2 Marks Questions

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

$\text{P}_{4}\text{O}_{6}+\text{6H}_{2}\text{O}\rightarrow\text{4H}_{3}\text{PO}_{3}.$

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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).

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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

$CH_3MgBr, H_3O^+$
$Cl_2, P$

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Question 42 Marks

Write the structures of the following:

$BrF_3.$
$XeF_4.$

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Question 52 Marks

Draw the structures of the following:

$XeF_{4.}$
$BrF_{5.}$

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Question 62 Marks

Draw the structures of the following:

$XeO_3$
$H2SO_4$

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Question 72 Marks

Complete the following equations:

$C +$ conc.$ H_2SO_4 →$
$XeF_2 + H_2O →$

Answer

$C + 2H_2SO_4($conc.$) → CO_2 + 2 SO_2 + 2 H_2O$
$2XeF_2+ 2H_2O → 2Xe + 4HF + O_2$

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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

$2\text{F}\text{e}^{3+} + \text{SO}_{2} + 2\text{H}_{2}\text{O}\rightarrow 2\text{Fe}^{2+} + \text{SO}_{4}^{ 2-} + 4\text{H}^{+}$
$\text{XeF}_{4} + \text{SbF}_{5}\rightarrow [\text{ XeF}_{3}]^{+} [\text{SbF}_{6}]^{-}.$

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Question 92 Marks

Draw the structures of the following :
(a) $H_2S_2O_8$
(b) $ClF_3$

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Question 102 Marks

Draw the structures of the following:

$XeF_4$
$HClO_4$

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Question 112 Marks

Complete the following equations:

$\mathrm{Ag}+\mathrm{PCl}_5 \rightarrow$
$\mathrm{CaF}_2+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow$

Answer

$2 \mathrm{Ag}+\mathrm{PCl}_5 \rightarrow 2 \mathrm{AgCl}+\mathrm{PCl}_3$.
$\mathrm{CaF}_2+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CaSO}_4+2 \mathrm{HF}$.

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Question 122 Marks

Draw the structures of the following:

$\mathrm{H}_2 \mathrm{SO}_3$
$\mathrm{HClO}_3$

Answer

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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.
$\mathrm{XeF}_4+\mathrm{O}_2 \mathrm{~F}_2 \rightarrow \mathrm{XeF}_6+\mathrm{O}_2$.

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Question 142 Marks

Draw the structures of the following:

$\mathrm{XeF}_2$
$\mathrm{BrF}_3$

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Question 152 Marks

Draw the structure of the following molecules:

$ \mathrm{XeOF}_4 $
$ \mathrm{H}_3 \mathrm{PO}_3$

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Question 162 Marks

Complete the following reaction equations:

$C_6H_5N_2Cl + H_3PO_2 + H_2O\rightarrow$
$C_6H_5NH_2 + Br_2 (aq.) \rightarrow$

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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.

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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.

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Question 192 Marks

Draw the structural formulae of the following compounds:

$H_4P_2O_{5.}$
$XeF_{4.}$

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Question 202 Marks

Complete the following, chemical reaction equations:

$I_2 + HNO_3→$

(conc.)

$HgCl_2 + PH_3→$

Answer

$I_2 + 10HNO_3 \rightarrow 2HIO_3 + 10NO_2 + 4H_2O.$
$3HgCl_2 + PH_3 \rightarrow Hg_3P_2 + 6HCl.$

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Question 212 Marks

Complete the following chemical reaction equations:

$XeF_2 + H_2O →$
$PH_3 + HgCl_2 →$

Answer

$2XeF_2 + 2H_2O → 2Xe + 4HF + O_2.$
$2PH_3 + 3HgCl_2 → Hg_3P_2 + 6HCl.$

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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.

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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).

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Question 242 Marks

Write the structures of the following:

$H_2SO_3$
$XeF_4$

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Question 252 Marks

Complete the following chemical equations :

$F_2 + 2Cl^– \longrightarrow$
$2XeF_2 + 2H_2O \longrightarrow$

Answer

$F_2 + 2Cl^- \longrightarrow 2F^-+ Cl_2$
$2XeF_2 + 2H_2O \longrightarrow 2Xe + 4HF + O_2$

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Question 262 Marks

What happens when

$HCl$ is added to $MnO_2?$
$PCl_5$ is heated?

Write the equations involved.

Answer

$MnO_2 + 4HCl \longrightarrow MnCl_2+ Cl_2+ 2H_2O$
$\text{PCl}_5\xrightarrow{\text{ }\text{ }\text{ }\text{ }\text{ }\triangle\text{ }\text{ }\text{ }\text{ }\text{ }}\text{PCl}_3+\text{Cl}_2$

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Question 272 Marks

Draw the structures of the following :

$H_4P_2O_7$
$XeOF_4$

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Question 282 Marks

Draw the structures of the following:

$H_3PO_2$
$XeF_4$

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Question 292 Marks

Complete the following reactions :

$\text{C}l_2+\text{H}_2\text{O}\xrightarrow{\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }}$
$\text{XeF}_6+3\text{H}_2\text{O}\xrightarrow{\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }}$

Answer

Complete chemical reactions are:

$Cl_2 + H_2O → [HCl + HOCl] → 2 HCl + [O]\ ($Nascent oxygen$)$
$XeF_6 + 3H_2O → XeO_3+ 6HF$

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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 $H_2SO_4.$

$SO_3+ H_2O → H_2SO_4$

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Question 312 Marks

Complete the following chemical equations:

$Ca_3P_2+ H_2O →$
$Cu + H_2SO_4(conc.)→$

Answer

$Ca_3P_2+ 6H_2O→ 3Ca(OH)_2+ 2PH_3$
$Cu + 2H_2SO_4→ CuSO_4+ 2H_2O + SO_2$

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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.

Answer

$HI$
$H_2O2S2Se2Te$

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Question 332 Marks

Complete the following reactions:

$NH_3+ 3Cl_2($excess$) →$
$XeF_6 + 2H_2O →$

Answer

$NH_3 + 3Cl_2 ($excess$) → NCl_3 + 3HCl$
$XeF_6 + 2H_2O → XeO_2F_2 + 4HF$

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Question 342 Marks

What happens when

$(NH_4)_2Cr_2O_7$ is heated?
$H_3PO_3$ is heated?

Write the equations.

Answer

$(NH_4)_2Cr_2O_7 → N_2 + 4H_2O + Cr_2O_3$
$4H_3PO_3 → 3H_3PO_4 + PH_3$

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Question 352 Marks

Draw the structures of the following:

$H_2S_2O_7$
$XeF_6$

Answer

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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.

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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.

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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.

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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.

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Question 402 Marks

Draw the structures of the following molecules:

$XeF_{4.}$
$BrF_{3.}$

Answer

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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.

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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

$\text{P}_{4}+\text{3NaOH}+\text{3H}_{2}\text{O}\rightarrow\text{PH}_{3}+\text{3NaOH}_{2}\text{PO}_{2}.$
$\text{2I}^{-}+\text{H}_{2}\text{O}+\text{O}_{3}\rightarrow\text{2OH}^{-}+\text{I}_{2}+\text{O}_{2}.$

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Question 432 Marks

Draw the structures of the following molecules:

$SF_4.$
$XeF_4.$

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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.

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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

$\text{2Al}+\text{6HCl}+\text{12H}_2{O}\rightarrow\text{2[Al(H}_{2}\text{O})_{6}]\text{Cl}_{3}+3\text{H}_{2}$
$\text{Sn}+\text{2KOH}+\text{4H}_{2}\text{O}\rightarrow\text{K}_{2}\text{[Sn(OH)}_{6}]+\text{2H}_{2}$

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Question 462 Marks

Write the structures of the following species: