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.
AnswerCorrect option: B. $\mathrm{H}_2 \mathrm{O}_2$ is oxidised.
Oxidation half reactions is,
$\mathrm{H}_2 \mathrm{O}_2 \rightarrow 2 \mathrm{H}^{+}+\mathrm{O}_2+2 \mathrm{e}^{-}$
View full question & answer→MCQ 21 Mark
What is the oxidation number of lithium in $\ce{LiCl}\ ?$
AnswerOxidation number of $\ce{Li}$ in $\ce{LiCl } : x − 1 = 0$
$\Rightarrow x = +1$
View full question & answer→MCQ 31 Mark
In $\mathrm{Ni}(\mathrm{CO})_4,$ the oxidation state of $\ce{Ni}$ is?
AnswerIn 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.
View full question & answer→MCQ 41 Mark
Which of the following elements does not show disproportionation tendency?
- A
$\ce{Cl}$
- B
$\ce{Br}$
- ✓
$\ce{F}$
- D
$\ce{I}$
AnswerCorrect 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.
View full question & answer→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
$2\text{K}+\text{F}_2\xrightarrow{ \ \ \ \ \ \ \ }2\text{KF}$
- ✓
$\text{BaCl}_2+\text{H}_2\text{SO}_4\xrightarrow{ \ \ \ \ \ \ \ \ }\text{BaSO}_4+2\text{HCl}$
AnswerCorrect 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.
View full question & answer→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
$\text{Zn}+\text{Cu}^{2+}\xrightarrow{ \ \ \ \ \ \ \ }$
AnswerCorrect option: A. $\text{Cu}+\text{Mg}^{2+}\xrightarrow{ \ \ \ \ \ \ \ }$
Will not take place because $'\text{Cu}\ '$ is less reactive than $\text{Mg}.$
View full question & answer→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
$\ce{ICl HIO_4,HI, I_2}$
AnswerCorrect option: A. $\ce{HlO_4, ICl, I_2, Hl}$
$\mathrm{HI}(-1), \mathrm{I}_2(\mathrm{O}), \mathrm{ICl}(+1), \mathrm{HIO}_4(+7)$
View full question & answer→MCQ 81 Mark
The formation of nitrous oxide from nitrogen and oxygen is the example for
- A
- B
Chemical combination of one element and one compound.
- C
Chemical combination of two compounds.
- ✓
Chemical combination of two elements.
AnswerCorrect 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.
$\text{N}_2\text{O}\rightleftarrows\text{N}_2+\text{O}.$
$\text{N}_2\text{O}\rightleftarrows\text{N}_2+\text{O}_2$
View full question & answer→MCQ 91 Mark
Oxidation number of $C$ in $\text{HNC}$ is $ ..........$
AnswerOxidation 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 $x,$
$1 + (−3) + x = 0$
$x = +2$
View full question & answer→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
$\text{Hg}_2\text{CrO}_4+2\text{OH}^-\xrightarrow{ \ \ \ \ \ \ \ \ }\text{Hg}_2\text{O}+\text{CrO}_4^{2-}+\text{H}_2\text{O}$
- ✓
$3\text{Br}_2+6\text{OH}^-\xrightarrow{ \ \ \ \ \ \ \ }5\text{Br}^-+\text{BrO}_3^-+3\text{H}_2\text{O}$
AnswerCorrect 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 $-1$.
View full question & answer→MCQ 111 Mark
Plumbous ion is represented as :
- ✓
$ \mathrm{Pb}^{+2}$
- B
$ \mathrm{~Pb}^{+4}$
- C
$ \mathrm{~Pb}^{+3} $
- D
$\mathrm{~Pb}^{+1}$
AnswerCorrect option: A. $ \mathrm{Pb}^{+2}$
$\mathrm{~Pb}^{2+}=$ Plumbous ion
$\mathrm{~Pb}^{4+}=$ Plumbic ion
View full question & answer→MCQ 121 Mark
What is the oxidation state of $\text{Mn}$ in the compound $\ce{K_2MnO_4}\ ?$
View full question & answer→MCQ 131 Mark
In which of the following, the highest oxidation state is not possible?
AnswerCorrect 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.
View full question & answer→MCQ 141 Mark
What is the oxidation number of $O$ in a diatomic molecule $\ce{(O_2)}\ ?$
AnswerThe oxidation state of any element in its native state is zero.
View full question & answer→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 :
View full question & answer→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.
AnswerCorrect option: D. Displacement of hydrogen.
View full question & answer→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
$2\text{F}_2+2\text{OH}^-\xrightarrow{ \ \ \ \ \ \ \ \ }2\text{F}^-+\text{OF}_2+\text{H}_2\text{O}$
- ✓
$2\text{NO}_2+2\text{OH}^-\xrightarrow{ \ \ \ \ \ \ \ \ }\text{NO}^-_2+\text{NO}^-_3+\text{H}_2\text{O}$
AnswerCorrect 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,
- $\stackrel{-4+1}{\text{CH}}_4+\stackrel{0}{\text{2O}}_2\xrightarrow{ \ \ \ \ \ \ \ \ }\stackrel{+4-2}{\text{CO}}_2+\stackrel{+1-2}{\text{2H}}_2\text{O}$
- $\stackrel{-4+1}{\text{CH}}_4+\stackrel{0}{\text{4Cl}}_2\xrightarrow{ \ \ \ \ \ \ \ \ }\stackrel{+4-1}{\text{CCl}}_4+\stackrel{+1-1}{\text{4HCl}}\text{}$
- $2\stackrel{0}{\text{F}}_2+\stackrel{-2+1}{\text{2OH}^-}\xrightarrow{ \ \ \ \ \ \ \ \ }\stackrel{-1}{\text{2F}^-+}\stackrel{+2-1}{\text{OF}_2}+\stackrel{+1-2}{\text{H}_2\text{O}}$
- $\stackrel{+4-2}{\text{2NO}_2}+\stackrel{-2+1}{\text{2OH}^-}\xrightarrow{ \ \ \ \ \ \ \ \ }\stackrel{+3-2}{\text{NO}^-_2}+\stackrel{+5-2}{\text{NO}^-_3}+\stackrel{+1-2}{\text{H}_2\text{O}}$
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 $\mathrm{NO}^{-} 2$. View full question & answer→MCQ 181 Mark
Which among the following shows maximum oxidation state?
Answer
| Metal |
Maximum Oxidation state |
| $V$ |
$+3$ |
| $Cr$ |
$+6$ |
| $Fe$ |
$+3$ |
| $Mn$ |
$+7$ |
View full question & answer→MCQ 191 Mark
What is the oxidation state of central atom in $\ce{Ca[PtCl_4]}\ ?$
Answer$\mathrm{Ca}\left[\mathrm{PtCl}_4\right] \Leftrightarrow \mathrm{Ca}^{+2}[\mathrm{PtCl} 4]^{2-}$
$\text { Take }[\mathrm{PtCl} 4]^{2-} \text {. Central atom }=\mathrm{Pt} \text {. }$
Let $x$ be the oxidation no. of $\ce{Pt},$
$\ce{Cl}$ oxidation no. is $−1.$
$x + 4(−1) = −2$
$x − 4 = −2$
$x = 2$
View full question & answer→MCQ 201 Mark
The oxidation state of the underlined element in the given compound is : $\ce{BaCl_2}$
Answer$\ce{BaCl_2}$
$\Rightarrow x + (−2) = 0$
$\Rightarrow x = 2$
As chlorine needs only one electron to get octet.
View full question & answer→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 $+6$
AnswerCorrect 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
$2(+1) + x + 2(−1) + 3(−2) = 0, x = +6$
View full question & answer→MCQ 221 Mark
In which of the following compounds, is the oxidation number of sulphur is the least?
AnswerCorrect 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 $\ce{H_2SO_4}$ are $+4,+6,+2.5$ and $+6$ respectively.
View full question & answer→MCQ 231 Mark
In the electrochemical series, metals are arranged in order of their tendency to :
- A
Release halogens from their salts.
- B
- C
- ✓
AnswerIn 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$).$
View full question & answer→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 $−2$
AnswerCorrect option: C. $0, +1$ and $+2$
View full question & answer→MCQ 251 Mark
Oxygen has an oxidation state of $+2$ in.
- A
$\ce{H_2O_2}$
- ✓
$\ce{OF_2}$
- C
$\ce{SO_2}$
- D
$\ce{H_2O}$
AnswerCorrect option: B. $\ce{OF_2}$
Oxidation state of oxygen is always $-2$ except in peroxides,superoxides and when it reacts with fluorine.
In $\ce{H_2O_2},$ 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 $\ce{H_2O}$, oxidation state of oxygen is $-2.$
View full question & answer→MCQ 261 Mark
Metals exhibit $........$ oxidation states in their compounds.
AnswerAs 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 $\ce{Na}^+$
View full question & answer→MCQ 271 Mark
Oxidation state of nitrogen in $\ce{NH_2OH}$ is :
AnswerLet $x$ be the oxidation state of $N$ in $\ce{NH_2OH}$.
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, $x = −1$.
View full question & answer→MCQ 281 Mark
The oxidation number of chromium in $\ce{CrO_5}$ is :
AnswerThe oxidation number of chromium in chromium pentaoxide is $6.$
View full question & answer→MCQ 291 Mark
The lowest possible oxidation number for hydrogen is :
MCQ 301 Mark
The sum of oxidation number of all the atoms in a neutral molecule must be zero.
AnswerThe 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 $+1.$
View full question & answer→MCQ 311 Mark
Which titrant is used in the Iodometric titration which involves $\ce{I_2} ?$
AnswerCorrect 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$).$
View full question & answer→MCQ 321 Mark
In $\ce{MgCl_2},$ the oxidation number of chlorine is :
AnswerIn $\ce{MgCl_2},$ oxidation number of $\ce{Cl}$ is :
$\Rightarrow 2 + 2x = 0$
$\Rightarrow x = −1.$
View full question & answer→MCQ 331 Mark
The oxidation state of $\ce{Cr}$ in $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is :
Answer$\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$
Let the oxidation state of $\ce{Cr}$ is $x.$
$2(+1) + 2x + 7(−2) = 0$
$+2 + 2x − 14 = 0$
$2x − 12 = 0$
$2x = 12$
$x = +6$
View full question & answer→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 :
Answer$\mathrm{CH}_2=\mathrm{CH}_2+\mathrm{H}-\mathrm{H} \rightarrow \mathrm{H}_3 \mathrm{C}-\mathrm{CH}_3$
$($Addition of hydrogen$)$
Because of the addition of hydrogen, there occurs reduction of ethylene.
View full question & answer→MCQ 351 Mark
Tailing of mercury is $ .......... $ redox change.
AnswerWhen ozone is passed through mercury, mercurous oxide $\ce{(Hg_2O)}$ 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.
View full question & answer→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.
AnswerCorrect 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.
View full question & answer→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 $+1$
AnswerCorrect 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 $-1 .$
View full question & answer→MCQ 381 Mark
AnswerHeat is released in exothermic reactions.
View full question & answer→MCQ 391 Mark
When $P$ reacts with caustic soda, the products are $\ce{PH_3}$ and $\ce{NaH_2PO_2}$. The reaction is an example of.
- A
- B
- ✓
Both oxidation and reduction.
- D
AnswerCorrect 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 $\ce{NaH_2PO_2}$ it is present in $(+1)$ oxidation state.
View full question & answer→MCQ 401 Mark
The rods of transition metals such as copper and zinc where potential difference is generated, are termed as :
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
$\text{X > Y > Z}$
AnswerCorrect 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.
View full question & answer→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
AnswerCorrect option: C. One substance is reduced and other is oxidised.
In a redox reaction both oxidation and reduction is happening together.
View full question & answer→MCQ 431 Mark
What is the oxidation number of $\ce{Si}$ in the compound $\ce{CaSiO_3} \ ?$
Answer$\ce{CaSiO_3}$
Total charge present $= 0.$
Oxidation no. of Oxygen is $−2.$
Oxidation no. od Calcium is $+2.$
Let, oxidation no. of Slilicon be $X.$
So, $[+2] + x + 3[−2] = 0$
$2 + x − 6 = 0$
$x = 4$
View full question & answer→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 :
AnswerIn $\mathrm{Na}_2\mathrm{~S}_4 \mathrm{O}_6,$ 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 $5 − 0 = 0.$
View full question & answer→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
$\text{III > II > IV > I}$
AnswerCorrect option: C. $\text{I > IV > II > III}$
View full question & answer→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 $\ \text{Cl}^-$
AnswerCorrect 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 $\text{Cr}^{3+}$
View full question & answer→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 $x + y + z$
AnswerThe balanced redox reaction is : $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+\mathrm{xH}_2 \mathrm{SO}_4+\mathrm{3SO}_2 \rightarrow \mathrm{K}_2 \mathrm{SO}_4+\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3+\mathrm{zH}_2 \mathrm{O}$
$x = 1, y = 3, z = 1$
View full question & answer→MCQ 481 Mark
Which of the following compounds we use in our laboratory as a standard solution $($titrant$) \ ?$
AnswerA 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.
View full question & answer→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
- ✓
Both oxidised and reduced.
- C
- D
AnswerCorrect option: B. Both oxidised and reduced.
View full question & answer→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?
AnswerLet oxidation state of $\text{Sn}$ is $x$ and we know oxidation of $\text{Cl}$ is $−1,$ so $x + 6(−1) = −2, x = +4.$
View full question & answer→