Question 12 Marks
Write the chemical equations involved in the preparation of the following:
- $XeF_4.$
- $H_3PO_3.$
Answer
- $\text{Xe}+\text{2F}_{2}\xrightarrow{873K 7bar}\text{XeF}_{4}.$
- $\text{PCl}_{3}+\text{3H}_{2}\text{O}\rightarrow\text{H}_{3}\text{PO}_{3}+\text{3HCl}.$
Alternate answer
$\text{P}_{4}\text{O}_{6}+\text{6H}_{2}\text{O}\rightarrow\text{4H}_{3}\text{PO}_{3}.$
View full question & answer→Question 22 Marks
Account for the following:
- $N_2$ has higher bond dissociation energy than NO.
- $N_2$ and CO both have same bond order but CO is more reactive than $N_2.$
Answer
- $N_2$ have 6 more electrons in a bonding atomic orbital so it’s bond order is 3, whereas $N_{2+}$ has only 5 in bonding so it's bond order is 2.5.
- Bond dissociation energy is directly proportional to Bond order of a molecule so $N_2$ has more bond dissociation energy.
- Because of higher electronegativity difference, CO is polar and therefore more reactive or any other suitable reason (polarity or heteronuclear nature).
View full question & answer→Question 32 Marks
Name the reagents used in the following reactions:
- $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{CO}-\text{CH}_3\xrightarrow{{ \ \ \ \ ? \ \ \ \ }}\text{CH}_3-\text{C}-\text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH}$
- $\text{CH}_3-\text{COOH}\xrightarrow{\ \ \ \ \ ? \ \ \ \ \ }\text{ClCH}_2-\text{COOH}$
Answer
- $CH_3MgBr, H_3O^+$
- $Cl_2, P$
View full question & answer→Question 42 Marks
Write the structures of the following:
- $BrF_3.$
- $XeF_4.$
View full question & answer→Question 52 Marks
Draw the structures of the following:
- $XeF_{4.}$
- $BrF_{5.}$
View full question & answer→Question 62 Marks
Draw the structures of the following:
- $XeO_3$
- $H2SO_4$
View full question & answer→Question 72 Marks
Complete the following equations:
- $C +$ conc.$ H_2SO_4 →$
- $XeF_2 + H_2O →$
Answer
- $C + 2H_2SO_4($conc.$) → CO_2 + 2 SO_2 + 2 H_2O$
- $2XeF_2+ 2H_2O → 2Xe + 4HF + O_2$
View full question & answer→Question 82 Marks
What happens when:
- $SO_2$ gas is passed through an aqueous solution $Fe^{3+}$ salt?
- $XeF_4$ reacts with $SbF_5?$
Answer
- $2\text{F}\text{e}^{3+} + \text{SO}_{2} + 2\text{H}_{2}\text{O}\rightarrow 2\text{Fe}^{2+} + \text{SO}_{4}^{ 2-} + 4\text{H}^{+}$
- $\text{XeF}_{4} + \text{SbF}_{5}\rightarrow [\text{ XeF}_{3}]^{+} [\text{SbF}_{6}]^{-}.$
View full question & answer→Question 92 Marks
Draw the structures of the following :
(a) $H_2S_2O_8$
(b) $ClF_3$
View full question & answer→Question 102 Marks
Draw the structures of the following:
- $XeF_4$
- $HClO_4$
View full question & answer→Question 112 Marks
Complete the following equations:
- $\mathrm{Ag}+\mathrm{PCl}_5 \rightarrow$
- $\mathrm{CaF}_2+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow$
Answer
- $2 \mathrm{Ag}+\mathrm{PCl}_5 \rightarrow 2 \mathrm{AgCl}+\mathrm{PCl}_3$.
- $\mathrm{CaF}_2+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CaSO}_4+2 \mathrm{HF}$.
View full question & answer→Question 122 Marks
Draw the structures of the following:
- $\mathrm{H}_2 \mathrm{SO}_3$
- $\mathrm{HClO}_3$
View full question & answer→Question 132 Marks
Complete the following equations:
- $\mathrm{P}_4+\mathrm{H}_2 \mathrm{O} \rightarrow$
- $\mathrm{XeF}_4+\mathrm{O}_2 \mathrm{~F}_2 \rightarrow$
Answer
- $\mathrm{P}_4+\mathrm{H}_2 \mathrm{O} \rightarrow$ no reaction.
- $\mathrm{XeF}_4+\mathrm{O}_2 \mathrm{~F}_2 \rightarrow \mathrm{XeF}_6+\mathrm{O}_2$.
View full question & answer→Question 142 Marks
Draw the structures of the following:
- $\mathrm{XeF}_2$
- $\mathrm{BrF}_3$
View full question & answer→Question 152 Marks
Draw the structure of the following molecules:
- $ \mathrm{XeOF}_4 $
- $ \mathrm{H}_3 \mathrm{PO}_3$
View full question & answer→Question 162 Marks
Complete the following reaction equations:
- $C_6H_5N_2Cl + H_3PO_2 + H_2O\rightarrow$
- $C_6H_5NH_2 + Br_2 (aq.) \rightarrow$
View full question & answer→Question 172 Marks
Explain the following facts giving appropriate reason in each case:
- $NF_3$ is an exothermic compound where as $NCl_3$ is not.
- All the bonds in $SF_4$ are not equivalent.
Answer
- Because bond energy of $F_2$ is lower than that of $Cl_2$ and therefore F forms stronger bond with $N$ with the release of energy.
- $SF_4$ has trigonal bipyramidal structure with one $l.p.$ Due to $l.p-b.p$ repulsion two axial $S-F$ bonds are longer than two $S-F$ equatorial bonds.
View full question & answer→Question 182 Marks
Describe the principle controlling each of the following processes:
- Vapour phase refining of titanium metal.
- Froth flotation method of concentration of a sulphide ore.
Answer
- In this method the titanium metal is heated with $I_2$ to form a volatile compound $TiI_4$ which on further heating at higher temperature decomposes to give pure titanium metal.
- This method is based upon the fact that the surface of the sulphide ores is preferentially wetted by oil while that of gangue is wetted by water.
View full question & answer→Question 192 Marks
Draw the structural formulae of the following compounds:
- $H_4P_2O_{5.}$
- $XeF_{4.}$
View full question & answer→Question 202 Marks
Complete the following, chemical reaction equations:
- $I_2 + HNO_3→$
(conc.)
- $HgCl_2 + PH_3→$
Answer
- $I_2 + 10HNO_3 \rightarrow 2HIO_3 + 10NO_2 + 4H_2O.$
- $3HgCl_2 + PH_3 \rightarrow Hg_3P_2 + 6HCl.$
View full question & answer→Question 212 Marks
Complete the following chemical reaction equations:
- $XeF_2 + H_2O →$
- $PH_3 + HgCl_2 →$
Answer
- $2XeF_2 + 2H_2O → 2Xe + 4HF + O_2.$
- $2PH_3 + 3HgCl_2 → Hg_3P_2 + 6HCl.$
View full question & answer→Question 222 Marks
Answer the following:
- Which neutral molecule would be isoelectronic with $CIO^–?$
- Of $Bi(V)$ and $Sb(V)$ which may be a stronger oxidising agent and why?
Answer
- $ClF.$
- $Bi\ (V)$, due to greater stability of its lower oxidation state effect.
View full question & answer→Question 232 Marks
How would you account for the following:
- Sulphur hexafluoride is less reactive than sulphur tetrafluoride.
- Of the noble gases only xenon forms known chemical compounds.
Answer
- Due to sterically protected sulphur atom.
- Because Xe is with the lowest ionization enthalpy among the rest (except Rn which is radioactive).
View full question & answer→Question 242 Marks
Write the structures of the following:
- $H_2SO_3$
- $XeF_4$
View full question & answer→Question 252 Marks
Complete the following chemical equations :
- $F_2 + 2Cl^– \longrightarrow$
- $2XeF_2 + 2H_2O \longrightarrow$
Answer
- $F_2 + 2Cl^- \longrightarrow 2F^-+ Cl_2$
- $2XeF_2 + 2H_2O \longrightarrow 2Xe + 4HF + O_2$
View full question & answer→Question 262 Marks
What happens when
- $HCl$ is added to $MnO_2?$
- $PCl_5$ is heated?
Write the equations involved. Answer
- $MnO_2 + 4HCl \longrightarrow MnCl_2+ Cl_2+ 2H_2O$
- $\text{PCl}_5\xrightarrow{\text{ }\text{ }\text{ }\text{ }\text{ }\triangle\text{ }\text{ }\text{ }\text{ }\text{ }}\text{PCl}_3+\text{Cl}_2$
View full question & answer→Question 272 Marks
Draw the structures of the following :
- $H_4P_2O_7$
- $XeOF_4$
View full question & answer→Question 282 Marks
Draw the structures of the following:
- $H_3PO_2$
- $XeF_4$
View full question & answer→Question 292 Marks
Complete the following reactions :
- $\text{C}l_2+\text{H}_2\text{O}\xrightarrow{\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }}$
- $\text{XeF}_6+3\text{H}_2\text{O}\xrightarrow{\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }}$
AnswerComplete chemical reactions are:
- $Cl_2 + H_2O → [HCl + HOCl] → 2 HCl + [O]\ ($Nascent oxygen$)$
- $XeF_6 + 3H_2O → XeO_3+ 6HF$
View full question & answer→Question 302 Marks
What happens when
- conc. $H_2SO_4$ is added to $Cu?$
- $SO_3$ is passed through water?
Write the equations. Answer
- When conc. $H_2SO_4$ is added to Cu then its leads to the formation of $CuSO_4, SO_2$ and $H_2O.$
$Cu + H_2SO_4 ($conc.$) → CuSO_4 + SO_2 + 2H_2O$
- When $SO_3$ is passed through water then its leads to the formation of $H_2SO_4.$
$SO_3+ H_2O → H_2SO_4$ View full question & answer→Question 312 Marks
Complete the following chemical equations:
- $Ca_3P_2+ H_2O →$
- $Cu + H_2SO_4(conc.)→$
Answer
- $Ca_3P_2+ 6H_2O→ 3Ca(OH)_2+ 2PH_3$
- $Cu + 2H_2SO_4→ CuSO_4+ 2H_2O + SO_2$
View full question & answer→Question 322 Marks
Arrange the following in the order of property indicated against each set:
- $HF,\ HCl,\ HBr,\ HI -$ increasing bond dissociation enthalpy.
- $H_2O, H_2S, H_2Se, H_2Te -$ increasing acidic character.
View full question & answer→Question 332 Marks
Complete the following reactions:
- $NH_3+ 3Cl_2($excess$) →$
- $XeF_6 + 2H_2O →$
Answer
- $NH_3 + 3Cl_2 ($excess$) → NCl_3 + 3HCl$
- $XeF_6 + 2H_2O → XeO_2F_2 + 4HF$
View full question & answer→Question 342 Marks
What happens when
- $(NH_4)_2Cr_2O_7$ is heated?
- $H_3PO_3$ is heated?
Write the equations. Answer
- $(NH_4)_2Cr_2O_7 → N_2 + 4H_2O + Cr_2O_3$
- $4H_3PO_3 → 3H_3PO_4 + PH_3$
View full question & answer→Question 352 Marks
Draw the structures of the following:
- $H_2S_2O_7$
- $XeF_6$
View full question & answer→Question 362 Marks
Explain the following giving an appropriate reason in each case.
- $O_2$ and $F_2$ both stabilise higher oxidation states of metals but $O_2$ exceeds $F_2$ in doing so.
- Structures of Xenon fluorides cannot be explained by Valence Bond approach.
Answer
- This is due to the ability of oxygen to form multiple bonds to metals.
- This is because the energy required for the promotion of electrons inXenon is very high./Energy factor does not favour VB approach.
View full question & answer→Question 372 Marks
State reasons for each of the following:
- The N–O bond in ${NO_2}^-$ is shorter than the $N–O$ bond in ${NO^-}_3.$
- $SF_6$ is kinetically an inert substance.
Answer
- In the resonance structure of these two species, in ${NO_2}^–, 2$ bonds are sharing a double bond while in ${NO_3}^–, 3$ bonds are sharing a double bond which means that bond in $NO_2$ will be shorter than in ${NO_3}^–.$
Alternate answer
In ${NO^–}_2$, bond order is 1.5 while in ${NO^–}_3,$ bond order is $1.33.$
- Because $SF6$ is stericallyprotected by six $F$ atoms $/$ co-ordinatively saturated.
View full question & answer→Question 382 Marks
State reasons for each of the following:
- All the $P-Cl$ bonds in $PCl_5$ molecule are not equivalent.
- Sulphur has greater tendency for catenation than oxygen.
Answer
- Because $PCl_5$ has a trigonal bipyramidal structure in which three $P-Cl$ bonds are equatorial and two $P-Cl$ bonds are axial.
- Because $S-S$ single bond is stronger than $O-O$ single bond.
View full question & answer→Question 392 Marks
Draw the structures of white phosphorus and red phosphorus. Which one of these two types of phosphorus is more reactive and why?
Answer
White phosphorus is more reactive due to its discrete tetrahedral structure and angular strain.
View full question & answer→Question 402 Marks
Draw the structures of the following molecules:
- $XeF_{4.}$
- $BrF_{3.}$
View full question & answer→Question 412 Marks
Explain the role of:
- Cryolite in the electrolytic reduction of alumina.
- Carbon monoxide in the purification of nickel.
Answer
- Cryolite lowers the melting point of alumina. Or Molten cryolite dissolves alumina.
- Carbon monoxide forms a volatile complex with nickel which on heating de-composes to gives pure nickle.
View full question & answer→Question 422 Marks
Complete the following chemical reaction equations:
- $\text{P}_{4(\text{s})}+\text{NaoH}_{(\text{aq})}+\text{H}_{2}\text{O}_{(\text{l})}\rightarrow$
- $\text{I}^{-}_{(\text{aq})}+\text{H}_{2}\text{O}_{(\text{l})}+{O}_{3(\text{g})}\rightarrow$
Answer
- $\text{P}_{4}+\text{3NaOH}+\text{3H}_{2}\text{O}\rightarrow\text{PH}_{3}+\text{3NaOH}_{2}\text{PO}_{2}.$
- $\text{2I}^{-}+\text{H}_{2}\text{O}+\text{O}_{3}\rightarrow\text{2OH}^{-}+\text{I}_{2}+\text{O}_{2}.$
View full question & answer→Question 432 Marks
Draw the structures of the following molecules:
- $SF_4.$
- $XeF_4.$
View full question & answer→Question 442 Marks
Assign a reason for each of the following statements:
- Ammonia is a stronger base than phosphine.
- Sulphur in vapour state exhibits a paramagnetic behaviour.
Answer
- The lone pair of electrons on $N$ atom in $NH_3$ is directed and not diffused/delocalized as it is in $PH_3$ due to larger size of $P.$
- $S_2$ molecule like $O_2,$ has two unpaired electrons in antibonding $\eth$* orbitals.
View full question & answer→Question 452 Marks
Write balanced chemical equations for the following reactions:
- Aluminium dissolves in aqueous hydrochloric acid.
- Tin reacts with a hot alkali solution.
Answer
- $\text{2Al}+\text{6HCl}+\text{12H}_2{O}\rightarrow\text{2[Al(H}_{2}\text{O})_{6}]\text{Cl}_{3}+3\text{H}_{2}$
- $\text{Sn}+\text{2KOH}+\text{4H}_{2}\text{O}\rightarrow\text{K}_{2}\text{[Sn(OH)}_{6}]+\text{2H}_{2}$
View full question & answer→Question 462 Marks
Write the structures of the following species:
- $H_3PO_2.$
- $H_2SO_5.$
View full question & answer→Question 472 Marks
- Draw the structure of $XeF_2$ molecule.
- Write the outer electronic configuration of $Cr$ atom $(Z=24).$
Answer
-
- $3d^54s^1.$
View full question & answer→Question 482 Marks
Write balanced chemical equations for the following reactions:
- $Ca_3P_2 + H_2O \rightarrow$
- $XeF_6 + 3H_2O \rightarrow$
Answer
- $Ca_3P_2 + 6H_2O \rightarrow 2PH_3 + 3Ca(OH)_2.$
- $XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF.$
View full question & answer→Question 492 Marks
Among the hydrides of Group-15 elements, which have the,
- Lowest boiling point?
- Maximum basic character?
- Highest bond angle?
- Maximum reducing character?
Answer
- $PH_3$ will have the lowest boiling point.
- Ammonia has the maximum basic character.
- $NH_3$ will have the highest bond angle.
- $BiH_3$ will have the maximum reducing character.
View full question & answer→Question 502 Marks
Draw the shape of the following molecules:
- $H_2S_2O_{8.}$
- $XeF_2.$
View full question & answer→Question 512 Marks
Draw the shape of the following molecules:
- $XeOF_4.$
- $BrF_3.$
View full question & answer→Question 522 Marks
Arrange the following in order of property indicated for each set:
- $H_2O,\ H_2S,\ H_2Se,\ H_2Te$ -Increasing acidic character.
- $HF,\ HCl,\ HBr,\ HI $-Decreasing bond enthalpy.
Answer
- The increasing order of acidic character is $H_2O < H_2S < H_2Se < H_2Te < h < h$
- The decreasing order of bond enthalpy is $HF > HCl > HBr > HI$
View full question & answer→Question 532 Marks
Write balanced chemical equations for the following processes:
- $XeF_2$ undergoes hydrolysis.
- $MnO_2$ is heated with conc. $HCl.$
Answer
- $2\text{XeF}_2(\text{s})\ +\ 2\text{H}_2\text{O}(\text{l})\xrightarrow{\ \ \ \ \ \ \ \ \ \ }2\text{Xe(g)}\ +\ 4\text{HF(aq)}\ +\ \text{O}_2(\text{g})$
- $4\text{HCl}\ +\ \text{MnO}_2\xrightarrow{\ \ \ \ \ \ \ \ \ \ }\text{Cl}_2+\text{MnCl}_2\ +\ 2\text{H}_2\text{O}$
View full question & answer→Question 542 Marks
Give reason for the following:
Noble gases form compounds with fluorine and oxygen only.
AnswerFluorine and oxygen are the most electronegative elements and hence are very reactive. So, they form compounds with noble gases, particularly xenon.
View full question & answer→Question 552 Marks
Out of $H_2O$ and $H_2S,$ which one has higher bond angle and why?
AnswerBond angle of $H_2O$ is larger, because oxygen is more electronegative than sulphur therefore bond pair electron of $O–H$ bond will be closer to oxygen and there will be more bond-pair bond -pair repulsion between bond pairs of two $O–H$ bonds.
View full question & answer→Question 562 Marks
Account for the following:
$PF_5$ is known but $NF_5$ is not known.
AnswerP has vacant 3d-orbitals in its valence shell while N does not have. As a result, P can form additional bonds to give $PF_5$ while N cannot extend its covalency beyond three and hence it forms only $NF_3$ but not $NF_5.$
View full question & answer→Question 572 Marks
Give reason for the following:
$NCl_3$ gets readily hydrolysed while $NF_3$ does not.
AnswerIn $NCl_3,$ Cl has vacant d-orbitals to accept the lone pair of electrons donated by O-atom of $H_2O$ molecule but in $NF_3,\ F$ does not have d-orbitals.
$NCl_3 + 3H_2O → NH_3 + 3HOCl$
$NF_3 + H_2O →$ No reaction
View full question & answer→MCQ 582 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: $N_2$ is less reactive than $P_4.$
Reason: Nitrogen has more electron gain enthalpy than phosphorus.
- 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.
$N_2$ is less reactive than $P_4$ molecule this is so, because nitrogen has very high bond dissociation enthalpy because of triple bond between two nitrogen atom which is not the case with phosphorus.
View full question & answer→Question 592 Marks
Account for the following:
Both $NO$ and $ClO_2$ are odd electron species but $NO$ dimerises while $ClO_2$ does not.
AnswerIn $NO,$ the odd electron on $N$ is attracted by only one $O-$atom but in $ClO_2,$ the odd electron on $Cl$ is attracted by two $O-$atoms. Thus, the odd electron on $N$ in $NO$ is localised while the odd electron on $Cl$ in $ClO_2$ is delocalised. Consequently, $NO$ has a tendency to dimerise but $ClO_2$ does not.
View full question & answer→Question 602 Marks
Sea is the greatest source of some halogens. Comment.
AnswerSea water contains chlorides, bromides, and iodides of $Na,\ K,\ Mg,$ and $Ca$.
However, it primarily contains $NaCl.$ The deposits of dried up sea beds contain sodium chloride and carnallite, $KCl.MgCl_2.6H_2O.$
Marine life also contains iodine in their systems.
For example, sea weeds contain upto $0.5%$ iodine as sodium iodide. Thus, sea is the greatest source of halogens.
View full question & answer→Question 612 Marks
Account for the following:
Bismuth is a strong oxidising agent in the pentavalent state.
AnswerAs the inert pair effect is very prominent in Bi, its $+5$ oxidation state is less stable than its $+3$ oxidation state. In other words, bismuth in the pentavalent state can easily accept two electrons and thus gets reduced to trivalent bismuth.
$Bi^{5+} + 2e^- → Bi^{3+}$
Thus, it acts as a strong oxidising agent.
View full question & answer→Question 622 Marks
Answer the following question:
Why is $\text{K}_{\text{a}_2}<<\text{K}_{\text{a}_1}$ for $\text{H}_2\text{SO}_4$ in water?
Answer$\text{K}_{\text{a}_2}<<\text{K}_{\text{a}_1},$ because $\text{HSO}_{4}-$ ion has much less tendency to donate a proton to $H_2O$ as compared to $H_2SO_4.$
View full question & answer→Question 632 Marks
Give reason for the following:
Elemental nitrogen exists as a diatomic molecule whereas elemental phosphorus is a tetraatomic molecule.
AnswerNitrogen because of its small size and high electronegativity forms $\text{p}\pi-\text{p}\pi$ multiple bonds. Thus, it exists as a diatomic molecule having a triple bond between the two $N-$atoms. Phosphorus due to its larger size and lower electronegativity usually does not form $\text{p}\pi-\text{p}\pi$ multiple bonds with itself. Instead it prefers to form $P-P$ single bonds and hence it exists as tetrahedral $P_4$ molecules.
View full question & answer→Question 642 Marks
Why is helium used in diving apparatus?
AnswerAir contains a large amount of nitrogen and the solubility of gases in liquids increases with increase in pressure. When sea divers dive deep into the sea, large amount of nitrogen dissolves in their blood. When they come back to the surface, solubility of nitrogen decreases and it separates from the blood and forms small air bubbles. This leads to a dangerous medical condition called bends. Therefore, air in oxygen cylinders used for diving is diluted with helium gas. This is done as He is sparingly less soluble in blood.
View full question & answer→Question 652 Marks
Why do noble gases have comparatively large atomic sizes?
AnswerNoble gas are stable so they cannot form molecules In case noble gas from molecule the force of attraction that acting is vander waal force.
On other hand, other element form covalent bond and it is well known that vander waal radii is larger than covalent radii. Thus noble gases have comparatively large atomic sizes.
View full question & answer→Question 662 Marks
How is the presence of $SO_2$ detected?
Answer$SO_2$ is a colourless and pungent smelling gas. It can be detected with the help of potassium permanganate solution. When $SO_2$ is passed through an acidified potassium permanganate solution, it decolonizes the solution as it reduces. $\text{MnO}^-_4 \ \text{ions to} \text{Mn}^{2+} \ \text{ions}$$5\text{SO}_2+2\text{MnO}_4^-+2\text{H}_2\text{O}\rightarrow5\text{SO}^{2-}_4+4\text{H}^++2\text{Mn}^{2+}$
View full question & answer→Question 672 Marks
Account for the following:
Sulphur exhibits greater tendency for catenation than selenium.
AnswerAs we move from S to Se, the atomic size increases and hence the strength of E-E bond decreases. Thus, S-S bond is much stronger than Se-Se bond. Consequently, S shows greater tendency for catenation than selenium.
View full question & answer→Question 682 Marks
Give the resonating structures of $NO_2$ and $N_2O_5.$
View full question & answer→Question 692 Marks
Assign a reason for the following:
Sulphur hexafluoride is used as a gaseous electrical insulator.
Answer$SF_6$ is a colourless, odourless and non-toxic gas at room temperature. It is thermally stable and chemically inert. Because of its inertness and high tendency to suppress internal discharges, it is used as a gaseous electrical insulator in high voltage generators and switch gears.
View full question & answer→Question 702 Marks
Give two examples to show the anomalous behaviour of fluorine.
Answer
- Oxidation state: Fluorine shows oxidation state of – 1 only. It does not show any positive oxidation state. Other halogens show oxidation states such as +1, +3, +5, +7 also.
- Extra-ordinary reactivity: Fluorine is extraordinary most reactive element. This is due to F—F bond energy is very low as compared to that of other halogen molecules.
View full question & answer→Question 712 Marks
Phosphorus forms a number of oxoacids. Out of these oxoacids phosphinic acid has strong reducing property. Write its structure and also write a reaction showing its reducing behaviour.
AnswerPhosphinic acid is hypophosphorus acid $(H_3PO_2)$ which acts as a reducing agent due to possession of P – H bonds.

Its reducing properties are:
$4\text{AgNO}_3+\text{H}_3\text{PO}_2+2\text{H}_2\text{O}\xrightarrow{\ \ \ \ \ \ \ \ \ \ }4\text{Ag}+\text{H}_3\text{PO}_4+4\text{HNO}_3$
$2\text{HgCl}_2+\text{H}_3\text{PO}_2+2\text{H}_2\text{O}\xrightarrow{\ \ \ \ \ \ \ \ \ \ }2\text{Hg}+\text{H}_3\text{PO}_4+4\text{HCl}$
$\text{H}_3\text{PO}_2+2\text{H}_2\text{O}+2\text{Cl}_2\xrightarrow{\ \ \ \ \ \ \ \ \ \ }\text{H}_3\text{PO}_4+4\text{HCl}$ View full question & answer→Question 722 Marks
Give reason for the following:
ICl is more reactive than $I_2.$
AnswerICl is more reactive than $I_2$ because I-Cl bond is weaker than I-I bond. Consequently, ICl breaks easily to form halogen atoms which readily bring about the reactions.
View full question & answer→MCQ 732 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: $HI$ cannot be prepared by the reaction of $KI$ with concentrated $\ce{H_2SO_4}$
Reason: $HI$ has lowest $H-X$ bond strength among halogen acids.
- A
Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.
- ✓
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: B. Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion.
$HI$ cannot be prepared by the reaction of $KI$ with concentrated $\ce{H_2SO_4}$ because HI formed is converted to $I_2.$
View full question & answer→Question 742 Marks
Mention three areas in which $H_2SO_4$ plays an important role.
Answer
-
It is used in the manufacture of fertilizers such as $(NH_4)^2$ $SO_4 $, calcium superphosphate.
-
It is used as an electrolyte in storage batteries.
-
It is used in petroleum refining, detergent industry and in the manufacture of paints, pigments and dyes.
View full question & answer→MCQ 752 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: $SF_6$ cannot be hydrolysed but $SF_4$ can be.
Reason: Six $F$ atoms in $SF_6$ prevent the attack of $H_2O$ on sulphur atom of $SF_6.$
- ✓
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.
$SF_6$ is sterically protected due to presence of six $F$ atoms around $S$ which prevents the attack of $H_2O$ on $SF_6.$
View full question & answer→Question 762 Marks
Why is nitric oxide paramagnetic in gaseous state but the solid obtained on cooling it is diamagnetic?
Answer
Nitric oxide (NO) in the gaseous state is monomer having one unpaired electron as it is paramagnetic in nature.

However, in solid state it forms a dimer, $N_2O_2$ which is diamagnetic in nature.

View full question & answer→Question 772 Marks
Answer the following question: Complete the following chemical equations.
- $\text{NH}_4\text{Cl(aq)}+\text{NaNO}_2\text{(aq)}\rightarrow$
- $\text{P}_4+3\text{NaOH}+3\text{H}_2\text{O}\rightarrow$
Answer
- $\text{NH}_4\text{Cl(aq)}+\text{NaNO}_2\text{(aq)}\rightarrow\text{N}_2\text{(g)}+2\text{H}_2\text{O(l)}+\text{NaCl(aq)}$
- $\text{P}_4+3\text{NaOH}+3\text{H}_2\text{O}\ \ \xrightarrow{\ \ \ \ \ }\ \ \ \ \text{PH}_3\ \ +\ \ 3\text{NaH}_2\text{PO}_2\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{Phosphine}$
View full question & answer→Question 782 Marks
Assign appropriate reasons for the following statement:
Perchloric acid is a stronger acid than sulphuric acid.
AnswerThe oxidation state of Cl in perchloric acid is +7 while that of S in sulphuric acid is +6. Greater the oxidation state of central atom, more readily the O-H bond breaks and hence stronger is the acid.
View full question & answer→Question 792 Marks
Explain the following:
$SF_6$ is inert towards hydrolysis.
AnswerIn $SF_6,$ S atom is sterically protected by six F atoms and does not allow water molecules to attack the S atom. Further, F does not have d-orbitals to accept the electrons donated by $H_2O$ molecules. Due to these reasons, $SF_6$ is kinetically an inert substance.
View full question & answer→Question 802 Marks
Give reason for the following:
Among the hydrides of Group $16$, water shows unusual physical properties.
AnswerBecause of high electronegativity of $O,$ the $O-H$ in $H_2O$ forms strong intermolecular H-bonds. Thus, water exists as an associated molecule while other hydrides of Group 16 do not form H-bonds and hence exist as discrete molecules. Hence, water shows unusual physical properties, i.e., high boiling point, high thermal stability and weaker acidic character as compared to other hydrides of Group $16.$
View full question & answer→Question 812 Marks
Answer the following question:
Write balanced chemical equations for the following:
- Complete hydrolysis of $XeF_6.$
- Disproportionation reaction of orthophosphorus acid.
Answer
- $\text{XeF}_6+3\text{H}_2\text{O}\rightarrow\text{XeO}_3+6\text{HF}$
- $4\text{H}_3\text{PO}_3\xrightarrow{\ \ \triangle\ \ }3\text{H}_3\text{PO}_45+\text{PH}_3$
View full question & answer→Question 822 Marks
The HNH angle value is higher than HPH, HAsH and HSbH angles. Why? [Hint: Can be explained on the basis of $sp_3$ hybridisation in NH3 and only s–p bonding between hydrogen and other elements of the group].
View full question & answer→Question 832 Marks
Account for the following observation:
Among the halogens, $F_2$ is the strongest oxidising agent.
AnswerThis is due to the:
- Low enthalpy of dissociation of $F-F$ bond.
- High hydration enthalpy of $F^-.$
View full question & answer→Question 842 Marks
Write main differences between the properties of white phosphorus and red phosphorus.
Answer
| White phosphorus |
Red phosphorus |
|
It is a soft and waxy solid.
It possesses a garlic smell
|
It is a hard and crystalline solid, without any smell.
|
|
It is poisonous.
|
It is non-poisonous.
|
|
It is insoluble in water but soluble in carbon
disulphide.
|
It is insoluble in both water and carbon disulphide.
|
|
It undergoes spontaneous
combustion in air.
|
It is relatively less reactive.
|
|
In both solid and vapour
states, it exists as a $P_4$
molecule.
|
It exists as a chain of tetrahedral $P_4$ units. 
|
View full question & answer→Question 852 Marks
Which aerosols deplete ozone?
AnswerFreons or chlorofluorocarbons (CFCs) are aerosols that accelerate the depletion of ozone. In the presence of ultraviolet radiations, molecules of CFCs break down to form chlorine-free radicals that combine with ozone to form oxygen.
View full question & answer→Question 862 Marks
Answer the following question:
Why is HF acid stored in wax coated glass bottles?
AnswerHF does not attack wax but reacts with glass. It dissolves $SiO_2$ present in glass forming hydrofluorosilicic acid.
$SiO_2 + 6HF → H_2SiF_6 + 2H_2O$
View full question & answer→Question 872 Marks
Why is $K_{a2} << K_{a1}$ for $H_2SO_4$ in water?
Answer$\text{H}_2\text{SO}_{4(\text{aq})}+\text{H}_2\text{O}_{(\text{I})}\rightarrow\text{H}_3\text{O}^+_{(\text{aq})}+\text{HSO}_{4(\text{aq})}^-; \ \text{K}_{\text{a}_1}>10$$\text{H}\text{SO}_{4(\text{aq})}+\text{H}_2\text{O}_{(\text{I})}\rightarrow\text{H}_3\text{O}^+_{(\text{aq})}+\text{SO}_{4(\text{aq})}^-; \ \text{K}_{\text{a}_2}=1.2\times10^{-2}$
It can be noticed that $\text{K}_{\text{a}_1}>>\text{K}_{\text{a}_2}$ This is because a neutral $H_2SO_4$ has a much higher tendency to lose a proton than the negatively charged . Thus, the former is a much stronger acid than the latter.
View full question & answer→Question 882 Marks
Account for the following:
Chlorine water has both oxidising and bleaching properties.
AnswerChlorine water produces nascent oxygen which is responsible for bleaching action and oxidation.
$Cl_2 + H_2O → 2HCl + [O]$
View full question & answer→Question 892 Marks
Sulphur dioxide is passed into aqueous solution of $Fe\ (III)$ salt?
Answer$SO_2$ acts as a reducing agent and hence reduces an aqueous solutuin of $Fe\ (III)$ salt to $Fe\ (II)$ salt.
View full question & answer→Question 902 Marks
How can you prepare $Cl_2$ from $HCl$ and $HCl$ from $Cl_2?$ Write reactions only.
Answer$\text{MnO}_2+4\text{HCl}\rightarrow\text{MnCl}_2+\text{Cl}_2+2\text{H}_2\text{O}\\\text{Oxidising}\\\text{agent}$We can also used $KMnO_4, K_2Cr_2O_7,$ etc., in place of $MnO_2.$
$\text{H}_2+\text{Cl}_2\xrightarrow[]{\text{Diffiused sunlight}}2\text{HCl}$
View full question & answer→Question 912 Marks
Draw the structure of the following: $HOClO_2$
View full question & answer→Question 922 Marks
Draw the structure of the following: $XeO_3$
View full question & answer→Question 932 Marks
Assign appropriate reasons for the following statement:
Metal fluorides are more ionic in nature than metal chlorides.
AnswerAccording to Fajan’s rule, a bigger anion is more easily polarised than a smaller anion. As a result, same metal cation can polarise a bigger $Cl^-$ ion more easily than the smaller $F^-$ ion. In other words, for the same metal, the metal fluoride is more ionic than metal chloride. So, in general, we can easily say that metal fluorides are more ionic than metal chlorides.
View full question & answer→Question 942 Marks
Mention the conditions required to maximise the yield of ammonia.
AnswerAmmonia is prepared using the Haber’s process. The yield of ammonia can be maximized under the following conditions:
- High pressure $(200\ atm).$
- A temperature of $700\ K.$
- Use of a catalyst such as iron oxide mixed with small amounts of $K_2O$ and $Al_2O_3.$
View full question & answer→Question 952 Marks
Explain the following:
$PCl_5$ is ionic in nature in the solid state.
AnswerThis is because in solid state $PCl_5$ exists as $[PCl_4]^+\ [PCl_6]^-$ in which cation is tetrahedral and the anion is octahedral. On melting, these ions become free to move and hence it conducts electricity.
View full question & answer→Question 962 Marks
Give reason for the following:
Neon is generally used for warning signals.
AnswerNeon lights are visible from long distances even in fog and mist and hence neon is generally used for warning signals.
View full question & answer→Question 972 Marks
Nitrogen exists as diatomic molecule and phosphorus as $P_4$. Why?
AnswerNitrogen owing to its small size has a tendency to form $\text{p}\pi-\text{p}\pi$ multiple bonds with itself.
Nitrogen thus forms a very stable diatomic molecule,$N_2$ On moving down a group, the tendency to form $\text{p}\pi-\text{p}\pi$ bonds decreases (because of the large size of heavier elements).
Therefore, phosphorus (like other heavier metals) exists in the $P_4$ state.
View full question & answer→Question 982 Marks
Explain why ozone is thermodynamically less stable than oxygen.
AnswerOzone is thermodynamically unstable with respect to oxygen since its decomposition into oxygen results in the liberation of heat ($\Delta\text{H}$ is negative) and an increase in entropy ($\Delta\text{S}$ is positive). These two effects reinforce each other, resulting in large negative Gibbs energy change ($\Delta\text{G}$) for its conversion into oxygen.
View full question & answer→Question 992 Marks
How is ammonia manufactured industrially?
AnswerAmmonia is manufactured industrially by Haber's process.$\text{N}_2(\text{g})+3\text{H}_2(\text{g})\rightleftharpoons2\text{NH}_3(\text{g}) \ \Delta_\text{f}\text{H}^\circ=-46.1\text{KJ mol}^{-1}$
A mixture of dry nitrogen and hydrogen gases in the ratio of $1:3$ by volume is compressed to about $200$ to $300$ atm and passed over iron catalyst at a temperature of about $723$ k to $773$ k. The iron catalyst is mixed with aluminiumoxide $(Al_2O_3)$ and potassium oxide $(K_2O)$ which act as promoters. Ammonia being formed is continuously removed by liquefying it. The optimum conditions for the production of ammonia are a pressure of about $200$ atm, a temperature of $\sim 700$ k and use of a catalyst such as iron oxide with small amounts of $K_2O$ and $Al_2,\ O_3$ to increase the rate of attainment of equilibrium.
View full question & answer→Question 1002 Marks
How are $XeO_3$ and $XeOF_4$ prepared?
AnswerPreparation of $XeO_3$: It is prepared by the hydrolysis of $XeF_4$ and $XeF_6$ under controlled pH of medium.$6\text{XeF}_4+12\text{H}_2\text{O}\rightarrow4\text{Xe}^-+2\text{XeO}_3+24\text{HF}+3\text{O}_2$
$\text{XeF}_4+3\text{H}_2\text{O}\rightarrow\text{XeO}_3+6\text{HF}$
Preparation of $XeOF_4$ Partial hydrolysis of $XeF_6$ gives $XeOF_4.\text{XeF}_6+\text{H}_2\text{O}\rightarrow\text{XeO}_4+2\text{HF}$
View full question & answer→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.
View full question & answer→Question 1192 Marks
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 $Cl^-.$
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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$\text{CU}+2\text{HNO}_3(\text{conc.})\rightarrow\text{Cu}(\text{NO}_3)_2+2\text{H}_2\text{O}+2\text{NO}_2$
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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.
$2\text{I}^-+\text{H}_2\text{O}+\text{O}_3\rightarrow2\text{OH}^-+\text{I}_2+\text{O}_2\\\text{Iodine} \ \ \ \ \ \ \ \ \ \ \text{Ozone} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Iodine}$
$\text{I}_2+2\text{Na}_2\text{S}_2\text{O}_3\rightarrow2\text{Na}_2\text{S}_4\text{O}_6+2\text{NaI}\\ \ \ \ \ \ \ \ \ \text{Sodium} \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Sodium} \\\ \ \ \ \ \ \ \text{thiosulphate} \ \ \ \ \text{tetrathionate}$
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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.$\text{Na}_2\text{SO}_3+\text{S}\ \ \xrightarrow{\ \ \text{Heat}\ \ }\ \ \text{Na}_2\text{S}_2\text{O}_3$
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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.
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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.$\text{AgCl}\ \ \ \ \ +\ \ \ 2\text{NH}_3\ \ \ \xrightarrow{\ \ \ \ \ }\ \ \ [\text{Ag(NH}_3)_2]\text{Cl}\\^\text{Silver chloride}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{Diam mine e (I) chloride}$
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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.
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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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