Case study (4 Marks)

Question 14 Marks

Read the passage given below and answer the following questions:
Interhalogen compounds are formed when halogen group elements react with each other. These are the compounds which consist of two or more different elements of group$-17$. A halogen with large size and low electronegativity reacts with an element of group$-17$ with small size and high electronegativity. As the ratio of radius of larger and smaller halogen increases, the number of atoms in a molecule also increases.
The following questions are multiple choice questions. Choose the most appropriate answer:

The stability of interhalogen compounds follows the order.

$IF_3 > BrF_3 > CIF_3$
$CIF_3 > BrF_3 > IF_3$
$BrF_3 > IF_3 > CIF_3$
$CIF_3 > IF_3 > BrF_3$

Identify the correct match from the following.

$[ICI_2]^- - $ bent.
$IF_7 $- pentagonal bipyramidal.
$CIF_3 $- trigonal planar.
$[BrF_4]^-$ - square pyramidal.

In $XA_5$, the central atom has $($both $X$ and $A$ are halogens$).$

$5$ bond pairs and no lone pairs.
$5$ bond pairs and one lone pair.
$6$ bond pairs and no lone pairs.
$4$ bond pairs and one lone pair.

In the known interbalogen compounds, the maximum number of atoms are,

Which of the following is not the characteristic of interhalogen compounds.

They are more reactive than halogens.
They are quite unstable but none of them is explosive.
They are covalent in nature.
They have low boiling points and are highly volatile.

Answer

(a) $IF_3 > BrF_3 > CIF_3$

Explanation:

Thermal stability decreases as the size difference or the electronegativity difference between the two halogen atoms decreases.

(b) $IF_7 - $ pentagonal bipyramidal.

Explanation:

$[ICI_2]^- -$ linear, $CIF_3 - T-$shaped,

$[BrF_4]^- -$ Square planar.

(b) 5 bond pairs and one lone pair.

Explanation:

It has square pyramidal shape and has 5 bond pairs and one lone pair.

Explanation:

In $IF_7,$ iodine is the least electronegative halogen, so its highest oxidation number $(+7)$ is more stable than those of the lighter member of the group.

(d) They have low boiling points and are highly volatile.

Explanation:

Some interhalogens are solids and are not volatile.

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Question 24 Marks

Read the passage given below and answer the following questions:
Ozone is an unstable, dark blue diamagnetic gas. It absorbs the UV radiation strongly, thus protecting the people on earth from the harmful UV-radiation from the sun. The use of chlorofluorocarbon (CFC) in aerosol and refrigerators and their subsequent escape into the atmosphere, is blamed for making holes in the ozone layer over the Antarctica. Ozone acts as a strong oxidising agent in acidic and alkaline medium. for this property, ozone is used as a germicide and disinfectant for sterilizing water. It is also used in laboratory for the ozonolysis of organic compounds and in industry for the manufacture of potassium permanganate, artificial silk, etc.
The following questions are multiple choice questions. Choose the most appropriate answer:

Which of the following statements is not correct for ozone?

lt oxidises lead sulphide.
It oxidises potassium iodide.
It oxidises mercury.
It cannot act as bleaching agent in dry state.

Ozone gives carbonyl compounds with.

Alkyl chloride
Alkanes
Alkenes followed by decomposition with $\frac{\text{Zn}}{\text{H}_2\text{O}}.$
Alcohols followed by decomposition with $\frac{\text{Zn}}{\text{H}_2\text{O}}.$

Ozone reacts with moist iodine gives.

$HI$
$HIO_3$
$I_2O_5$
$I_2O_4$

Ozone acts as an oxidising agent due to.

Iiberation of nascent oxygen.
Iiberation of nascent oxygen.
Both $(a)$ and $(b).$
Both $(a)$ and $(b).$

The colour of ozone molecule is:

White.
Blue.
Pale green.
Pale yellow.

Answer

(d) It cannot act as bleaching agent in dry state.
(c) Alkenes followed by decomposition with $\frac{\text{Zn}}{\text{H}_2\text{O}}.$

Explanation:
Ozone reacts with alkenes to form ozonides which on hydrolysis or reduction gives carbonyl compounds.

(b) $HIO_3$

Explanation:
$I_2 + 5O_3 + H_2O \rightarrow 2HIO_3 + 5O_2$

(a) Iiberation of nascent oxygen.
(b) Blue.

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Question 34 Marks

Read the passage given below and answer the following questions:
The halogen elements show great resemblances to one another in their chemical behaviour and properties of their compounds with other elements. There is, however, a progressive change in properties from F through $Cl, Br,$ and I to At. F is most reactive among the halogens and in fact, from all other elements and it has certain other properties that set it apart from the other halogens.
In these questions (Q.No. i - iv), a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.

Assertion and reason both are correct statements and reason is correct explanation for assertion.
Assertion and reason both are correct statements but reason is not correct explanation for assertion.
Assertion is correct statement but reason is wrong statement.
Assertion is wrong statement but reason is correct statement.

Assertion: $F_2 $ has high reactivity.

Reason: $F_2$ has low bond dissociation enthalpy.

Assertion: The bond between $F - F$ is weaker than between $Cl - Cl.$

Reason: Atomic size of $F$ is smaller than that of $Cl.$

Assertion: Fluoride does not show oxidation number greater than zero.

Reason: The halogens chlorine, bromine and iodine can show positive oxidation state of $+1, +3$ and $+7.$

Assertion: F atom has less negative electron affinity than Cl atom.

Reason: Additional electrons are repelled more effectively by $3p-$electrons in $Cl$ than by $2p-$electrons in Fatom.

Assertion: Fluorine is strongest oxidising agent in halogens.

Reason: It displaces other halogens from its aqueous solution.

Answer

(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

Explanation:

Fluorine is most electronegative element and has low bond dissociation enthalpy.

(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

Explanation:

In $F - F$ bond, due to smaller size of fluorine, $e^- - e^-$ repulsion occur, so its bond strength is less.

(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

Explanation:

Fluorine is most electronegative element and can not exhibit any $+ve$ oxidation state. Other halogens have vacant $d-$orbitals.

Explanation:

Additional electrons are repelled more effectively by $2$p-electrons in F than by $3$p-electrons in Cl atom.

(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

Explanation:

The electrode potential of $F_2$ is maximum while that of $I_2$ is minimum.

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Question 44 Marks

Read the passage given below and answer the following questions:
Under the normal conditions, noble gases are monoatomic and have closed shell electronic configuration. Lighter noble gases have low boiling points due to weak dispersion forces between the atoms and the absence of other interatomic interactions. Xenon, one of the important noble gas, forms a series of compounds with fluorine with oxidation number $+2, +4$ and $+6.$ All xenon fluorides are strong oxidising agents. $XeF_4$ reacts violently with water to give $XeO_3.$ The geometry of xenon compounds can be deduced by considering the total number of electron pairs in their valence shell.
The following questions are multiple choice questions. Choose the most appropriate answer:

Among noble gases (from He to Xe) only xenon reacts with fluorine to form stable xenon fluorides because xenon.

Has the largest size.
Has the lowest ionisation enthalpy.
Has the highest heat ofvapourisation.
Is the most readily available noble gas.

The structure of $XeO_3$ is:

Square planar.
Pyramidal.
Linear.
T-shaped.

$XeF_6$ is expected to be.

Oxidising agent.
Reducing agent.
Unreactive.
Strongly basic.

In the preparation of compound of xenon, Bartlett had taken $\text{O}_2^+\text{PtF}_6^-$ as a base compound. This is because,

Both $O_2$ and $Xe$ have same size.
Both $Xe$ and $O_2$ have same electron gain enthalpy.
Both have almost same ionisation enthalpy.
Both $Xe$ and $O_2$ are gases.

The oxidation state of xenon in $XeO_3$ is:

$+4$
$+2$
$+8$
$+6$

Answer

(b) Has the lowest ionisation enthalpy.
(b) Pyramidal.

Explanation:

(a) Oxidising agent.

Explanation:

All xenon fluorides are strong oxidising agents.

$XeF_6 + H_2O \rightarrow XeO_3 + 6HF$

(c) Both have almost same ionisation enthalpy.
(d) $+6$

Explanation:

$XeO_3 \Rightarrow x + (-2) \times 3 = 0 \Rightarrow x = +6$

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Question 54 Marks

Read the passage given below and answer the following questions:
All the elements of group $16$ have $ns^2np^4$ configuration in their outermost shell. Therefore, the atoms of these elements try to gain or share two electrons to achieve noble gas configuration. Sulphur and other elements of group $16$ are less electronegative than oxygen, so, they cannot accept electrons easily. By sharing of two electrons with other elements, these elements acquire $ns^2np^6$ configuration and exhibit $+2$ oxidation state. Except oxygen, group $16$ elements have vacant d-orbitals in their valence shell to which electrons Call be promoted from P- and s-orbitals of the same shell As a result, they Call show $+4$ and $+6$ oxidation states also.
The following questions are multiple choice questions. Choose the most appropriate answer:

Oxygen shows $+2$ oxidation state in.

$OF_2$
$H_2O$
$CI_2O$
$H_2O_2$

Like sulphur, oxygen is not able to show $+4$ and $+6$ oxidation states because?

Oxygen is a gas while sulphur is a solid.
Sulphur has high ionisation enthalpy as compared to oxygen.
Oxygen has nod-orbitals in its valence shell.
Oxygen has high electron affmity as compared to sulphur.

Compounds of sulphur with $+4$ oxidation state acts as a/ an.

Oxidising agent.
Reducing agent.
Both oxidising as well as reducing agen.
Cannot be predicted.

Oxidation state of sulphur in $Na_2S_4O_6$ is:

$\frac{7}{2}$
$\frac{5}{2}$
$\frac{1}{2}$
$\frac{3}{2}$

The oxidation states of sulphur in $S_8, SO_3$ and $H_2S$ are respectively.

$0, +6$ and $-2$
$+6, 0$ and $-2$
$-2, 0$ and $+6$
$+2, +6$ and $-2$

Answer

(a) $OF_2$

Explanation:

As fluorine is more electronegative than oxygen, so, oxygen exhibits $+2$ oxidation state in $OF_2$.

Explanation:

In $SO_2$, sulphur having $+4$ oxidation state, so it can Jose its two more electrons to attain $+6$ oxidation state. It can gain electrons to attain its lowest oxidation state of $-2$. Therefore, it can behave as both reducing and oxidising agent.

(b) $\frac{5}{2}$

Explanation:

$\text{Na}_2\text{S}_4\text{O}_6\Rightarrow2(+1)+4\text{x}+6(-2)=0\Rightarrow\text{x}=\frac{5}{2}$

(a) $0, +6$ and $-2$

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Question 64 Marks

Read the passage given below and answer the following questions:
Noble gases are inert gases with general electronic configuration of $ns^2np^6$. These are monoatomic, colourless, odourless and tasteless gases. The first compound of noble gases was obtained by the reaction of Xe with $PtF_6.$ A large number of compounds of Xe and fluorine have been prepared till now. The structure of these compounds can be explained on the basis of VSEPR theory as well as concept of hybridisation. The compounds of krypton are fewer. Only the diftuoride of krypton $(KrF_2)$ has been studied in detail. Compounds of radon have not isolated but only identified by radio tracer technique. However, no true compounds of helium, neon or argon are yet known.
The following questions are multiple choice questions. Choose the most appropriate answer:

The formula of the compound when $Xe$ and $PtF_6$ are mixed, is:

$XeF_6$
$XeF_4$
$Xe_2PtF_6$
$Xe^+[PtF_6]^-$

The shape and hybridisation of some xenon oxy-fluoride and fluoride compounds are given below. Find the incorrect one.

$XeOF_2 - T-$shape $- sy^3d$
$XeOF_4 -$ square pyramidal $- sp^3d^2$
$XeF_2 -$ linear $- sp^3d$
$XeF_6 -$ square planar $- dsp^2$

Which of the following is not formed by $Xe?$

$XeF_5$
$XeF$
$XeF_3$
All of these.

The number oflone pairs and bond pairs of electrons around $Xe$ in $XeOF_4$ respectively are,

$0$ and $5$
$1$ and $5$
$1$ and $4$
$2$ and $3$

Which of the following compounds has more than one lone pair of electrons around central atom?

$XeO_3$
$XeF_2$
$XeOF_4$
$XeO_2F_2$

Answer

(d) $Xe^+[PtF_6]^-$
(d) $XeF_6 -$ square planar $- dsp^2$

Explanation:

$XeF_6$ has $sp^3d^3$ hybridisation and distorted octahedral shape.

(d) All of these.

Explanation:

$Xe$ has completely filled $5$p-orbital. As a result, when it undergoes bonding with an odd number $(1, 3$ or $5)$ of fluorine atoms, it leaves behind one unpaired electron. This causes the molecule to become unstable. As a result, $XeF, XeF_3$ and $XeF_5$ do not exist.

(b) $1$ and $5$

Explanation:

(b) $XeF_2$

Explanation:

$XeF_2$ has 3 lone pairs on Xe atom.

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Question 74 Marks

Read the passage given below and answer the following questions:
Nitric acid reacts with most of the metals (except noble metals like gold and platinum) and non-metals. Towards its reaction with metals. $HNO_3$ acts as an acid as well as an oxidising agent. Like other acids, $HNO_3$ liberate nascent hydrogen from metals which further reduces the nitric acid into number of products like $NO, NO_2, N_{2O}$ or $NH_3.$ The different stages of reduction of nitric acid are:
$\ \ _{+5}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _{+4}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _{+2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _{+1}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _{-3}\\\text{HNO}_3\xrightarrow{\ \ \ \ +\text{e}\ \ \ \ }\text{NO}_2\xrightarrow{\ \ \ \ +2\text{e}^-}\text{NO}\xrightarrow[\text{NaOH}]{\ \ \ \ +\text{e}^-}\text{N}_2\text{O}\xrightarrow{\ \ \ \ +4\text{e}^-}\text{NH}_3$
The product of the reduction of $HNO_3$ depends upon the nature of the metal, concentration of nitric acid and temperature.
The following questions are multiple choice questions. Choose the most appropriate answer:

Which of the following reactions is used to prepare laughing gas?

$Pb +$ dil. $HNO_3 \rightarrow$
$Hg +$ dil. $HNO_3 \rightarrow$
$Zn +$ dil. $HNO_3 \rightarrow$
$Cu +$ dil. $HNO_3 \rightarrow$

Gold and platinum does not dissolve in $HNO_3$ but soluble in $1 : 3$ mixture of $HNO_3$ and $HCl$ due to the formation of respectively.

$Au(NO_3)_2, [Pt(NO_3)_2]$
$H[AuCl_4J, H_2[PtCl_6]$
$[AuCl_6]^{2-}, [PtCl_2)^{2-}$
$[Au(NO_3)_4], [Pt(NO_3)_6]^{2-}$

Identify B in the following reaction.

$\text{Cu}+\text{HNO}_{3{\text{(conc.)}}}\xrightarrow{\ \ \ \ }\ \ \ \ \ \ (\text{A})\ \ \ \ \ \ \ +\ \ \ \ \ \ \ (\text{B})+\text{H}_2\text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\text{Deep blue colour}}\ \ \ \ \ \ \ \text{Gas}$

$NO_2$
$N_2$
$NO$
$N_2O$

In which of the following reactions $HNO_3$ will not act as an oxidising agent?

$HNO_3 + H_2SO_4 \rightarrow$
$HNO_3 + FeS0_4 + H_2S0_4 \rightarrow$
$KI + HNO_3 \rightarrow$
$Au + HNO_3\rightarrow$

When dil. $HNO_3$ reacts with Hg, which gas will liberate?

$N_2$
$O_2$
$NO$
$NO_2$

Answer

Explanation:

$4\text{Zn}+10\text{HNO}_3\xrightarrow{}4\text{Zn}(\text{NO}_3)_2+5\text{H}_2\text{O}+\text{N}_2\text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Laughing gas}$

(b) $H[AuCl_4J, H_2[PtCl_6]$

Explanation:

$\text{Au}+3\text{[CI]}\xrightarrow{}\text{AuCl}_3\xrightarrow{\text{HCl}}\text{H}[\text{AuCl}_4]\\\ \ \ \ \text{Form aqua regia}$

$\text{Pt}+4\text{[CI]}\xrightarrow{}\text{PtCl}_4\xrightarrow{\text{HCl}}\text{H}_2[\text{PtCl}_6]\\\ \ \ \ \text{Form aqua regia}$

(a) $NO_2$

Explanation:

$\text{Cu}+4\text{HNO}_{3\text{(conc.)}}\xrightarrow{}\text{Cu}(\text{NO}_3)_2+2\text{NO}_2+2\text{H}_2\text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(A)}\ \ \ \ \ \ \ \ \ \ \ \ \text{(B)}$

(a) $HNO_3 + H2SO_4 \rightarrow$

Explanation:

$\text{HNO}_3+\text{H}_2\text{SO}_4\xrightarrow{}\text{NO}_2^++\text{HSO}_4^-+\text{H}_2\text{O}$

In this reaction $HNO_3$ is acting as $OH^-$ donor and $H_2SO_4$ as $H^+$ donor. This is not a redox reaction.

Explanation:

$6\text{Hg}+8\text{HNO}_{3\text{(dil.)}}\xrightarrow{}3\text{Hg}_2(\text{NO}_3)_2+2\text{NO}+4\text{H}_2\text{O}$

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Question 84 Marks

Read the passage given below and answer the following questions:
Chlorine is a greenish yellow gas with pungent and suffocating odour. With dry slaked lime, it gives bleaching powder. Bleaching powder is a mixture of calcium hypochlorite and basic calcium chloride:
$[Ca(OCl)_2· CaCl_2· Ca(OH)_2· 2H_2O].$
The amount of chlorine obtained from a sample of bleaching powder by the treatment with excess of dilute acids or $CO_2$ is called available chlorine. Chlorine is a powerful bleaching agent. Bleaching effect of chlorine is permanent.
The following questions are multiple choice questions. Choose the most appropriate answer:

Chlorine gas reacts with _____ to form bleaching powder.

$Ca(OH)_2$
$CaCl_2$
$CaSO_4$
dry $CaO$

Chlorine reacts with cold and dilute alkali to form:

Chloride
Hypochlorite
Chlorate
Both $(a)$ and $(b)$

Which of the following is produced on the reaction of bleaching powder with a few drops of cone. $HCl?$

Hypochlorous acid
Oxygen
Chlorine
Calcium oxide

Chlorine is used as a bleaching agent. The bleaching action is due to.

Oxidation
Chlorination
Hydrogenation
Reduction

Bleaching powder contains a salt of an oxoacid as one of its components. The anhydride of that oxoacid is:

$Cl_2O$
$Cl_2O_7$
$ClO_2$
$Cl_2O_6$

Answer

(a) $Ca(OH)_2$
(d) Both $(a)$ and $(b)$

Explanation:

In cold, chlorine reacts with dilute alkalies to form chlorides and hypochlorites.

(c) Chlorine
(a) Oxidation

Explanation:

$Cl_2 + H_2O \rightarrow 2HCl + [O]$

(a) $Cl_2O$

Explanation:

Bleaching powder contains $OCl^-$ ion, hence the oxoacid is $HOCl$. Anhydride of $HOCl$ is $Cl_2O.$

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Question 94 Marks

Read the passage given below and answer the following questions:
All the elements of group $16$ form hydrides: $H_2O, H_2S, H_2Se, H_2Te$ and $H_2Po.$ All these hydrides have angular structure which involves $sp^3$ hybridisation of the central atom. All hydrides are volatile. The volatility increases from $H_2O$ to $H_2S$ and then decreases. All hydrides are weakly acidic in character. The increase in acidic character from $H_2O$ to $H_2Te$ is a resultof thedecrease in the $1 H-E ($where$ E = O, S, Se, Te, Po)$ bond dissociation enthalpy from $H_2O$ to $H_2Te.$ All the hydrides except water are reducing agents. The reducing property of these hydrides increases from $H_2S$ to $H_2Te.$
In these questions $(Q.No. i - iv),$ a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.

Assertion and reason both are correct statements and reason is correct explanation for assertion.
Assertion and reason both are correct statements but reason is not correct explanation for assertion.
Assertion is correct statement but reason is wrong statement.
Assertion is wrong statement but reason is correct statement.

Assertion: Water has high boiling point.

Reason: Water molecules are associated with hydrogen bonding.

Assertion: $H_2Te$ has less acidic character than $H_2S.$

Reason: Bond dissociation enthalpy of $H-Te$ is less than $H-S.$

Assertion: Reducing nature of hydrides of group-16 elements increases as the atomic number of central atom increases.

Reason: Due to strong force of attraction of $H-E$ bond.

Assertion: $H_2O$ is the only hydrides of the chakogens which is liquid.

Reason: In ice each $O-$atom is surrounded by $4H-$atoms.

Assertion: The thermal stability of the hydrides decreases as: $H_2O > H_2S > H_2Se > H_2Te.$

Reason: Due to increase in the size of central atom on going down the group.

Answer

(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

Explanation:

The high boiling point of water is due to the association of $H_2O$ molecules through hydrogen bonding.

(d) Assertion is wrong statement but reason is correct statement.

Explanation:

$H_2Te$ is more acidic than $H_2S.$

Explanation:

Due to weakening of $H-E$ bond as the bond length increases with increase of size of E-atom.

(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

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Chemistry Case study (4 Marks)

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