ is reduced from $+6$ to $+3$ as :
$\mathrm{Cr}_2 \mathrm{O}_7^{-2}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{+3}+7 \mathrm{H}_2 \mathrm{O}$
and $\mathrm{SO}_2$ is oxidised to $\mathrm{SO}_3$ as :
$\mathrm{SO}_2+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{SO}_3+2 \mathrm{H}^{+}+\mathrm{e}^{-}$
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MCQ 1041 Mark
In alkaline medium, the reaction of hydrogen peroxide with potassium permanganate produces a compound in which the oxidation state of $\text{Mn}$ is :
  • A
    $0$
  • B
    $+2$
  • C
    $+3$
  • ✓
    $+4$
Answer
Correct option: D.
$+4$
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MCQ 1051 Mark
In Daniell cell, electrons flow from :
  • A
    Ccathode to anode.
  • ✓
    Canode to cathode.
  • C
    Copper to zinc.
  • D
    $\text{SO}^-_2$ to $\text{Cu}^{2+}$
Answer
Correct option: B.
Canode to cathode.
In Daniell cell, electrons flow from anode to cathode and current flows from cathode to anode.
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MCQ 1061 Mark
In the reaction $,4 \mathrm{Na}+\mathrm{O}_2 \rightarrow 2 \mathrm{Na}_2 \mathrm{O},$ sodium acts as a/an :
  • A
    Oxidising agent.
  • ✓
    Reducing agent.
  • C
    Complexing agent.
  • D
    None of these.
Answer
Correct option: B.
Reducing agent.
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MCQ 1071 Mark
$ ? + \mathrm{O}_2 \rightarrow 2 \mathrm{~K}_2 \mathrm{O}$
  • A
    $K$
  • B
    $K_2$​
  • C
    $2K$
  • ✓
    $4K$
Answer
Correct option: D.
$4K$
$ ? +\mathrm{O}_2 \rightarrow 2 \mathrm{~K}_2 \mathrm{O}$
Since in the product side there are $4$ atoms of potassium. so, to have a balanced equation, reactant side should also have $4K.$
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MCQ 1081 Mark
Oxidation state of nitrogen is not an integer in :
  • A
    Hydroxyl amine $\left(\mathrm{NH}_2 \mathrm{OH}\right)$
  • B
    Ammonia $\left(\mathrm{NH}_3\right)$
  • C
    Hydrazine $\left(\mathrm{N}_2 \mathrm{H}_4\right)$
  • ✓
    Hydrazoic acid $\left(\mathrm{N}_3 \mathrm{H}\right)$
Answer
Correct option: D.
Hydrazoic acid $\left(\mathrm{N}_3 \mathrm{H}\right)$
In $\ce{N_3H}​,$ the oxidation state of $N$ is $-\frac{1}{3}$​. Hence, it is not an integer.
The oxidation state of $N$ in $\mathrm{NH}_2 \mathrm{OH}, \mathrm{NH}_3$ and $\mathrm{N}_2 \mathrm{H}_4$ are $-1,-3$ and $-2$ respectively.
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MCQ 1091 Mark
A mole of $\ce{N_2​H_4}$​ loses $10$ mol of electrons to form a new compound $Y$. Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in $Y\ ?\ ($There is no change in the oxidation number of hydrogen$)$.
  • A
    $−1$
  • B
    $−3$
  • ✓
    $+3$
  • D
    $+5$
Answer
Correct option: C.
$+3$
As given,
$\mathrm{N}_2 \mathrm{H}_4 \rightarrow 2 \mathrm{y}+10 \mathrm{e}^{-}$
Let $x$ be the oxidation state of $N$ in $\mathrm{N}_2 \mathrm{H}_4$.
Since, the overall charge on the complex is $0,$ the sum of oxidation states of all elements in it should be equal to $0.$
$2x + 4 = 0$
$2y = 2x,$ by replacing $y,$ we get
$2x + 4 = 10$
or $, x = 3$
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MCQ 1101 Mark
The oxidation number of $\text{Mn}$ is $+8$ in the compound:
  • A
    $ \mathrm{K}_2 \mathrm{MnO}_4$
  • ✓
    $ \mathrm{MnO}_4 $
  • C
    $ \mathrm{KMnO}_4$
  • D
    $ \mathrm{Mn}_2 \mathrm{O}_7 $
Answer
Correct option: B.
$ \mathrm{MnO}_4 $
In $ \mathrm{MnO}_4 $
manganese oxidation number is $+8.$
Oxidation number of oxygen is $−2. \mathrm{MnO}_4 $
​​​​​​​ is neutral molecule so total oxidation number of molecule is zero.
Oxidation number of $\text{Mn} + 4(−2) = 0$
Oxidation number of $\text{Mn} = +8.$
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MCQ 1111 Mark
The oxidation number of Phosphorus in $\mathrm{Mg}_2 \mathrm{P}_2 \mathrm{O}_7$​ is :
  • A
    $+3$
  • B
    $2$
  • ✓
    $+5$
  • D
    $-3$
Answer
Correct option: C.
$+5$
$\text{Mg}$ has o.n. $= +2$ and $O$ has o.n. $= -2$
$( 2 \times 2 ) + 2x + ( 7 \times -2) = 0$
$4 + 2x - 14 = 0$
$2x = 10$
$x = +5\ ($o.n. of $P)$
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MCQ 1121 Mark
Consider the following chemical reaction $\text{MnO}_4^-\text{(aq)}+\text{I}^-\text{(aq)}\xrightarrow{\ \ \ \ \ \ \ }\text{MnO}_2\text{(s)}+\text{I}_2\text{(s)}$ Which of the following reactions is an oxidation half $-$ reaction?
  • A
    $\text{MnO}_4^- \text{(aq)}\xrightarrow{\ \ \ \ \ \ \ \ \ \ } \text{MnO}_2\text{(s) }$
  • ✓
    $\mathrm{I}^{-}(\mathrm{aq}) \rightarrow \mathrm{I}_2(\mathrm{~s})$.
  • C
    Both $(a)$ and $(b).$
  • D
    None of the above.
Answer
Correct option: B.
$\mathrm{I}^{-}(\mathrm{aq}) \rightarrow \mathrm{I}_2(\mathrm{~s})$.
$\text{MnO}^-_4\text{(aq)}+\text{I}^-\text{(aq)}\xrightarrow{\ \ \ \ \ \ \ }\text{MnO}_2+\text{I}_2\text{(s)}$
Oxidation half reaction : $\text{I}^-\text{(aq)}\text{}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ }\text{I}_2\text{(s)}$
Reduction half reaction : $\text{MnO}^-_4\text{(aq)}\xrightarrow{\ \ \ \ \ \ \ }\text{MnO}_2\text{(s)}$
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MCQ 1131 Mark
$\text{xKMnO}_4+\text{NH}_3\xrightarrow{ \ \ \ \ \ \ \ \ \ \ }\text{yKNO}_3$
$+\text{MnO}_3+\text{KOH}+\text{H}_2\text{O}$
  • A
    $x = 4, y = 6$
  • B
    $x = 8, y = 6$
  • C
    $x = 3, y = 8$
  • ✓
    $x = 8, y = 3$
Answer
Correct option: D.
$x = 8, y = 3$
$8\text{KMnO}_4+3\text{NH}_3\xrightarrow{ \ \ \ \ \ \ \ \ \ \ }3\text{KNO}_3$
$+8\text{MnO}_3+5\text{KOH}+2\text{H}_2\text{O}$
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MCQ 1141 Mark
Which of the following metal displacement reaction will not take place and why?
  • ✓
    $\text{Cu}+\text{Mg}^{2+}\xrightarrow{ \ \ \ \ \ }$
  • B
    $\text{Mg}+\text{Cu}^{2+}\xrightarrow{ \ \ \ \ \ }$
  • C
    $\text{Cu}+\text{Ag}^+\xrightarrow{ \ \ \ \ \ }$
  • D
    $\text{Zn}+\text{Cu}^{2+}\xrightarrow{ \ \ \ \ \ }$
Answer
Correct option: A.
$\text{Cu}+\text{Mg}^{2+}\xrightarrow{ \ \ \ \ \ }$
Will not take place because $'\text{Cu}\ '$ is less reactive than $\text{Mg}.$
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MCQ 1151 Mark
Oxidation state of $\ce{Fe}$ in Sodium Nitroprusside is :
  • A
    $+3$
  • B
    $+4$
  • ✓
    $+2$
  • D
    $+1$
Answer
Correct option: C.
$+2$
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MCQ 1161 Mark
The oxidation state of $\text{Cr}$ in $\text{CrO}_5$, is:
  • A
    $+10$
  • B
    $+5$
  • ✓
    $+6$
  • D
    $+3$
Answer
Correct option: C.
$+6$

There are two Peroxide linkage $\text{O}^{2-}_2$
$\because$ Oxidation state is $x - 1 - 1 - 1 - 1 - 2 = 0$
$x = +6$
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MCQ 1171 Mark
Oxidation number of $S$ in $\mathrm{S}_2 \mathrm{O}_3{ }^{2-}$ is :
  • A
    $-2$
  • ✓
    $+2$
  • C
    $+6$
  • D
    $0$
Answer
Correct option: B.
$+2$
Let Oxidation number of $S$ in $\mathrm{S}_2 \mathrm{O}_3{ }^{2-}$ be $x.$
Thus,
$2\text{x} + (-2 \times 3) = -2 $
$2\text{x} -6 = -2$
$2\text{x} = -2 + 6$
$2\text{x} = 4$
$\text{x}=\frac{4}{2}$
$\text{x}=2$
So the oxidation state of sulfur is $+2.$
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MCQ 1181 Mark
The oxidation state of the metal atom is zero in :
  • A
    Ferrous sulphate.
  • B
    Ferric sulphate.
  • C
    Rust.
  • ✓
    Iron carbonyl.
Answer
Correct option: D.
Iron carbonyl.
In iron carbonyl, the oxidation state of metal is zero. In all other given compounds, metal has positive oxidation state.
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MCQ 1191 Mark
If a reaction is carried out in acidic medium then which is used to balance the equation?
  • ✓
    $\mathrm{H}^{+}$ ions.
  • B
    $\mathrm{OH}^{-}$ ions.
  • C
    $\mathrm{H}^{-}$ ions.
  • D
    $\mathrm{O}^{2-}$ ions.
Answer
Correct option: A.
$\mathrm{H}^{+}$ ions.
If a reaction is carried out in acidic medium, $\mathrm{H}^{+}$ ions are used to balance the equation. If it is carried out in basic medium, $\text{OH}^-$ ions are used.
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MCQ 1201 Mark
When a piece of magnesium ribbon is placed in a beaker of copper sulfate solution, which one of the following is most likely to happen?
  • A
    A vigorous reaction immediately takes place, with the release of a lot of heat.
  • B
    The blue colour of the copper sulfate solution gets darker.
  • ✓
    The blue colour of the copper sulfate solution gets lighter.
  • D
    The magnesium catches fire.
Answer
Correct option: C.
The blue colour of the copper sulfate solution gets lighter.
$\mathrm{Mg}+\mathrm{CuSO}_4 \rightarrow \mathrm{MgSO}_4+\mathrm{Cu}$
$($Blue$)$
By adding Magnesium to $\mathrm{CuSO}_4$​, it turns to $\mathrm{MgSO}_4$​.
Because $\text{Mg}$ is above $\text{Cu}$ in reactivity series.
So $\text{Mg}$ displaces $\text{Cu}$ in $\mathrm{CuSO}_4$​.
As reaction processes, $\mathrm{CuSO}_4$ content decreases.
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MCQ 1211 Mark
Find the oxidation number of $V$ in $\mathrm{Rb}_4 \mathrm{Na}\left[\mathrm{HV}_{10} \mathrm{O}_{28}\right].$
  • ✓
    $+5$
  • B
    $+2$
  • C
    $-5$
  • D
    None of these
Answer
Correct option: A.
$+5$
$H(1) + 1 + 1 + 10x − 2(28) = 0$
$6 + 10x − 56 = 0$
$10x = 50$
$x = +5$
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MCQ 1221 Mark
A balanced chemical equation is in accordance with :
  • A
    Avogadro's Law.
  • B
    Law of constant proportion.
  • ✓
    Law of conservation of mass.
  • D
    Law of gaseous volumes.
Answer
Correct option: C.
Law of conservation of mass.
The Law of Conservation Of Mass states that matter can be changed from one form to another and mixtures can be prepared or pure substances can be decomposed but the total amount of mass remains constant. In a chemical equation the number of moles of reactants is equal to the number of moles of products irrespective of the type of reaction the total moles in equation remains constant.
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MCQ 1231 Mark
Which one of the following substances is a good oxidising agent?
  • A
    Coke.
  • B
    Water.
  • ✓
    Hydrogen peroxide.
  • D
    Sulphur dioxide.
Answer
Correct option: C.
Hydrogen peroxide.
A good oxidizing agent is one which can readily oxidize other chemical species and reduce itself.
Therefore, the compound which is good oxidizing agent must have initial oxidation as high, so that it can reduce its oxidation state and get reduced and oxidize others.
Coke : Reducing agent
Hydrogen peroxide $ \ce{(H_2​O_2)}$ : because oxidation state of oxygen is $-1$ in hydrogen peroxide.
So, it oxidizes to $0$ acts as a reducing agent it reduces to $-2$ acts as an oxidizing agent.
But, generally, it acts as an oxidizing agent because a stable oxidation state of oxygen is $-2\ ($oxide form$)$
$\ce{H_2​O}$ : Reducing agent.
$\ce{SO_2}$ : sulfur in the $+4$ oxidation state, sulfur dioxide is a reducing agent.
It is oxidized by halogens to give the sulfuryl halides, such as sulfuryl chloride.
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MCQ 1241 Mark
Number of moles of $\text{MnO}_4^-$ required to oxidise one mole of ferrous oxalate completely in acidic medium will be :
  • ✓
    $0.6$ moles.
  • B
    $0.4$ moles.
  • C
    $7.5$ moles.
  • D
    $0.2$ moles.
Answer
Correct option: A.
$0.6$ moles.
$[\text{MnO}^-_4+8\text{H}+5\text{C}^-\xrightarrow{ \ \ \ \ \ \ \ }\text{Mn}^{2+}+3\text{H}_2\text{O}]\times3$
$\text{Fe}^{2+}\xrightarrow{ \ \ \ \ \ \ \ \ \ \ }\text{Fe}^{3+}+\text{e}^-$
$\ \text{COO}^- \xrightarrow{ \ \ \ \ \ \ \ \ }2\text{CO}_2+2\text{e}^-$
$\ |$
$\text{COO}^- $
$[\text{Fe}^{2+}+\ \text{COO}^- \xrightarrow{ \ \ \ \ \ \ \ \ }\text{Fe}^{3+}+2\text{CO}_2+3\text{e}]\times5$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{COO}^- $
$24\text{H}^+3\text{MnO}_4^-+5\text{Fe}\text{C}_2\text{O}_4$
$\xrightarrow{ \ \ \ \ \ \ \ \ \ \ \ }3\text{Mn}^{2+}+5\text{Fe}^{3+}+5\text{CO}_2+12\text{H}_2\text{O}$
$5$ moles of $ \mathrm{FeC}_2 \mathrm{O}_4 $ is getting oxidising by $ 3 $ mole of $ \mathrm{KMnO}_4$
$1$ mole of $ \mathrm{FeC}_2 \mathrm{O}_4$ is getting oxidising by $\frac{3}{5}=0.6\text{ moles}$
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MCQ 1251 Mark
Choose the correct explanation regarding half $-$ reaction such as $\text{Cr}_2\text{O}^{2-}_7\xrightarrow{\ \ \ \ \ \ \ \ }\text{Cr}^{3+}$ from the following.
  • A
    It is oxidation half $-$ reaction.
  • B
    Chromium being oxidized.
  • C
    $\text{Cr}_2\text{O}^{2-}_7$ is a good reducing agent.
  • ✓
    Chromium being reduced.
Answer
Correct option: D.
Chromium being reduced.
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MCQ 1261 Mark
In the reaction, $2 \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3+\mathrm{I}_2 \rightarrow \mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6+2 \mathrm{NaI}, \mathrm{I}_2$ acts as :
  • ✓
    Oxidising agent.
  • B
    Reducing agent.
  • C
    Oxidising as well as reducing agent.
  • D
    None of the above.
Answer
Correct option: A.
Oxidising agent.
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MCQ 1271 Mark
In $\mathrm{FeCr}_2 \mathrm{O}_4$ the oxidation numbers of $\text{Fe}$ and $\text{Cr}$ are :
  • ✓
    $+2$ and $+3$
  • B
    $0$ and $+2$
  • C
    $+2$ and $+6$
  • D
    $+3$ and $+6$
Answer
Correct option: A.
$+2$ and $+3$
$\text{FeCr}_2\text{​O}_4​$
$\text{Fe: x} + 6 + (4 \times −2) = 0$
$\Rightarrow \text{x}=8−6$
$\Rightarrow \text{x}=+2.$
$\text{Cr}:2+2\text{x}+(4\times −2)=0$
$\Rightarrow 2\text{x}=8−2$
$\Rightarrow \text{x}=\frac{6}{2}$
$\Rightarrow \text{x}=+3.$
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MCQ 1281 Mark
Identify which of the following reactions is/ are combustion reactions :
  • A
    $\text{C}+\text{O}_2\xrightarrow{\ \ \Delta\ \ }\text{CO}_2$
  • B
    $\text{H}_2+\text{Cl}_2\xrightarrow{\ \ \Delta \ \ }2\text{HCl}$
  • C
    $\text{CH}_4+\text{O}_2\xrightarrow{\ \ \ \Delta\ \ \ }\text{CO}_2+\text{H}_2\text{O}$
  • ✓
    Both $(a)$ and $(c)$
Answer
Correct option: D.
Both $(a)$ and $(c)$
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MCQ 1291 Mark
Water molecule is formed by the reaction, $2 \mathrm{H}_2+\mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}_2$ What does happen in this reaction?
  • ✓
    Electrons are transferred from $H$ to $O-$ atom.
  • B
    Electrons are transferred from $0$ to $H-$ atom.
  • C
    Electrons are accepted by $H$ from $O-$ atom.
  • D
    Electrons are donated by $O$ to $H-$ atom.
Answer
Correct option: A.
Electrons are transferred from $H$ to $O-$ atom.
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MCQ 1301 Mark
The oxidation number of sulphur in $\mathrm{H}_2 \mathrm{SO}_4$ is:
  • A
    $4$
  • B
    $2$
  • ✓
    $6$
  • D
    Zero
Answer
Correct option: C.
$6$
For a compound $\mathrm{H}_2 \mathrm{SO}_4$,
As net charge on compound is $0$
Charge on: $H = 1$
$S = x$
$O = −2$
$1(2) + x + (−2) (4) = 0$
$2 + x − 8 = 0$
$x = 6$
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MCQ 1311 Mark
What is the sum of oxidation number of various elements in $\mathrm{HCO}_3{ }^{\ominus}​ ($bicarbonate$)$ ion?
  • ✓
    $−1$
  • B
    $0$
  • C
    $+1$
  • D
    None of these
Answer
Correct option: A.
$−1$
The sum of oxidation number of various elements in $\text{HCO}^-_3 ($bicarbonate$)$ ion is $1 + 4 + 3(−2) = −1.$
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MCQ 1321 Mark
From the given species such as $Li, K, Ca$ and $Na$, which of the following is the strongest reducing agent?
  • A
    $Na$
  • ✓
    $Li$
  • C
    $Ca$
  • D
    $K$
Answer
Correct option: B.
$Li$
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MCQ 1331 Mark
The oxidation number of $Cl$ in $\mathrm{CaOCl}_2$ is:
  • ✓
    $−1$ and $+1$
  • B
    $+2$
  • C
    $−2$
  • D
    None
Answer
Correct option: A.
$−1$ and $+1$
In the given compound $\mathrm{CaOCl}_2$ , one $Cl$ is directly attached to $Ca$ and other to $O$.
$ \mathrm{OCl}^{\ominus}:-2+\mathrm{x}=-1 $
$\Rightarrow \mathrm{x}=+1$
$ \mathrm{Cl}^{\ominus}:-1 $
Oxidation states of $Cl$ are $−1$ and $+1$.
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MCQ 1341 Mark
It is found that $V$ forms a double salt isomorphous with Mohr's salt. The oxidation number of $V$ in this compound is $.........$
  • A
    $3$
  • ✓
    $+2$
  • C
    $+4$
  • D
    $−4$
Answer
Correct option: B.
$+2$
Double salt of $V$ isomorphous with Mohr's salt is $\left(\mathrm{NH}_4\right) 2 \mathrm{~V}\left(\mathrm{SO}_4\right)_2 \cdot 6 \mathrm{H}_2 \mathrm{O}$.
When double salt dissolves in water it dissociates to give $\mathrm{NH}_4{ }^{+}, \mathrm{V}^{+2}$ and $\mathrm{SO} 4-2$.
So oxidation state of vanadium is $+2$.
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MCQ 1351 Mark
The oxidation number of chlorine in $\ce{NaClO_3}$ is:
  • A
    $−5$
  • B
    $+2$
  • ✓
    $+5$
  • D
    $−3$
Answer
Correct option: C.
$+5$
The overall charge of molecule is zero $($because it's not an ion$)$. Since sodium is a $1A$ family member, you can assume that the charge is $+1$. The charge of oxygen is almost always $−2$ so you can assume that as well.
$(+1) +$ oxidation of $Cl + 3(−2) = 0$
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MCQ 1361 Mark
The lowest possible oxidation state of nitrogen is $−3$ as in $N^{-3}$.
  • A
    True
  • B
    False
  • C
    Ambiguous
  • ✓
    None of these
Answer
Correct option: D.
None of these
The lowest possible oxidation state of nitrogen is $−3$ as in $N^{-3}$.
Nitrogen can form compounds in which oxidation state ranges from $−3$ to $+5$.
Ammonia, $\ce{NH_3}$ and magnesium nitride, $\ce{Mg_3​N_2}$ have $N$ in $−3$ oxidation state.
$N$ has $5$ valence electrons. It accepts $3$ electrons to complete its octet.
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MCQ 1371 Mark
In the following reaction, $2 \mathrm{Mg}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{Mgo}(\mathrm{s})$ with respect to $Mg$, the process is called :
  • ✓
    Oxidation.
  • B
    Reduction.
  • C
    Redox reaction.
  • D
    None of these.
Answer
Correct option: A.
Oxidation.
Oxidation is a process, which involves addition of oxygen$/$ electronegative element to a substance or removal of hydrogen$/$ electropositive element from a substance. e.g.,

Addition of oxygen $=$ oxidation
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MCQ 1381 Mark
Which of the following arrangements represent increasing oxidation number of the central atom?
  • ✓
    $\text{CrO}^{-}_2,\text{ClO}^-_3,\text{CrO}^{2-}_4,\text{MnO}^-_4$
  • B
    $\text{ClO}^-_3,\text{CrO}^{2-}_4,\text{MnO}^-_4,\text{CrO}^{-}_2$
  • C
    $\ce{{CrO}^-_2,{ClO}^-_3,{MnO}^-_4,{CrO}^{2-}_4}$
  • D
    $\ce{{CrO}^{2-}_4,{MnO}^-_4,{CrO}_2^-,{ClO}^-_3}$
Answer
Correct option: A.
$\text{CrO}^{-}_2,\text{ClO}^-_3,\text{CrO}^{2-}_4,\text{MnO}^-_4$
Writing the $O.N$. of $Cr, C$l and $Mn$ each species in the four set of ions,
we have,
  1. $\stackrel{+3}{\text{Cr}}\text{O}^-_2,\stackrel{+5}{\text{Cl}}\text{O}^-_3,\stackrel{+6}{\text{Cr}}\text{O}^{2-}_4,\stackrel{+7}{\text{Mn}}\text{O}^-_4$
  2. $\stackrel{+5}{\text{Cl}}\text{O}^-_3,\stackrel{+6}{\text{Cr}}\text{O}^{2-}_4,\stackrel{+7}{\text{Mn}}\text{O}^-_4,\stackrel{+3}{\text{Cr}}\text{O}^-_2$
  3. $\stackrel{+3}{\text{Cr}}\text{O}^-_2,\stackrel{+5}{\text{Cl}}\text{O}^-_3,\stackrel{+7}{\text{Mn}}\text{O}^-_4,\stackrel{+6}{\text{Cr}}\text{O}^{2-}_4$
  4. $\stackrel{+6}{\text{Cr}}\text{O}^{2-}_4,\stackrel{+7}{\text{Mn}}\text{O}^-_4,\stackrel{+3}{\text{Cr}}\text{O}^-_2,\stackrel{+5}{\text{Cl}}\text{O}^{3-}_3$
Only in arrangement $(a)$, the $O.N$. of central atom increases from left to right. Therefore, option $(a)$ is correct.
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MCQ 1391 Mark
$\mathrm{H}_2 \mathrm{SO}_5+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O}_2$ Oxidation number of sulphur in $\mathrm{H}_2 \mathrm{SO}_5$ in the above reaction is:
  • ✓
    $6$
  • B
    $4$
  • C
    $5$
  • D
    $2$
Answer
Correct option: A.
$6$
Formation of $\mathrm{H}_2 \mathrm{O}_2$ indicates that there is one peroxy linkage in $\mathrm{H}_2 \mathrm{SO}_5$​. The oxidation state of $H = +1.$
Therefore, $x + 2 − 2 − 6 = 0$ or, $x = +6$
Hence, $x = 6.$
Thus, oxidation number of sulphur $= +6.$
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MCQ 1401 Mark
In the balanced chemical equation : $\text{IO}_3+\text{aI}^-+\text{bH}^+\xrightarrow{ \ \ \ \ \ }\text{cH}_2\text{O}+\text{dI}_2\ a, b, c, d$ respectively are:
  • ✓
    $5, 6, 3, 3$
  • B
    $5, 3, 6, 3$
  • C
    $3, 5, 3, 6$
  • D
    $5, 6, 5, 5$
Answer
Correct option: A.
$5, 6, 3, 3$
$\text{IO}_3+\text{aI}^-+\text{bH}^+\xrightarrow{ \ \ \ \ \ \ \ }\text{cH}_2\text{O}+\text{dI}_2$
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MCQ 1411 Mark
Negative $ER$ indicates that redox couple is:
  • A
    Weaker reducing agent than $H^+/ H$, couple.
  • ✓
    Stronger reducing agent than $H^+/ H,$ couple.
  • C
    Stronger oxidising agent than $H^+/ H,$ couple.
  • D
    None of the above.
Answer
Correct option: B.
Stronger reducing agent than $H^+/ H,$ couple.
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MCQ 1421 Mark
Identify the pair of binary corresponds in which nitrogen exhibits the lowest and the highest oxidation state.
  • A
    $\mathrm{NH}_3, \mathrm{NO}_2$
  • ✓
    $\mathrm{N}_3 \mathrm{H}, \mathrm{N}_2 \mathrm{O}_5$
    ​
  • C
    $\mathrm{N}_2, \mathrm{HNO}_3$
    ​
  • D
    $\mathrm{N}_2 \mathrm{O}, \mathrm{N}_2 \mathrm{O}_5$
Answer
Correct option: B.
$\mathrm{N}_3 \mathrm{H}, \mathrm{N}_2 \mathrm{O}_5$
​
Oxidation of nitrogen in $N_3​H$ is $-\frac{1}{3},$ which is the lowest possible.
Oxidation of nitrogen in $N_2​O_5$​ is $+5$ which is the highest possible.
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MCQ 1431 Mark
When $KMnO_4$​ is reduced with oxalic acid in acidic solution, the oxidation number of $Mn$ changes from.
  • ✓
    $7$ to $2$
  • B
    $7$ to $4$
  • C
    $7$ to $6$
  • D
    $6$ to $2$
Answer
Correct option: A.
$7$ to $2$
In acidic medium it reacts as follows : $\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{+2}+4 \mathrm{H}_2 \mathrm{O}$
So oxidation state of $Mn$ changes from $+7$ to $+2.$
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MCQ 1441 Mark
The oxidation state of manganese in $\mathrm{K}_2 \mathrm{MnO}_4$​ is:
  • A
    $+5$
  • B
    $+6$
  • C
    $+4$
  • ✓
    $+7$
Answer
Correct option: D.
$+7$
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MCQ 1451 Mark
The oxidation number of chlorine is maximum in $.........$
  • A
    $\ce{HOCl}$
  • B
    $ \mathrm{Cl}_2 \mathrm{O}_6 $
    ​
  • ✓
    $ \mathrm{KClO}_4 $
    ​
  • D
    $\mathrm{NaClO}_3 $
Answer
Correct option: C.
$ \mathrm{KClO}_4 $
​
The oxidation numbers of chlorine in $\ce{HOCl},\   \mathrm{Cl}_2 \mathrm{O}_6  ,\  \mathrm{KClO}_4 $
and $\ce{NaClO3}$​ are $+1, +6, +7$ and $+5$ respectively. Hence, the oxidation number of chlorine is maximum in $ \mathrm{KClO}_4 $
​​​​​​​​.
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MCQ 1461 Mark
In which of the following compounds, an element exhibits two different oxidation states:
  • A
    $\mathrm{NH}_2 \mathrm{OH} $
  • ✓
    $ \mathrm{NH}_4 \mathrm{NO}_3 $
  • C
    $ \mathrm{~N}_2 \mathrm{H}_4 $
  • D
    $ \mathrm{~N}_3 \mathrm{H}$
Answer
Correct option: B.
$ \mathrm{NH}_4 \mathrm{NO}_3 $
$ \mathrm{NH}_4 \mathrm{NO}_3 $ is an ionic compound consisting of $\mathrm{NH}_4{ }^{+}$ and $\mathrm{NO}_3{ }^{-}$ ion.
The oxidation number of $N$ in two species is different as shown below:
In $NH$;,
Let the oxidation state of $N$ in $\mathrm{NH}_4{ }^{+}$ be $x$.
$x + 4 \times (+l) = +l$
$x = -3$
In $\mathrm{NO}_3{ }^{-}$
Let the oxidation state of N in $\mathrm{NO}_3{ }^{-}$ be $y$,
$y + 3 \times (-2) = -1$
$y = +5$
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MCQ 1471 Mark
The oxidation number of sulphur in $\mathrm{S}_8, \mathrm{~S}_2 \mathrm{~F}_2$ and $\mathrm{H}_2 \mathrm{S}$ respectively are $.........$
  • ✓
    $0, +1$ and $−2$
  • B
    $+2, +1$ and $−2$
  • C
    $0, +1$ and $+2$
  • D
    $−2, +1$ and $−2$
Answer
Correct option: A.
$0, +1$ and $−2$
Oxidation number of sulphur in $S_8$​ is $0$.
Oxidation number of $F$ is $-1$ so oxidation number of sulphur in $\mathrm{S}_2 \mathrm{F}_2$ is $+1$.
Oxidation number of $H$ is $+1$ so oxidation number of sulphur in $\mathrm{H}_2 \mathrm{S}$ is $-2$.
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MCQ 1481 Mark
The oxidation state of $C$ in diamond is:
  • ✓
    $0$
  • B
    $+1$
  • C
    $−1$
  • D
    $+2$
Answer
Correct option: A.
$0$
Carbon in diamond is in elemental state, so the oxidation state of $C$ in diamond is zero.
The structure of diamond has been shown below:
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MCQ 1491 Mark
Using the standard electrode potential, find out the pair between which redox reaction is not feasible: $\text{E}^{\ominus}\text{values:}$ $\text{Fe}^{3+}/\text{ Fe}^{2+}=+0.77;\text{I}_2/\text{I}^{-}=+0.54;$$\text{Cu}^{2+}/\text{Cu}=+0.34;\text{Ag}^+/\text{Ag}=+0.80\text{V}$
  • A
    $\text{Fe}^{3+}$ and $\text{ I}^-$
  • B
    $\ce{Ag}$ and $\ce{Cu}$
  • C
    $\text{Fe}^{3+}$ and $\text{Cu}$
  • ✓
    $\text{Ag}$ and $\ce{{Fe}^{3+}}$
Answer
Correct option: D.
$\text{Ag}$ and $\ce{{Fe}^{3+}}$
Using the standard electrode potential, find out the pair between which redox reaction is not feasible.
  1. $2\text{Fe}^{3+}+2\text{e}^-\xrightarrow[]{\ \ \ \ \ \ \ \ \ \ \ }2\text{Fe}^{2+};\ \text{E}^0=+0.77\text{V}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\text{I}^-\xrightarrow[]{\ \ \ \ \ \ \ \ \ \ \ }\text{I}_2+2\text{e}^-;\text{E}^0=-0.54\text{V (sign of E}^{\text{0}}\text{is reversed})\\\underline{\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\\ \ \ \ \ \ \ 2\text{Fe}^{3+}2\text{I}^-\xrightarrow[]{\ \ \ \ \ \ \ \ \ \ \ \\ }2\text{Fe}^{2+}+\text{I}_2;\text{E}^0_{\text{cell}}=+0.23\text{V}\\ \text{This reaction is feasible since E}^0_\text{cell}=+0.23\text{V}$
  2. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Cu}\xrightarrow[]{\ \ \ \ \ \ \ \ \ \ \ }\text{Cu}^{2+}+2\text{e}^-;\text{E}^0=-0.34\text{V(sing of E}^0\text{is reversed)}\\\ 2\text{Ag}^++2\text{e}^-\xrightarrow[]{\ \ \ \ \ \ \ \ \ \ \ }2\text{Ag};\text{E}^0=+0.80\text{V}\\\underline{\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\\ \ \text{Cu}+2\text{Ag}^+\xrightarrow[]{\ \ \ \ \ \ \ \ \ \ \ }\text{Cu}^{2+}+2\text{Ag;}\text{E}^0=+0.46\text{V} \\ \text{This reaction is feasible since E}^0_\text{cell}=+\text{Ve.}$
  3. $2\text{Fe}^{3+}+2\text{e}^-\xrightarrow[]{\ \ \ \ \ \ \ \ \ \ \ }2\text{Fe}^{2+};\ \text{E}^0=+0.77\text{V}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Cu}\xrightarrow[]{\ \ \ \ \ \ \ \ \ \ \ }\text{Cu}^{2+}+2\text{e}^-;\text{E}^0=-0.34\text{V (sign of E}^{\text{0}}\text{is reversed})\\\underline{\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\\ \ \ \ \ \ \ 2\text{Fe}^{3+}\text{Cu}\xrightarrow[]{\ \ \ \ \ \ \ \ \ \ \ \\ }2\text{Fe}^{2+}+\text{Cu}^{2+};\text{E}^0=+0.43\text{V}\\ \text{This reaction is feasible since E}^0_\text{cell}=+0.43\text{V}$
  4. $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Ag}\xrightarrow[]{\ \ \ \ \ \ \ \ \ \ \ }\text{Ag}^+\text{e}^-;\text{E}^0=-0.80\text{V(sing of E}^0\text{is reversed}\\\ \text{Fe}^{3+}+2\text{e}^-\xrightarrow[]{\ \ \ \ \ \ \ \ \ \ \ }\text{Fe}^{2+};\text{E}^0+0.77\text{V}\\\underline{\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\\ \ \ \ \text{Ag}+\text{Fe}^{3+}\xrightarrow[]{\ \ \ \ \ \ \ \ \ \ \ \\ }\text{Ag}^+\text{Fe}^{2+};\text{E}^0=-0.03\text{V}\\ \text{This reaction is not feasible since E}^0_\text{cell}=-\text{ve.}$
Thus, option $(d)$ is correct.
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MCQ 1501 Mark
Oxidation number of $Cl$ in $\ce{CaOCl_2}$ is $.........$
  • ✓
    $−1$ and $+1$
  • B
    $+2$
  • C
    $−2$
  • D
    None of these
Answer
Correct option: A.
$−1$ and $+1$
In $\ce{Ca(OCl)Cl}$, two $Cl$ atoms are in different oxidation state i.e., one $Cl^-$ in $-1$ oxidation state and other as $\ce{OCl^-}$ in $+1$ oxidation state.
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MCQ 1511 Mark
Which are of the following can act as oxidising as well reducing agent?
  • A
    $H_2$
  • B
    $I_2$
  • C
    $\ce{H_2O_2}$
  • ✓
    All of these
Answer
Correct option: D.
All of these
All of them can act as oxidising as well as reducing agent because their oxidation state can increase as well as decrease.
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MCQ 1521 Mark
he ratio of oxygen atom having $−2$ and $−1$ oxidation numbers in $\text{S}_2\text{O}^{2-}_8$​ is $.........$
  • A
    $1$
  • B
    $2$
  • ✓
    $3$
  • D
    $4$
Answer
Correct option: C.
$3$
The ratio of oxygen atom having $−2$ and $−1$ oxidation numbers in $\text{S}_2\text{O}^{2-}_8$​ is three as only one peroxy linkage is present.
So, we can see from the structure below that the number of oxygen atoms having $−2$ oxidation state is $6$ while those having $−1$ oxidation state is $2$.
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MCQ 1531 Mark
$\mathrm{KMnO}_4$ and $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$​ are $.........$ agents and $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_7$​ is $.........$ agent.
  • A
    Reducing, oxydising.
  • ✓
    Oxydising, reducing.
  • C
    Reducing, reducing.
  • D
    Oxydising, oxydising.
Answer
Correct option: B.
Oxydising, reducing.
An oxidizing agent $($oxidant, oxidizer$)$ is a substance that has the ability to oxidize other substances $($cause them to lose electrons$)$.
$\mathrm{KMnO}_4$, $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$​ 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.
$\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ is a reducing agent.
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MCQ 1541 Mark
Which of the following statement$(s)$ is$/$ are not true about the following decomposition reaction :
$2\text{KClO}_3\xrightarrow{ \ \ \ \ \ \ \ }2\text{KCl}+3\text{O}_2$
  • A
    Potassium is undergoing oxidation.
  • B
    Chlorine is undergoing oxidation.
  • C
    Oxygen is reduced.
  • ✓
    None of the species are undergoing oxidation or reduction.
Answer
Correct option: D.
None of the species are undergoing oxidation or reduction.
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MCQ 1551 Mark
A half cell reaction is one which:
  • A
    Involves only half a mole of electrolyte.
  • B
    Goes only half way to completion.
  • ✓
    Takes place at one electrode.
  • D
    Consumes half a unit of electricity.
Answer
Correct option: C.
Takes place at one electrode.
In an electrochemical cell, a redox reaction occurs. Oxidation occurs at one electrode and reduction occurs at another electrode. The reaction occuring at each electrode is known as half cell reaction.
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MCQ 1561 Mark
The oxidation state of phosphorus is maximum in $.........$
  • A
    Phospine $\left(\mathrm{PH}_3\right)$.
  • B
    Diphosphine $\left(\mathrm{P}_2 \mathrm{H}_4\right)$.
  • ✓
    Metaphosphoric acid $\left(\mathrm{HPO}_3\right)$.
  • D
    Phosphorus acid $\left(\mathrm{H}_3 \mathrm{PO}_3\right)$.
Answer
Correct option: C.
Metaphosphoric acid $\left(\mathrm{HPO}_3\right)$.
The oxidation states of phosphorus in phosphine, diphosphine, metaphosphoric acid and phosphorus acid are $−3, −2, +5$ and $+3$ respectively. Hence, it is maximum in metaphosphoric acid.
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MCQ 1571 Mark
Oxidation number of carbon in $\ce{(CN)_2}$​ is $.........$
  • A
    $1$
  • B
    $2$
  • ✓
    $3$
  • D
    $4$
Answer
Correct option: C.
$3$
Let $x$ be the oxidation number of $C$ in $\ce{(CN)_2}$​.
Since the overall
Charge on the compound is $0$, the sum of oxidation states of all elements in it should be equal to $0$.
Therefore, $2(x + (−3)) = 0$ or, $x = +3$
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MCQ 1581 Mark
The oxidation number of carbon in $\mathrm{CH}_2 \mathrm{Cl}_2$​ is
  • ✓
    $0$
  • B
    $2$
  • C
    $3$
  • D
    $5$
Answer
Correct option: A.
$0$
Let $x$ be the oxidation state of $C$ in $\mathrm{CH}_2 \mathrm{Cl}_2$​.
Since the overall charge on the complex is $0$, the sum of oxidation states of all elements in it should be equal to $0$.
Therefore, $x + 2 + 2(−1) = 0$
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MCQ 1591 Mark
Thiosulphate reacts differently with iodine and bromine in the reactions given below: $\ce{2 {S}_2{O}^{2-}_3 + {I}\xrightarrow{ \ \ \ \ \ \ \ } {S}_4{O}^{2-}_6 + 2 {I}^{-}{S}_2{O}^{2-}_3 + 2{Br}_2 + 5 {H}_2{O}\xrightarrow{ \ \ \ \ \ \ \ }2{SO}^{2-}_4 + 2 {Br}^{-} + 10 {H}^{+}}$ Which of the following statements justifies the above dual behaviour of thiosulphate?
  • ✓
    Bromine is a stronger oxidant than iodine.
  • B
    Bromine is a weaker oxidant than iodine.
  • C
    Thiosulphate undergoes oxidation by bromine and reduction by iodine in these reactions.
  • D
    Bromine undergoes oxidation and iodine undergoes reduction in these reactions.
Answer
Correct option: A.
Bromine is a stronger oxidant than iodine.
$\ce{2{S}_2{O}^{2-}_3({aq}) + {I}_2 {(s)}\xrightarrow{ \ \ \ \ \ \ \ \ \ } {S}_4{O}^{2-}_6({aq}) + 2{I}^{-}({aq})$
${S}_2{O}^{2-}_3({aq}) + 2{Br}({l}) + 5{H}_2 {O}({l})\xrightarrow{ \ \ \ \ \ \ \ \ \ }2{SO}^{2-}_4({Aq}) + 4 {Br}^{-}({aq}) + 10 {H}^{+}({aq})}$
Bromine being stronger oxidizing agent than $I_2$, it oxidises S of $\mathrm{S}_2  \mathrm{O}^{2-}{ }_3$ to $\mathrm{SO}^{2-}{ }_4$ whereas $\mathrm{I}_2$ oxidises it only into $\mathrm{S}_4 \mathrm{O}^{2-}{ }_6$ ion.
 
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MCQ 1601 Mark
The oxidation number of an element in a compound is evaluated on the basis of certain rules. Which of the following rules is not correct in this respect?
  • ✓
    The oxidation number of hydrogen is always $+1$.
  • B
    The algebraic sum of all the oxidation numbers in a compound is zero.
  • C
    An element in the free or the uncombined state bears oxidation number zero.
  • D
    In all its compounds, the oxidation number of fluorine is $– 1$.
Answer
Correct option: A.
The oxidation number of hydrogen is always $+1$.
In ionic hydrides hydrogen exist in $-1$ oxidation state because the hydrogen acquires negative charge in the presence of its companion.
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MCQ 1611 Mark
The oxidation state of chromium in the final product formed in the reaction between $KI$ and acidified potassium dichromate solution is:
  • A
    $+4$
  • B
    $+6$
  • C
    $+2$
  • ✓
    $+3$
Answer
Correct option: D.
$+3$
The balanced reaction equation for reaction between potassium iodide and acidified potassium dichromate is :
$\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+7 \mathrm{H}_2 \mathrm{SO}_4+6 \mathrm{KI} \rightarrow \mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3+3 \mathrm{I}_2+7 \mathrm{H}_2 \mathrm{O}+4 \mathrm{~K}_2 \mathrm{SO}_4$
In this reaction, potassium dichromate gives a chromium sulfate. The oxidation state of chromium in chromium sulfate can be calculated as:
Consider, the oxidation state of $Cr$ be $X$ in $\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3$
$\therefore  2\text{X} + 3(−2) = 0$
$∴ 2\text{X}−6=0$
$∴ 2\text{X}=6$
$\therefore X=\frac{6}{2}$
$∴X=+3$
Therefore, the oxidation state of $Cr$ in the final product is $+3$.
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MCQ 1621 Mark
In which one of the following compounds does oxygen have an oxidation number of $+2$ ?
  • ✓
    $ \mathrm{F}_2 \mathrm{O} $
  • B
    $ \mathrm{Cl}_2 \mathrm{O} $
  • C
    $ \mathrm{Na}_2 \mathrm{O}_2 $
  • D
    $ \mathrm{Na}_2 \mathrm{O} $
Answer
Correct option: A.
$ \mathrm{F}_2 \mathrm{O} $
Flourine is more electronegative than oxygen which in turn is more electronegative than $Cl$ and $Na$.
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MCQ 1631 Mark
In which of the following, iron is present in its lowest oxidation state?
  • A
    $ \mathrm{FeSO}_4 .7 \mathrm{H}_2 \mathrm{O} $
  • ✓
    $ \mathrm{Fe}(\mathrm{CO})_5$
    ​
  • C
    $ \mathrm{Fe}_{0.94} \mathrm{O} $
  • D
    $ \mathrm{Fe}_3 \mathrm{O}_4 $
Answer
Correct option: B.
$ \mathrm{Fe}(\mathrm{CO})_5$
​
The oxidation state of iron in iron pentacarbonyl is $0$.
In all other compounds, iron has a positive oxidation state.
Thus, $ \mathrm{Fe}(\mathrm{CO})_5$​ has the lowest oxidation state of iron.
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MCQ 1641 Mark
In which of the following compounds the oxidation number of carbon is not zero?
  • A
    $\text{HCHO}$
  • B
    $\mathrm{CH}_3 \mathrm{COOH} $
  • C
    $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} $
  • ✓
    $ \mathrm{CH}_3 \mathrm{CHO}$
Answer
Correct option: D.
$ \mathrm{CH}_3 \mathrm{CHO}$
Oxidation number of $C$ in $ \mathrm{CH}_3 \mathrm{CHO}$ is $-1$, in other it is zero.
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MCQ 1651 Mark
The largest oxidation number exhibited by an element depends on its outer electronic configuration. With which of the following outer electronic configurations the element will exhibit largest oxidation number?
  • A
    $3 d^1 4 s^2 $
  • B
    $ 3 d ^34 s ^2 $
  • C
    $ 3 d^5 4 s^1 $
  • ✓
    $ 3 d^5 4 s^2$
Answer
Correct option: D.
$ 3 d^5 4 s^2$
Highest $O.N$. of any transition element $= (n – 1)d$ electrons $+ns$ electrons. Therefore, larger the number of electrons in the $3d$ orbitals, higher is the maximum $O.N$.
  1. $ 3 d^1 4 s^2=3 $
  2. $ 3 d^2 4 s^2=3+2=5 $
  3. $ 3 d^5 4 s^1=5+1=6 $
  4. $ 3 d^5 4 s^2=5+2=7$
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MCQ 1661 Mark
The oxidation number of nitrogen atoms in $\mathrm{NH}_4 \mathrm{NO}_3$ are:
  • A
    $+3, +3$
  • B
    $+3, −3$
  • ✓
    $−3, +5$
  • D
    $−5, +3$
Answer
Correct option: C.
$−3, +5$
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MCQ 1671 Mark
An element in its native state will have the oxidation number :
  • A
    $+1$
  • B
    $2$
  • ✓
    $0$
  • D
    $−1$
Answer
Correct option: C.
$0$
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MCQ 1681 Mark
What is the oxidation number of $Mg$ and $N$ in magnesium nitride?
  • ✓
    $Mg = +2; N = −3$
  • B
    $Mg = +1; N = −3$
  • C
    $Mg = +2; N = −2$
  • D
    None of these
Answer
Correct option: A.
$Mg = +2; N = −3$
In magnesium nitride $(\mathrm{Mg}_3 \mathrm{~N}_2)$, the oxidation number of $Mg$ and $N$ are $+2$ and $−3$ respectively.
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MCQ 1691 Mark
Identify redox reaction among the following:
  • A
    Acid$-$base neutralization.
  • B
    Precipitation reaction.
  • ✓
    Metal displacement reaction.
  • D
    All the above.
Answer
Correct option: C.
Metal displacement reaction.
Metal displacement reactions are redox reactions in which both oxidation and reduction takes place.Example; $Mg + \mathrm{ZnSO}_4 \rightarrow \mathrm{MgSO}_4+\mathrm{Zn}$
Here $Mg$ is undergoing oxidation with an increase in oxidation number from $0$ to $+2$ and $Zn$ is undergoing reduction with a decrease in oxidation number.
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MCQ 1701 Mark
Which of the following is not a redox reaction?
  • ✓
    $\text{CaCO}_3\xrightarrow{ \ \ \ \ \ \ \ }\text{CaO}+\text{CO}_2$
  • B
    $2\text{H}_2+\text{O}_2\xrightarrow{ \ \ \ \ \ \ }2\text{H}_2\text{O}$
  • C
    $2\text{Na}+2\text{H}_2\text{O}\xrightarrow{ \ \ \ \ \ \ \ }2\text{NaOH}+\text{H}_2$
  • D
    $\text{MnCl}_3\xrightarrow{ \ \ \ \ \ \ }\text{MnCl}_2+\frac{1}{2}\text{Cl}_2$
Answer
Correct option: A.
$\text{CaCO}_3\xrightarrow{ \ \ \ \ \ \ \ }\text{CaO}+\text{CO}_2$
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MCQ 1711 Mark
When magnesium and iron react with steam, they produce:
  • ✓
    $\ce{H_2}$
  • B
    $\ce{O_2}$
  • C
    $\ce{CO_2}$
  • D
    None of these.
Answer
Correct option: A.
$\ce{H_2}$
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