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Multiple Choice Questions
One and More than One Correct
1. The species present in solution when $\text{CO}_2$ is dissolved in water are [JEE Main 2006]
a. $\text{CO}_2, \text{H}_2\text{CO}_3, \text{HCO}_3^-, \text{CO}_3^{2-}$
b. $\text{H}_2\text{CO}_3, \text{CO}_3^{2-}$
c. $\text{CO}_2, \text{HCO}_3^-$
d. $\text{CO}_2, \text{H}_2\text{CO}_3$
Answer: (a)
Solution:
$$\text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^- \rightleftharpoons \text{H}^+ + \text{CO}_3^{2-}$$2. $\text{MgSO}_4$ on reaction with $\text{NH}_4\text{OH}$ and $\text{Na}_2\text{HPO}_4$ forms a white crystalline precipitate. What is its formula? [JEE Main 2006]
a. $\text{Mg}(\text{NH}_4)\text{PO}_4$
b. $\text{Mg}_3(\text{PO}_4)_2$
c. $\text{MgCl}_2 \cdot \text{MgSO}_4$
d. $\text{MgSO}_4$
Answer: (a)
Solution:
Test of $\text{Mg}^{2+}$ ion
$$\text{Mg}^{2+} + \text{NH}_4\text{OH} + \text{Na}_2\text{HPO}_4 \longrightarrow \text{Mg}(\text{NH}_4)\text{PO}_4 \downarrow$$3. Among the following, the paramagnetic compound is [JEE 2007 P-I]
a. $\text{Na}_2\text{O}_2$
b. $\text{O}_3$
c. $\text{N}_2\text{O}$
d. $\text{KO}_2$
Answer: (d)
Solution:
$\text{Na}_2\text{O}_2$: $\text{O}_2^{2-} = 1s^2 \sigma^*1s^2 \sigma 2s^2 \sigma^*2s^2 \sigma 2p_z^2 \pi 2p_x^2 \pi 2p_y^2 \pi^*2p_x^2 \pi^*2p_y^2$
Number of unpaired electrons $= 0$.
$\text{O}_3$: Number of unpaired electrons $= 0$.
$\text{N}_2\text{O}$: $\text{N}\equiv\text{N}\rightarrow\text{O}$, Number of unpaired electrons $= 0$.
$\text{KO}_2$: $\text{O}_2^- = 1s^2 \sigma^*1s^2 \sigma 2s^2 \sigma^*2s^2 \sigma 2p_z^2 \pi 2p_x^2 \pi 2p_y^2 \pi^*2p_x^2 \pi^*2p_y^1$
Number of unpaired electrons $= 1$
Thus $\text{KO}_2$ is paramagnetic. Hence, (d) is correct.
4. The percentage of p-character in the orbitals forming $\text{P}-\text{P}$ bonds in $\text{P}_4$ is [JEE 2007 P-I]
a. 25
b. 33
c. 50
d. 75
Answer: (d)
Solution:
$\text{P}$ is $sp^3$ hybridised in $\text{P}_4$. Hence (d) is correct.
5. Aqueous solutions of $\text{Na}_2\text{S}_2\text{O}_3$ on reaction with $\text{Cl}_2$ gives [JEE 2008 P-I]
a. $\text{Na}_2\text{S}_4\text{O}_6$
b. $\text{NaHSO}_4$
c. $\text{NaCl}$
d. $\text{NaOH}$
Answer: (b)
Solution:
$$\text{Na}_2\text{S}_2\text{O}_3 + 4\text{Cl}_2 + 5\text{H}_2\text{O} \longrightarrow 2\text{NaHSO}_4 + 6\text{HCl}$$6. The reaction of $\text{P}_4$ with X leads selectively to $\text{P}_4\text{O}_6$. The X is [JEE 2009 P-I]
a. Dry $\text{O}_2$
b. A mixture of $\text{O}_2$ and $\text{N}_2$
c. Moist $\text{O}_2$
d. $\text{O}_2$ in the presence of aqueous $\text{NaOH}$ ($\text{N}_2$ is used to retard the further oxidation.)
Answer: (b)
Solution:
$$\text{P}_4 + 3\text{O}_2 \xrightarrow{\text{N}_2} \text{P}_4\text{O}_6 \text{ (exclusively)}$$7. The compound(s) formed upon combustion of sodium metal in excess air is(are) [JEE 2009 P-I]
a. $\text{Na}_2\text{O}_2$
b. $\text{Na}_2\text{O}$
c. $\text{NaO}_2$
d. $\text{NaOH}$
Answer: (a, b)
Solution:
In dry air
$$\text{Na} + \text{O}_2(\text{g}) \longrightarrow \text{Na}_2\text{O}_2 \text{ (excess)}$$ $$\text{Na} + \text{O}_2(\text{g}) \longrightarrow \text{Na}_2\text{O}$$8. The nitrogen oxide(s) that contain(s) $\text{N}-\text{N}$ bond(s) is(are) [JEE 2009 P-II]
a. $\text{N}_2\text{O}$
b. $\text{N}_2\text{O}_3$
c. $\text{N}_2\text{O}_4$
d. $\text{N}_2\text{O}_5$
Answer: (a, b, c)
Solution:
$\text{N}_2\text{O}$: $\text{N} \equiv \text{N} \rightarrow \text{O}$
$\text{N}_2\text{O}_3$: $\text{O}=\text{N}-\text{N}(=\text{O})\rightarrow\text{O}$
$\text{N}_2\text{O}_4$: $\text{O}_2\text{N}-\text{NO}_2$
$\text{N}_2\text{O}_5$: $\text{O}_2\text{N}-\text{O}-\text{NO}_2$
9. In the reaction $2\text{X} + \text{B}_2\text{H}_6 \rightarrow [\text{BH}_2(\text{X})_2]^+ [\text{BH}_4]^-$ the amine (s) X is (are) [JEE 2009 P-II]
a. $\text{NH}_3$
b. $\text{CH}_3\text{NH}_2$
c. $(\text{CH}_3)_2\text{NH}$
d. $(\text{CH}_3)_3\text{N}$
Answer: (a, b, c)
Solution:
Due to bulkiness of trimethylamine, it does not react.
10. The reagent(s) used for softening the temporary hardness of water is(are) [JEE 2010 P-I]
a. $\text{Ca}_3(\text{PO}_4)_2$
b. $\text{Ca}(\text{OH})_2$
c. $\text{Na}_2\text{CO}_3$
d. $\text{NaOCl}$
Answer: (b, c, d)
Solution:
$$\text{Ca}(\text{HCO}_3)_2 + \text{Ca}(\text{OH})_2 \longrightarrow 2\text{CaCO}_3 \downarrow + 2\text{H}_2\text{O} \text{ [Clarke's method]}$$ $$\text{NaOCl} + \text{H}_2\text{O} \rightleftharpoons \text{HOCl} + \text{NaOH}$$ $$\text{OH}^- + \text{HCO}_3^- \longrightarrow \text{CO}_3^{2-} + \text{H}_2\text{O}$$ $$\text{Ca}(\text{HCO}_3)_2 + \text{Na}_2\text{CO}_3 \longrightarrow \text{CaCO}_3 \downarrow + 2\text{NaHCO}_3$$11. Extraction of metal from the ore cassiterite involves [JEE 2011 P-I]
a. carbon reduction of an oxide ore
b. self-reduction of a sulphide ore
c. removal of copper impurity
d. removal of iron impurity
Answer: (a, d)
Solution:
Cassiterite ($\text{SnO}_2$) contains impurity of $\text{FeWO}_4$.
$$\text{SnO}_2 + 2\text{C} \longrightarrow \text{Sn} + 2\text{CO}$$12. Extra pure $\text{N}_2$ can be obtained by heating [JEE 2011 P-I]
a. $\text{NH}_3$ with $\text{CuO}$
b. $\text{NH}_4\text{NO}_3$
c. $(\text{NH}_4)_2\text{Cr}_2\text{O}_7$
d. $\text{Ba}(\text{N}_3)_2$
Answer: (d)
Solution:
$$\text{Ba}(\text{N}_3)_2 \xrightarrow{\Delta} \text{Ba} + 3\text{N}_2$$13. Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen? [JEE 2012 P-I]
a. $\text{HNO}_3, \text{NO}, \text{NH}_4\text{Cl}, \text{N}_2$
b. $\text{HNO}_3, \text{NO}, \text{N}_2, \text{NH}_4\text{Cl}$
c. $\text{HNO}_3, \text{NH}_4\text{Cl}, \text{NO}, \text{N}_2$
d. $\text{NO}, \text{HNO}_3, \text{NH}_4\text{Cl}, \text{N}_2$
Answer: (b)
Solution:
$\text{HNO}_3 \longrightarrow +5$
$\text{NO} \longrightarrow +2$
$\text{N}_2 \longrightarrow 0$
$\text{NH}_4\text{Cl} \longrightarrow -3$
14. With respect to graphite and diamond, which of the statement(s) given below is (are) correct? [JEE 2012 P-II]
a. Graphite is harder than diamond
b. Graphite has higher electrical conductivity than diamond
c. Graphite has higher thermal conductivity than diamond
d. Graphite has higher C–C bond order than diamond
Answer: (b, c, d)
Solution:
Graphite is $sp^2$ hybridized and contains van der waal’s force. While diamond is $sp^3$ hybridised.
15. The reaction of white phosphorus with aqueous $\text{NaOH}$ gives phosphine along with another phosphorus containing compound. The reaction type; the oxidation states of phosphorus in phosphine and the other product are respectively
a. redox reaction; –3 and –5
b. redox reaction; +3 and +5
c. disproportionation reaction; –3 and +5
d. disproportionation reaction; –3 and +3
Answer: None of these
Solution:
$$\text{P}_4 \text{ (white)} + \text{NaOH} + 3\text{H}_2\text{O} \longrightarrow \text{PH}_3 + 3\text{NaH}_2\text{PO}_2$$$\text{PH}_3$ has $-3$ and $\text{NaH}_2\text{PO}_2$ has $+1$ O.N.
16. The shape of $\text{XeO}_2\text{F}_2$ molecule is [JEE 2012 P-II]
a. trigonal bipyramidal
b. square planar
c. tetrahedral
d. see-saw
Answer: (d)
Solution:
Shape of $\text{XeO}_2\text{F}_2$ is see-saw.
Hybridisation: $\frac{1}{2}(V + M - C + A) = \frac{1}{2}(8 + 2 - 0 + 0) = 5 (sp^3d)$
17. The correct statement(s) about $\text{O}_3$ is(are) [JEE Adv. 2013 P-II]
a. O–O bond lengths are equal.
b. Thermal decomposition of $\text{O}_3$ is endothermic.
c. $\text{O}_3$ is diamagnetic in nature.
d. $\text{O}_3$ has a bent structure.
Answer: (a, c, d)
18. Which of the following exists as covalent crystals in the solid state? [JEE Main 2013]
a. Iodine
b. Silicon
c. Sulphur
d. Phosphorus
Answer: (b)
Solution:
Silicon exists as covalent crystals in the solid state.
19. Concentrated nitric acid, upon long standing, turns yellow-brown due to the formation of [JEE Adv. 2013 P-I]
a. NO
b. $\text{NO}_2$
c. $\text{N}_2\text{O}$
d. $\text{N}_2\text{O}_4$
Answer: (b)
Solution:
$$4\text{HNO}_3 \longrightarrow 2\text{H}_2\text{O} + 4\text{NO}_2 + \text{O}_2$$$\text{NO}_2$ remains dissolved in nitric acid colouring it yellow or even red at higher temperature.
20. The carbon-based reduction method is NOT used for the extraction of [JEE Adv. 2013 P-II]
a. tin from $\text{SnO}_2$
b. iron from $\text{Fe}_2\text{O}_3$
c. aluminium from $\text{Al}_2\text{O}_3$
d. magnesium from $\text{MgCO}_3\cdot\text{CaCO}_3$
Answer: (c, d)
Solution:
$\text{Fe}_2\text{O}_3$ and $\text{SnO}_2$ undergoes C reduction. Hence (c) and (d) are correct.
21. The correct statement(s) for orthoboric acid is/are [JEE Adv. 2014 P-I]
a. It behaves as a weak acid in water due to self ionization
b. Acidity of its aqueous solution increases upon addition of ethylene glycol
c. It has a three dimensional structure due to hydrogen bonding.
d. It is a weak electrolyte in water
Answer: (b, d)
Solution:
(A) $\text{H}_3\text{BO}_3$ is a weak monobasic Lewis acid.
$$\text{H}_3\text{BO}_3 + \text{H}_2\text{O} \rightleftharpoons \text{B}(\text{OH})_4^- + \text{H}^+ \quad \ldots \text{(i)}$$(B) Equilibrium (i) is shifted in forward direction by the addition of syn-diols like ethylene glycol which forms a stable complex with $[\text{B}(\text{OH})_4]^-$.
(C) It has a planar sheet like structure due to hydrogen bonding.
(D) $\text{H}_3\text{BO}_3$ is a weak electrolyte in water.
22. The pair(s) of reagents that yield paramagnetic species is/are [JEE Adv. 2014 P-I]
a. Na and excess of $\text{NH}_3$
b. K and excess of $\text{O}_2$
c. Cu and dilute $\text{HNO}_3$
d. $\text{O}_2$ and 2-ethylanthraquinol
Answer: (a, b, c)
Solution:
(a) Sodium (Na) when dissolved in excess liquid ammonia, forms a blue coloured paramagnetic solution.
(b) $\text{K} + \text{O}_2 \longrightarrow \text{KO}_2 \text{ (potassium superoxide)}$ and $\text{KO}_2$ is paramagnetic.
(c) $3\text{Cu} + 8\text{HNO}_3\text{ (dilute)} \longrightarrow 3\text{Cu}(\text{NO}_3)_2 + 2\text{NO} + 4\text{H}_2\text{O}$
Where “$\text{NO}$” is paramagnetic.
(d) $\text{2-ethylanthraquinol} + \text{O}_2\text{ (air)} \longrightarrow \text{H}_2\text{O}_2 + \text{2-ethylanthraquinone}$
Where $\text{H}_2\text{O}_2$ is diamagnetic.
23. Under ambient conditions, the total number of gases released as products in the final step of the reaction scheme shown below is [JEE Adv. 2014 P-II]
$\text{XeF}_6 \xrightarrow[\text{slow disproportionation}]{\text{Complete Hydrolysis in OH}^- / \text{H}_2\text{O}} \text{P} + \text{other product } \text{Q} \xrightarrow{\text{Products}}$
a. 0
b. 1
c. 2
d. 3
Answer: (c)
Solution:
$$\text{XeF}_6 \xrightarrow[\text{OH}^-/\text{H}_2\text{O}]{\text{3H}_2\text{O}} \text{XeO}_3 + 3\text{H}_2\text{F}_2 \xrightarrow{\text{OH}^-/\text{H}_2\text{O}} \text{HXeO}_4^-$$ $$\text{HXeO}_4^- \xrightarrow[\text{disproportionation}]{\text{OH}^- / \text{H}_2\text{O}} \text{XeO}_6^{4-}(\text{s}) + \text{Xe}(\text{g}) + \text{H}_2\text{O}(\text{l}) + \text{O}_2(\text{g})$$24. The product formed in the reaction of $\text{SOCl}_2$ with white phosphorous is [JEE Adv. 2014 P-II]
a. $\text{PCl}_3$
b. $\text{SO}_2\text{Cl}_2$
c. $\text{SCl}_2$
d. $\text{POCl}_3$
Answer: (a)
Solution:
$$\text{P}_4(\text{s}) + 8\text{SOCl}_2(\text{l}) \longrightarrow 4\text{PCl}_3(\text{l}) + 4\text{SO}_2(\text{g}) + 2\text{S}_2\text{Cl}_2(\text{l})$$25. Hydrogen peroxide in its reaction with $\text{KIO}_4$ and $\text{NH}_2\text{OH}$ respectively, is acting as a [JEE Adv. 2014 P-II]
a. reducing agent, oxidising agent
b. reducing agent, reducing agent
c. oxidising agent, oxidising agent
d. oxidising agent, reducing agent
Answer: (a)
Solution:
$$\text{KIO}_4 + \text{H}_2\text{O}_2 \longrightarrow \text{KIO}_3 + \text{H}_2\text{O} + \text{O}_2$$ $$\text{NH}_2\text{OH} + 3\text{H}_2\text{O}_2 \longrightarrow \text{HNO}_3 + 4\text{H}_2\text{O}$$26. In the context of the Hall-Heroult process for the extraction of $\text{Al}$, which of the following statements is false? [JEE Main 2015]
a. CO and $\text{CO}_2$ are produced in this process
b. $\text{Al}_2\text{O}_3$ is mixed with $\text{CaF}_2$ which lowers the melting point of the mixture and brings conductivity
c. $\text{Al}^{3+}$ is reduced at the cathode to form $\text{Al}$
d. $\text{Na}_3\text{AlF}_6$ serves as the electrolyte
Answer: (d)
27. Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy? [JEE Main 2015]
a. $\text{CaSO}_4$
b. $\text{BeSO}_4$
c. $\text{BaSO}_4$
d. $\text{SrSO}_4$
Answer: (b)
Solution:
$\text{BaSO}_4$ is least soluble.
$\text{BeSO}_4$ is most soluble.
28. Which among the following is the most reactive? [JEE Main 2015]
a. $\text{Cl}_2$
b. $\text{Br}_2$
c. $\text{I}_2$
d. $\text{ICl}$
Answer: (d)
Solution:
The interhalogen compounds are generally more reactive than halogens (except $\text{F}_2$).
29. Which one has the highest boiling point? [JEE Main 2015]
a. He
b. Ne
c. Kr
d. Xe
Answer: (d)
Solution:
Xe has the highest boiling point.
30. The correct statement(s) regarding, (i) $\text{HClO}$, (ii) $\text{HClO}_2$, (iii) $\text{HClO}_3$ and (iv) $\text{HClO}_4$ is (are) [JEE Adv. 2015 P-II]
a. The number of $\text{Cl}=\text{O}$ bonds in (ii) and (iii) together is two
b. The number of lone pairs of electrons on $\text{Cl}$ in (ii) and (iii) together is three
c. The hybridization of $\text{Cl}$ in (iv) is $sp^3$
d. Amongst (i) to (iv), the strongest acid is (i)
Answer: (b, c)
Solution:
(I) $\text{H}-\text{O}-\text{Cl}$
(II) $\text{H}-\text{O}-\text{Cl}=\text{O}$
(III) $\text{H}-\text{O}-\text{Cl}(=\text{O})_2$
(IV) $\text{H}-\text{O}-\text{Cl}(=\text{O})_3$
Multiple Choice Questions
One and More than One Correct
31. The species present in solution when $\text{CO}_2$ is dissolved in water are [JEE Main 2006]
a. $\text{H}_2\text{CO}_3, \text{HCO}_3^-, \text{CO}_3^{2-}$
b. $\text{CO}_2, \text{H}_2\text{CO}_3, \text{HCO}_3^-, \text{CO}_3^{2-}$
c. $\text{H}_2\text{CO}_3, \text{CO}_3^{2-}$
d. $\text{CO}_2, \text{HCO}_3^-$
Answer: (a)
Solution:
$$\text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^- \rightleftharpoons \text{H}^+ + \text{CO}_3^{2-}$$32. $\text{MgSO}_4$ on reaction with $\text{NH}_4\text{OH}$ and $\text{Na}_2\text{HPO}_4$ forms a white crystalline precipitate. What is its formula? [JEE Main 2006]
a. $\text{Mg(NH}_4)\text{PO}_4$
b. $\text{Mg}_3(\text{PO}_4)_2$
c. $\text{MgCl}_2 \cdot \text{MgSO}_4$
d. $\text{MgSO}_4$
Answer: (a)
Solution:
Test of $\text{Mg}^{2+}$ ion:
$$\text{Mg}^{2+} + \text{NH}_4\text{OH} + \text{Na}_2\text{HPO}_4 \longrightarrow \text{Mg(NH}_4)\text{PO}_4 \downarrow$$33. Among the following, the paramagnetic compound is [JEE 2007 P-I]
a. $\text{Na}_2\text{O}_2$
b. $\text{O}_3$
c. $\text{N}_2\text{O}$
d. $\text{KO}_2$
Answer: (d)
Solution:
$\text{Na}_2\text{O}_2 \implies \text{O}_2^{2-} \implies \text{Number of unpaired electrons} = 0$.
$\text{O}_3 \implies \text{Number of unpaired electrons} = 0$.
$\text{N}_2\text{O} \implies \text{Number of unpaired electrons} = 0$.
$\text{KO}_2 \implies \text{O}_2^- \implies \text{Number of unpaired electrons} = 1$.
Thus $\text{KO}_2$ is paramagnetic.
34. The percentage of p-character in the orbitals forming $\text{P}-\text{P}$ bonds in $\text{P}_4$ is [JEE 2007 P-I]
a. 25
b. 33
c. 50
d. 75
Answer: (d)
Solution:
$\text{P}$ is $sp^3$ hybridised in $\text{P}_4$. Hence % p-character is $75\%$.
35. Aqueous solutions of $\text{Na}_2\text{S}_2\text{O}_3$ on reaction with $\text{Cl}_2$ gives [JEE 2008 P-I]
a. $\text{Na}_2\text{S}_4\text{O}_6$
b. $\text{NaHSO}_4$
c. $\text{NaCl}$
d. $\text{NaOH}$
Answer: (b)
Solution:
$$\text{Na}_2\text{S}_2\text{O}_3 + 4\text{Cl}_2 + 5\text{H}_2\text{O} \longrightarrow 2\text{NaHSO}_4 + 8\text{HCl}$$36. The reaction of $\text{P}_4$ with X leads selectively to $\text{P}_4\text{O}_6$. The X is [JEE 2009 P-I]
a. Dry $\text{O}_2$
b. A mixture of $\text{O}_2$ and $\text{N}_2$
c. Moist $\text{O}_2$
d. $\text{O}_2$ in the presence of aqueous $\text{NaOH}$ ($\text{N}_2$ is used to retard the further oxidation.)
Answer: (b)
Solution:
$$\text{P}_4 + 3\text{O}_2 \xrightarrow{\text{N}_2} \text{P}_4\text{O}_6\text{ (exclusively)}$$37. The compound(s) formed upon combustion of sodium metal in excess air is(are) [JEE 2009 P-I]
a. $\text{Na}_2\text{O}_2$
b. $\text{Na}_2\text{O}$
c. $\text{NaO}_2$
d. $\text{NaOH}$
Answer: (a, b)
Solution:
In dry air, sodium metal on combustion forms $\text{Na}_2\text{O}_2$ along with small amounts of $\text{Na}_2\text{O}$.
38. The nitrogen oxide(s) that contain(s) $\text{N}-\text{N}$ bond(s) is(are) [JEE 2009 P-II]
a. $\text{N}_2\text{O}$
b. $\text{N}_2\text{O}_3$
c. $\text{N}_2\text{O}_4$
d. $\text{N}_2\text{O}_5$
Answer: (a, b, c)
Solution:
$\text{N}_2\text{O} \implies \text{N}\equiv\text{N}\rightarrow\text{O}$
$\text{N}_2\text{O}_3 \implies \text{O}=\text{N}-\text{N}(=\text{O})\rightarrow\text{O}$
$\text{N}_2\text{O}_4 \implies \text{O}_2\text{N}-\text{NO}_2$
39. In the reaction $2\text{X} + \text{B}_2\text{H}_6 \longrightarrow [\text{BH}_2(\text{X})_2]^+[\text{BH}_4]^-$, the amine (s) X is (are) [JEE 2009 P-II]
a. $\text{NH}_3$
b. $\text{CH}_3\text{NH}_2$
c. $(\text{CH}_3)_2\text{NH}$
d. $(\text{CH}_3)_3\text{N}$
Answer: (a, b, c)
Solution:
Due to bulkiness of trimethylamine $(\text{CH}_3)_3\text{N}$, it does not react to form this unsymmetrical cleavage complex.
40. The reagent(s) used for softening the temporary hardness of water is(are) [JEE 2010 P-I]
a. $\text{Ca}_3(\text{PO}_4)_2$
b. $\text{Ca(OH)}_2$
c. $\text{Na}_2\text{CO}_3$
d. $\text{NaOCl}$
Answer: (b, c, d)
Solution:
$$\text{Ca(HCO}_3)_2 + \text{Ca(OH)}_2 \longrightarrow 2\text{CaCO}_3 \downarrow + 2\text{H}_2\text{O}\text{ [Clarke's method]}$$ $$\text{NaOCl} + \text{H}_2\text{O} \rightleftharpoons \text{HOCl} + \text{NaOH}$$ $$2\text{OH}^- + \text{HCO}_3^- \longrightarrow \text{CO}_3^{2-} + \text{H}_2\text{O}$$ $$\text{Ca(HCO}_3)_2 + \text{Na}_2\text{CO}_3 \longrightarrow \text{CaCO}_3 \downarrow + 2\text{NaHCO}_3$$41. Extraction of metal from the ore cassiterite involves [JEE 2011 P-I]
a. carbon reduction of an oxide ore
b. self-reduction of a sulphide ore
c. removal of copper impurity
d. removal of iron impurity
Answer: (a, d)
Solution:
Cassiterite ($\text{SnO}_2$) contains impurity of $\text{FeWO}_4$.
$$\text{SnO}_2 + 2\text{C} \longrightarrow \text{Sn} + 2\text{CO}$$42. Extra pure $\text{N}_2$ can be obtained by heating [JEE 2011 P-I]
a. $\text{NH}_3$ with $\text{CuO}$
b. $\text{NH}_4\text{NO}_3$
c. $(\text{NH}_4)_2\text{Cr}_2\text{O}_7$
d. $\text{Ba(N}_3)_2$
Answer: (d)
Solution:
$$\text{Ba(N}_3)_2 \xrightarrow{\Delta} \text{Ba} + 3\text{N}_2$$43. Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen? [JEE 2012 P-I]
a. $\text{HNO}_3, \text{NO}, \text{NH}_4\text{Cl}, \text{N}_2$
b. $\text{HNO}_3, \text{NO}, \text{N}_2, \text{NH}_4\text{Cl}$
c. $\text{HNO}_3, \text{NH}_4\text{Cl}, \text{NO}, \text{N}_2$
d. $\text{NO}, \text{HNO}_3, \text{NH}_4\text{Cl}, \text{N}_2$
Answer: (b)
Solution:
$\text{HNO}_3 \longrightarrow +5$
$\text{NO} \longrightarrow +2$
$\text{N}_2 \longrightarrow 0$
$\text{NH}_4\text{Cl} \longrightarrow -3$
44. With respect to graphite and diamond, which of the statement(s) given below is (are) correct? [JEE 2012 P-II]
a. Graphite is harder than diamond
b. Graphite has higher electrical conductivity than diamond
c. Graphite has higher thermal conductivity than diamond
d. Graphite has higher C–C bond order than diamond
Answer: (b, c, d)
Solution:
Graphite is $sp^2$ hybridized and contains van der Waals forces between layers, having delocalized $\pi$-electrons. Diamond is $sp^3$ hybridized.
45. The reaction of white phosphorus with aqueous $\text{NaOH}$ gives phosphine along with another phosphorus containing compound. The reaction type; the oxidation states of phosphorus in phosphine and the other product are respectively [JEE 2012 P-II]
a. redox reaction; –3 and –5
b. redox reaction; +3 and +5
c. disproportionation reaction; –3 and +5
d. disproportionation reaction; –3 and +1
Answer: None of these
Solution:
$$\text{P}_4\text{ (white)} + 3\text{NaOH} + 3\text{H}_2\text{O} \longrightarrow \text{PH}_3 + 3\text{NaH}_2\text{PO}_2$$$\text{PH}_3$ has $-3$ and $\text{NaH}_2\text{PO}_2$ has $+1$ O.N.
46. The shape of $\text{XeO}_2\text{F}_2$ molecule is [JEE 2012 P-II]
a. trigonal bipyramidal
b. square planar
c. tetrahedral
d. see-saw
Answer: (d)
Solution:
Shape of $\text{XeO}_2\text{F}_2$ is see-saw ($sp^3d$ hybridization with 1 lone pair).
$$H = \frac{1}{2}(V + M - C + A) = \frac{1}{2}(8 + 2 - 0 + 0) = 5\text{ }(sp^3d)$$47. The correct statement(s) about $\text{O}_3$ is(are) [JEE Adv. 2013 P-II]
a. O–O bond lengths are equal.
b. Thermal decomposition of $\text{O}_3$ is endothermic.
c. $\text{O}_3$ is diamagnetic in nature.
d. $\text{O}_3$ has a bent structure.
Answer: (a, c, d)
48. Which of the following exists as covalent crystals in the solid state? [JEE Main 2013]
a. Iodine
b. Silicon
c. Sulphur
d. Phosphorus
Answer: (b)
Solution:
Silicon exists as covalent crystals in the solid state.
49. Concentrated nitric acid, upon long standing, turns yellow-brown due to the formation of [JEE Adv. 2013 P-I]
a. $\text{NO}$
b. $\text{NO}_2$
c. $\text{N}_2\text{O}$
d. $\text{N}_2\text{O}_4$
Answer: (b)
Solution:
$$4\text{HNO}_3 \longrightarrow 2\text{H}_2\text{O} + 4\text{NO}_2 + \text{O}_2$$$\text{NO}_2$ remains dissolved in nitric acid colouring it yellow or even red at higher temperature.
50. The carbon-based reduction method is NOT used for the extraction of [JEE Adv. 2013 P-II]
a. tin from $\text{SnO}_2$
b. iron from $\text{Fe}_2\text{O}_3$
c. aluminium from $\text{Al}_2\text{O}_3$
d. magnesium from $\text{MgCO}_3 \cdot \text{CaCO}_3$
Answer: (c, d)
Solution:
$\text{Fe}_2\text{O}_3$ and $\text{SnO}_2$ undergo carbon reduction. Aluminium and Magnesium are extracted by electrolytic reduction. Hence (c) and (d) are correct.
51. The correct statement(s) for orthoboric acid is/are [JEE Adv. 2014 P-I]
a. It behaves as a weak acid in water due to self ionization
b. Acidity of its aqueous solution increases upon addition of ethylene glycol
c. It has a three dimensional structure due to hydrogen bonding.
d. It is a weak electrolyte in water
Answer: (b, d)
Solution:
(A) $\text{H}_3\text{BO}_3$ is a weak monobasic Lewis acid:
$$\text{H}_3\text{BO}_3 + \text{H}_2\text{O} \rightleftharpoons \text{B(OH)}_4^- + \text{H}^+$$(B) Equilibrium is shifted in forward direction by the addition of cis-1,2-diols like ethylene glycol which forms a stable chelate complex with $[\text{B(OH)}_4]^-$.
(C) It has a two-dimensional planar sheet-like structure due to hydrogen bonding.
(D) $\text{H}_3\text{BO}_3$ is a weak electrolyte in water.
52. The pair(s) of reagents that yield paramagnetic species is/are [JEE Adv. 2014 P-I]
a. Na and excess of $\text{NH}_3$
b. K and excess of $\text{O}_2$
c. Cu and dilute $\text{HNO}_3$
d. $\text{O}_2$ and 2-ethylanthraquinol
Answer: (a, b, c)
Solution:
(a) Sodium ($\text{Na}$) when dissolved in excess liquid ammonia, forms a blue coloured paramagnetic solution due to ammoniated electrons.
(b) $\text{K} + \text{O}_2 \longrightarrow \text{KO}_2$ (potassium superoxide), which contains $\text{O}_2^-$ and is paramagnetic.
(c) $3\text{Cu} + 8\text{HNO}_3\text{ (dilute)} \longrightarrow 3\text{Cu(NO}_3)_2 + 2\text{NO} + 4\text{H}_2\text{O}$, where $\text{NO}$ is paramagnetic.
(d) 2-ethylanthraquinol $+\text{ O}_2 \longrightarrow \text{2-ethylanthraquinone} + \text{H}_2\text{O}_2$, where products are diamagnetic.
53. Under ambient conditions, the total number of gases released as products in the final step of the reaction scheme shown below is [JEE Adv. 2014 P-II]
$\text{XeF}_6 \xrightarrow{\text{Complete Hydrolysis}} P + \text{other product}$
$P + \text{OH}^- / \text{H}_2\text{O} \xrightarrow{\text{slow disproportionation}} Q$
$Q \xrightarrow{\text{Complete disproportionation in OH}^- / \text{H}_2\text{O}} \text{Products}$
a. 0
b. 1
c. 2
d. 3
Answer: (c)
Solution:
$$\text{XeF}_6 + 3\text{H}_2\text{O} \longrightarrow \text{XeO}_3 + 6\text{HF}$$ $$\text{XeO}_3 + \text{OH}^- \longrightarrow \text{HXeO}_4^-$$ $$2\text{HXeO}_4^- + 2\text{OH}^- \longrightarrow \text{XeO}_6^{4-} + \text{Xe}(g) + 2\text{H}_2\text{O} + \text{O}_2(g)$$The gases released in the final step are $\text{Xe}(g)$ and $\text{O}_2(g)$. Total number of gases = 2.
54. The product formed in the reaction of $\text{SOCl}_2$ with white phosphorous is [JEE Adv. 2014 P-II]
a. $\text{PCl}_3$
b. $\text{SO}_2\text{Cl}_2$
c. $\text{SCl}_2$
d. $\text{POCl}_3$
Answer: (a)
Solution:
$$\text{P}_4(s) + 8\text{SOCl}_2(l) \longrightarrow 4\text{PCl}_3(l) + 4\text{SO}_2(g) + 2\text{S}_2\text{Cl}_2(l)$$55. Hydrogen peroxide in its reaction with $\text{KIO}_4$ and $\text{NH}_2\text{OH}$ respectively, is acting as a [JEE Adv. 2014 P-II]
a. reducing agent, oxidising agent
b. reducing agent, reducing agent
c. oxidising agent, oxidising agent
d. oxidising agent, reducing agent
Answer: (a)
Solution:
$$\text{KIO}_4 + \text{H}_2\text{O}_2 \longrightarrow \text{KIO}_3 + \text{H}_2\text{O} + \text{O}_2\text{ (Reducing agent)}$$ $$2\text{NH}_2\text{OH} + 3\text{H}_2\text{O}_2 \longrightarrow \text{HNO}_3 + 4\text{H}_2\text{O}\text{ (Oxidising agent)}$$56. In the context of the Hall-Heroult process for the extraction of $\text{Al}$, which of the following statements is false? [JEE Main 2015]
a. $\text{CO}$ and $\text{CO}_2$ are produced in this process
b. $\text{Al}_2\text{O}_3$ is mixed with $\text{CaF}_2$ which lowers the melting point of the mixture and brings conductivity
c. $\text{Al}^{3+}$ is reduced at the cathode to form $\text{Al}$
d. $\text{Na}_3\text{AlF}_6$ serves as the electrolyte
Answer: (d)
Solution:
$\text{Al}_2\text{O}_3$ is the electrolyte dissolved in molten cryolite ($\text{Na}_3\text{AlF}_6$) and fluorspar ($\text{CaF}_2$).
57. Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy? [JEE Main 2015]
a. $\text{CaSO}_4$
b. $\text{BeSO}_4$
c. $\text{BaSO}_4$
d. $\text{SrSO}_4$
Answer: (b)
Solution:
$\text{BaSO}_4$ is least soluble, while $\text{BeSO}_4$ is most soluble because hydration energy of $\text{Be}^{2+}$ exceeds its lattice energy due to small size.
58. Which among the following is the most reactive? [JEE Main 2015]
a. $\text{Cl}_2$
b. $\text{Br}_2$
c. $\text{I}_2$
d. $\text{ICl}$
Answer: (d)
Solution:
The interhalogen compounds are generally more reactive than halogens (except $\text{F}_2$).
59. Which one has the highest boiling point? [JEE Main 2015]
a. He
b. Ne
c. Kr
d. Xe
Answer: (d)
Solution:
Xe has the largest atomic size and polarizability, hence strongest dispersion forces and highest boiling point.
60. The correct statement(s) regarding, (i) $\text{HClO}$, (ii) $\text{HClO}_2$, (iii) $\text{HClO}_3$ and (iv) $\text{HClO}_4$ is (are) [JEE Adv. 2015 P-II]
a. The number of $\text{Cl}=\text{O}$ bonds in (ii) and (iii) together is two
b. The number of lone pairs of electrons on Cl in (ii) and (iii) together is three
c. The hybridization of Cl in (iv) is $sp^3$
d. Amongst (i) to (iv), the strongest acid is (i)
Answer: (b, c)
Solution:
(I) $\text{HClO} \implies \text{H}-\text{O}-\text{Cl}$ (3 lone pairs on Cl, 0 $\text{Cl}=\text{O}$ bonds)
(II) $\text{HClO}_2 \implies \text{H}-\text{O}-\text{Cl}=\text{O}$ (2 lone pairs on Cl, 1 $\text{Cl}=\text{O}$ bond)
(III) $\text{HClO}_3 \implies \text{H}-\text{O}-\text{Cl}(=\text{O})_2$ (1 lone pair on Cl, 2 $\text{Cl}=\text{O}$ bonds)
(IV) $\text{HClO}_4 \implies \text{H}-\text{O}-\text{Cl}(=\text{O})_3$ (0 lone pairs on Cl, 3 $\text{Cl}=\text{O}$ bonds, $sp^3$ hybridized)
Number of lone pairs in (ii) and (iii) together $= 2 + 1 = 3$.
Matrix Match/Column Match
59. Match the extraction processes listed in Column I with metals listed in Column II: [JEE 2006]
Column I
(A) Self reduction
(B) Carbon reduction
(C) Complex formation and displacement by metal
(D) Decomposition of iodide
Column II
1. Lead
2. Silver
3. Copper
4. Boron
a. A$\rightarrow$1, 3; B$\rightarrow$1, 3; C$\rightarrow$2; D$\rightarrow$4
b. A$\rightarrow$1, 2; B$\rightarrow$1, 2; C$\rightarrow$1, 2, 3; D$\rightarrow$1, 2, 4
c. A$\rightarrow$1, 2, 3; B$\rightarrow$1, 2; C$\rightarrow$1, 2, 3; D$\rightarrow$1, 2, 3, 4
d. A$\rightarrow$1, 2, 3; B$\rightarrow$1, 2, 4; C$\rightarrow$1, 3, 2; D$\rightarrow$1, 2, 3
Answer: (a)
Solution:
A$\rightarrow$1, 3; B$\rightarrow$1, 3; C$\rightarrow$2; D$\rightarrow$4
60. Match the following: [JEE 2006]
Column I
(A) $\text{Bi}^{3+} \longrightarrow (\text{BiO})^+$
(B) $[\text{AlO}_2]^- \longrightarrow \text{Al(OH)}_3$
(C) $\text{SiO}_4^{4-} \longrightarrow \text{Si}_2\text{O}_7^{6-}$
(D) $(\text{B}_4\text{O}_7)^{2-} \longrightarrow [\text{B(OH)}_4]^-$
Column II
1. Heat
2. Hydrolysis
3. Acidification
4. Dilution by water
a. A$\rightarrow$1; B$\rightarrow$3; C$\rightarrow$4; D$\rightarrow$2
b. A$\rightarrow$1; B$\rightarrow$2; C$\rightarrow$3; D$\rightarrow$4
c. A$\rightarrow$3; B$\rightarrow$1; C$\rightarrow$2; D$\rightarrow$4
d. A$\rightarrow$2; B$\rightarrow$3; C$\rightarrow$1; D$\rightarrow$2, 3
Answer: (d)
Solution:
A$\rightarrow$2; B$\rightarrow$3; C$\rightarrow$1; D$\rightarrow$2, 3
61. Match the conversions in Column I with the type(s) of reaction(s) given in Column II. [JEE 2008 P-II]
Column I
(A) $\text{PbS} \rightarrow \text{PbO}$
(B) $\text{CaCO}_3 \rightarrow \text{CaO}$
(C) $\text{ZnS} \rightarrow \text{Zn}$
(D) $\text{Cu}_2\text{S} \rightarrow \text{Cu}$
Column II
1. Roasting
2. Calcination
3. Carbon reduction
4. Self reduction
a. A$\rightarrow$1, 3, 4; B$\rightarrow$1, 3; C$\rightarrow$3, 4; D$\rightarrow$1, 2
b. A$\rightarrow$1, 2; B$\rightarrow$1, 2; C$\rightarrow$3, 4; D$\rightarrow$1, 2, 3, 4
c. A$\rightarrow$1; B$\rightarrow$2; C$\rightarrow$1, 3; D$\rightarrow$1, 4
d. A$\rightarrow$1, 3; B$\rightarrow$1, 2, 3, 4; C$\rightarrow$2, 1; D$\rightarrow$1, 2
Answer: (c)
Solution:
A$\rightarrow$1; B$\rightarrow$2; C$\rightarrow$1, 3; D$\rightarrow$1, 4
62. All the compounds listed in Column I react with water. Match the result of the respective reactions with the appropriate options listed in Column II. [JEE 2010 P-II]
Column I
(A) $(\text{CH}_3)_2\text{SiCl}_2$
(B) $\text{XeF}_4$
(C) $\text{Cl}_2$
(D) $\text{VCl}_5$
Column II
1. Hydrogen halide formation
2. Redox reaction
3. Reacts with glass
4. Polymerization
5. $\text{O}_2$ formation
a. A$\rightarrow$1, 2, 3; B$\rightarrow$1, 3; C$\rightarrow$1, 2, 3, 4; D$\rightarrow$1, 2
b. A$\rightarrow$1, 4; B$\rightarrow$1, 2, 3, 5; C$\rightarrow$1, 2, 3; D$\rightarrow$1
c. A$\rightarrow$1, 3; B$\rightarrow$1, 2, 3, 4; C$\rightarrow$1, 2, 3; D$\rightarrow$1, 2, 4
d. A$\rightarrow$1, 2, 3, 4; B$\rightarrow$2, 3, 4; C$\rightarrow$2, 3; D$\rightarrow$1, 2, 3, 4, 5
Answer: (b)
Solution:
A$\rightarrow$1, 4; B$\rightarrow$1, 2, 3, 5; C$\rightarrow$1, 2, 3; D$\rightarrow$1
(A) $(\text{CH}_3)_2\text{SiCl}_2 + 2\text{H}_2\text{O} \rightarrow (\text{CH}_3)_2\text{Si(OH)}_2 + 2\text{HCl}$; $(\text{CH}_3)_2\text{Si(OH)}_2$ undergoes polymerisation to form silicones.
(B) $3\text{XeF}_4 + 6\text{H}_2\text{O} \rightarrow \text{XeO}_3 + 2\text{Xe} + 12\text{HF} + \frac{3}{2}\text{O}_2$. Hydrofluoric acid ($\text{HF}$) reacts with glass.
(C) $\text{Cl}_2 + \text{H}_2\text{O} \rightarrow \text{HCl} + \text{HOCl}$ (Redox reaction, produces $\text{HCl}$ which reacts with glass).
(D) $\text{VCl}_5 + \text{H}_2\text{O} \rightarrow \text{VOCl}_3 + 2\text{HCl}$.
63. The unbalanced chemical reactions given in Column I show missing reagent or condition (?) which are provided in Column II. [JEE Adv. 2013 P-II]
Column I
(A) $\text{PbO}_2 + \text{H}_2\text{SO}_4 \xrightarrow{?} \text{PbSO}_4 + \text{O}_2 + \text{other product}$
(B) $\text{Na}_2\text{S}_2\text{O}_3 + \text{H}_2\text{O} + \text{Cl}_2 \xrightarrow{?} \text{NaHSO}_4 + \text{other product}$
(C) $\text{N}_2\text{H}_4 \xrightarrow{?} \text{N}_2 + \text{other product}$
(D) $\text{XeF}_2 \xrightarrow{?} \text{Xe} + \text{other product}$
Column II
1. NO
2. $\text{I}_2$
3. Warm
4. $\text{Cl}_2$
a. A$\rightarrow$1; B$\rightarrow$3; C$\rightarrow$4; D$\rightarrow$2
b. A$\rightarrow$3; B$\rightarrow$4; C$\rightarrow$2; D$\rightarrow$1
c. A$\rightarrow$3; B$\rightarrow$1; C$\rightarrow$2; D$\rightarrow$4
d. A$\rightarrow$3; B$\rightarrow$4; C$\rightarrow$1; D$\rightarrow$2
Answer: (b)
Solution:
A$\rightarrow$3; B$\rightarrow$4; C$\rightarrow$2; D$\rightarrow$1
(A) $\text{PbO}_2 + \text{H}_2\text{SO}_4 \xrightarrow{\text{Warm}} \text{PbSO}_4 + \text{H}_2\text{O} + \frac{1}{2}\text{O}_2$
(B) $\text{Na}_2\text{S}_2\text{O}_3 + 5\text{H}_2\text{O} + 4\text{Cl}_2 \rightarrow 2\text{NaHSO}_4 + 8\text{HCl}$
(C) $\text{N}_2\text{H}_4 + 2\text{I}_2 \rightarrow \text{N}_2 + 4\text{HI}$
(D) $\text{XeF}_2 + 2\text{NO} \rightarrow \text{Xe} + 2\text{NOF}$
Statement/Assertion and Reason
65. Statement-I: Boron always forms covalent bond. And
Statement-II: The small size of $\text{B}^{3+}$ favours formation of covalent bond. [JEE 2007 P-I]
a. Statement-I: is True, Statement-II is True; Statement-II: is correct explanation for Statement-I
b. Statement-I is True, Statement-II: is True; Statement-II is not a correct explanation for Statement-I
c. Statement-I is True, Statement-II is False
d. Statement-I is False, Statement-II is True
Answer: (a)
Solution:
Due to high sum of first three ionization enthalpies and small size of $\text{B}^{3+}$, boron always forms covalent compounds.
66. Statement-I: In water, orthoboric acid behaves as a weak monobasic acid. And
Statement-II: In water, orthoboric acid acts as a proton donor. [JEE 2007 P-I]
a. Statement-I: is True, Statement-II is True; Statement-II: is correct explanation for Statement-I
b. Statement-I is True, Statement-II: is True; Statement-II is not a correct explanation for Statement-I
c. Statement-I is True, Statement-II is False
d. Statement-I is False, Statement-II is True
Answer: (c)
Solution:
Orthoboric acid behaves as a weak monobasic acid because it acts as a Lewis acid by accepting $\text{OH}^-$ from water molecule, releasing a proton ($\text{H}^+$) from water, rather than donating its own proton.
67. Statement-I: Alkali metals dissolve in liquid ammonia to give blue solution. And
Statement-II: Alkali metals in liquid ammonia give solvated species of the type $[\text{M(NH}_3)_n]^+$ ($\text{M} =$ alkali metals). [JEE 2007 P-II]
a. Statement-I: is True, Statement-II is True; Statement-II: is correct explanation for Statement-I
b. Statement-I is True, Statement-II: is True; Statement-II is not a correct explanation for Statement-I
c. Statement-I is True, Statement-II is False
d. Statement-I is False, Statement-II is True
Answer: (b)
Solution:
The deep blue colour of the solution is due to ammoniated electrons $[\text{e(NH}_3)_y]^-$, which absorb light in the visible region, and not due to the cation $[\text{M(NH}_3)_x]^+$.
68. Statement-I: $\text{Pb}^{4+}$ compounds are stronger oxidizing agents than $\text{Sn}^{4+}$ compounds. And
Statement-II: The higher oxidation states for the group 14 elements are more stable for the heavier members of the group due to ‘inert pair effect’. [JEE 2008 P-I]
a. Statement-I: is True, Statement-II is True; Statement-II: is correct explanation for Statement-I
b. Statement-I is True, Statement-II: is True; Statement-II is not a correct explanation for Statement-I
c. Statement-I is True, Statement-II is False
d. Statement-I is False, Statement-II is True
Answer: (c)
Solution:
Statement-I is True ($\text{Pb}^{2+}$ is more stable than $\text{Pb}^{4+}$ due to inert pair effect, making $\text{Pb}^{4+}$ a strong oxidizing agent). Statement-II is False because lower oxidation states are more stable for heavier members, not higher oxidation states.
69. Statement-I: Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen.
Statement-II: The reaction between nitrogen and oxygen requires high temperature. [JEE Main 2015]
a. Statement-I: is True, Statement-II is True; Statement-II: is correct explanation for Statement-I
b. Statement-I is True, Statement-II: is True; Statement-II is not a correct explanation for Statement-I
c. Statement-I is True, Statement-II is False
d. Statement-I is False, Statement-II is True
Answer: (a)
Solution:
The triple bond in $\text{N}_2$ ($\text{N}\equiv\text{N}$) has a very high bond dissociation energy, so its reaction with oxygen is endothermic and requires very high temperature.
Paragraph
Paragraph for Question Nos. 70 to 72
The noble gases have closed-shell electronic configuration and are monoatomic gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions. The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers +2, +4 and +6. $\text{XeF}_4$ reacts violently with water to give $\text{XeO}_3$. The compounds of xenon exhibit rich stereochemistry and their geometries can be deduced considering the total number of electron pairs in the valence shell. [JEE 2007 P-I]
70. Argon is used in arc welding because of its
a. low reactivity with metal
b. ability to lower the melting point of metal
c. flammability
d. high calorific value
Answer: (a)
71. The structure of $\text{XeO}_3$ is
a. linear
b. planar
c. pyramidal
d. T-shaped
Answer: (c)
Solution:
$\text{XeO}_3$ has $sp^3$ hybridization with 1 lone pair, leading to a trigonal pyramidal structure.
72. $\text{XeF}_4$ and $\text{XeF}_6$ are expected to be
a. oxidizing
b. reducing
c. unreactive
d. strongly basic
Answer: (a)
Solution:
Fluorides of Xenon are strong fluorinating and oxidizing agents.
Paragraph for Question Nos. 73 to 75
There are some deposits of nitrates and phosphates in earth’s crust. Nitrates are more soluble in water. Nitrates are difficult to reduce under the laboratory conditions but microbes do it easily. Ammonia forms large number of complexes with transition metal ions. Hybridization easily explains the ease of sigma donation capability of $\text{NH}_3$ and $\text{PH}_3$. Phosphine is a flammable gas and is prepared from white phosphorus. [JEE 2008 P-I]
73. Among the following, the correct statement is
a. Phosphates have no biological significance in humans
b. Between nitrates and phosphates, phosphates are less abundant in earth’s crust
c. Between nitrates and phosphates, nitrates are less abundant in earth’s crust
d. Oxidation of nitrates is possible in soil
Answer: (c)
Solution:
Nitrates are highly soluble in water and get washed away, making them less abundant in the earth's crust compared to insoluble phosphates.
74. Among the following, the correct statement is
a. Between $\text{NH}_3$ and $\text{PH}_3$, $\text{NH}_3$ is better electron donor because the lone pair of electrons occupies spherical ‘s’ orbital and is less directional
b. Between $\text{NH}_3$ and $\text{PH}_3$, $\text{NH}_3$ is better electron donor because the lone pair of electrons occupies $sp^3$ orbital and is more directional
c. Between $\text{NH}_3$ and $\text{PH}_3$, $\text{NH}_3$ is a better electron donor because the lone pair of electrons occupies $sp^2$ orbital and is more directional
d. Between $\text{NH}_3$ and $\text{PH}_3$, $\text{NH}_3$ is better electron donor because the lone pair of electrons occupies spherical ‘s’ orbital and is less directional
Answer: (b)
Solution:
In $\text{NH}_3$, nitrogen is $sp^3$ hybridized, so the lone pair is in a directional $sp^3$ orbital. In $\text{PH}_3$, according to Drago's rule, hybridization is almost absent and the lone pair resides predominantly in an unhybridized, spherical pure $s$-orbital which is less directional.
75. White phosphorus on reaction with $\text{NaOH}$ gives $\text{PH}_3$, as one of the products. This is a
a. dimerization reaction
b. disproportionation reaction
c. condensation reaction
d. precipitation reaction
Answer: (b)
Solution:
$$\text{P}_4 + 3\text{NaOH} + 3\text{H}_2\text{O} \longrightarrow \text{PH}_3 + 3\text{NaH}_2\text{PO}_2$$Phosphorus undergoes oxidation (from $0$ to $+1$ in $\text{NaH}_2\text{PO}_2$) and reduction (from $0$ to $-3$ in $\text{PH}_3$), so it is a disproportionation reaction.
Paragraph for Question Nos. 76 and 77
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]
76. 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:
$$\text{Milli-equivalents of Bleach} = \text{Milli-equivalents of }\text{Na}_2\text{S}_2\text{O}_3$$ $$25 \times N = 48 \times 0.25 = 12$$ $$N = \frac{12}{25} = 0.48\text{ N}$$Since n-factor for $\text{OCl}^- \rightarrow \text{Cl}^-$ is 2:
$$\text{Molarity} = \frac{N}{2} = \frac{0.48}{2} = 0.24\text{ M}$$77. 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. $\text{Cl}_2\text{O}_7$
c. $\text{ClO}_2$
d. $\text{Cl}_2\text{O}_6$
Answer: (a)
Solution:
Bleaching powder contains $\text{Ca(OCl)}_2$, salt of hypochlorous acid ($\text{HOCl}$). The anhydride of $\text{HOCl}$ is $\text{Cl}_2\text{O}$.
$$2\text{HOCl} \xrightarrow{-\text{H}_2\text{O}} \text{Cl}_2\text{O}$$Paragraph for Question Nos. 78 and 79
The reactions of $\text{Cl}_2$ gas with cold-dilute and hot-concentrated $\text{NaOH}$ in water give sodium salts of two (different) oxoacids of chlorine, P and Q, respectively. The $\text{Cl}_2$ gas reacts with $\text{SO}_2$ gas, in presence of charcoal, to give a product R. R reacts with white phosphorus to give a compound S. On hydrolysis, S gives an oxoacid of phosphorus, T. [JEE Adv. 2013 P-II]
78. P and Q, respectively, are the sodium salts of
a. hypochlorous and chloric acids
b. hypochlorous and chlorous acids
c. chloric and perchloric acids
d. chloric and hypochlorous acids
Answer: (a)
Solution:
Cold & dilute: $\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO}\text{ (P: Sodium hypochlorite)} + \text{H}_2\text{O}$
Hot & conc.: $3\text{Cl}_2 + 6\text{NaOH} \rightarrow 5\text{NaCl} + \text{NaClO}_3\text{ (Q: Sodium chlorate)} + 3\text{H}_2\text{O}$
79. R, S and T, respectively, are
a. $\text{SO}_2\text{Cl}_2, \text{PCl}_5, \text{H}_3\text{PO}_4$
b. $\text{SO}_2\text{Cl}_2, \text{PCl}_3, \text{H}_3\text{PO}_3$
c. $\text{SOCl}_2, \text{PCl}_3, \text{H}_3\text{PO}_2$
d. $\text{SOCl}_2, \text{PCl}_5, \text{H}_3\text{PO}_4$
Answer: (a)
Solution:
$\text{Cl}_2 + \text{SO}_2 \xrightarrow{\text{charcoal}} \text{SO}_2\text{Cl}_2\text{ (R)}$
$\text{P}_4 + 10\text{SO}_2\text{Cl}_2 \rightarrow 4\text{PCl}_5\text{ (S)} + 10\text{SO}_2$
$\text{PCl}_5 + 4\text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4\text{ (T)} + 5\text{HCl}$
Paragraph for Question Nos. 80 and 81
Upon heating $\text{KClO}_3$ in the presence of catalytic amount of $\text{MnO}_2$, a gas W is formed. Excess amount of W reacts with white phosphorus to give X. The reaction of X with pure $\text{HNO}_3$ gives Y and Z. [JEE Adv. 2017 P-II]
80. W and X are, respectively
a. $\text{O}_3$ and $\text{P}_4\text{O}_6$
b. $\text{O}_2$ and $\text{P}_4\text{O}_6$
c. $\text{O}_2$ and $\text{P}_4\text{O}_{10}$
d. $\text{O}_3$ and $\text{P}_4\text{O}_{10}$
Answer: (c)
Solution:
$$2\text{KClO}_3 \xrightarrow[\Delta]{\text{MnO}_2} 2\text{KCl} + 3\text{O}_2\text{ (W)}$$ $$\text{P}_4 + 5\text{O}_2\text{ (excess)} \rightarrow \text{P}_4\text{O}_{10}\text{ (X)}$$81. Y and Z are, respectively
a. $\text{N}_2\text{O}_5$ and $\text{HPO}_3$
b. $\text{N}_2\text{O}_3$ and $\text{H}_3\text{PO}_4$
c. $\text{N}_2\text{O}_4$ and $\text{HPO}_3$
d. $\text{N}_2\text{O}_4$ and $\text{H}_3\text{PO}_3$
Answer: (a)
Solution:
$$\text{P}_4\text{O}_{10} + 4\text{HNO}_3 \rightarrow 2\text{N}_2\text{O}_5\text{ (Y)} + 4\text{HPO}_3\text{ (Z)}$$Integer and Subjective
82. The coordination number of Al in the crystalline state of $\text{AlCl}_3$ is_______. [JEE 2009 P-II]
Answer: 6
Solution:
In the solid crystalline state, $\text{AlCl}_3$ forms a 6-coordinate layer lattice structure where each $\text{Al}^{3+}$ is surrounded octahedrally by six $\text{Cl}^-$ ions.
83. The value of n in the molecular formula $\text{Be}_3\text{Al}_2\text{Si}_n\text{O}_{18}$ is_______. [JEE 2010 P-I]
Answer: 6
Solution:
Beryl formula: $\text{Be}_3\text{Al}_2\text{Si}_6\text{O}_{18}$. Thus, $n = 6$.
84. The total number of diprotic acids among the following is_______. [JEE 2010 P-II]
$\text{H}_3\text{PO}_4, \text{H}_2\text{SO}_4, \text{H}_3\text{PO}_3, \text{H}_2\text{CO}_3, \text{H}_2\text{S}_2\text{O}_7, \text{H}_3\text{BO}_3, \text{H}_3\text{PO}_2, \text{H}_2\text{CrO}_4, \text{H}_2\text{SO}_3$
Answer: 6
Solution:
Diprotic acids (containing 2 ionizable $\text{H}^+$): $\text{H}_2\text{SO}_4, \text{H}_3\text{PO}_3, \text{H}_2\text{CO}_3, \text{H}_2\text{S}_2\text{O}_7, \text{H}_2\text{CrO}_4, \text{H}_2\text{SO}_3$.
Total = 6.
85. Reaction of $\text{Br}_2$ with $\text{Na}_2\text{CO}_3$ in aqueous solution gives sodium bromide and sodium bromate with evolution of $\text{CO}_2$ gas. The number of sodium bromide molecules involved in the balanced chemical equation is _______. [JEE 2011 Paper -I]
Answer: 5
Solution:
$$3\text{Br}_2 + 3\text{Na}_2\text{CO}_3 \rightarrow 5\text{NaBr} + \text{NaBrO}_3 + 3\text{CO}_2$$Number of $\text{NaBr}$ molecules involved = 5.
86. The difference in the oxidation numbers of the two types of sulphur atoms in $\text{Na}_2\text{S}_4\text{O}_6$ is_______. [JEE 2011 P-I]
Answer: 5
Solution:
Structure of tetrathionate ion ($\text{S}_4\text{O}_6^{2-}$):
The two central S-atoms have oxidation number $0$.
The two terminal S-atoms have oxidation number $+5$.
Difference $= 5 - 0 = 5$.
87. Among the following, the number of compounds that can react with $\text{PCl}_5$ to give $\text{POCl}_3$ is _______. ($\text{O}_2, \text{CO}_2, \text{SO}_2, \text{H}_2\text{O}, \text{H}_2\text{SO}_4, \text{P}_4\text{O}_{10}$)
[JEE 2011 P-II]
Answer: 5
Solution:
Compounds reacting with $\text{PCl}_5$ to give $\text{POCl}_3$ are: $\text{SO}_2, \text{H}_2\text{O}, \text{H}_2\text{SO}_4, \text{P}_4\text{O}_{10}, \text{O}_2$. Total = 5.
88. A list of species having the formula $\text{XZ}_4$ is given below. Defining shape on the basis of the location of X and Z atoms, the total number of species having a square planar shape is_______. [JEE Adv. 2014 P-I]
$[\text{XeF}_4, \text{SF}_4, \text{SiF}_4, \text{BF}_4^-, \text{BrF}_4^-, [\text{Cu(NH}_3)_4]^{2+}, [\text{FeCl}_4]^-, [\text{CoCl}_4]^{2-}, [\text{PtCl}_4]^{2-}]$
Answer: 4
Solution:
Square planar species are: $\text{XeF}_4, \text{BrF}_4^-, [\text{Cu(NH}_3)_4]^{2+}, [\text{PtCl}_4]^{2-}$. Total = 4.
89. Three moles of $\text{B}_2\text{H}_6$ are completely reacted with methanol. The number of moles of boron containing product formed is_______. [JEE Adv. 2015 P-II]
Answer: 6
Solution:
$$\text{B}_2\text{H}_6 + 6\text{CH}_3\text{OH} \rightarrow 2\text{B(OCH}_3)_3 + 6\text{H}_2$$1 mole of $\text{B}_2\text{H}_6$ yields 2 moles of $\text{B(OCH}_3)_3$.
So, 3 moles of $\text{B}_2\text{H}_6$ yield $3 \times 2 = 6$ moles of trimethyl borate.
90. (a) Calculate the amount of Calcium oxide required when it reacts with $852\text{ gm}$ of $\text{P}_4\text{O}_{10}$. [JEE 2005]
(b) Write the structure of $\text{P}_4\text{O}_{10}$.
Solution:
(a) Reaction:
$$6\text{CaO} + \text{P}_4\text{O}_{10} \longrightarrow 2\text{Ca}_3(\text{PO}_4)_2$$Molar mass of $\text{P}_4\text{O}_{10} = 284\text{ g mol}^{-1}$
Moles of $\text{P}_4\text{O}_{10} = \frac{852}{284} = 3\text{ mol}$
Moles of $\text{CaO}$ required $= 6 \times 3 = 18\text{ mol}$
Mass of $\text{CaO} = 18 \times 56 = 1008\text{ g}$
91. Predict whether the following molecules are isostructural or not. Justify your answer. (i) $\text{NMe}_3$; (ii) $\text{N(SiMe}_3)_3$ [JEE 2005]
Solution:
No, they are not isostructural.
(i) $\text{NMe}_3$ has a pyramidal geometry ($sp^3$ hybridized N with a lone pair).
(ii) $\text{N(SiMe}_3)_3$ has a planar geometry ($sp^2$ hybridized N) due to $p\pi-d\pi$ back-bonding of the lone pair on nitrogen into the vacant d-orbitals of silicon.
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