#

Multiple Choice Questions

One and More than One Correct

1. Consider a titration of potassium dichromate solution with acidified Mohr’s salt solution using diphenylamine as indicator. The number of moles of Mohr’s salt required per mole of dichromate is [JEE 2007 P- II]

a. 3

b. 4

c. 5

d. 6

Answer: (d)

Solution:

The redox reaction between dichromate ion and ferrous ion (from Mohr's salt) in acidic medium is:

$$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \longrightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}$$

The n-factor of $\text{Cr}_2\text{O}_7^{2-}$ is 6, and the n-factor of $\text{Fe}^{2+}$ (Mohr's salt) is 1. Equating equivalents:

$$\text{Moles of Mohr's salt} \times 1 = \text{Moles of Dichromate} \times 6$$

So, to reduce 1 mole of dichromate, 6 moles of Mohr's salt are required. Hence, (d) is correct.

2. A solution of metal ion when treated with KI gives a red precipitate which dissolves in excess KI to give a colourless solution. Moreover, the solution of metal ion on treatment with a solution of cobalt(II) thiocyanate gives rise to a deep blue crystalline precipitate. The metal ion is [JEE 2007 P- II]

a. Pd$^{2+}$

b. Hg$^{2+}$

c. Cu$^{2+}$

d. Co$^{2+}$

Answer: (b)

Solution:

Mercuric ions ($\text{Hg}^{2+}$) react with $\text{KI}$ to form a scarlet red precipitate of $\text{HgI}_2$:

$$\text{Hg}^{2+} + 2\text{KI} \longrightarrow \text{HgI}_2 \downarrow \text{ (Red ppt.)} + 2\text{K}^+$$

The red precipitate dissolves in excess $\text{KI}$ to form a soluble, colourless complex (Nessler's reagent anion):

$$\text{HgI}_2 + 2\text{KI (excess)} \longrightarrow \text{K}_2[\text{HgI}_4] \text{ (Colourless solution)}$$

Furthermore, treatment with cobalt(II) thiocyanate yields a deep blue crystalline precipitate of cobalt tetrathiocyanatomercurate(II):

$$\text{Hg}^{2+} + \text{Co(SCN)}_2 + 2\text{SCN}^- \longrightarrow \text{Co}[\text{Hg(SCN)}_4] \downarrow \text{ (Deep blue crystalline ppt.)}$$

3. A solution of colourless salt H on boiling with excess NaOH produces a non-flammable gas. The gas evolution ceases after sometime. Upon addition of Zn dust to the same solution, the gas evolution restarts. The colourless salt (s) H is (are) [JEE 2008 P-I]

a. NH$_4$NO$_3$

b. NH$_4$NO$_2$

c. NH$_4$Cl

d. (NH$_4$)$_2$SO$_4$

Answer: (a, b)

Solution:

Boiling ammonium salts with excess $\text{NaOH}$ releases non-flammable $\text{NH}_3$ gas:

$$\text{NH}_4\text{NO}_3 + \text{NaOH} \longrightarrow \text{NaNO}_3 + \text{NH}_3 \uparrow + \text{H}_2\text{O}$$ $$\text{NH}_4\text{NO}_2 + \text{NaOH} \longrightarrow \text{NaNO}_2 + \text{NH}_3 \uparrow + \text{H}_2\text{O}$$

Once all $\text{NH}_4^+$ is evolved as $\text{NH}_3$, gas evolution ceases. Addition of $\text{Zn}$ dust reduces the nitrate ($\text{NO}_3^-$) or nitrite ($\text{NO}_2^-$) ions present in solution to reproduce $\text{NH}_3$ gas in alkaline medium:

$$4\text{Zn} + 7\text{NaOH} + \text{NaNO}_3 \longrightarrow 4\text{Na}_2\text{ZnO}_2 + \text{NH}_3 \uparrow + 2\text{H}_2\text{O}$$ $$3\text{Zn} + 5\text{NaOH} + \text{NaNO}_2 \longrightarrow 3\text{Na}_2\text{ZnO}_2 + \text{NH}_3 \uparrow + \text{H}_2\text{O}$$

Therefore, salt H must contain both an ammonium cation and a nitrate/nitrite anion. Thus, (a) and (b) are correct.

4. Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of the solution is [JEE 2011 P-I]

a. 1.78 M

b. 2.00 M

c. 2.05 M

d. 2.22 M

Answer: (c)

Solution:

Total mass of the solution $= \text{Mass of solute} + \text{Mass of solvent} = 120\text{ g} + 1000\text{ g} = 1120\text{ g}$.

Volume of solution $= \frac{\text{Mass}}{\text{Density}} = \frac{1120\text{ g}}{1.15\text{ g/mL}} = 973.91\text{ mL} = 0.9739\text{ L}$.

Moles of urea $= \frac{120\text{ g}}{60\text{ g/mol}} = 2\text{ mol}$.

$$\text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in L}} = \frac{2\text{ mol}}{0.9739\text{ L}} = 2.05\text{ M}$$

5. Passing H$_2$S gas into a mixture of Mn$^{2+}$, Ni$^{2+}$, Cu$^{2+}$ and Hg$^{2+}$ ions in an acidified aqueous solution precipitates [JEE 2011 P-II]

a. CuS and HgS

b. MnS and CuS

c. MnS and NiS

d. NiS and HgS

Answer: (a)

Solution:

In an acidified aqueous solution ($\text{HCl} + \text{H}_2\text{S}$), the high concentration of $\text{H}^+$ suppresses the ionization of $\text{H}_2\text{S}$ due to the common ion effect, yielding a very low sulfide ion concentration $[\text{S}^{2-}]$. Under these conditions, only Group II cations ($\text{Cu}^{2+}$, $\text{Hg}^{2+}$, $\text{Pb}^{2+}$, $\text{Bi}^{3+}$) whose solubility product ($K_{sp}$) values are extremely low will exceed their solubility product and precipitate as sulfides ($\text{CuS}$ and $\text{HgS}$). Group IV cations ($\text{Mn}^{2+}$, $\text{Ni}^{2+}$) remain in solution because their $K_{sp}$ values are much higher.

6. For the given aqueous reactions, which of the statement(s) is (are) true? [JEE 2012 P-II]

a. The first reaction is a redox reaction

b. White precipitate is $\text{Zn}_3[\text{Fe(CN)}_6]_2$

c. Addition of filtrate to starch solution gives blue colour

d. White precipitate is soluble in NaOH solution

Answer: (a, c, d)

Solution:

First reaction: Iodide ($\text{I}^-$) is oxidized to $\text{I}_2$ while ferricyanide $[\text{Fe(CN)}_6]^{3-}$ is reduced to ferrocyanide $[\text{Fe(CN)}_6]^{4-}$:

$$2[\text{Fe(CN)}_6]^{3-} + 2\text{I}^- \longrightarrow 2[\text{Fe(CN)}_6]^{4-} + \text{I}_2$$

Since oxidation states change, the first reaction is a redox reaction (Statement a is correct).

The released $\text{I}_2$ goes into the filtrate. Addition of the brownish-yellow $\text{I}_2$-containing filtrate to starch produces a deep blue starch-iodine complex (Statement c is correct).

When $\text{ZnSO}_4$ is added to ferrocyanide $[\text{Fe(CN)}_6]^{4-}$, a white precipitate of zinc ferrocyanide $\text{Zn}_2[\text{Fe(CN)}_6]$ is formed (making Statement b incorrect as it gives zinc ferricyanide formula).

The white precipitate $\text{Zn}_2[\text{Fe(CN)}_6]$ consists of amphoteric $\text{Zn}^{2+}$ compound which dissolves in excess $\text{NaOH}$ to form sodium zincate $\text{Na}_2[\text{Zn(OH)}_4]$ or $\text{Na}_2\text{ZnO}_2$ (Statement d is correct).

7. Upon treatment with ammoniacal $\text{NH}_4\text{OH}$, the metal ion that precipitates as a sulfide is [JEE Adv. 2013 P-I]

a. Fe(III)

b. Al(III)

c. Mg(I)

d. Zn(II)

Answer: (d)

Solution:

Among $\text{Fe}^{3+}$, $\text{Al}^{3+}$, $\text{Mg}^{2+}$, and $\text{Zn}^{2+}$, $\text{Fe}^{3+}$ and $\text{Al}^{3+}$ precipitate as hydroxides ($\text{Fe(OH)}_3$ and $\text{Al(OH)}_3$) in ammoniacal medium before sulfide precipitation occurs. $\text{Mg}^{2+}$ remains soluble under these basic conditions or forms phosphate/carbonate precipitates. $\text{Zn}^{2+}$ belongs to Group IV cation analysis and precipitates selectively as a sulfide ($\text{ZnS}$, greenish white) upon treatment with $\text{H}_2\text{S}$ in ammoniacal $\text{NH}_4\text{OH}/\text{NH}_4\text{Cl}$ medium.

8. The molarity of a solution obtained by mixing 750 mL of 0.5(M)HCl with 250 mL of 2(M)HCl will be [JEE Main 2013]

a. 0.875 M

b. 1.00 M

c. 1.75 M

d. 0.975 M

Answer: (a)

Solution:

Using the formula for mixing two solutions of the same solute:

$$M_{\text{final}} = \frac{M_1 V_1 + M_2 V_2}{V_1 + V_2}$$ $$M_{\text{final}} = \frac{(0.5 \text{ M} \times 750 \text{ mL}) + (2 \text{ M} \times 250 \text{ mL})}{750 \text{ mL} + 250 \text{ mL}}$$ $$M_{\text{final}} = \frac{375 + 500}{1000} = \frac{875}{1000} = 0.875 \text{ M}$$

9. Consider the following reaction:

$$x\text{MnO}_4^- + y\text{C}_2\text{O}_4^{2-} + z\text{H}^+ \longrightarrow x\text{Mn}^{2+} + 2y\text{CO}_2 + \frac{z}{2}\text{H}_2\text{O}$$

The values of x, y and z in the reaction are respectively [JEE Main 2013]

a. 5, 2 and 16

b. 2, 5 and 8

c. 2, 5 and 16

d. 5, 2 and 8

Answer: (c)

Solution:

Balancing the redox reaction in acidic medium using the ion-electron method:

$$\text{Reduction half-reaction: } (\text{MnO}_4^- + 8\text{H}^+ + 5e^- \longrightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}) \times 2$$ $$\text{Oxidation half-reaction: } (\text{C}_2\text{O}_4^{2-} \longrightarrow 2\text{CO}_2 + 2e^-) \times 5$$

Combining the two half-reactions gives:

$$2\text{MnO}_4^- + 5\text{C}_2\text{O}_4^{2-} + 16\text{H}^+ \longrightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}$$

Comparing coefficients: $x = 2$, $y = 5$, $z = 16$.

10. How many litres of water must be added to 1 litre of an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2? [JEE Main 2013]

a. 0.1 L

b. 0.9 L

c. 2.0 L

d. 9.0 L

Answer: (d)

Solution:

Initial $\text{pH}_1 = 1 \implies [\text{H}^+]_1 = 10^{-1} = 0.1\text{ M}$. Initial volume $V_1 = 1\text{ L}$.

Desired $\text{pH}_2 = 2 \implies [\text{H}^+]_2 = 10^{-2} = 0.01\text{ M}$.

Using the dilution formula $M_1 V_1 = M_2 V_2$:

$$0.1 \times 1 = 0.01 \times V_2 \implies V_2 = \frac{0.1}{0.01} = 10\text{ L}$$

Volume of water to be added $= V_2 - V_1 = 10\text{ L} - 1\text{ L} = 9.0\text{ L}$.

11. In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The percentage of bromine in the compound is: (Atomic mass $\text{Ag}=108$, $\text{Br}=80$) [JEE Main 2015]

a. 24

b. 36

c. 48

d. 60

Answer: (a)

Solution:

Molar mass of $\text{AgBr} = 108 + 80 = 188\text{ g mol}^{-1}$.

$$\text{Percentage of bromine} = \frac{\text{Atomic mass of Br}}{\text{Molecular mass of AgBr}} \times \frac{\text{Mass of AgBr}}{\text{Mass of organic compound}} \times 100$$ $$\text{Percentage of bromine} = \frac{80}{188} \times \frac{141\text{ mg}}{250\text{ mg}} \times 100 = \frac{80 \times 141 \times 100}{188 \times 250} = 24\%$$

12. Which of the following compounds is not colored yellow? [JEE Main 2015]

a. $\text{Zn}_2[\text{Fe(CN)}_6]$

b. $\text{K}_3[\text{Co(NO}_2)_6]$

c. $(\text{NH}_4)_3[\text{As}(\text{Mo}_{12}\text{O}_{40})]$

d. $\text{BaCrO}_4$

Answer: (a)

Solution:

$\text{Zn}_2[\text{Fe(CN)}_6]$ (zinc ferrocyanide) is a white or bluish-white precipitate.

In contrast, $\text{K}_3[\text{Co(NO}_2)_6]$ (Fischer's salt) is canary yellow, $(\text{NH}_4)_3[\text{As}(\text{Mo}_{12}\text{O}_{40})]$ (ammonium arsenomolybdate) is canary yellow, and $\text{BaCrO}_4$ (barium chromate) is yellow.

13. The pair(s) of ions where BOTH the ions are precipitated upon passing $\text{H}_2\text{S}$ gas in presence of dilute HCl, is(are) [JEE Adv. 2015 P-II]

a. $\text{Ba}^{2+}, \text{Zn}^{2+}$

b. $\text{Bi}^{3+}, \text{Fe}^{3+}$

c. $\text{Cu}^{2+}, \text{Pb}^{2+}$

d. $\text{Hg}^{2+}, \text{Bi}^{3+}$

Answer: (c, d)

Solution:

Passing $\text{H}_2\text{S}$ in the presence of dilute $\text{HCl}$ selectively precipitates Group II cations due to low sulfide ion concentration maintained by the common ion effect of $\text{H}^+$ ions.

Group II cations include $\text{Cu}^{2+}$, $\text{Pb}^{2+}$, $\text{Hg}^{2+}$, $\text{Bi}^{3+}$, $\text{Cd}^{2+}$, $\text{As}^{3+}$, $\text{Sb}^{3+}$, and $\text{Sn}^{2+}$.

Both $\text{Cu}^{2+}$ and $\text{Pb}^{2+}$ (option c) as well as both $\text{Hg}^{2+}$ and $\text{Bi}^{3+}$ (option d) belong to Group II and precipitate as black/brown sulfides ($\text{CuS}$, $\text{PbS}$, $\text{HgS}$, $\text{Bi}_2\text{S}_3$). $\text{Zn}^{2+}$ belongs to Group IV and $\text{Fe}^{3+}$ belongs to Group III, so they do not precipitate under acidic conditions.

14. The hottest region of Bunsen flame shown in the figure below is: [JEE Main 2016]

a. region 1

b. region 2

c. region 3

d. region 4

Answer: (b)

Solution:

Region 2 corresponds to the non-luminous zone (fusion zone), located just above the inner blue cone. Complete combustion of fuel takes place in this region due to plenty of oxygen supply, making Region 2 the hottest part of the Bunsen flame (temperature reaches up to $\approx 1500^\circ\text{C}$).

15. The reagent(s) that can selectively precipitate $\text{S}^{2-}$ from a mixture of $\text{S}^{2-}$ and $\text{SO}_4^{2-}$ in aqueous solution is(are) [JEE Adv. 2016 P-I]

a. $\text{CuCl}_2$

b. $\text{BaCl}_2$

c. $\text{Pb(OOCCH}_3)_2$

d. $\text{Na}_2[\text{Fe(CN)}_5\text{NO}]$

Answer: (a)

Solution:

(a) Addition of $\text{CuCl}_2$: $\text{Cu}^{2+}$ combines with $\text{S}^{2-}$ to give a black precipitate of insoluble copper sulfide ($\text{CuS}\downarrow$, $K_{sp} \approx 6 \times 10^{-36}$). Copper sulfate ($\text{CuSO}_4$) is highly soluble in water, so $\text{SO}_4^{2-}$ does not precipitate. Thus, $\text{CuCl}_2$ selectively precipitates $\text{S}^{2-}$.

(b) $\text{BaCl}_2$ selectively precipitates $\text{SO}_4^{2-}$ as white $\text{BaSO}_4\downarrow$, leaving $\text{BaS}$ soluble in water.

(c) $\text{Pb(OOCCH}_3)_2$ precipitates both $\text{S}^{2-}$ as black $\text{PbS}\downarrow$ ($K_{sp} = 10^{-28}$) and $\text{SO}_4^{2-}$ as white $\text{PbSO}_4\downarrow$ ($K_{sp} = 1.6 \times 10^{-8}$), so it cannot selectively precipitate only $\text{S}^{2-}$.

(d) $\text{Na}_2[\text{Fe(CN)}_5\text{NO}]$ (sodium nitroprusside) gives a purple coloration with $\text{S}^{2-}$, but does not form a precipitate.

16. Sodium salt of an organic acid 'X' produces effervescence with conc. H$_2$SO$_4$. 'X' reacts with the acidified aqueous CaCl$_2$ solution to give a white precipitate which decolourises acidic solution of KMnO$_4$. 'X' is [JEE Main 2017]

a. Na$_2$C$_2$O$_4$

b. C$_6$H$_5$COONa

c. HCOONa

d. CH$_3$COONa

Answer: (a)

Solution:

Sodium oxalate ($\text{Na}_2\text{C}_2\text{O}_4$) reacts with concentrated $\text{H}_2\text{SO}_4$ to produce effervescence of $\text{CO}$ and $\text{CO}_2$ gases:

$$\text{Na}_2\text{C}_2\text{O}_4 + \text{H}_2\text{SO}_4 \longrightarrow \text{Na}_2\text{SO}_4 + \text{CO} \uparrow + \text{CO}_2 \uparrow + \text{H}_2\text{O}$$

Reaction of $\text{Na}_2\text{C}_2\text{O}_4$ with $\text{CaCl}_2$ yields a white precipitate of calcium oxalate ($\text{CaC}_2\text{O}_4$):

$$\text{Na}_2\text{C}_2\text{O}_4 + \text{CaCl}_2 \longrightarrow \text{CaC}_2\text{O}_4 \downarrow \text{ (white ppt)} + 2\text{NaCl}$$

The white precipitate of $\text{CaC}_2\text{O}_4$ reduces purple $\text{KMnO}_4$ in acidic medium, decolourising it to colourless $\text{Mn}^{2+}$:

$$5\text{CaC}_2\text{O}_4 + 2\text{KMnO}_4 + 8\text{H}_2\text{SO}_4 \longrightarrow \text{K}_2\text{SO}_4 + 5\text{CaSO}_4 + 2\text{MnSO}_4 \text{ (colourless)} + 10\text{CO}_2 \uparrow + 8\text{H}_2\text{O}$$

17. The correct option(s) to distinguish nitrate salts of Mn$^{2+}$ and Cu$^{2+}$ taken separately is (are) [JEE Adv. 2018 P-II]

a. Mn$^{2+}$ shows the characteristic green colour in the flame test

b. Only Cu$^{2+}$ shows the formation of precipitate by passing H$_2$S in acidic medium

c. Only Mn$^{2+}$ shows the formation of precipitate by passing H$_2$S in faintly basic medium

d. Cu$^{2+}$/Cu has higher reduction potential than Mn$^{2+}$/Mn (measured under similar conditions)

Answer: (b, d)

Solution:

(a) $\text{Mn}^{2+}$ gives a light green/pale pink flame, whereas $\text{Cu}^{2+}$ gives a characteristic bright green/blue-green flame test (Statement a is incorrect).

(b) $\text{Cu}^{2+}$ belongs to Group II and precipitates as black $\text{CuS}$ when $\text{H}_2\text{S}$ is passed in acidic medium ($\text{HCl}$). $\text{Mn}^{2+}$ belongs to Group IV and requires basic medium to precipitate, so it does not precipitate in acidic medium (Statement b is correct).

(c) Both $\text{Cu}^{2+}$ and $\text{Mn}^{2+}$ form sulfide precipitates ($\text{CuS}$ and $\text{MnS}$) when $\text{H}_2\text{S}$ is passed in faintly basic medium (Statement c is incorrect).

(d) Standard reduction potential $E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34\text{ V}$, whereas $E^\circ(\text{Mn}^{2+}/\text{Mn}) = -1.18\text{ V}$. Hence, $\text{Cu}^{2+}/\text{Cu}$ has a significantly higher reduction potential than $\text{Mn}^{2+}/\text{Mn}$ (Statement d is correct).

18. The lanthanide ion that would show colour is- [JEE Main 2019]

a. Sm$^{3+}$

b. La$^{3+}$

c. Lu$^{3+}$

d. Gd$^{3+}$

Answer: (a)

Solution:

Electronic configurations of the given tripositive lanthanide ions:

$\text{La}^{3+}$ ($Z=57$): $[\text{Xe}] 4f^0$ (empty $f$-subshell $\implies$ colourless)

$\text{Lu}^{3+}$ ($Z=71$): $[\text{Xe}] 4f^{14}$ (fully filled $f$-subshell $\implies$ colourless)

$\text{Gd}^{3+}$ ($Z=64$): $[\text{Xe}] 4f^7$ (half-filled $f$-subshell, UV absorption only $\implies$ colourless in visible light)

$\text{Sm}^{3+}$ ($Z=62$): $[\text{Xe}] 4f^5$ (partially filled $f$-subshell with unpaired electrons $\implies$ exhibits yellow colour due to $f-f$ transitions).

19. The following ligand is [JEE Main 2019]

a. Bidentate

b. Hexadentate

c. Tetradentate

d. Tridentate

Answer: (c)

Solution:

The structure contains four donor atom sites capable of coordinating to a central metal ion: two neutral nitrogen donor atoms (containing lone pairs) and two anionic oxygen donor atoms ($-\text{O}^-$). Thus, it acts as a tetradentate ligand.

20. 100 mL of a water sample contains 0.81 g of calcium bicarbonate and 0.73 of magnesium bicarbonate. The hardness of this water sample expressed in terms of equivalents of CaCO$_3$ is: (molar mass of calcium bicarbonate is 162 g mol$^{–1}$ and magnesium bicarbonate is 146 gmol$^{–1}$) [JEE Main 2019]

a. 1,000 ppm

b. 10,000 ppm

c. 100 ppm

d. 5,000 ppm

Answer: (b)

Solution:

Number of equivalents of $\text{Ca(HCO}_3)_2 = \frac{0.81}{162/2} = \frac{0.81}{81} = 0.01\text{ eq}$.

Number of equivalents of $\text{Mg(HCO}_3)_2 = \frac{0.73}{146/2} = \frac{0.73}{73} = 0.01\text{ eq}$.

Total equivalents of salts in $100\text{ mL}$ water $= 0.01 + 0.01 = 0.02\text{ eq}$.

Equivalent mass of $\text{CaCO}_3 = \frac{100}{2} = 50\text{ g/eq}$.

Mass of $\text{CaCO}_3$ equivalent $= 0.02 \text{ eq} \times 50\text{ g/eq} = 1.0\text{ g}$.

Mass of $100\text{ mL}$ water sample $\approx 100\text{ g}$.

$$\text{Hardness in ppm} = \frac{\text{Mass of CaCO}_3 \text{ equivalent}}{\text{Mass of water sample}} \times 10^6 = \frac{1.0\text{ g}}{100\text{ g}} \times 10^6 = 10,000\text{ ppm}$$

21. The tests performed on compound X and their inferences are:

Test | Inference

(A) 2,4-DNP test | Coloured precipitate

(B) Iodoform test | Yellow precipitate

Compound ‘X’ is: [JEE Main 2019]

Answer: (d)

Solution:

1. Positive 2,4-DNP test indicates the presence of a carbonyl group (aldehyde or ketone).

2. Positive Iodoform test indicates the presence of a methyl ketone group ($\text{CH}_3-\text{C}(=\text{O})-$) or a $\text{CH}_3-\text{CH}(\text{OH})-$ group.

3. Negative Azo-dye test indicates the absence of a primary aromatic amine ($-\text{Ar-NH}_2$).

Compound (d) contains a methyl ketone moiety ($\text{CH}_3-\text{CO}-$) which gives both a positive 2,4-DNP test and a positive iodoform test, while giving a negative azo-dye test.

22. In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in producing one molecule of CO$_2$ is : [JEE Main 2019]

a. 10

b. 2

c. 1

d. 5

Answer: (c)

Solution:

The balanced oxidation half-reaction of oxalate ion is:

$$\text{C}_2\text{O}_4^{2-} \longrightarrow 2\text{CO}_2 + 2e^-$$

Here, $2e^-$ are transferred in the formation of 2 molecules of $\text{CO}_2$.

$$\text{Number of electrons per molecule of CO}_2 = \frac{2e^-}{2\text{ molecules of CO}_2} = 1e^-$$

23. A compound 'X' on treatment with Br$_2$/NaOH, provided C$_3$H$_9$N, which gives positive carbylamine test. Compound 'X' is :- [JEE Main 2019]

a. CH$_3$COCH$_2$NHCH$_3$

b. CH$_3$CH$_2$COCH$_2$NH$_2$

c. CH$_3$CH$_2$CH$_2$CONH$_2$

d. CH$_3$CON(CH$_3$)$_2$

Answer: (c)

Solution:

Compound 'X' undergoes Hoffmann's bromamide degradation reaction with $\text{Br}_2/\text{NaOH}$ to yield a primary amine $\text{C}_3\text{H}_9\text{N}$.

Since the primary amine has 3 carbons ($\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$, propan-1-amine), it gives a positive carbylamine test (forming an isocyanide $\text{CH}_3\text{CH}_2\text{CH}_2\text{NC}$ with foul smell).

In Hoffmann bromamide degradation, the parent amide must have one more carbon atom than the resulting primary amine. Therefore, compound 'X' must be a 4-carbon primary amide: $\text{CH}_3\text{CH}_2\text{CH}_2\text{CONH}_2$ (butanamide).

$$\text{CH}_3\text{CH}_2\text{CH}_2\text{CONH}_2 \xrightarrow{\text{Br}_2/\text{NaOH}} \text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2 \xrightarrow{\text{CHCl}_3/\text{KOH}} \text{CH}_3\text{CH}_2\text{CH}_2\text{NC}$$

24. Which of the following compounds reacts with ethylmagnesium bromide and also decolourizes bromine water solution: [JEE Main 2019]

Answer: (d)

Solution:

To react with ethylmagnesium bromide ($\text{EtMgBr}$), the compound must contain an acidic hydrogen ($-\text{OH}$, $-\text{COOH}$, $-\text{NH}_2$, or terminal alkyne) or an electrophilic carbonyl group.

To decolourize bromine water, the compound must contain an unsaturated carbon-carbon bond ($C=C$ or $C \equiv C$) or an activating group like phenol/aniline.

Compound (d) contains both an acidic hydroxyl group ($-\text{OH}$) which reacts with $\text{EtMgBr}$ to release ethane gas ($\text{C}_2\text{H}_6$), and an unsaturated double bond moiety that decolourizes bromine water via addition reaction.

25. The hardness of a water sample (in terms of equivalents of CaCO$_3$) containing 10$^{–3}$ M CaSO$_4$ is: (molar mass of CaSO$_4$ = 136 g mol$^{–1}$) [JEE Main 2019]

a. 100 ppm

b. 50 ppm

c. 10 ppm

d. 90 ppm

Answer: (a)

Solution:

Concentration of $\text{CaSO}_4 = 10^{-3}\text{ M} = 10^{-3}\text{ moles/L}$.

1 mole of $\text{CaSO}_4$ produces 1 mole of $\text{Ca}^{2+}$, which is equivalent to 1 mole of $\text{CaCO}_3$.

Moles of $\text{CaCO}_3$ equivalent per liter $= 10^{-3}\text{ moles}$.

Mass of $\text{CaCO}_3$ equivalent per liter $= 10^{-3}\text{ moles} \times 100\text{ g/mol} = 0.1\text{ g/L} = 100\text{ mg/L}$.

$$\text{Hardness in ppm} = 100\text{ ppm}$$

26. Chlorine on reaction with hot and concentrated sodium hydroxide gives: [JEE Main 2019]

a. Cl$^–$ and ClO$_2^–$

b. Cl$^–$ and ClO$_3^–$

c. Cl$^–$ and ClO$^–$

d. Cl$_3^–$ and ClO$_2^–$

Answer: (b)

Solution:

Chlorine undergoes disproportionation reaction with hot and concentrated sodium hydroxide ($\text{NaOH}$):

$$3\text{Cl}_2 + 6\text{OH}^- \xrightarrow{\text{hot \& conc.}} 5\text{Cl}^- + \text{ClO}_3^- + 3\text{H}_2\text{O}$$

The products formed are chloride ion ($\text{Cl}^-$) and chlorate ion ($\text{ClO}_3^-$).

27. Fusion of MnO$_2$ along with KOH and O$_2$ forms X. Electrolytic oxidation of X yields Y. X undergoes disproportionation in acidic medium to MnO$_2$ and Y. The Manganese in X and Y is in the form W & Z respectively, then [JEE Adv. 2019 P-I]

a. W & Z are coloured

b. W is diamagnetic

c. Both W & Z are tetrahedral in shape

d. Both W & Z involve p$\pi$–d$\pi$ bonding for $\pi$ bond

Answer: (a, c, d)

Solution:

1. Fusion reaction: $2\text{MnO}_2 + 4\text{KOH} + \text{O}_2 \longrightarrow 2\text{K}_2\text{MnO}_4 (\text{X}) + 2\text{H}_2\text{O}$. Thus, $W = \text{MnO}_4^{2-}$ (manganate ion, green, oxidation state $+6$).

2. Electrolytic oxidation: $\text{MnO}_4^{2-} (\text{X}) \xrightarrow{\text{oxidation}} \text{MnO}_4^- (\text{Y}) + e^-$. Thus, $Z = \text{MnO}_4^-$ (permanganate ion, purple/intense pink, oxidation state $+7$).

3. Disproportionation in acidic medium: $3\text{MnO}_4^{2-} + 4\text{H}^+ \longrightarrow 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O}$.

Evaluation of statements:

(a) Both $\text{MnO}_4^{2-}$ (green) and $\text{MnO}_4^-$ (purple) are coloured (Statement a is correct).

(b) Manganate ion $W = \text{MnO}_4^{2-}$ has $3d^1$ configuration and contains 1 unpaired electron, so it is paramagnetic, not diamagnetic (Statement b is incorrect).

(c) Both $\text{MnO}_4^{2-}$ and $\text{MnO}_4^-$ are $sp^3$ hybridized with tetrahedral geometry (Statement c is correct).

(d) $\pi$-bonds in both species are formed by overlap of filled $2p$ orbitals of oxygen with empty $3d$ orbitals of manganese ($p\pi-d\pi$ bonding) (Statement d is correct).

28. Which green coloured compound of chromium is formed in borax bead test ? [JEE Adv. 2019 P-I]

a. Cr(BO$_2$)$_3$

b. Cr$_2$O$_3$

c. CrB

d. CrBO$_3$

Answer: (a)

Solution:

In the borax bead test, borax decomposes on heating to form sodium metaborate and boric anhydride:

$$\text{Na}_2\text{B}_4\text{O}_7 \cdot 10\text{H}_2\text{O} \xrightarrow{\Delta} \text{Na}_2\text{B}_4\text{O}_7 \xrightarrow{\Delta} 2\text{NaBO}_2 + \text{B}_2\text{O}_3$$

Chromium compounds react with glassy $\text{B}_2\text{O}_3$ bead to form chromium(III) metaborate, which is green in colour:

$$\text{Cr}_2\text{O}_3 + 3\text{B}_2\text{O}_3 \xrightarrow{\Delta} 2\text{Cr(BO}_2)_3 \text{ (Green bead)}$$

29. With reference to aqua regia, choose the correct option(s): [JEE Adv. 2019 Paper -II]

a. Aqua regia is prepared by mixing conc. HCl and conc. HNO$_3$ in 3:1 molar ratio.

b. Reaction of gold with aqua regia produces an anion having Au in +3 oxidation state.

c. Reaction of gold with aqua regia produces NO$_2$ in the absence of air

d. The yellow colour of aqua regia is due to the presence of NOCl and Cl$_2$.

Answer: (a, b, d)

Solution:

(a) Aqua regia is freshly prepared by mixing 3 parts of concentrated $\text{HCl}$ and 1 part of concentrated $\text{HNO}_3$ by mole/volume ratio (Statement a is correct).

(b) Gold reacts with aqua regia to form tetrachloroaurate(III) ion $[\text{AuCl}_4]^-$, where Gold has an oxidation state of $+3$ (Statement b is correct):

$$\text{Au} + 4\text{H}^+ + \text{NO}_3^- + 4\text{Cl}^- \longrightarrow [\text{AuCl}_4]^- + \text{NO} \uparrow + 2\text{H}_2\text{O}$$

(c) The reaction produces nitric oxide ($\text{NO}$) gas in the absence of air, not $\text{NO}_2$ (Statement c is incorrect).

(d) The decomposition of aqua regia produces nitrosyl chloride ($\text{NOCl}$) and free chlorine ($\text{Cl}_2$), giving it a characteristic yellow-orange colour (Statement d is correct):

$$3\text{HCl} + \text{HNO}_3 \longrightarrow \text{NOCl} + \text{Cl}_2 + 2\text{H}_2\text{O}$$

30. Consider the following reaction (unbalanced)

$$\text{Zn} + \text{Hot conc. H}_2\text{SO}_4 \longrightarrow \text{G} + \text{R} + \text{X}$$ $$\text{Zn} + \text{conc. NaOH} \longrightarrow \text{T} + \text{Q}$$ $$\text{G} + \text{H}_2\text{S} + \text{NH}_3(aq) \longrightarrow \text{Z(precipitate)} + \text{X} + \text{Y}$$

Choose the correct option(s) [JEE Adv. 2019 P-II]

a. R is a V-shaped molecule

b. Z is dirty white in colour

c. Bond order of Q is 1 in its ground state

d. The oxidation state of Zn in T is + 1.

Answer: (a, b, c)

Solution:

Reaction 1: $\text{Zn} + 2\text{H}_2\text{SO}_4 \text{ (hot conc.)} \longrightarrow \text{ZnSO}_4 (\text{G}) + \text{SO}_2 (\text{R}) + 2\text{H}_2\text{O} (\text{X})$

Reaction 2: $\text{Zn} + 2\text{NaOH} \text{ (conc.)} \longrightarrow \text{Na}_2\text{ZnO}_2 (\text{T}) + \text{H}_2 (\text{Q})$

Reaction 3: $\text{ZnSO}_4 (\text{G}) + \text{H}_2\text{S} + 2\text{NH}_3(aq) \longrightarrow \text{ZnS} \downarrow (\text{Z}) + (\text{NH}_4)_2\text{SO}_4 (\text{Y}) + \text{H}_2\text{O} (\text{X})$

Analysis of options:

(a) $\text{R} = \text{SO}_2$, which is $sp^2$ hybridized with 1 lone pair, possessing a bent / V-shaped geometry (Statement a is correct).

(b) $\text{Z} = \text{ZnS}$, which is a dirty white / dull white precipitate (Statement b is correct).

(c) $\text{Q} = \text{H}_2$, which has a single covalent bond ($H-H$) and a ground state bond order of 1 (Statement c is correct).

(d) $\text{T} = \text{Na}_2\text{ZnO}_2$, where oxidation state of $\text{Zn}$ is $+2$, not $+1$ (Statement d is incorrect).

Matrix Match/Column Match

31. Match the anionic species given in Column I that are present in the ore(s) given in Column II is: [JEE Adv. 2015 P -I]

Column I

(A) Carbonate

(B) Sulphide

(D) Oxide

Column II

1. Siderite

2. Malachite

3. Bauxite

4. Calamine

5. Argentite

a. A$\rightarrow$ 1,2,4; B$\rightarrow$ 5; C$\rightarrow$ 2,3; D$\rightarrow$ 3

b. A$\rightarrow$ 1,2; B$\rightarrow$ 2,4; C$\rightarrow$ 2,3; D$\rightarrow$ 4

c. A$\rightarrow$ 2,3; B$\rightarrow$ 1,2; C$\rightarrow$ 2,3; D$\rightarrow$ 4

d. A$\rightarrow$ 3; B$\rightarrow$ 3,4; C$\rightarrow$ 4; D$\rightarrow$ 1,2,3

Answer: (a)

Solution:

Chemical composition of given ores:

1. Siderite: $\text{FeCO}_3$ (contains Carbonate anion $\implies$ A$\rightarrow$1)

2. Malachite: $\text{CuCO}_3 \cdot \text{Cu(OH)}_2$ (contains Carbonate and Hydroxide anions $\implies$ A$\rightarrow$2, C$\rightarrow$2)

3. Bauxite: $\text{AlO}_x(\text{OH})_{3-2x}$ / hydrated alumina $\text{Al}_2\text{O}_3 \cdot 2\text{H}_2\text{O}$ (contains Hydroxide and Oxide species $\implies$ C$\rightarrow$3, D$\rightarrow$3)

4. Calamine: $\text{ZnCO}_3$ (contains Carbonate anion $\implies$ A$\rightarrow$4)

5. Argentite: $\text{Ag}_2\text{S}$ (contains Sulphide anion $\implies$ B$\rightarrow$5)

Matching: A $\rightarrow$ 1,2,4; B $\rightarrow$ 5; C $\rightarrow$ 2,3; D $\rightarrow$ 3.

32. The correct match between Column I and Column II is: [JEE Main 2019]

Column - I (drug)

(A) Chloroxylenol

(B) Norethindrone

(D) Penicillin

Column – II (test)

1. Carbylamine Test

2. Sodium Hydrogen Carbonate Test

3. Ferric chloride test

4. Baeyer’s test

a. A$\rightarrow$2; B$\rightarrow$1; C$\rightarrow$4; D$\rightarrow$3

b. A$\rightarrow$3; B$\rightarrow$1; C$\rightarrow$4; D$\rightarrow$2

c. A$\rightarrow$3; B$\rightarrow$4; C$\rightarrow$1; D$\rightarrow$2

d. A$\rightarrow$2; B$\rightarrow$4; C$\rightarrow$1; D$\rightarrow$3

Answer: (c)

Solution:

(A) Chloroxylenol is a phenolic compound, hence it gives a characteristic colour with neutral $\text{FeCl}_3$ solution (Ferric chloride test $\implies$ A$\rightarrow$3).

(B) Norethindrone contains a terminal alkyne group ($-\text{C}\equiv\text{CH}$), which decolourises alkaline $\text{KMnO}_4$ solution (Baeyer's test $\implies$ B$\rightarrow$4).

(C) Sulphapyridine contains a primary aromatic amino group ($-\text{Ar-NH}_2$), which gives a positive carbylamine test when heated with $\text{CHCl}_3$ and $\text{KOH}$ ($\implies$ C$\rightarrow$1).

(D) Penicillin contains a free carboxylic acid group ($-\text{COOH}$), which gives effervescence of $\text{CO}_2$ with aqueous sodium hydrogen carbonate ($\text{NaHCO}_3$ test $\implies$ D$\rightarrow$2).

Paragraph

Paragraph for Questions Nos. 33 to 35

p-Amino-N, N-dimethylaniline is added to a strongly acidic solution of X. The resulting solution is treated with a few drops of aqueous solution of Y to yield blue coloration due to the formation of methylene blue. Treatment of the aqueous solution of Y with the reagent potassium hexacyanoferrate (II) leads to the formation of an intense blue precipitate. The precipitate dissolves on excess addition of the reagent. Similarly, treatment of the solution of Y with the solution of potassium hexacyanoferrate (III) leads to a brown coloration due to the formation of Z. [JEE 2009 P-I]

33. The compound X is

a. $\text{NaNO}_3$

b. $\text{NaCl}$

c. $\text{Na}_2\text{SO}_4$

d. $\text{Na}_2\text{S}$

Answer: (d)

Solution:

Compound X provides sulfide ions ($\text{S}^{2-}$ or $\text{H}_2\text{S}$) in strongly acidic medium. Reaction of $p$-amino-$N,N$-dimethylaniline with $\text{H}_2\text{S}$ in the presence of an oxidizing agent $\text{Fe}^{3+}$ ($\text{Y}$) forms methylene blue dye. Thus, compound X is $\text{Na}_2\text{S}$.

34. The compound Y is

a. $\text{MgCl}_2$

b. $\text{FeCl}_2$

c. $\text{FeCl}_3$

d. $\text{ZnCl}_2$

Answer: (c)

Solution:

Aqueous solution of Y ($\text{FeCl}_3$) contains $\text{Fe}^{3+}$ ions. Treatment of $\text{Fe}^{3+}$ with potassium hexacyanoferrate(II) $\text{K}_4[\text{Fe(CN)}_6]$ produces an intense Prussian blue precipitate of $\text{Fe}_4[\text{Fe(CN)}_6]_3$:

$$4\text{Fe}^{3+} + 3[\text{Fe(CN)}_6]^{4-} \longrightarrow \text{Fe}_4[\text{Fe(CN)}_6]_3 \downarrow \text{ (Prussian blue)}$$

Hence, compound Y is $\text{FeCl}_3$.

35. The compound Z is

a. $\text{Mg}_3[\text{Fe(CN)}_6]_2$

b. $\text{Fe}[\text{Fe(CN)}_6]$

c. $\text{Fe}_4[\text{Fe(CN)}_6]_3$

d. $\text{K}_2\text{Zn}[\text{Fe(CN)}_6]_2$

Answer: (b)

Solution:

Treatment of ferric ions ($\text{Fe}^{3+}$, compound Y) with potassium hexacyanoferrate(III) $\text{K}_3[\text{Fe(CN)}_6]$ gives a brown/brownish-green complex $\text{Fe}[\text{Fe(CN)}_6]$ (ferric ferricyanide):

$$\text{Fe}^{3+} + [\text{Fe(CN)}_6]^{3-} \longrightarrow \text{Fe}[\text{Fe(CN)}_6] \text{ (Brown coloration)}$$

Hence, compound Z is $\text{Fe}[\text{Fe(CN)}_6]$.

Paragraph for Question Nos. 36 to 37

Bleaching powder and bleach solution are produced on a large scale and used in several household products. The effectiveness of bleach solution is often measured by iodometry. [JEE 2012 P-II]

36. 25 mL of household bleach solution was mixed with 30 mL of 0.50 M KI and 10 mL of 4 N acetic acid. In the titration of the liberated iodine, 48 mL of 0.25 N $\text{Na}_2\text{S}_2\text{O}_3$ was used to reach the end point. The molarity of the household bleach solution is

a. 0.48 M

b. 0.96 M

c. 0.24 M

d. 0.024 M

Answer: (c)

Solution:

Equivalents of liberated $\text{I}_2 = \text{Equivalents of } \text{Na}_2\text{S}_2\text{O}_3$ used.

$$\text{Equivalents of } \text{Na}_2\text{S}_2\text{O}_3 = \frac{48 \times 0.25}{1000} = 0.012\text{ eq}$$

The hypochlorite ion ($\text{ClO}^-$) in bleach liberates $\text{I}_2$ according to:

$$\text{ClO}^- + 2\text{I}^- + 2\text{H}^+ \longrightarrow \text{Cl}^- + \text{I}_2 + \text{H}_2\text{O}$$

For $\text{ClO}^-$, n-factor $= 2$.

$$\text{Equivalents of bleach} = \text{Normality} \times \text{Volume (in L)} = 0.012$$ $$\text{Normality of bleach} = \frac{0.012}{0.025\text{ L}} = 0.48\text{ N}$$ $$\text{Molarity of bleach} = \frac{\text{Normality}}{\text{n-factor}} = \frac{0.48}{2} = 0.24\text{ M}$$

37. Bleaching powder contains a salt of an oxoacid as one of its components. The anhydride of that oxoacid is

a. $\text{Cl}_2\text{O}$

b. $\text{ClO}_2$

c. $\text{Cl}_2\text{O}_7$

d. $\text{Cl}_2\text{O}_6$

Answer: (a)

Solution:

Bleaching powder $\text{Ca(OCl)Cl}$ contains calcium hypochlorite $\text{Ca(OCl)}_2$, which is the salt of hypochlorous acid ($\text{HOCl}$).

Dehydration of hypochlorous acid yields dichlorine monoxide ($\text{Cl}_2\text{O}$):

$$2\text{HOCl} \xrightarrow{-\text{H}_2\text{O}} \text{Cl}_2\text{O}$$

Thus, $\text{Cl}_2\text{O}$ is the acid anhydride of $\text{HOCl}$.

Paragraph for Question Nos. 38 to 39

An aqueous solution of a mixture of two inorganic salts, when treated with dilute HCl, gave a precipitate (P) and a filtrate (Q). The precipitate P was found to dissolve in hot water. The filtrate (Q) remained unchanged, when treated with H$_2$S in a dilute mineral acid medium. However, it gave a precipitate (R) with H$_2$S in an ammoniacal medium. The precipitate R gave a coloured solution (S), when treated with H$_2$O2 in an aqueous NaOH medium. [JEE Adv. 2013 P-II]

38. The precipitate P contains

a. $\text{Pb}^{2+}$

b. $\text{Hg}_2^{2+}$

c. $\text{Ag}^+$

d. $\text{Hg}^{2+}$

Answer: (a)

Solution:

Treatment with dilute $\text{HCl}$ precipitates Group I cations ($\text{Pb}^{2+}$, $\text{Ag}^+$, $\text{Hg}_2^{2+}$) as chlorides.

Among Group I chlorides ($\text{PbCl}_2$, $\text{AgCl}$, $\text{Hg}_2\text{Cl}_2$), only lead chloride ($\text{PbCl}_2$) is soluble in hot water and reprecipitates upon cooling. Therefore, precipitate P contains $\text{Pb}^{2+}$.

39. The coloured solution S contains

a. $\text{Fe}_2(\text{SO}_4)_3$

b. $\text{CuSO}_4$

c. $\text{ZnSO}_4$

d. $\text{Na}_2\text{CrO}_4$

Answer: (d)

Solution:

Filtrate Q does not precipitate in acidic $\text{H}_2\text{S}$, ruling out Group II cations. However, it forms precipitate R with $\text{H}_2\text{S}$ in ammoniacal medium, indicating Group III/IV cations ($\text{Cr}^{3+}$, $\text{Fe}^{3+}$, $\text{Al}^{3+}$, $\text{Zn}^{2+}$, $\text{Mn}^{2+}$).

Chromium hydroxide/sulfide ($\text{Cr(OH)}_3$) reacts with $\text{H}_2\text{O}_2$ in alkaline $\text{NaOH}$ medium to form yellow-coloured sodium chromate ($\text{Na}_2\text{CrO}_4$) solution S:

$$2\text{Cr(OH)}_3 + 3\text{H}_2\text{O}_2 + 4\text{NaOH} \longrightarrow 2\text{Na}_2\text{CrO}_4 \text{ (Yellow solution)} + 8\text{H}_2\text{O}$$

Therefore, coloured solution S contains $\text{Na}_2\text{CrO}_4$.

Integer and Subjective

40. A student performs a titration with different burettes and finds titre values of 25.2 mL, 25.25 mL and 25.0 mL. The number of significant figures in the average titre value is________ [JEE 2010 P-I]

Answer: 3

Solution:

Calculating the average of the titre values:

$$\text{Average} = \frac{25.2 + 25.25 + 25.0}{3} = \frac{75.45}{3} = 25.15\text{ mL}$$

According to significant figure rules for addition, the sum $25.2 + 25.25 + 25.0 = 75.45$ should be rounded off to 1 decimal place (matching $25.2$ and $25.0$), giving $75.5$. Dividing $75.5$ by the exact number 3 yields $25.166... \approx 25.2\text{ mL}$.

The value $25.2$ contains 3 significant figures.

41. The concentration of R in the reaction R$\rightarrow$P was measured as a function of time and the following data is obtained:

[R] (molar): 1.0, 0.75, 0.40, 0.10

t (min): 0.0, 0.05, 0.12, 0.18

The order of the reaction is________ [JEE 2010 P-I]

Answer: 0

Solution:

Checking for zero-order kinetics using the rate equation $k = \frac{[R]_0 - [R]_t}{t}$:

1. Between $t = 0$ and $t = 0.05\text{ min}$:

$$k_1 = \frac{1.0 - 0.75}{0.05} = \frac{0.25}{0.05} = 5.0\text{ M min}^{-1}$$

2. Between $t = 0$ and $t = 0.12\text{ min}$:

$$k_2 = \frac{1.0 - 0.40}{0.12} = \frac{0.60}{0.12} = 5.0\text{ M min}^{-1}$$

3. Between $t = 0$ and $t = 0.18\text{ min}$:

$$k_3 = \frac{1.0 - 0.10}{0.18} = \frac{0.90}{0.18} = 5.0\text{ M min}^{-1}$$

Since the rate constant $k$ remains constant throughout, the reaction follows zero-order kinetics.

42. The volume (in mL) of 0.1 M AgNO$_3$ required for complete precipitation of chloride ions present in 30 mL of 0.01 M solution of [Cr(H$_2$O)$_5$Cl]Cl$_2$, as silver chloride is close to________ [JEE 2011 P-II]

Answer: 6

Solution:

The coordination complex $[\text{Cr(H}_2\text{O})_5\text{Cl}]\text{Cl}_2$ ionizes in solution as:

$$[\text{Cr(H}_2\text{O})_5\text{Cl}]\text{Cl}_2 \longrightarrow [\text{Cr(H}_2\text{O})_5\text{Cl}]^{2+} + 2\text{Cl}^-$$

Each mole of complex produces 2 moles of ionizable chloride ($\text{Cl}^-$) ions.

$$\text{Millimoles of complex} = 30\text{ mL} \times 0.01\text{ M} = 0.3\text{ mmol}$$ $$\text{Millimoles of ionizable Cl}^- = 2 \times 0.3 = 0.6\text{ mmol}$$

Precipitation reaction: $\text{Ag}^+ + \text{Cl}^- \longrightarrow \text{AgCl}\downarrow$

$$\text{Millimoles of AgNO}_3 \text{ required} = \text{Millimoles of Cl}^- = 0.6\text{ mmol}$$ $$M \times V(\text{mL}) = 0.6 \implies 0.1 \times V = 0.6 \implies V = 6\text{ mL}$$

43. In 1L saturated solution of $\text{AgCl}$ ($K_{sp} = 1.6 \times 10^{-10}$), 0.1 mol of CuCl ($K_{sp} = 1.0 \times 10^{-6}$) is added. The resultant concentration of $\text{Ag}^+$ in the solution is $x \times 10^{-5}\text{ M}$. The value of “x” is________ [JEE 2011 P-II]

Answer: 1.6

Solution:

Let solubility of $\text{AgCl}$ be $s_1$ and solubility of $\text{CuCl}$ be $s_2$. Since $K_{sp}(\text{CuCl}) = 1.0 \times 10^{-6} \gg K_{sp}(\text{AgCl}) = 1.6 \times 10^{-10}$, the chloride ion concentration $[\text{Cl}^-]$ is predominantly governed by $\text{CuCl}$.

$$[\text{Cl}^-] \approx \sqrt{K_{sp}(\text{CuCl})} = \sqrt{1.0 \times 10^{-6}} = 10^{-3}\text{ M}$$

Now, applying solubility product expression for $\text{AgCl}$:

$$K_{sp}(\text{AgCl}) = [\text{Ag}^+][\text{Cl}^-]$$ $$1.6 \times 10^{-10} = [\text{Ag}^+] \times 10^{-3}$$ $$[\text{Ag}^+] = 1.6 \times 10^{-7}\text{ M}$$

Rewriting in terms of $x \times 10^{-5}\text{ M}$:

$$[\text{Ag}^+] = 0.016 \times 10^{-5}\text{ M} \implies x = 1.6 \times 10^{-2} \text{ or directly } x = 1.6 \text{ (when expressed as } 1.6 \times 10^{-7}\text{ M)}.$$

44. 29.2% (w/w) HCl stock solution has a density of 1.25 g mL$^{–1}$. The molecular weight of HCl is 36.5 g mol$^{–1}$. The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is________ [JEE 2012 P-I]

Answer: 8

Solution:

First, calculate molarity of stock $\text{HCl}$ solution:

$$\text{Molarity} = \frac{\% (w/w) \times \text{density} \times 10}{\text{Molar Mass}} = \frac{29.2 \times 1.25 \times 10}{36.5} = 10\text{ M}$$

Now, using dilution equation $M_1 V_1 = M_2 V_2$ to prepare $200\text{ mL}$ of $0.4\text{ M HCl}$:

$$10\text{ M} \times V_1 = 0.4\text{ M} \times 200\text{ mL}$$ $$V_1 = \frac{0.4 \times 200}{10} = 8\text{ mL}$$

45. Consider the following list of reagents: K$_2$Cr$_2$O$_7$, alkaline KMnO$_4$, CuSO$_4$, H$_2$O$_2$, Cl$_2$, O$_3$, FeCl$_3$, HNO$_3$ and Na$_2$S$_2$O$_3$. The total number of reagents that can oxidise aqueous iodide to iodine is________ [JEE Adv. 2014 P-I]

Answer: 7

Solution:

Reagents capable of oxidizing aqueous iodide ($\text{I}^-$) to iodine ($\text{I}_2$):

1. Acidified $\text{K}_2\text{Cr}_2\text{O}_7$ ($\text{Cr}_2\text{O}_7^{2-} + 6\text{I}^- + 14\text{H}^+ \longrightarrow 2\text{Cr}^{3+} + 3\text{I}_2 + 7\text{H}_2\text{O}$)

2. $\text{CuSO}_4$ ($2\text{Cu}^{2+} + 4\text{I}^- \longrightarrow \text{Cu}_2\text{I}_2\downarrow + \text{I}_2$)

3. $\text{H}_2\text{O}_2$ ($\text{H}_2\text{O}_2 + 2\text{I}^- + 2\text{H}^+ \longrightarrow \text{I}_2 + 2\text{H}_2\text{O}$)

4. $\text{Cl}_2$ ($\text{Cl}_2 + 2\text{I}^- \longrightarrow 2\text{Cl}^- + \text{I}_2$)

5. $\text{O}_3$ ($\text{O}_3 + 2\text{I}^- + \text{H}_2\text{O} \longrightarrow \text{O}_2 + \text{I}_2 + 2\text{OH}^-$)

6. $\text{FeCl}_3$ ($2\text{Fe}^{3+} + 2\text{I}^- \longrightarrow 2\text{Fe}^{2+} + \text{I}_2$)

7. $\text{HNO}_3$ ($2\text{NO}_3^- + 8\text{H}^+ + 6\text{I}^- \longrightarrow 2\text{NO} + 3\text{I}_2 + 4\text{H}_2\text{O}$)

Note: Alkaline $\text{KMnO}_4$ oxidizes $\text{I}^-$ further to iodate ($\text{IO}_3^-$), not $\text{I}_2$. $\text{Na}_2\text{S}_2\text{O}_3$ acts as a reducing agent. Therefore, 7 reagents can oxidise $\text{I}^-$ to $\text{I}_2$.

46. Among PbS, CuS, HgS, MnS, Ag$_2$S, NiS, CoS, Bi$_2$S$_3$ and SnS$_2$, the total number of black coloured sulphides is____ [JEE Adv. 2014 P-I]

Answer: 6

Solution:

Examining the colors of the given metal sulfides:

1. $\text{PbS}$: Black

2. $\text{CuS}$: Black

3. $\text{HgS}$: Black

4. $\text{Ag}_2\text{S}$: Black

5. $\text{NiS}$: Black

6. $\text{CoS}$: Black

7. $\text{MnS}$: Buff / Pink

8. $\text{Bi}_2\text{S}_3$: Dark brown (precipitate appears blackish-brown)

9. $\text{SnS}_2$: Yellow

The total number of strictly black coloured sulfides is 6 ($\text{PbS}, \text{CuS}, \text{HgS}, \text{Ag}_2\text{S}, \text{NiS}, \text{CoS}$).