Question 11 Mark
Justify that the following reactions are redox reactions:$4BCl_3(g) + 3LiAlH_4(s) → 2B_2H_6(g) + 3LiCl(s) + 3AlCl_3(s)$
Answer$4BCl_{3(g)} + 3LiAlH_{4(s)} → 2B_2H_{6(g)} + 3LiCl_{(s)} + 3AlCl_{3(s)}$The oxidation number of each element in the given reaction can be represented as:
$\stackrel{{+3\ -1}}{4\ \ \ \ \ \hbox{BCl}_{3(\text{g})}}+\stackrel{{+1\ \ +3\ -1}}{3\ \ \ \ \hbox{LiAlH}_{4(\text{s})}}\ \rightarrow\stackrel{{-3}}{2\ \ \hbox{B}_2}\stackrel{{+1}}{\ \ \ \ \ \hbox{H}_{6(\text{g})}}+\stackrel{{+1}}{3\hbox{Li}}\stackrel{{-1}}{\ \ \ \ \hbox{Cl}_{(\text{s})}}+\stackrel{{+3}}{3\hbox{Al}}\stackrel{{-1}}{\ \ \ \ \ \ \hbox{Cl}_{3(\text{s})}}$
In this reaction, the oxidation number of B decreases from +3 in $BCl_3$ to –3 in $B_2H_6.$ i.e., $BCl_3$ is reduced to $B_2H_6$. Also, the oxidation number of H increases from $–1 in LiAlH_4 to +1 in B_2H_6 i.e., LiAlH_4$ is oxidized to $B_2H_6.$ Hence, the given reaction is a redox reaction.
View full question & answer→Question 21 Mark
Write formula for the following compound:
Nickel(II) sulphate.
AnswerNickel(II) sulphate:
$\text{NiSO}_4$
View full question & answer→Question 31 Mark
Write formula for the following compound:
Iron(III) sulphate.
AnswerIron(III) sulphate:
$\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3$
View full question & answer→Question 41 Mark
Consider the elements:
Cs, Ne, I and F
Identify the element that exhibits only negative oxidation state.
AnswerF. Fluorine being the most electronegative element shows only a -ve oxidation state of -1.
View full question & answer→Question 51 Mark
Write formula for the following compound:
Tin(IV) oxide.
AnswerTin(IV) oxide:
$\text{SnO}_2$
View full question & answer→Question 61 Mark
Refer to the periodic table given in your book and now answer the following questions:
Select the possible non metals that can show disproportionation reaction.
AnswerIn disproportionation reactions, one of the reacting substances always contains an element that can exist in at least three oxidation states.
P, Cl, and S can show disproportionation reactions as these elements can exist in three or more oxidation states.
View full question & answer→Question 71 Mark
Write formula for the following compound:
Chromium(III) oxide.
AnswerChromium(III) oxide:
$\mathrm{Cr}_2 \mathrm{O}_3$
View full question & answer→Question 81 Mark
Justify that the following reactions are redox reactions:
$\mathrm{Fe}_2 \mathrm{O}_{3(\mathrm{~s})}+3 \mathrm{CO}_{(\mathrm{g})} \rightarrow 2 \mathrm{Fe}_{(\mathrm{s})}+3 \mathrm{CO}_{2(\mathrm{~g})}$
Answer$\mathrm{Fe}_2 \mathrm{O}_{3(\mathrm{~s})}+3 \mathrm{CO}_{(\mathrm{g})} \rightarrow 2 \mathrm{Fe}_{(\mathrm{s})}+3 \mathrm{CO}_{2(\mathrm{~g})}$
Let us write the oxidation number of each element in the given reaction as:
$\stackrel{+3}{\mathrm{Fe}}_2 \quad \stackrel{-2}{\mathrm{O}}_{3(\mathrm{~s})}+3 \stackrel{+2-2}{\mathrm{CO}}_{(\mathrm{g})} \rightarrow 2 \stackrel{0}{\mathrm{~F} \mathrm{e}_{(\mathrm{s})}}+3 \stackrel{+4-2}{\mathrm{CO}}_{2(\mathrm{~g})}$
Here, the oxidation number of Fe decreases from +3 in $\mathrm{Fe}_2 \mathrm{O}_3$ to 0 in Fe i.e., $\mathrm{Fe}_2 \mathrm{O}_3$ is reduced to Fe . On the other hand, the oxidation number of C increases from +2 in CO to +4 in $\mathrm{CO}_2$ i.e., CO is oxidized to $\mathrm{CO}_2$. Hence, the given reaction is a redox reaction.
View full question & answer→Question 91 Mark
Depict the galvanic cell in which the reaction $\mathrm{Zn}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$ takes place, Further show:
Individual reaction at each electrode.
AnswerThe galvanic cell corresponding to the given redox reaction can be represented as:
$\text{Zn}|\text{Zn}^{2+}_{(\text{aq})}||\text{Ag}^{+}_{(\text{aq})}|\text{Ag}$
The reaction taking place at Zn electrode can be represented as:
$\text{Zn}_{(\text{s})}\rightarrow\text{Zn}^{2+}_{(\text{aq})}+2\text{e}^-$
And the reaction taking place at Ag electrode can be represented as:
$\text{Ag}^+_{(\text{aq})}+\text{e}^-\rightarrow\text{Ag}_{(\text{s})}.$
View full question & answer→Question 101 Mark
Consider the elements: Cs, Ne, I and FIdentify the element that exhibits only postive oxidation state.
AnswerCs. Alkali metals because of the presence of a single electron in the valence shell, exhibit an oxidation state of +1.
View full question & answer→Question 111 Mark
Refer to the periodic table given in your book and now answer the following questions:
Select three metals that can show disproportionation reaction.
AnswerIn disproportionation reactions, one of the reacting substances always contains an element that can exist in at least three oxidation states.
Mn, Cu, and Ga can show disproportionation reactions as these elements can exist in three or more oxidation states.
View full question & answer→Question 121 Mark
Consider the elements:
Cs, Ne, I and F
Identify the element which exhibits neither the negative nor does the positive oxidation state.
AnswerNe. It is an inert gas (with high ionization enthalpy and high positive electron gain enthalpy) and hence it neither exhibits -ve nor +ve oxidation states.
View full question & answer→Question 131 Mark
Write formula for the following compound:
Mercury(II) chloride.
AnswerMercury(II) chloride:
$\mathrm{HgCl}_2$
View full question & answer→Question 141 Mark
Justify that the following reactions are redox reactions:$2K(s) + F_2(g) → 2K^+F^-(s)$
Answer$2K_{(s)} + F_{2(g)} → 2K^+F^-_{(s)}$The oxidation number of each element in the given reaction can be represented as:
$\stackrel{{0}}{2\ \ \text{K}_{(\text{s})}}+\stackrel{{0}}{\ \ \ \ \ \text{F}_{2(\text{g})}}\ \rightarrow\stackrel{{+1}}{2\ \ \text{K}^+}\stackrel{{-1}}{\ \ \ \ \text{F}^-_{(\text{s})}}$ In this reaction, the oxidation number of K increases from 0 in K to +1 in KF i.e., K is oxidized to KF. On the other hand, the oxidation number of F decreases from 0 in $F_2$ to -1 in KF i.e., $F_2$ is reduced to KF. Hence, the above reaction is a redox reaction.
View full question & answer→Question 151 Mark
Write formula for the following compound:
Thallium(I) sulphate.
AnswerThallium(I) sulphate:
$\mathrm{Tl}_2\mathrm{SO}_4$
View full question & answer→Question 161 Mark
Justify that the following reactions are redox reactions:$CuO(s) + H_2(g) → Cu(s) + H_2O(g)$
Answer$CuO_{(s)} + H_{2(g)} → Cu_{(s)} + H_2O_{(g}$ Let us write the oxidation number of each element involved in the given reaction as:
$\stackrel{{+2}\ -2}{\hbox{Cu O}}_{(\text{s})}\ +\stackrel{{0}}{\ \ \ \ \ \ \hbox{H}_{2(\text{g})}}\ \rightarrow\stackrel{0}{\ \ \ \ \hbox{ Cu}_{(\text{s})}}+\stackrel{+1\ \ \ \ -2}{\ \ \ \hbox{H}_2\ \ \text{O}_{(\text{g})}}$ Here, the oxidation number of Cu decreases from +2 in CuO to 0in Cu i.e., CuO is reduced to Cu . Also, the oxidation number of H increases from 0 in $H _2$ to +1 in $H _2 O$ i.e., $H _2$ is oxidized to $H _2 O$. Hence, this reaction is a redox reaction.
View full question & answer→Question 171 Mark
Predict the products of electrolysis in the following:
An aqueous solution $\mathrm{AgNO}_3$ with platinum electrodes.
AnswerPt cannot be oxidized easily. Hence, at the anode, oxidation of water occurs to liberate $\mathrm{O}_2$. At the cathode, $\mathrm{Ag}^{+}$ions are reduced and get deposited.
View full question & answer→Question 181 Mark
Justify that the following reactions are redox reactions:$4NH_3(g) + 5O_2(g) → 4NO(g) + 6H_2O(g)$
Answer$4NH_{3(g)} + 5O_{2(g)} → 4NO_{(g)} + 6H_2O_{(g)}$
The oxidation number of each element in the given reaction can be represented as:
$\stackrel{{-3\ \ \ +1}}{\ \ \ \ 4\text{N H}_{3(\text{g})}}+\stackrel{{0}}{\ \ \ \ \ 5\text{O}_{2(\text{g})}}\ \rightarrow\stackrel{{+2-2}}{\ \ 4\text{NO}_{(\text{g})}}+\stackrel{+1\ \ \ \ \ -2}{ \ \ \ 6\text{H}_2\ \ \text{O}_{(\text{g})}}$
Here, the oxidation number of N increases from -3 in $NH _3$ to +2 in NO . On the other hand, the oxidation number of $O _2$ decreases from 0 in $O _2$ to -2 in NO and $H _2 O$ i.e., $O _2$ is reduced. Hence, the given reaction is a redox reaction.
View full question & answer→Question 191 Mark
Depict the galvanic cell in which the reaction $\mathrm{Zn}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$ takes place, Further show:
Which of the electrode is negatively charged.
AnswerThe galvanic cell corresponding to the given redox reaction can be represented as:
$\text{Zn}|\text{Zn}^{2+}_{(\text{aq})}||\text{Ag}^{+}_{(\text{aq})}|\text{Ag}$
Zn electrode is negatively charged because at this electrode, Zn oxidizes to $\text{Zn}^{2+}$ and the leaving electrons accumulate on this electrode.
View full question & answer→Question 201 Mark
Depict the galvanic cell in which the reaction $\mathrm{Zn}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$ takes place, Further show:The carriers of the current in the cell.
AnswerThe galvanic cell corresponding to the given redox reaction can be represented as:
$\text{Zn}|\text{Zn}^{2+}_{(\text{aq})}||\text{Ag}^{+}_{(\text{aq})}|\text{Ag}$
Ions are the carriers of current in the cell.
View full question & answer→Question 211 Mark
Write the name of cell in which chemical energy is converted into electrical energy.
Question 221 Mark
Why are positive ions called cations, whereas negative ions are called anions?
AnswerPositive ions are called cations because they are attracted towards cathode whereas negative ions are called anions because they are attracted towards anode.
View full question & answer→Question 231 Mark
Calculate oxidation number of O in $\mathrm{KO}_2 \mathrm{~Na}$ in $\mathrm{Na}_2 \mathrm{O}_2$
Answer$\stackrel{2+1}{\mathrm{KO}_2}$
$+1+2 \mathrm{x}=0$
$2 x=-1$
$\mathrm{x}=-\frac{1}{2}$
$\mathrm{Na}_2 \mathrm{O}_2$
$+2+2 \mathrm{x}=0$
$2 x=-2$
$x=-1$
View full question & answer→Question 241 Mark
At what concenration of $\mathrm{Cu}^{2+}(\mathrm{aq})$ will electrode potential become equal to its standard electrode potential?
Answer$1 \mathrm{M}\left(1 \mathrm{~mol} \mathrm{~L}^{-1}\right)$.
View full question & answer→Question 251 Mark
Out of aluminium and silver vessel, which one will be more suitable to store 1 M HCl solution and why?
$\text{E}^{\circ}_{\text{Al}^{3+}|\text{Al}}=-1.66\text{V, E}^{\circ}_{\text{Ag}^+|\text{Ag}}=+0.80\text{V}$
AnswerSince, reduction potential of silver is more than that of hydrogen $\Big(\text{E}^{\circ}_{\text{H}^+|\text{H}_2},\text{Pt}=0\Big)$ silver vessel will be suitable to store 1M HCl. On the other hand, $\text{E}^\circ_{\text{H}^+|\text{H}_2},$ is less than that of hydrogen $\text{E}^\circ_{\text{H}^+|\text{H}_2}$ so hydrogen will get liberated if stored in aluminium vessel.
View full question & answer→Question 261 Mark
Write formula for the following compound:
Nickel(II) sulphate.
AnswerNickel(II) sulphate:
$\mathrm{NiSO}_4$
View full question & answer→Question 271 Mark
Find the value of n in:
$4\text{MnO}^-_4+8\text{H}^++\text{ne}^-\xrightarrow{\ \ \ \ \ \ }\text{Mn}^{2+}+4\text{H}_2\text{O}$
Answer$\text{MnO}^-_4+8\text{H}^++\text{ne}^-\xrightarrow{\ \ \ \ \ }\text{Mn}^{2+}+4\text{H}_2\text{O}$
$-1+8+n=+2$
$-1-2+8+n=0$
$n=-5 \text { or } 5 e^{-}$
View full question & answer→Question 281 Mark
Does the oxidation number of an element in any molecule or any polyatomic ion represent the actual charge on it?
AnswerNo, The oxidation number of an element in any species is an apparent charge on the atom which it appears to have acquired when all other atoms in the species are removed as ions.
View full question & answer→Question 291 Mark
Name the indicator used in redox titration involving $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ as an oxidising agent.
AnswerDiphenyl amine is used as indicator which gives dark blue colour at end point.
View full question & answer→Question 301 Mark
Write formula for the following compound:
Iron(III) sulphate.
AnswerIron(III) sulphate:
$\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3$
View full question & answer→Question 311 Mark
Write the oxidation and reduction reactions separately from the following redox reaction.
$2\text{Fe}+2\text{H}_2\text{O}+\text{O}_2\xrightarrow{ \ \ \ \\ \ \ }2\text{Fe(OH)}_2$
Answer$\text{Fe}\xrightarrow{ \ \ \ \ \ \ }\text{Fe}^{2+}+2\text{e}^-\text{(Oxidation)}$
$\frac{1}{2}\text{O}_2+\text{H}_2\text{O}+2\text{e}^-\xrightarrow{ \ \ \ \ \ }2\text{OH}^-\text{(Reduction)}$
View full question & answer→Question 321 Mark
Consider the elements:
Cs, Ne, I and F
Identify the element that exhibits only negative oxidation state.
AnswerF. Fluorine being the most electronegative element shows only a -ve oxidation state of -1.
View full question & answer→Question 331 Mark
Can $\mathrm{Fe}^{3+}$ oxidise $\mathrm{Br}^{-}$to $\mathrm{Br}^2$ at 1 M concentrations?
$\mathrm{E}^{\circ}\left(\mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+}\right)=0.77 \mathrm{~V}$ and $\mathrm{E}^{\circ}\left(\mathrm{Br} \mid \mathrm{Br}^{-}\right)=1.09 \mathrm{~V}$
Answer$\mathrm{E}^{\circ}\left(\mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+}\right)$ is lower than that of $\mathrm{E}^{\circ}\left(\mathrm{Br}^{-} \mid \mathrm{Br}^{-}\right)$.
Therefore, $\mathrm{Fe}^{2+}$ can reduce $\mathrm{Br}^2$ but $\mathrm{Br}^{-}$cannot reduce $\mathrm{Fe}^{3+}$.
Thus, $\mathrm{Fe}^{3+}$ cannot oxidise $\mathrm{Br}^{-}$to $\mathrm{Br}^2$.
View full question & answer→Question 341 Mark
Write formula for the following compound:
Tin(IV) oxide.
AnswerTin(IV) oxide:
$\mathrm{SnO}_2$
View full question & answer→Question 351 Mark
Refer to the periodic table given in your book and now answer the following questions:
Select the possible non metals that can show disproportionation reaction.
AnswerIn disproportionation reactions, one of the reacting substances always contains an element that can exist in at least three oxidation states.
P, Cl, and S can show disproportionation reactions as these elements can exist in three or more oxidation states.
View full question & answer→Question 361 Mark
Write formula for the following compound:
Chromium(III) oxide.
AnswerChromium(III) oxide:
$\mathrm{Cr}_2\mathrm{O}_3$
View full question & answer→Question 371 Mark
What is the oxidation number of S in $\mathrm{Na}_2 \mathrm{S}_4 \mathrm{O}_6$ and $\mathrm{Na}_2 \mathrm{SO}_3$ ?
View full question & answer→Question 381 Mark
$\mathrm{Br}_2+2 \mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_2+2 \mathrm{Br}^{-}$,
will this reaction take place or not?$\text{E}^0_\frac{\text{Br}_2}{\text{Br}^-}=+1.09\text{V}$
$\text{E}^0_\frac{\text{Cl}_2}{\text{Cl}^-}=+1.36\text{V}$
Answer$\text{E}^0_\text{cell}=\text{E}^0_\frac{\text{Br}_2}{\text{Br}^-}-\text{E}^0_\frac{\text{Cl}_2}{\text{Cl}^-}$
$=1.09\text{V}-1.36$
$=-0.27\text{V}$
Since $\text{E}^0$ cell is -ve, reaction will not take place.
View full question & answer→Question 391 Mark
Justify that the following reactions are redox reactions:$Fe_2O_3(s) + 3CO(g) → 2Fe(s) + 3CO_2(g)$
Answer$Fe_2O_{3(s)} + 3CO_{(g)} → 2Fe_{(s)} + 3CO_{2(g)}$ Let us write the oxidation number of each element in the given reaction as:
$\stackrel{{+3}}{\ \ \ \ \hbox{Fe}_2}\stackrel{{-2}}{\ \ \ \ \ \hbox{O}_{3(\text{s})}}+\stackrel{{+2-2}}{3\ \ \ \hbox{CO}_{(\text{g})}}\ \rightarrow\stackrel{{0}}{\ \ \ \ \hbox{2Fe}_{(\text{s})}}+\stackrel{{+4-2}}{3\ \ \ \ \ \hbox{CO}_{2(\text{g})}}$
Here, the oxidation number of Fe decreases from +3 in $Fe _2 O _3$ to 0 in Fe i.e., $Fe _2 O _3$ is reduced to Fe . On the other hand, the oxidation number of C increases from +2 in CO to +4 in $CO _2$ i.e., CO is oxidized to $CO _2$. Hence, the given reaction is a redox reaction.
View full question & answer→Question 401 Mark
A freshly cut apple is almost white but turns brown after some time, why?
AnswerApple contains $\mathrm{Fe}^{2+}$ which get oxidised to $\mathrm{Fe}^{3+}$ which is brown in colours. Apple turns brown due to oxidation of $\mathrm{Fe}^{2+}$ to $\mathrm{Fe}^{3+}$
View full question & answer→Question 411 Mark
How to find strength of $\mathrm{KMnO}_4$ by titrating it with Mohr's salt in acidic medium?
Answer$5 \mathrm{M}_1 \mathrm{V}_1=\mathrm{M}_2 \mathrm{V}_2$ is used because in $\mathrm{KMnO}_4, \mathrm{Mn}^{7+}$ changes $\left(\mathrm{KMnO}_4\right)$ (Mohr's salt) to $\mathrm{Mn}^{2+}$ by gaining 5 electrons, therefore we have $5 \mathrm{M}_1 \mathrm{V}_1$ but in Mohr's salt $\left(\mathrm{FeSO}_4\left(\mathrm{NH}_4\right) \cdot 6 \mathrm{H}_2 \mathrm{O}, \mathrm{Fe}^{2+}\right.$ loses one electrons to form $\mathrm{Fe}^{2+}$ therefore $\mathrm{M}_2 \mathrm{V}_2$ is used.
View full question & answer→Question 421 Mark
Out of Zn and Cu vessel which one will be more suitable to store 1M HCl?
$\text{E}^\circ_\frac{\text{Zn}^{2+}}{\text{Zn}}=-0.76\text{V}$
$\text{E}^\circ_\frac{\text{Cu}^{2+}}{\text{Cu}}=+0.34\text{V}$
Answer'Cu' vessel is more suitable because Cu is less reactive than hydrogen due to higher value of reduction potential where 'Zn' is more reactive than hydrogen, will displace $\text{H}_2$ from IM HCl.
View full question & answer→Question 431 Mark
$\text{Fe}_2\text{O}_3+3\text{CO}\text{(g)}\xrightarrow{ \ \ \ \ \ \ \ }2\text{Fe}\text{(s)}+3\text{CO}_2\text{(g)}$
AnswerSubstance reduced is $\text{Fe}_2\text{O}_3$.
View full question & answer→Question 441 Mark
Can the following reaction,
$\text{Cr}_{2}\text{O}^{2-}_7 + \text{H}_{2}\text{O}\rightleftharpoons 2\text{CrO}^{2-}_{4} + 2\text{H}^{+}$
be regarded as a redox reaction?
AnswerIn this reaction, oxidation number of Cr in $\text{Cr}_{2}\text{O}^{2-}_{4}$ is +6 and oxidation number of Cr in $\text{Cr}_{2}\text{O}^{2-}_{4}$is + 6. Since, during the reaction, the oxidation number of Cr has neither decreased nor increased, therefore, the above reaction is not a redox reaction.
View full question & answer→Question 451 Mark
Identify oxidant and reductant in the reaction:
$\text{I}_2\text{(aq)}+2\text{S}_2\text{O}_3^{2-}\xrightarrow{ \ \ \ \ \ }2\text{I}^-+\text{S}_4\text{O}_6^{2-}$
AnswerI, is oxidant, $\text{S}_2\text{O}_3^{2-}$ is reductant.
View full question & answer→Question 461 Mark
What is the relationship between standard oxidation potential and standard reduction potential?
AnswerBoth are equal in magnitude but opposite in sign.
View full question & answer→Question 471 Mark
What are spectator ions? Give one example.
AnswerSpectator ions are ions that stay unaffected during a chemical reaction. They appear both as reactant and as product in an ionic equation, e.g. in the following ionic equation, the sodium and nitrate ions are spectator ions.
$\text{Ag}^+\text{(aq)}+\text{NO}^-_3\text{(aq)}+\text{Na}^+\text{(aq)}+\text{Cl}^-\text{(aq)}\\\xrightarrow{\ \ \ \ \ \ \ \ }\text{AgCl(s)}+\text{Na}^+\text{(aq)}+\text{NO}^-_3\text{(aq)}$
View full question & answer→Question 481 Mark
What is oxidation state of Cr in $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_3$
AnswerLet oxidation state of Cr be ' x ', H is $+1, \mathrm{O}$ is -2 ,
$\mathrm{Cl}=-1$
$\mathrm{x}+12-12-3=0$
$\mathrm{x}=3$
View full question & answer→Question 491 Mark
What is the relationship between direction of current and flow of electrons by convention?
AnswerThe current flows from cathode to anode, whereas electrons flow from anode to cathode.
View full question & answer→Question 501 Mark
What happens when $\text{Cu}^{2+}$ is added KI solution? Indicator used in this titration?
Answer$2\text{Cu}^{2+}+4\text{I}^-\text{(aq)}\xrightarrow{ \ \ \ \ \ }\text{Cu}_2\text{I}_2\text{(s)}+\text{I}_2\text{(aq)}$
Starch is used as indicator which gives blue colour with $\text{I}_2$.
View full question & answer→