5 Marks Questions

Question 15 Marks

Give reasons for each of the following:

$\text{SiF}_{6}^{2-}$ is known but $\text{SiCl}_{6}^{2-}$ is not known.
Sulphur in vapour state exhibits paramagnetic behaviour.
$PbO_2$ is a stronger oxidizing agent than $SnO_2$.
$H_3PO_2$ acts as a monobasic acid.
Bond dissociation energy of $F_2$ is less than that of $Cl_2$.

Answer

It is due to the small size of Fluorine (steric repulsion will be less in $\text{SiF}_{6}^{-2}$) Silicon cannot hold Chlorine atoms because of its larger size.
In vapour state sulphur partly exists as $S_2$ molecule and $S_2$ molecule like $O_2$ has two unpaired electrons and hence exhibits paramagnetism.
Because $Pb^{2+}$ is more stable than $Pb^{4+}$ due to inert pair effect whereas $Sn^{4+}$ is more stable than $Sn^{2+}$.
Due to the presence of only one ionizable –OH group in $H_3PO_2$.

$F_2$ is expected to have more bond dissociation energy due to its small bond length as compared to $Cl_2$. But actually the dissociation energy of $F_2$ is less because of greater repulsion in the non bonded electron pairs in $F_2$ molecule than in $Cl_2$.

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Question 25 Marks

Account for the following:

Thermal stability of water is much higher than that of $H_2S$.
Anhydrous aluminium chloride acts as a catalyst.
White phosphorus is more reactive than red phosphorus.

Draw the structures of (i) $H_3PO_3$ and (ii) $XeOF_4$.

Answer

Because of stronger hydrogen bonds in $H_2O$, water has high stability as compared to hydrogen sulphide.
Because $AlCl_3$ is a strong lewis acid/or is an electron deficient compound.
Because white phosphorus is a discrete $P_4$ molecule whereas red phosphorous is polymeric.

Phosphorous Aci.

$XeOF_4$

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Question 35 Marks

Account for the following:

Bond angle in $NH_4^+$is greater than that in $NH_3$.
Reducing character decreases from $SO_2$ to $TeO_2$.
$HClO_4$ is a stronger acid than $HClO$.

Draw the structures of the following :

$H_2S_2O_8$
$XeOF_4$

Answer

Due to lone pair of electron on nitrogen in $NH_3$.
Due to inert pair effect/Stability of higher oxidation state decreases down the group from S to Te/Stability of lower oxidation state increases down the group.
$ClO_4^–$ is more stable than $ClO^–/ClO_4^–$ is weak conjugate base than $ClO^–$.

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Question 45 Marks

Which poisonous gas is evolved when white phosphorus is heated with conc. $NaOH$ solution? Write the chemical equation.
Write the formula of first noble gas compound prepared by $N$. Bartlett. What inspired $N$. Bartlett to prepare this compound?
Fluorine is a stronger oxidizing agent than chlorine. Why?
Write one use of chlorine gas.
Complete the following equation: $CaF_2 + H_2SO_4 \xrightarrow{\text{ }\ \ \ \ \ \ \ \ \ \ \ \ }$

Answer

$\text{PH}_{3}\ \ \ \ \ \ \ \ \ \text{P}_{4}+\text{3NaOH}+\text{3H}_{2}\text{O}\xrightarrow{\text{ }\ \ \ \ \ \ \ \ \ \ \ \ }\text{3NaH}_{2}\text{PO}_{2}+\text{PH}_{3}$
$Xe+[PtF_6]^–,$ Approximately same molecular size of $Xe\ \&\ O_2/$Comparable ionisation energies of $Xe\ \&\ O_2$
It is due to

Low enthalpy of dissociation of $F-F$ bond
High hydration enthalpy of $F^–.$

For bleaching wood pulp $($required for manufacture of paper and rayon$),$ cotton and textiles.
In the metallurgy $($extraction$)$ of gold and platinum.
In the manufacture of dyes, drugs and organic compounds such as $CHCl_3, CCl_4, DDT,$ refrigerants $(CCl_2F_2, $ freon$), $ and bleaching powder.
In the preparation of poisonous gases such as phosgene $(COCl_2)$, tear gas $(CCl_3NO_2)$, mustard gas $(ClCH_2CH_2SCH_2CH_2Cl),$ etc. Mustard gas was used by Germany in World War I.
In sterilizing drinking water.

$\text{CaF}_{2}+\text{H}_{2}\text{SO}_{4}\xrightarrow{\text{ }\ \ \ \ \ \ \ \ \ \ \ \ }\text{CaSO}_{4}+\text{2HF}$

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Question 55 Marks

Complete the following chemical equations:

$Cu + HNO_3 (dilute) →$
$XeF_4 + O_2F_2 \rightarrow $

Explain the following observations:

Phosphorus has greater tendency for catenation than nitrogen.
Oxygen is a gas but sulphur a solid.
The halogens are coloured. Why?

Answer

a.$3Cu + 8HNO_{3 (dilute)}$ $\rightarrow$ $3Cu(NO_3)_2 + 2NO + 4H_2O$.
$XeF_4 + O_2F_2$ $\rightarrow$ $XeF_6 + O_2$.
b.Because P-P single bond is stronger than N-N single bond.
Due to smaller size of oxygen it forms p$\pi$-p$\pi$ bonds and form $O_2 (O=O)$ which is absent in sulphur due to which it acquires a stable ring structure.
Because halogens absorb radiations in the visible region. This results in the excitation of valence electrons to a higher energy region.

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Question 65 Marks

Draw the molecular structures of the following compounds:

$N_2O_5$.
$XeOF_4$.

Explain the following observations:

Sulphur has a greater tendency for catenation than oxygen.
ICl is more reactive than $I_2$.
Despite lower value of its electron gain enthalpy with negative sign, fluorine $(F_2)$ is a stronger oxidising agent than $Cl_2$.

Answer

Because S-S single bond is stronger than O-O single bond.
Because I-Cl bond has lower bond dissociation enthalpy than I-I bond.
Because of lower bond dissociation enthalpy and high hydration enthalpy of Fluorine.

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Question 75 Marks

Draw the structures of the following molecules:

$(HPO_3)_3$.
$BrF_3$.

Complete the following chemical equations:

$HgCl_2 + PH_3 \rightarrow $
$SO_3 + H_2SO_4 \rightarrow $
$XeF_4 + H_2O \rightarrow $

Answer

$3HgCl_2 + 2PH_3$ $\rightarrow$ $Hg3P_2+ 6HCl.$
$SO_3 + H_2SO_4$ $\rightarrow$ $H_2S_2O_7.$
$6XeF_4 + 12H_2O$ $\rightarrow$ $4Xe + 2XeO_3 + 24HF + 3O_2$.

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Question 85 Marks

What happens when

Chlorine gas is passed through a hot concentrated solution of $NaOH$?
Sulphur dioxide gas is passed through an aqueous solution of a Fe (III) salt?

Answer the following:

What is the basicity of $H_3PO_3$ and why?
Why does fluorine not play the role of a central atomin interhalogen compounds?
Why do noble gases have very low boiling points?

Answer

$3Cl_2 + 6NaOH$ $\rightarrow$ $5NaCl + NaClO_3 +3H_2O$.
$2Fe^{3+} + SO_2 + 2H_2O$ $\rightarrow$ $2Fe^{2+} + SO_4^{2–} + 4H^+$.

Two, due to presence of two P-OH bonds.
Due to high electronegativityof fluorine.
There are no interatomic forces exceptweak dispersion forces.

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Question 95 Marks

Draw the structures of the following:

$H_2S_2O_7$.
$HClO_3$.

Explain the following observations:

In the structure of $HNO_3$, the N–O bond.

$(121\ pm)$ is shorter than the N–OH bond $(140\ pm)$.

All the P–Cl bonds in $PCl_5$ are not equivalent.
ICl is more reactive than $I_2$.

Answer

Because of resonance, N-O bond length is the average of single and double bond whereas N-OH bond has purely single bond.
Because the three P-CI bonds are equatorial and two P-Cl bonds are axial.
Because I-Cl bond has lower bond dissociation enthalpy than Cl-Cl bond.

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Question 105 Marks

Draw the structures of the following:

$H_2S_2O_8$.
$HClO_4$.

How would you account for the following:

$NH_3$ is a stronger base than $PH_3$.
Sulphur, has a greater tendency for ‘catenation than oxygen'.
$F_2$ is a stronger oxidising agent than $Cl_2$.

Answer

The lone pair of electrons on N atom in $NH_3$ is directed and not diffused or delocalized as it is in $PH_3$ due to larger size of P or due to availability of d-orbitals in P.
Because S-S single bond is stronger than O-O single bond.
Because of large electron-electron repulsion among the lone pairs in $F_3$ than that of $Cl_2$.

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Question 115 Marks

Assign reasons for the following:

Sulphur vapour is paramagnetic.
Ammonia $(NH_3)$ has greater affinity for protons than phosphine $(PH_3)$.
The negative value of electron gain enthalpy of fluorine is less than that of chlorine.
$SF_6$ is much less reactive than $SF_4$.
Of the noble gases only xenon is known to form well-established chemical compounds.

Answer

Because sulplhur vapour like $O_2$ contains two unpaired electrons in antibonding orbitals.
Because of small size and high elctron density on N.
Because of small size and high elctron repulsoion in F.
Because $SF_6$ is sterically protected by six F atoms.
Because of low lonisation enthalpy of Xe (comparable to that of oxygen )/compared to that of other noble gases.

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Question 125 Marks

Describe the favourable conditions for the manufacture of (i) ammonia by Haber’s process, and (ii) sulphuric acid by contact process.
Draw the structures of the following:

$PCl_5 (g)$.
$S_8 (g)$.
$CIF_3 (g)$.

Answer

Favourable conditions for the manufacture of

$NH_3$ by Haber’s process are low temperature and high pressure.
$H_2SO_4$ by contact process are low temperature and high pressure.

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Question 135 Marks

Assign reasons for the following:

The acid strengths of acids increase in the order.

$HF < HCl < HBr < HI.$

The lower oxidation state becomes more stable with increasing atomic number in Group $13.$
$H_3PO_2$ behaves as a monoprotic acid.

Draw the structures of the following compounds:

$SF_4.$
$XeF_2.$

Answer

1. Due to decrease in bond energy from HF to HI.
2. Due to inert pair effect/the energy required to unpair the $ns^2$ electrons is not compensated by the energy released in forming the two additional bonds.
3. Due to the presence of one ionizable hydrogen atom.

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Question 145 Marks

Assign reasons for the following:

i. $PbO _2$ is a stronger oxidising agent than $SnO _2$.
ii. In solid state $PCl _5$ behaves as an ionic species.
iii. Aluminium chloride $\left( AlCl _3\right)$ is very often used as a catalyst.

What is the structural difference between orthosilicates and pyrosilicates?

Answer

a. i. Because $Pb ^{4+}$ changes to more stable $Pb ^{2+}$ whereas $Sn ^{4+}$ is more stable than $Sn ^{2+}$ due to inert pair effect.
ii. Due to the formation of $\left[ PCl _4\right]^{+}\left[ PCl _6\right]^{-}$.
iii. Because $AlCl _3$ is a lewis acid/electron deficient.
b. Orthosilicates contain discrete $SiO _4$ tetrahedral units to form $Si _2 O _7^{-6}$ whereas in pyrosilicates two tetrahedral $SiO _4$ units are joined sharing a common oxygen atom.

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Question 155 Marks

Account for the following:

$BiH_3$ is the strongest reducing agent in Group $15$ elements hydrides.
$Cl_2$ acts as a bleaching agent.
Noble gases have very low boiling points.

Draw the structures of the following:

$H_4P_2O_7$
$XeOF_4$

Answer

Thermal stability of hydrides decreases down the group/Bond dissociation enthalpy decreases down the group.
Because $Cl_2$ in presence of moisture liberates nascent oxygen.
Interatomic interactions are weak.

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Question 165 Marks

Although nitrogen and chlorine have nearly same electronegativity yet nitrogen forms hydrogen bonding while chlorine does not. Why?
What happens when $F_2$ reacts with water?
Write the name of the gas evolved when $Ca_3P_2$ is dissolved in water.
Write the formula of a noble gas species which is isostructural with $IBr_2$.
Complete the equation:

$[Fe(H_2O)_6]^{2+}$ + NO $\longrightarrow$

Answer

Size of Nitrogen is smaller than Chlorine.
$2F_2 + 2H_2O$ $\longrightarrow$ $4HF + O_2 / HF$ and $O_2$ are produced.
$PH_3$/Phosphine.
$XeF_2$
$[Fe(H_2O)_6]^{2+} + NO^-$$\longrightarrow$ $[Fe(H_2O)_5(NO)]^{2+} + H_2O$

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Question 175 Marks

Compare the oxidizing action of $F_2$ and $Cl_2$ by considering parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
Write the conditions to maximize the yield of $H_2SO_4$ by contact process.
Arrange the following in the increasing order of property mentioned:

$H_3PO_3, H_3PO_4, H_3PO_2$ (Reducing character).
$NH_3, PH_3, AsH_3, SbH3, BiH_3$ (Base strength).

Answer

$F_2$ is the stronger oxidising agent than chlorine.

Low enthalpy of dissociation of F-F bond.
Less negative electron gain enthalpy of F.
High hydration enthalpy of F- ion.

Low temperature, high pressure and presence of catalyst.

$H_3PO_43PO_33PO_2.$
$BiH_33333.$

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Question 185 Marks

Account for the following:
1. Acidic character increases from HF to HI.
2. There is large difference between the melting and boiling points of oxygen and sulphur.
3. Nitrogen does not form pentahalide.

Draw the structures of the following:

$ClF_3$
$XeF_4$

Answer

Due to decrease in bond dissociation enthalpy from HF to HI, there is an increase in acidic character observed.
Oxygen exists as diatomic $O_2$ molecule while sulphur as polyatomic $S_8$.
Due to non-availability of d orbitals.

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Question 195 Marks

Which allotrope of phosphorus is more reactive and why?
How the supersonic jet aeroplanes are responsible for the depletion of ozone layers?
$F_2$ has lower bond dissociation enthalpy than $Cl_2$. Why?
Which noble gas is used in filling balloons for meteorological observations?
Complete the equation:

$XeF_2 + PF_5$ $\xrightarrow{\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }}$

Answer

White Phosphorus because it is less stable due to angular strain.
Nitrogen oxides emitted by supersonic jet planes are responsible for depletion of ozone layer.

Or $\text{NO+O}_{3}\rightarrow\text{NO}_{2}+\text{O}_{2}$

Due to small size of F, large inter electronic repulsion/electron- electron repulsion among the lone pairs of fluorine.
Helium.
$\text{XeF}_{2}+\text{PF}_{5}\rightarrow\text{[XeF]}^{+}\text{[PF}_{6}]^{-}$

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Question 205 Marks

Account for the following:

Ozone is thermodynamically unstable.
Solid $PCl_5$ is ionic in nature.
Fluorine forms only one oxoacid HOF.

Draw the structure of

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

Answer

Endothermic compound/decomposition of ozone is exothermic in nature and $\Delta $G is negative/decomposition of ozone is spontaneous.
Exists as $[PCl_4]^+[PCl_6]^-$
Shows only $-1$ oxidation state/most electronegative element/absence of d-orbitals.

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Question 215 Marks

Give reasons for the following:

Bond enthalpy of $f_2$ is lower than that of $Cl_2$.
$PH_3$ has lower boiling point than $NH_3$.

Draw the structures of the following molecules:

$BrF_{3.}$
$(HPO_3)_{3.}$
$XeF_{4.}$

Answer

Because of smaller size of F-atom/shorter bond length, the electron – electron repulsion among the lone pairs is greater in $F_2$ than $Cl_2$.
Due to hydrogen bonding in $NH3$.

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Question 225 Marks

Account for the following:

Helium is used in diving apparatus.
Fluorine does not exhibit positive oxidation state.
Oxygen shows catenation behavior less than sulphur.

Draw the structure of the following molecules.

$XeF_2$.
$H_2S_2O_8$.

Answer

Because of its low solubility in blood.
Because of its highest electronegativity.
Because O-O single bond is weaker than S-S single bond.

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Question 235 Marks

Complete the following chemical reaction equations:

$P_4 + SO_2Cl_2 \rightarrow $
$XeF_6 + H_2O \rightarrow $

Predict the shape and the asked angle ($90o$ or more or less) in each of the following cases:

$SO^{2–}_3$ and the angle $O–S–O$.
$ClF_3$ and the angle $F–Cl–F$.
$XeF_2$ and the angle $F–Xe–F$.

Answer

$ \text{P}_{4}+\text{10SO}_{2}\text{Cl}_{2}\rightarrow\text{4PCl}_{5}+\text{10SO}_{2} $.
$\text{XeF}_{6}+\text{H}_{2}\text{O}\rightarrow\text{XeOF}_{4}+\text{2HF}$.

Alternate Answer

$\text{XeF}_{6}+\text{3H}_{2}\text{O}\rightarrow\text{XeOF}_{3}+\text{6HF}$.

The angle O-S-O is greater than $90^o$

The angle F-Cl-F is lesser than $90^o$

The angle F-Xe-F is greater than $90^o$

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Question 245 Marks

Complete the following chemical equations:

$NaOH + Cl_2 \rightarrow $

(hot and conc.)

$XeF_4 + O_2F_2 \rightarrow $

Draw the structures of the following molecules:

$H_3PO_{2.}$
$H_2S_2O_{7.}$
$XeOF_{4.}$

Answer

$3Cl_2 + 6NaOH \rightarrow 5NaCl + NaClO_3 +3H_2O.$
$XeF_4+ O_2F_2 \rightarrow XeF_6 + O_{2.}$

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Question 255 Marks

Account for the following:

The acidic strength decreases in the order $HCl > H_2S > PH_3$.
Tendency to form pentahalides decreases down the group in group $15$ of the periodic table.

Complete the following chemical equations:

$P_4+ SO_2Cl_2 \rightarrow $
$XeF_2 + H_2O\rightarrow $
$I_2 + HNO_3 \rightarrow $

(conc.)

Answer

Because of increase in bond dissociation enthalpy from H-Cl bond to H-P bond/Because of decrease in electronegativity from to Cl to P.
Because of the energy factor (inert pair effect), stability of $+3$ oxidation state increases than that of $+5$ oxidation state.

$P_2 + 10SO_2Cl_2\rightarrow 4PCl_5 + 10SO_{2.}$

Alternate Answer
$P_4 + 8SO_2Cl_2 \rightarrow 4PCl_3 + 4SO_2 + 2S_2Cl_{2.}$

$XeF_2 – 2H_2O \rightarrow 2Xe + 4HF + O_{2.}$
$I_2 + 10 HNO_3 (conc) \rightarrow 2HIO_3 + 10 NO_2 + 4H_2O.$

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Question 265 Marks

Explain the following:

$NF_3$ is an exothermic compound whereas $NCl_3$ is not.
$F_2$ is most reactive of all the four common halogens.

Complete the following chemical equations:

$C + H_2SO_4 (conc.)\rightarrow $
$P_4 + NaOH + H_2O\rightarrow $
$Cl_2 + F_2 \rightarrow $

(excess)

Answer

Becuase bond energy of $F_2$ is lower than that of $Cl_2$ and N-F bond is smaller & stronger than $N-Cl$ bond.
Because of low bond dissociation enthalpy of $F-F$ bond.

$C+2H_2SO_4$(conc)$\rightarrow$ $CO_2 + 2SO_2 + 2H_2O.$
$P_4 + 3NaOH + 3H_2O$$\rightarrow$ $PH_3 + 3NaH_2PO_{2.}$
$Cl_2 + 3F_2$$\rightarrow$ $2ClF_3$.

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Question 275 Marks

Complete the following chemical equations:

$NaOH(aq) + Cl_2(g)\rightarrow $

(Hot and conc.)

$XeF_6(s) + H_2O(l)\rightarrow $

How would you account for the following?

The value of electron gain enthalpy with negative sign for sulphur is higher than that for oxygen.
$NF_3$ is an exothermic compound but $NCl_3$ is endothermic compound.
$ClF_3$ molecule has a T-shaped structure and not a trigonal planar one.

Answer

$6NaOH + 3Cl_2 \rightarrow 5NaCl + NaClO_3 + 3H_2O$
$XeF_6 + H_2O \rightarrow XeOF_4 + 2HF.$

Alternate Answer

$XeF_6 + 2H_2O \rightarrow XeO_2F_2 + 4HF.$

Alternate Answer

$XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF.$

Becuase of larger size of sulphur atom than oxygen atom.
Becuase bond energy of $F_2$ is lower than $Cl_2$ and N-F bond is smaller & stronger than $N-Cl$ bond.
Becuase it has $sp^3d$ hybridization.

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Question 285 Marks

Complete the following chemical reaction equations:

$P_4 + SO_2Cl_2\rightarrow $
$XeF_4 + H_2O \rightarrow $

Explain the following observations giving appropriate reasons:

The stability of $+5$ oxidation state decreases down the group in group $15$ of the periodic table.
Solid phosphorus pentachloride behaves as an ionic compound.
Halogens are strong oxidising agents.

Answer

$P4 + 10SO_2Cl_2 \rightarrow _4PCl_5 + 10SO_{2.}$
$6XeF_4 + 12H_2O \rightarrow 2XeO_3 + 4Xe + 24HF + 3O_{2.}$

Becuase down the group, $+3$ oxidation state becomes more & more stable due to higher energy involved to unpair the s electrons/due to inert pair effect.
Due to the formation of $[PCl_4]^+ [PCl_6]^-$
Becuase they readily accept an electron.

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Question 295 Marks

Give reasons for the following:

Molten aluminium bromide is poor conductor of electricity.
Nitric oxide becomes brown when released in air.
$PCl_5$ is ionic in nature in the solid state.
Ammonia acts as a ligand.
Sulphur disappears when boiled with an aqueous solution of sodium sulphite.

Answer

Because molten aluminium bromide is predominantly a covalent compound.
Due to oxidation of $NO$.

Alternate Answer

Due to the formation of $NO_2$ which is brown in colour.

Alternate Answer

$2NO (g) + O_2 (g) \rightarrow 2NO_2 (g).$

colourless Brown

$PCl_5$ is ionic in solid state because it exists as $[PCl_4]^+ [PCl_6]^-_2$.
Because of the presence of lone pair of electrons on nitrogen.

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Question 305 Marks

When concentrated sulphuric acid was added to an unknown salt present in a test tube a brown gas (A) was evolved. This gas intensified when copper turnings were added to this test tube. On cooling, the gas (A) changed into a colourless solid (B).

Identify (A) and (B).
Write the structures of (A) and (B).
Why does gas (A) change to solid on cooling?

Arrange the following in the decreasing order of their reducing character:

HF, HCl, HBr, HI

Complete the following reaction:

$\text{XeF}_4+\text{SbF}_5\xrightarrow{\ \ \ \ \ \ \ \ }$

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Question 315 Marks

Give reasons:

$H_3PO_3$_ undergoes disproportionation reaction but $H_3PO_4$_ does not.
When $Cl_2$_ reacts with excess of $F_2, ClF_3$_ is formed and not $FCl_3.$
Dioxygen is a gas while Sulphur is a solid at room temperature.

Draw the structures of the following:

$XeF_4$
$HClO_3$

Answer

In $H _3 PO _3, P$ has +3 oxidation state, while in $H _3 PO _4$, P has +5 oxidation state-the maximum oxidation state. It does not exceed the oxidation state beyond this, and thus, it does not undergo a disproportionation reaction. The reaction for disproportionation of $H _3 PO _3$ is given as follows:

$4\text{H}_3\text{PO}_3\xrightarrow{\ \ \ \ \ \ \ }3\text{H}_3\text{PO}_4+\text{PH}_3$

Fluorine has a lack of d orbitals; hence, it cannot expand its octet to form a bond with three chlorine atoms. Chlorine, on the other hand, has empty d orbitals, and thus, it can expand its octet to bond with three fluorine atoms and forms $ClF_3.$
The size of the oxygen atom is smaller as compared to the size of the sulphur atom. Oxygen can effectively form $\text{p}\pi-\text{p}\pi$ bonds, and exists as an $O_2 (O=O)$ molecule due to its smaller size. Also, oxygen exists in the gaseous state because it has weak Van der Waal’s forces. In case of sulphur, it forms an $S_8$_ polyatomic molecule with a puckered structure. Hence, it is a solid at room temperature.

$XeF_4$

$HClO_3$

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Question 325 Marks

Give reasons for the following:

Sulphur in vapour state shows paramagnetic behaviour.
N-N bond is weaker than P-P bond.
Ozone is thermodynamically less stable than oxygen.

Write the name of gas released when Cu is added to

dilute $HNO_3$
conc. $HNO_3$

Answer

i. The paramagnetism in Sulphur in the vapour phase can be explained from the fact that in the vapour phase, $S _2$ is the dominant species and is therefore paramagnetic like $O _2$, which has two unpaired electrons. So like $O _2$ molecule, $S _2$ also has two unpaired electrons in the $\pi^* px$ and $\pi^*$ py orbital.
ii. The nitrogen atom is smaller in size than the Phosphorus atom. The bond length of $N - N$ bond is smaller than that of P-P bond. Due to this, the four non bonding electrons of the two nitrogen atoms repel each other making the bond weaker. This does not happen in the case of phosphorus because P-P bond length is more. So, the repulsion of non-bonding electrons is lesser than in N-N. So, the P-P sigma bond is stronger.
iii. Ozone is thermodynamically unstable with respect to oxygen since its decomposition into oxygen results in the liberation of heat ( $\Delta H$ is negative) and an increase in entropy ( $\Delta S$ is positive). These two effects reinforce each other, resulting in large negative Gibbs energy change $(\Delta G)$ for its conversion into oxygen.

Nitrogen monoxide:

$8\text{HNO}_3(\text{dil})+3\text{Cu}\xrightarrow{\ \ \ \ \ \ \ \ \ }3\text{Cu}\text{(NO}_3)_2+2\text{NO}+4\text{H}_2\text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{Nitrogen}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{Monoxide}$

Nitrogen dioxide:

$4\text{HNO}_3(\text{conc.})+\text{Cu}\xrightarrow{\ \ \ \ \ \ \ \ \ }\text{Cu}\text{(NO}_3)_2+2\text{NO}_2+2\text{H}_2\text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{Nitrogen}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{dioxide}$

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Question 335 Marks

Write the disproportionation reaction of $H_3PO_3$.
Draw the structure of $XeF_4$.

Account for the following:

Although Fluorine has less negative electron gain enthalpy yet $F_2$ is strong oxidizing agent.
Acidic character decreases from $N_2O_3$ to $Bi_2O_3$ in group $15$.

Write a chemical reaction to test sulphur dioxide gas. Write chemical equation involved.

Answer

$4\text{H}_3(\text{PO}_3)\xrightarrow{\ \ \ {\Delta}\ \ \ \ }3\text{H}_3(\text{PO}_4)+\text{PH}_3$
$\text{XeF}_4$

It has a square planar structure. Six electron pairs form an octahedron with two positions occupied by lone pairs.

The oxidising capability of a substance depends upon three factors: bond dissociation enthalpy, electron gain enthalpy and hydration energy. Although electron gain enthalpy of fluorine is less negative as compared to chlorine, fluorine is a stronger oxidizing agent than chlorine, because of low enthalpy of dissociation of F-F bond $(158.8/kJ\ mol^{-1})$ and high hydration enthalpy of $F-(515/kJ\ mol^{-1})$. Due to its small size, Fluorine's hydration energy is very high hence better the oxidizing power. Fluorine has a greater electron-electron repulsion among the lone pairs in the small-sized $F_2$ molecule where they are much closer to each other than in case of $Cl_2$, hence the enthalpy of dissociation of $F_2$ is lower than $Cl_2$ which makes it better oxidizing agent.
Acidic character of oxides of group $15$ elements decreases and basicity increases down the group. As we move down the group, the atomic size increases, electronegativity decreases and metallic character increases. Oxides of Nitrogen and phosphorous are acidic whereas those of As and Sb are amphoteric and that of Bismuth is basic.

Chemical reaction to test sulphur dioxide gas are:

$\text{K}_2\text{Cr O}_7+\text{H}_2\text{SO}_4+\text{SO}_2\xrightarrow[\text{green}]{\ \ \ \ \ \ \ \ \ \ \ \ }\text{K}_2\text{SO}_4+\text{Cr}_2(\text{SO}_4)_3+\text{H}_2\text{O}\\^\text{Orange}$

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Question 345 Marks

Match the formulas of oxides given in Column I with the type of oxide given in Column II and mark the correct option.

	Column I		Column II
$(A)$	$Pb_3O_4$	$(1)$	Neutral oxide
$(B)$	$N_2O$	$(2)$	Acidic oxide
$(C)$	$Mn_2O_7$	$(3)$	Basic oxide
$(D)$	$Bi_2O_3$	$(4)$	Mixed oxide

$A (1) B (2) C (3) D (4)$
$A (4) B (1) C (2) D (3)$
$A (3) B (2) C (4) D (1)$
$A (4) B (3) C (1) D (2)$

Answer

	Column I		Column II
$(A)$	$Pb_3O_4$	$(4)$	Mixed oxide
$(B)$	$N_2O$	$(1)$	Neutral oxide
$(C)$	$Mn_2O_7$	$(2)$	Acidic oxide
$(D)$	$Bi_2O_3$	$(3)$	Basic oxide

$A (4) B (1) C (2) D (3)$

Explanation:

	Formulas of the compound		Type of oxide
$(A)$	$Pb_3O_4(PbO.Pb_2O_3)$	$(4)$	Mixed oxide
$(B)$	$N_2O$	$(1)$	Neutral oxide
$(C)$	$Mn_2O_7$	$(2)$	Acidic oxide
$(D)$	$Bi_2O_3$	$(3)$	Basic oxide

$Mn_2O_7$ on dissolution in water produces acidic solution.
$Bi_2O_3$ on dissolution in water produces basic solution.

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Question 355 Marks

Match the compounds given in Column I with the hybridisation and shape given in Column II and mark the correct option.

	Column I		Column II
(A)	$XeF_6$	(1)	$sp^3d^3$– distorted octahedral
(B)	$XeO_3$	(2)	$sp^3d^2$ - square planar
(C)	$XeOF_4$	(3)	$sp^3$- pyramidal
(D)	$XeF_4$	(4)	$sp^3d^2$ - square pyramidal

A (1) B (3) C (4) D (2)
A (1) B (2) C (4) D (3)
A (4) B (3) C (1) D (2)
A (4) B (1) C (2) D (3)

Answer

	Column I		Column II
(A)	$XeF_6$	(1)	$sp^3d^3$ – distorted octahedral
(B)	$XeO_3$	(3)	$sp^3$ - pyramidal
(C)	$XeOF_4$	(4)	$sp^3d^2$ - square pyramidal
(D)	$XeF_4$	(2)	$sp^3d^2$ - square planar

A (1) B (3) C (4) D (2)

Explanation:

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Question 365 Marks

Answer the following question:
Account for the following:

$SnCl_4$ is more covalent than $SnCl_2$.
Tendency to form pentahalides decreases down the group in group $15$ of the periodic table.

Answer

The oxidation states of central atom $Sn$ in $SnCl_4$ and $SnCl_2$ are $+4$ and $+2$ respectively. $+4$ state of Sn has higher polarising power which, inturn, increase the covalent character of bond formed between the central atom and the other atoms.
This is due to inert pair effect. The stability of $+5$ oxidation state decreases down the group in group $15$.

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Question 375 Marks

On heating lead (II) nitrate gives a brown gas “A”. The gas “A” on cooling changes to colourless solid “B”. Solid “B” on heating with NO changes to a blue solid ‘C’. Identify ‘A’, ‘B’ and ‘C’ and also write reactions involved and draw the structures of ‘B’ and ‘C’.

Answer

$\text{Pb}(\text{NO}_3)_2\xrightarrow[673\text{K}]{\ \ \ \ \Delta\ \ \ \ }2\text{PbO}+4\text{NO}_2\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{\text{(A)}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{\text{(Brown colour)}}$

$2\text{NO}+\text{N}_2\text{O}_4\xrightarrow{\ \ \ \ \Delta250\text{K}\ \ \ \ }2\text{N}_2\text{O}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{\text{(C)}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{\text{(Blue Solid)}}$

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Question 385 Marks

Match the items of Columns I and II and mark the correct option.

	Column I		Column II
$(A)$	Its partial hydrolysis does not change oxidation state of central atom	$(1)$	$He$
$(B)$	It is used in modern diving apparatus	$(2)$	$XeF_6$
$(C)$	It is used to provide inert atmosphere for filling electrical bulbs	$(3)$	$XeF_4$
$(D)$	Its central atom is in $sp^3d^2$ hybridisation	$(4)$	$Ar$

$A (1) B (4) C (2) D (3)$
$A (1) B (2) C (3) D (4)$
$A (2) B (1) C (4) D (3)$
$A (1) B (3) C (2) D (4)$

Answer

	Column I		Column II
$(A)$	Its partial hydrolysis does not change oxidation state of central atom	$(2)$	$XeF_6$
$(B)$	It is used in modern diving apparatus	$(1)$	$He$
$(C)$	It is used to provide inert atmosphere for filling electrical bulbs	$(4)$	$Ar$
$(D)$	Its central atom is in $sp^3d^2$ hybridisation	$(3)$	$XeF_4$

$A (2) B (1) C (4) D (3)$

Explanation:

Partial hydrolysis of $XeF_6$ gives oxyfluorides, $XeOF_4$ and $XeO_2F_2.$

$XeF_6+ H_2O \rightarrow XeOF_4+ 2HF$
$XeF_6+ 2H_2O \rightarrow XeO_2F_2+ 4HF$
We can see that oxidation state of central atom Xe remains unchanged.

He is used in modern diving apparatus
Ar is used to provide inert atmosphere for filling electrical bulbs.
$XeF_4$ has $sp^3d^2$ hybridization $(4-$bond pair and $2-$lone pair$)$

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Question 395 Marks

Match the species given in Column I with the shape given in Column II and mark the correct option.

	Column I		Column II
(A)	$\text{SF}_4$	(1)	Tetrahedral
(B)	$\text{BrF}_3$	(2)	Pyramidal
(C)	$\text{BrO}_3^-$	(3)	Sea-saw shaped
(D)	$\text{NH}_4^+$	(4)	Bent T-shaped

A (3) B (2) C (1) D (4)
A (3) B (4) C (2) D (1)
A (1) B (2) C (3) D (4)
A (1) B (4) C (3) D (2)

Answer

	Column I		Column II
(A)	$\text{SF}_4$	(3)	Sea-saw shaped
(B)	$\text{BrF}_3$	(4)	Bent T-shaped
(C)	$\text{BrO}_3^-$	(2)	Pyramidal
(D)	$\text{NH}_4^+$	(1)	Tetrahedral

A (3) B (4) C (2) D (1)

Explanation:

Species: $\text{SF}_4$
Shape: Sea-saw shaped
Structure

Species: $\text{BrF}_3$
Shape: Bent T-shaped
Structure:

Species: $\text{BrO}_3^-$
Shape: Pyramidal
Structure:

Species: $\text{NH}_4^+$
Shape: Tetrahedral
Structure:

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Question 405 Marks

$P_4O_6$ reacts with water according to equation $P_4O_6 + 6H_2O \rightarrow 4H_3PO_3$. Calculate the volume of $0.1M$ $NaOH$ solution required to neutralise the acid formed by dissolving $1.1\ g$ of $P_4O_6$ in $H_2O$.

Answer

$\text{P}_4\text{O}_6+6\text{H}_2\text{O}\xrightarrow{\ \ \ \ \ \ \ \ \ \ }4\text{H}_3\text{PO}_3;\\\underline{4\text{H}_3\text{PO}_3+8\text{NaOH}\xrightarrow{\ \ \ \ \ \ \ \ \ \ }4\text{Na}_2\text{HPO}_3+8\text{H}_2\text{O}}\\\ \ \ \ \ \text{P}_4\text{O}_6+8\text{NaOH}\xrightarrow{\ \ \ \ \ \ \ \ \ \ }4\text{Na}_2\text{HPO}_3+2\text{H}_2\text{O}\\\ \ \ \ \ \ ^{1\text{ Mol}}\ \ \ \ \ \ \ \ \ \ ^{8\text{ Mol}}$
($H_3PO_3$ can be neutralised with NaOH)
Moles of $\text{P}_4\text{O}_6=\frac{1.1}{220}=0.005\text{ mol}$
Acid formed by $1\ mol$ of $P_4O_6$ require $NaOH = 8\ mol$
Acid formed by $0.005\ mol$ of $P_4O_6$ require $NaOH = 8 \times 0.005 = 0.04\ mol$
$0.1\ M$ $NaOH$ means $0.1\ mol$ of NaOH is present in 1000mL solution
$0.04\ M$ of $NaOH$ is present in solution $=\frac{1000}{0.1}\times0.04$
$=400\text{mL}$

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Question 415 Marks

An amorphous solid $“A”$ burns in air to form a gas $“B”$ which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified aqueous $KMnO_4$ solution and reduces $Fe^{3+}$ to $Fe^{2+}$. Identify the solid $“A”$ and the gas $“B”$ and write the reactions involved.

Answer

A is sulphur $(Sg)$ while B is sulphur dioxide $(SO_2)$.

$\ \ \ \ \ \ \ \ \text{S}_8+8\text{O}_2\xrightarrow{\ \ \ \ \ \ \ \ \ }8\text{SO}_2\\\ \ \ \ \ \ \ \ ^{\text{(A)}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{\text{(B)}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\text{ZnS}+3\text{O}_2\xrightarrow{\ \ \ \ \ \ \ \ \ \ }2\text{ZnO}+2\text{SO}_2\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{\text{(B)}}$

$\ \ \ \ \ \ \ \ 2\text{KMnO}_4+3\text{H}_2\text{SO}_4\xrightarrow{\ \ \ \ \ \ \ \ \ \ }\text{K}_2\text{SO}_4+2\text{MnSO}_4+3\text{H}_2\text{O}+5[\text{O}]\\\ \ \ \ \ \ \ \ \ \ \ [\text{SO}_2+[\text{O}]+\text{H}_2\text{O}\xrightarrow{\ \ \ \ \ \ \ \ \ \ }\text{H}_2\text{SO}_4]\times5$

$\overline{2\text{KMnO}_4+5\text{SO}_5+2\text{H}_2\text{O}\xrightarrow{ \ \ \ \ \ \ \ \ \ \ }\text{K}_2\text{SO}_4+2\text{MnSO}_4+2\text{H}_2\text{SO}_4}\\^{\overline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{SO}_2+2\text{H}_2\text{O}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ }\text{H}_2\text{SO}_4+2\text{H}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Fe}_2(\text{SO}_4)_3+2\text{H}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ }2\text{FeSO}_4+\text{H}_2\text{SO}_4$

$\ \ \ \ \ \ \ \ \overline{\text{Fe}_2(\text{SO}_4)_3+\text{SO}_2+2\text{H}_2\text{O}\xrightarrow{\ \ \ \ \ \ \ \ \ \ }2\text{FeSO}_4+2\text{H}_2\text{SO}_4}\\\ \ \ \ \ \ \ \ ^{\overline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }}$

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Question 425 Marks

Match the items of Columns I and II and mark the correct option.

	Column I		Column II
$(A)$	$H_2SO_4$	$(1)$	Highest electron gain enthalpy
$(B)$	$CCl_3NO_2$	$(2)$	Chalcogen
$(C)$	$Cl_2$	$(3)$	Tear gas
$(D)$	Sulphur	$(4)$	Storage batteries

$A (4) B (3) C (1) D (2)$
$A (3) B (4) C (1) D (2)$
$A (4) B (1) C (2) D (3)$
$A (2) B (1) C (3) D (4)$

Answer

	Column I		Column II
$(A)$	$H_2SO_4$	$(4)$	Storage batteries
$(B)$	$CCl_3NO_2$	$(3)$	Tear gas
$(C)$	$Cl_2$	$(1)$	Highest electron gain enthalpy
$(D)$	Sulphur	$(2)$	Chalcogen

$A (4) B (3) C (1) D (2)$

Explanation:

$H_2SO_4$ is used in Storage batteries.
$CCl_3NO_2$ is known as tear gas.
$Cl_2$ has highest electron gain enthalpy.
Sulphur is also called as chalcogen.

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Question 435 Marks

Answer the following question: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

Shorter the bond length, higher is the bond dissociation enthalpy of hydrogen halide. As the atomic size increases down the group the $E-H (E = F, Cl, Br, I)$ bond length increases and hence the bond dissociation enthalpy increases in the reverse order i.e.,

$HI < HBr < HCl < HF$

$H_2O < H_2S < H_2Se < H2Te$

The increase in acidic character from $H_2O$ to $H_2Te$ is due to decrease in bond enthalpy for dissociation of $H-E (E = O, S, Se, Te)$ bond down the group.

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Question 445 Marks

Account for the following:

When $NaBr$ is heated with conc. $H_2SO_4, Br_2$ is produced but when $NaCl$ is heated with conc. $H_2SO_4,$ $HCl$ is produced.
$H_2S$ acts only as a reducing agent but $SO_2$ acts both as a reducing agent as well as an oxidising agent.
The acid strength decreases in the order: $HCl > H_2S > PH_3.$

Answer

When $NaBr$ is heated with conc. $H_2SO_4, $ HBr is first produced which being a reducing agent reduces $H_2SO_4 $ to $SO_2$_ while HBr itself gets oxidised to $Br_2.$

$NaBr + H_2SO_4 \rightarrow NaHSO_4 + HBr$

$2HBr + H_2SO_4 \rightarrow 2H_2O + SO_2 + Br_2$

As a result, only $Br_2$ is produced.

Similarly, NaCl reacts with conc. $H_2SO_4$ to form HCl. Since HCl does not act as a reducing agent, it does not get oxidised to $Cl_2.$

$NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl$

$HCl + H_2SO_4 \rightarrow$ No action

As a result, only HCl is produced.

The minimum oxidation number $(O.N)$ of $S$ is $-2$ while its maximum $O.N$ is +6. In $SO_2,$ the O.N of sulphur is $+4$, hence, it cannot only increase its $O.N$ by losing electrons but also reduce its $O.N$ by gaining electrons. Thus, it acts both as a reducing agent as well as an oxidising agent.

In contrast, in $H_2S, S$ has an $O.N$ of $-2$. Thus, it can only increase its $O.N$ by losing electrons and hence acts only as a reducing agent.

Greater the polarity of the $H–A$ bond, more easily the bond break and hence greater is the acid strength. As the electronegativity of atom A decreases in the order; $Cl > S > P,$ therefore the polarity of the bond decreases in the order $HCl > H-S > H-P$ and hence the acid strength decreases in the same order $HCl > H_2S > PH_3.$

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Question 455 Marks

Account for the following: Give the formula and describe the structure of a noble gas species which is isostructural with $\text{IBr}^-_2.$

Answer

Total number of electron pairs around central atom I.
Therefore, according to VSEPR theory, $\text{IBr}^-_2,$ should be linear.

Now a noble gas compound having 10 electrons in the valence shell of central atom is $XeF^2 (8 + 2 \times 1 = 10)$. As it has $2$ bond pairs and $3$ lone pairs around Xe therefore like $\text{IBr}^-_2,$ XeF2 is also linear.

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Question 465 Marks

Answer the following question:
When conc. H2SO4 was added into an unknown salt present in a test tube, a brown gas (A) was evolved. This gas intensified when copper turnings were also added into this test tube. On cooling, the gas (A) changed into a colourless gas (B).

Identify the gases A and B.
Write the equations for the reactions involved.

Answer

$\text{A}=\text{NO}_2\text{(g)}, \ \text{B}= \text{N}_2\text{O}_4\text{(g)}$
$\text{MNO}_3+\text{H}_2\text{SO}_4\ \ \ \xrightarrow{\ \ \text{Heat}\ \ \ }\ \ \ \text{MHSO}_4+\text{HNO}_3$
$4\text{HNO}_3 \ \ \ \ \ \xrightarrow{\ \ \text{Heat}\ \ \ }\ \ \ \ \ \ \ \ 4\text{NO}_2\ \ \ \ +\ \ \ \ 2\text{H}_2\text{O}\ \ \ + \ \ \text{O}_2\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{Nitrogen dioxide}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{(Brown gas)}$
$\ \ \ \ \ \ \ \ \text{Cu}\ \ \ +\ \ \ 4\text{HNO}_3 \ \ \xrightarrow{\ \ \text{Heat}\ \ \ }\ \ \text{Cu(NO}_3)_2\ +\ 2\text{H}_2\text{O}\ 2\ +\ \text{NO}_2\\^\text{Copper turnings}$

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Question 475 Marks

Answer the following question:
Arrange the following in order of property indicated for each set:

$F_2, Cl_2, Br_2, I_2$-increasing bond dissociation enthalpy.
$HF, HCl, HBr, HI$-incre acid strength.
$NH_3, PH_3, AsH_3, SbH_3, BiH_3$-increasing base strength.asing acid strength.

Answer

$I_2 < F_2 < Br_2 < Cl_2$

Bond dissociation enthalpy decreases with increase in the size of the atom as we move from $Cl$ to $I$. The low F-F bond dissociation enthalpy is due to the fact that F atom is very small in size and hence the three lone pair of electrons on each F atom repel the bond pair of F-F bond very strongly.

As the size of atom increases from F to I, the bond dissociation enthalpy of H-X bond decreases from H-F to H-I. Therefore, the acid strength increases in the opposite order:

$HF < HCl < HBr < HI$.

As we move from $NH_3$ to $BiH_3$, the size of the central atom increases. Consequently, the electron density on the central atom decreases and hence the basic strength decreases as we move from $NH_3$ to $BiH_3$.

Therefore, the basic strength increases in the order:
$BiH_3 < SbH_3 < AsH_3 < PH_3 < NH_3.$

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Question 485 Marks

Discuss the general characteristics of Group $15$ elements with reference to their electronic configuration, oxidation state, atomic size, ionisation enthalpy and electronegativity.

Answer

Electronic configuration: The valence shell electronic configuration of these elements is $ns^2 np^3.$ The s orbital in these elements is completely filled and p orbitals are half filled, making their electronic configuration extra stable.
Atomic Size: Covalent and ionic (in a particular state) radii increase in size down the group. There is a considerable increase in covalent radius from $N$ to $P$. However, from As to Bi only a small increase in covalent radius is observed. This is due to the presence of completely filled d and f or f orbitals in heavier members.
Oxidation State: The common oxidation states of these elements are $–3, +3$ and $+5.$ The tendency to exhibit $–3$ oxidation state decreases down the group due to increase in size and metallic group. In the last member of the group, bismuth hardly forms any compound in $–3$ oxidation state. The stability of $+5$ oxidation state decreases down the group. The stability of $+5$ oxidation state decreases and that of $+3$ state increases (due to invert pair effect) down the group. Nitrogen exhibits $+1, +2, + 4$ oxidation states also when it reacts with oxygen. Phosphorus also shows $+1$ and $+4$ oxidation states in some oxo acids.
Ionization enthalpy: Ionization enthalpy decreases down the group due to gradual increase in atomic size. Because of the extra stable half filled p orbitals electronic configuration and smaller size, the ionization enthalpy of the group $15$ elements is much greater than that of group 14 elements in the corresponding periods. The order of successive ionization emthalpies are expected as $\triangle H_1, < \triangle H_2 < \triangle H_3.$
Electronegativity: The electronegativity value, in general, decreases down the group with increasing atomic size. However, amongst the heavier elements, the different is not that much pronounced.

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Question 495 Marks

Answer the following question:
$X_2$ is a greenish yellow gas with pungent smell and used in purification of water. On dissolving water it gives a solution which turns blue litmus red. When it is passed through NaBr solution $Br_2$ is obtained.

Identify the gas.
What are products obtained when $X_2$ reacts with ammonia? Give chemical equations.
What happens when $X_2$ reacts with cold and dilute NaOH solution? Write chemical equation and give the name of reaction.

Answer

$\text{X}_2=\text{Cl}_2$
$8\text{NH}_3\ +\ 3\text{Cl}_2\ \ \xrightarrow{\ \ \ \ }\ \ 6\text{NH}_4\text{Cl}+\text{N}_2\\^\text{(excess)}$

$\text{NH}_3+3\text{Cl}_2\ \xrightarrow{\ \ \ \ \ \ }\ \ \text{NCl}_3\ \ +\ \ 3\text{HCl}\\^\text{(excess)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{(oxidation)}$

$\ 2\text{NaOH}\ \ +\ \ \text{Cl}_2\ \ \xrightarrow{\ \ \ }\ \ \text{NaCl}+\text{NaOCl}+\text{H}_2\text{O},\\^{\text{(cold and dil.)}}$ Disproportionation reaction.

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Question 505 Marks

An element ‘A’ exists as a yellow solid in standard state. It forms a volatile hydride ‘B’ which is a foul smelling gas and is extensively used in qualitative analysis of salts. When treated with oxygen, ‘B’ forms an oxide ‘C’ which is a colourless, pungent smelling gas. This gas when passed through acidified $KMnO_4$ solution, decolourises it. ‘C’ gets oxidised to another oxide ‘D’ in the presence of a heterogeneous catalyst. Identify ‘A’, ‘B’, ‘C’, ‘D’ and also give the chemical equation of reaction of ‘C’ with acidified $KMnO_4$ solution and for conversion of ‘C’ to ‘D’.

Answer

$'A'$ = Sulphur, $'B'$ = $H_2S$ gas, $'C' = SO_2$ gas, $'D' = SO_3$ gas
Reactions are:

$\ \ \ \ 2\text{SO}_2\text{(g)}\ \ \ +\ \ \ \text{O}_2\text{(g)}\ \ \ \xrightarrow{\text{v}_2\text{O}_5\text{(a)}\ \ \ }\ \ 2\text{SO}_3(\text{g})\\^\text{Sulphur dioxide}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{Sulphur trioxide}\\ \ \ \ \ \ ^\text{(C)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{(D)}$

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Question 515 Marks

A translucent white waxy solid (A) on heating in an inert atmosphere is converted to its allotropic form (B). Allotrope (A) on reaction with very dilute aqueous KOH liberates a highly poisonous gas (C) having rotten fish smell. With excess of chlorine it forms (D) which hydrolyses to compound (E). Identify compounds (A) to (E).

Answer

A = White phosphorus.
B = Red phosphorus.
C = Phosphine $(PH_3)$.
D = Phosphorus pentachloride $(PCl_5)$.
E = Phosphoric acid $(H_3PO_4)$.
Reactions are:
$\ \ \ \ \text{P}_4\text{(s)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \xrightarrow[\text{inert gas}]{\triangle \ \ \ }\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{P}_4\text{s}\\^\text{White phosphorus}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{Red phosphorus}\\ \ \ \ \ \ \ \ \ ^\text{(A)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{(B)}$
$\ \ \ \ \ \ \ \text{P}_4\text{(g)}\ \ \ +\ \ \ 3\text{KOH(aq)}+3\text{H}_2\text{O}\ \ \xrightarrow{\ \ \ \ \ \triangle\ \ \ \ \ \ }3\text{KH}_2\text{PO}_2\text{(aq)}\ \ \ \ +\ \ \ \text{PH}_3\text{(g)}\\^\text{White plasphorus}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{Potassium}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{Phosphine}\\\ \ \ \ \ ^\text{(A)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{hypophosphite} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{(C)}$
$\text{PH}_3\text{(g)}+4\text{Cl}_2\text{(g)}\rightarrow\text{PCl}_5\text{(g)}+3\text{HCl(g)}\\ ^\text{Phosphine}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{Phosphorous}\\ \ \ \ \ ^\text{(C)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{pentachloride}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{(D)}$
$\ \ \ \ \ \ \ \ \ \text{PCl}_5\text{(g)}\ \ \ \ \ \ \ \ \ \ \ \ + \ \ \ \ \ \ 4\text{H}_2\text{O(l)}\ \ \ \xrightarrow{\ \ \ \ \ }\ \ \ \text{H}_3\text{PO}_4\text{(aq)}\ \ \ +\ \ \ 5\text{Hcl(aq)}\\ ^\text{Phosphorous pentachloride}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{Phosporic acid}\\ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{(D)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{(E)}$

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