Question 1012 Marks
Account for the following observation:
Acidity of oxo-acids of chlorine is $HOCl < HOClO < HOClO_2 < HOClO_3.$
AnswerOxygen is more electronegative than chlorine, therefore, dispersal of negative charge present on chlorine increases from $\text{ClO}^-\text{ to }\text{Cl}^-_4$ ion because number of oxygen atoms attached to chlorine is increasing. Therefore, stability of ions will increase in the order given below:$\text{ClO}^-<\text{ClO}^-_2<\text{ClO}^-_3<\text{ClO}^-_4$
This is due to increase in stability of conjugate base, acidic strength of corresponding acid increases in the following order:$\text{HClO}<\text{HClO}_2<\text{HClO}_3<\text{HClO}_4$
View full question & answer→Question 1022 Marks
Give reason for the following:
Nitric oxide becomes brown when released in air.
AnswerNitric oxide readily combines with $O_2$ of air to form nitrogen dioxide which is brown in colour.$\ \ \ 2\text{NO}\ \ \ \ +\ \ \ \text{O}_2\ \ \ \xrightarrow{\ \ \ \ \ }\ \ \ 2\text{NO}_2\\^\text{Nitric oxide}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{Nitrogen dioxide}\\^\text{(Colourless)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{(Brown)}$
View full question & answer→Question 1032 Marks
$PCl_5$ reacts with finely divided silver on heating and a white silver salt is obtained, which dissolves on adding excess aqueous $NH_3$ solution. Write the reactions involved to explain what happens.
Answer$PCl_5$ reacts with silver to form white silver salt $(AgCl).$ This dissolves in aqueous ammonia to form soluble complex.$\text{PCl}_5+2\text{Ag}\xrightarrow{\ \ \ \ \ \ \ \ \ }2\text{AgCl}+\text{PCl}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{\text{White ppt}}$
$\text{AgCl}+2\text{NH}_3(\text{aq})\xrightarrow{\ \ \ \ \ \ \ \ \ \ }[\text{Ag}(\text{NH}_3)]^+\text{Cl}^-\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{\text{Soluble complex}}$
View full question & answer→Question 1042 Marks
Account for the following:
On addition of ozone gas to $KI$ solution, violet vapours are obtained.
AnswerOzone gas acts as a strong oxidising agent, so it oxidises iodide ions to iodine.
$2I^-(aq) + H_2O(l) + O_3(g) → 2OH^-(aq) + I_2(g) + O_2(g)$
$I_2$ vapours evolved have violet colour.
View full question & answer→Question 1052 Marks
Give reason for the following:
Oxygen generally exhibits an oxidation state of -2 only whereas other members of its family show oxidation states of +2, +4 and +6 as well.
AnswerThe electronic configuration of oxygen is $1\text{S}^2\ 2\text{S}^2\ 2\text{P}^2_\text{x}\ 2\text{P}^1_\text{y}\ 2\text{P}^1_\text{z}$ i.e., it has two half-filled orbitals and there is no d-orbital available for excitation of electrons. Further, it is the most electronegative element of its family. Hence, it shows oxidation state of -2 only. Other elements like sulphur have d-orbitals available for excitation, thereby giving four and six half-filled orbitals. Moreover, they can combine with more electronegative elements. Hence, they show oxidation states of +2, +4 and +6 also.
View full question & answer→Question 1062 Marks
Give reason for the following:
$H_2O$ is a liquid and $H_2S$ is a gas.
AnswerDue to small size and high electronegativity of oxygen, molecules of water are associated through hydrogen bonding, resulting in its liquid state. On the other hand, $H_2S$ molecules are not associated through H-bonding. Hence, it is a gas.
View full question & answer→Question 1072 Marks
Why does nitrogen show catenation properties less than phosphorus?
AnswerCatenation is much more common in phosphorous compounds than in nitrogen compounds. This is because of the relative weakness of the N-N single bond as compared to the P-P single bond. Since nitrogen atom is smaller, there is greater repulsion of electron density of two nitrogen atoms, thereby weakening the N-N single bond.
View full question & answer→Question 1082 Marks
What happens when:
$XeF_6$ is partially hydrolysed?
Answer$XeF_6 + H_2O → XeOF_4 + 2HF$
$XeF_6 + 2H_2O → XeO_2F_2 + 4HF$
View full question & answer→MCQ 1092 Marks
Note: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Both rhombic and monoclinic sulphur exist as $S_8$ but oxygen exists as $O_2.$
Reason: Oxygen forms $\text{p}\pi-\text{p}\pi$ multiple bond due to small size and small bond length but $\text{p}\pi-\text{p}\pi$ bonding is not possible in sulphur.
- ✓
Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.
- B
Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion.
- C
Assertion is correct, but reason is wrong statement.
- D
Assertion is wrong but reason is correct statement.
AnswerCorrect option: A. Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.
$S$ exists as $S_8$ but oxygen exists as $O_2$ because oxygen forms $\text{p}\pi-\text{p}\pi$ multiple bonds which is not possible in $S.$
View full question & answer→Question 1102 Marks
Draw the structure of the following:
$H_2S_2O_8$
View full question & answer→Question 1112 Marks
On heating compound $(A)$ gives a gas $(B)$ which is a constituent of air. This gas when treated with $3$ mol of hydrogen $(H_2)$ in the presence of a catalyst gives another gas $(C)$ which is basic in nature. Gas $C$ on further oxidation in moist condition gives a compound $(D)$ which is a part of acid rain. Identify compounds $(A)$ to $(D)$ and also give necessary equations of all the steps involved.
Answer$\text{(A)}=\text{NH}_4\text{NO}_2,\text{(B)}=\text{N}_2,\text{(C)}=\text{NH}_3,\text{(D)}=\text{HNO}_3$$\text{NH}_4\text{NO}_2\xrightarrow{\ \ \ \ \text{Heat}\ \ \ \ }\text{N}_2+2\text{H}_2\text{O}\\\ \ \ \ \ \ ^{\text{(A)}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{\text{(B)}}$
$\text{N}_2+3\text{H}_2\xrightarrow{\ \ \ \text{Catalyst}\ \ \ }2\text{NH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{\text{(C)}}$
$4\text{NH}_3+5\text{O}_2\xrightarrow{\ \ \ \ \ \ \ \ \ \ }4\text{NO}+6\text{H}_2\text{O}$
$2\text{NO}+\text{O}_2\xrightarrow{\ \ \ \ \ \ \ \ \ \ }2\text{NO}_2$
$3\text{NO}_2+\text{H}_2\text{O}\xrightarrow{\ \ \ \ \ \ \ \ \ }2\text{HNO}_3+\text{NO}$
View full question & answer→Question 1122 Marks
Answer the following question:
$ClF_3$ exists but $FCl_3$ does not. Explain.
Answer
- $Cl$ has vacant d-orbitals and hence can show an oxidation state of $+3$ but $F$ has no d-orbitals, so, it cannot show positive oxidation states. Since $F$ can show only $-1$ oxidation state, $FCl_3$ does not exist.
- Because of bigger size, $Cl$ can accommodate three small $F$ atoms around it while F being smaller cannot accommodate three large sized $Cl$ atoms around it.
View full question & answer→Question 1132 Marks
Complete the following chemical reaction equation:$\text{p}_4(\text{s})+\text{NaOH(aq)}+\text{H}_2\text{O(l)}\rightarrow$
Answer$\text{p}_4(\text{s})+3\text{NaOH(aq)}+3\text{H}_2\text{O}\ \ \ \xrightarrow{\ \ \ \ \ \ \ } \ \ \ \text{PH}_3\ \ \ \ \ +\ \ \ \ \ 3\text{NaH}_2\text{PO}_2\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Phosphine}$
View full question & answer→Question 1142 Marks
Explain the following:
Out of noble gases only xenon is known to form established chemical compounds.
AnswerExcept radon which is radioactive, xenon has least ionization enthalpy among noble gases hence it forms compounds particularly with $O_2$ and $F_2.$
View full question & answer→Question 1152 Marks
Account for the following:
$H_3PO_2$ and $H_3PO_3$ act as good reducing agents while $H_3PO_4$ does not.
AnswerBoth $H_3PO_2$ and $H_3PO_3$ have $P-H$ bonds, so they act as reducing agents. $H_3PO_4,$ has no P-H bond but has $O-H$ bonds, so it cannot act as a reducing agent.
View full question & answer→Question 1162 Marks
Write the reactions of $F_2$ and $Cl_2$ with water.
Answer$2\text{F}_2(\text{g})+2\text{H}_2\text{O}(\text{l})\rightarrow4\text{H}^+(\text{aq})+4\text{F}^-(\text{aq})+\text{O}_2(\text{g})$$3\text{F}_2(\text{g})+3\text{H}_2\text{O}(\text{l})\rightarrow6\text{H}^+(\text{aq})+6\text{F}^-(\text{aq})+\text{O}_3(\text{g})$
$\text{Cl}_2(\text{g})+\text{H}_2\text{O}(\text{l})\rightarrow\text{HCl}(\text{aq})+\text{HOCl}(\text{aq})$
$F_2$ oxidisis water, whereas $Cl_2$ undergoes disproportion in water.
View full question & answer→Question 1172 Marks
Explain why $NH_3$ is basic while $BiH_3$ is only feebly basic.
AnswerNitrogen has a small size due to which the lone pair of electrons is concentrated in a small, region. This means that the charge density per unit volume is high. On moving down a group, the size of the central atom increases and the charge gets distributed over a large area decreasing the electron density. Hence, the electron donating capacity of group $15$ element hydrides decreases on moving down the group.
View full question & answer→Question 1182 Marks
Why is dioxygen a gas but sulphur a solid?
AnswerOxygen is smaller in size as compared to sulphur. Due to its smaller size, it can effectively form $\text{p}\pi-\text{p}\pi$ bonds and form $O_2(O==O)$ molecule. Also, the intermolecular forces in oxygen are weak van der Wall's, which cause it to exist as gas. On the other hand, sulphur does not form $M_2$ molecule but exists as a puckered structure held together by strong covalent bonds. Hence, it is a solid.
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Answer the following question:
Draw the structure of a noble gas species which is isostructural with $\text{BrO}^-_3.$
Answer$XeO_3$ is isostructural with $\text{BrO}^-_3.$
View full question & answer→Question 1202 Marks
Explain the following observations giving appropriate reason:
Ozone is thermodynamically unstable with respect to oxygen.
Answer$2O_3(g) → 3O_2(g)$
Ozone is thermodynamically unstable with respect to oxygen as its decomposition into oxygen results in the liberation of heat $(\triangle \text{H}=-\text{ve})$ and increase in entropy $(\triangle\text{S} =+\text{ve}).$ These two factors reinforce each other, resulting in large $-\text{ve}\triangle \text{G}(=\triangle \text{H}-\text{T}\triangle \text{S})$ for its conversion into oxygen.
View full question & answer→Question 1212 Marks
Knowing the electron gain enthalpy values for $O → O^–$ and $O → O^{_2–}$ as $–141$ and $702$ kJ mol$^{–1}$ respectively, how can you account for the formation of a large number of oxides having $O^{2–}$ species and not $O^–?$
(Hint: Consider lattice energy factor in the formation of compounds).
AnswerStability of an ionic compound depends on its lattice energy. More the lattice energy of a compound, more stable it will be.
Lattice energy is directly proportional to the charge carried by an ion. When a metal combines with oxygen, the lattice energy of the oxide involving $O^{2-}$ ion is much more than the oxide involving $O-$ion. Hence, the oxide having $O^{2-}$ ions are more stable than oxides having $O^-.$ Hence, we can say that formation of $O^{2-}$ is energetically more favourable than formation of $O^-.$
View full question & answer→MCQ 1222 Marks
Note: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: $\ce{HNO_3}$ makes iron passive.
Reason: $\ce{HNO_3}$ forms a protective layer of ferric nitrate on the surface of iron.
- A
Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.
- B
Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion.
- ✓
Assertion is correct, but reason is wrong statement.
- D
Assertion is wrong but reason is correct statement.
AnswerCorrect option: C. Assertion is correct, but reason is wrong statement.
Passivity is attained by formation of a thin film of oxide on iron.
View full question & answer→Question 1232 Marks
Explain the following observations giving appropriate reason:
Bleaching effect of chlorine is permanent.
Answer$\text{Cl}_2+\text{H}_2\text{O}\rightarrow [\text{HCl}+\text{HOCl}]\rightarrow2\text{HCl}+[\text{O}]$Coloured substance + [O] → Coloured substance.
As the bleaching action of chlorine is due to oxidation, therefore, it is permanent.
View full question & answer→MCQ 1242 Marks
Note: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: $\ce{NaCl}$ reacts with concentrated $\ce{H_2SO_4}$ to give colourless fumes with pungent smell. But on adding $\ce{MnO_2}$ the fumes become greenish yellow.
Reason: $\ce{MnO_2}$ oxidises $\ce{HCl}$ to chlorine gas which is greenish yellow.
- ✓
Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.
- B
Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion.
- C
Assertion is correct, but reason is wrong statement.
- D
Assertion is wrong but reason is correct statement.
AnswerCorrect option: A. Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.
$\ce{NaCl}$ reacts with concentrated $\ce{H_2SO_4}$ to give colourless fumes with pungent smell. But on adding $\ce{MnO_2}$ the fumes become greenish yellow. $\ce{MnO_2}$ oxidises $\ce{HCl}$ to chlorine gas which is greenish yellow.
$\text{NaCl}+\text{H}_2\text{SO}_4\xrightarrow{\ \ \ \ \ \ \ \ \ \ }\text{NaHSO}_4+\text{HCl}($fumes of $\ce{HCl}$ is colourless$)$
By heating manganese dioxide with concentrated hydrochloric acid.
$\text{MnO}_2+4\text{HCl}\xrightarrow{\ \ \ \ \ \ \ \ \ \ }\text{MnCl}_2+\text{Cl}_2+2\text{H}_2\text{O}$
View full question & answer→Question 1252 Marks
Give reason:
$NH_3$ has a higher proton affinity than $PH_3.$
AnswerWhen $NH_3$ or $PH_3$ accepts a proton, an additional $N-H$ or $P-H$ bond is formed.
$H_3N : + H^+ → NH + 4$
$H^3P : + H^+ → PH + 4$
Due to the bigger size of $P$ than $N$, $P-H$ bond thus formed is much weaker than the $N-H$ bond. Thus, $NH_3$ has higher proton affinity than $PH_3.$
View full question & answer→Question 1262 Marks
Complete the following chemical reaction equation:
$I^- (aq) + H_2O(l) + O_3(g) →$
Answer$2I^- (aq) + H_2O(l) + O_3(g) → 2OH^- (aq) + I_2(s) + O_2(g)$
View full question & answer→Question 1272 Marks
Explain the following:
Hydrogen fluoride is a weaker acid than hydrogen chloride in aqueous solution.
AnswerIt is due to:
- Higher $H-F$ bond dissociation energy than $H-Cl.$
- Stronger $H-$bonding of $F-$ion with $H_3O+$ than $Cl^-.$
View full question & answer→Question 1282 Marks
Illustrate how copper metal can give different products on reaction with $HNO_3.$
AnswerConcentrated nitric acid is strong oxidizing agent.
The product of oxidation depend on the concentration of acid, temperture and also the material undergoing oxidation.
Two condition aries : On heating with dilute nitric oxide $NO$ evolved or on reacting with concentrated nitric acid $NO_2$ is evolved.
reaction with dilute nitric acid$3\text{CU}+8\text{HNO}_3(\text{dil})\rightarrow3\text{Cu}(\text{NO}_3)_2+4\text{H}_2\text{O}+2\text{NO}$
Reaction with concentrated nitric acid$\text{CU}+2\text{HNO}_3(\text{conc.})\rightarrow\text{Cu}(\text{NO}_3)_2+2\text{H}_2\text{O}+2\text{NO}_2$
View full question & answer→Question 1292 Marks
How is $O_3$ estimated quantitatively?
AnswerQuantitatively, ozone can be estimated with the help of potassium iodide. When ozone is made to react with potassium iodide solution buffered with a borate buffer $($pH $9.2)$, iodine is liberated. This liberated iodine can be titrated against a standard solution of sodium thiosulphate using starch as an indicator. The reactions involved in the process are given below.
$2\text{I}^-+\text{H}_2\text{O}+\text{O}_3\rightarrow2\text{OH}^-+\text{I}_2+\text{O}_2\\\text{Iodine} \ \ \ \ \ \ \ \ \ \ \text{Ozone} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Iodine}$
$\text{I}_2+2\text{Na}_2\text{S}_2\text{O}_3\rightarrow2\text{Na}_2\text{S}_4\text{O}_6+2\text{NaI}\\ \ \ \ \ \ \ \ \ \text{Sodium} \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Sodium} \\\ \ \ \ \ \ \ \text{thiosulphate} \ \ \ \ \text{tetrathionate}$
View full question & answer→Question 1302 Marks
Account for the following:
Sulphur disappears when boiled with sodium sulphite.
AnswerWhen sodium sulphite is heated with sulphur, we get sodium thiosulphate which is soluble in water that is why sulphur disappears.$\text{Na}_2\text{SO}_3+\text{S}\ \ \xrightarrow{\ \ \text{Heat}\ \ }\ \ \text{Na}_2\text{S}_2\text{O}_3$
View full question & answer→Question 1312 Marks
List the uses of neon and argon gases.
AnswerNeon and argon is inert gas because they are more stable than other.
Uses of Neon:
- Neon is mainly used in fluorescent lamps of tubes for advertising purposes. These are known as neon signs and can be seen at long distances even when there is a fog. Neon actually produces an orange red glow in the tube and on mixing with the vapours of other gases, glows or signs of different colours can be obtained.
- It is used in filling sodium vapour lamps.
- It is used in safety devices for protecting certain electrical instruments (voltmeters, relays, rectifiers etc.)
Uses of Argon:
- It is used in metal filament electric lamps since it increases the life of the tungsten filament by retarding its vapourisation.
- A mixture of argon and mercury vapours is used in fluorescent tubes.
- It is used to create an inert atmosphere for welding and for carrying certain chemical reactions.
View full question & answer→Question 1322 Marks
Give reason for the following:
Ammonia acts as a ligand.
AnswerDue to the presence of lone pair of electrons on $N,\ NH_3$ acts as a ligand.$\text{AgCl}\ \ \ \ \ +\ \ \ 2\text{NH}_3\ \ \ \xrightarrow{\ \ \ \ \ }\ \ \ [\text{Ag(NH}_3)_2]\text{Cl}\\^\text{Silver chloride}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{Diam mine e (I) chloride}$
View full question & answer→Question 1332 Marks
Give reason:
NO (Nitric oxide) is paramagnetic in the gaseous state but diamagnetic in the liquid and solid states.
AnswerNO has an odd number of electrons (11 valence electrons) and hence is paramagnetic in the gaseous state. But in liquid and solid states, it exists as a symmetrical or asymmetrical dimer and hence is diamagnetic in these states.
View full question & answer→Question 1342 Marks
Explain the following observations giving appropriate reason:
The HEH bond angle of the hydrides of group $15$ elements decreases as we move down the group.
AnswerAs we move from $NH_3$ to $BiH_3,$ the size of the central atom goes on increasing and its electronegativity goes on decreasing. Due to this, the bond pair of electrons tend to lie away from the central atom. As a result, the repulsion between the pairs decreases as we move from $NH_3$ to $BiH_3.$ Consequently the bond angle decreases as we go down the group from $NH_3$ to $BiH_3.$
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