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Chapter 3 - p-Block Elements – II - 12th Chemistry Guide Samacheer Kalvi Solutions - Pdf Download [2024-2025]

Chapter 3 - p-Block Elements – II - 12th Chemistry Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

$ C while hydrogen fluoride and hydrogen chloride are stable at this temperature.
Reason (R) - Thermal stability of hydrogen halides decreases from fluoride to iodide.
(a) $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ explains $\mathrm{A}$
(b) A and R are correct but doesn't explains A
(c) $\mathrm{A}$ is correct but $\mathrm{R}$ is wrong
(d) $A$ is wrong but $R$ is correct
Answer:
(a) $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ explains $\mathrm{A}$
Question 4.
Assertion (A) - The bleaching of chlorine is temporary.
Reason (R) - Chlorine oxidises ferrous salts to ferric salts.
(a) $A$ and $R$ are correct and $R$ explains $A$
(b) $A$ and $R$ are correct but doesn't explains $A$
(c) $A$ is correct but $R$ is wrong
(d) $\mathrm{A}$ is wrong but $\mathrm{R}$ is correct
Answer:
(d) $\mathrm{A}$ is wrong but $\mathrm{R}$ is correct
Question 5.
Assertion (A) - Sulphuric acid is highly reactive.
Reason (R) - Sulphuric acid can act as strong acid and an oxidising agent.
(a) $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ explains $\mathrm{A}$
(b) $A$ and $R$ are correct but doesn't explains $A$
(c) $A$ is correct but $R$ is wrong
(d) $\mathrm{A}$ is wrong but $\mathrm{R}$ is correct
Answer:
(a) $\mathrm{A}$ and $\mathrm{R}$ are correct but doesn't explains $\mathrm{A}$
Question 6.
Assertion (A) - Sulphuric acid is a high boiling point and viscous liquid.
Reason $(\mathrm{R})$ - This is due to the association of molecules together through hydrogen bonding.
(a) $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ explains $\mathrm{A}$
(b) $A$ and $R$ are correct but doesn't explains $A$
(c) $\mathrm{A}$ is correct but $\mathrm{R}$ is wrong
(d) $\mathrm{A}$ is wrong but $R$ is correct
Answer:
(a) $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ explains $\mathrm{A}$

Question 7.
Assertion (A) - Monoclinic sulphur is less stable than rhomobic sulphur.
Reason (R) - Monoclinic sulphur is stable between $96^{\circ} \mathrm{C}-119^{\circ} \mathrm{C}$ and slowly changes into rhombic sulphur.
(a) $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ explains $\mathrm{A}$
(b) $\mathrm{A}$ and $\mathrm{R}$ are correct but doesn't explains $\mathrm{A}$
(c) $\mathrm{A}$ is correct but $\mathrm{R}$ is wrong
(d) $A$ is wrong but $R$ is correct
Answer:
(a) $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ explains $\mathrm{A}$
Question 8.
Assertion (A) - Nitrogen gas is chemically inert.
Reason (R) - Nitrogen has low bonding energy.
(a) $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ explains $\mathrm{A}$
(b) A and $\mathrm{R}$ are correct but doesn't explains $\mathrm{A}$
(c) $\mathrm{A}$ is correct but $\mathrm{R}$ is wrong
(d) $A$ is wrong but $R$ is correct
Answer:
(c) $\mathrm{A}$ is correct but $\mathrm{R}$ is wrong
V. Find the odd one out
Question 1.

(a) $\mathrm{NO}$
(b) $\mathrm{HNO}_3$
(C) $\mathrm{NO}_2$
(d) $\mathrm{N}_2 \mathrm{O}$
Answer:
(b) $\mathrm{HNO}_3$
Hint: $\mathrm{HNO}_3$ is acid and other are oxides.
Question 2.
(a) Nitrous acid
(b) Nitric acid
(c) Hyponitrous acid
(d) Pemitrous acid
Answer:
(d) Pemitrous acid
Hint: Pemitrous acid contains peroxide linkage others doesn't have peroxide linkage.
Question 3.
(a) White phosphorous
(b) Red phosphorous
(c) phosphorous pentaoxide
(d) black phosphorous
Answer:

(c) phosphorous pentaoxide
Hint: $\mathrm{P}_2 \mathrm{O}_5$ is a compound of phosphorous and others are allotropic form of phosphorous.
Question 4.
(a) $\mathrm{PH}_3$
(b) $\mathrm{HPO}_3$
(c) $\mathrm{H}_3 \mathrm{PO}_3$
(d) $\mathrm{H}_3 \mathrm{PO}_4$
Answer:
(a) $\mathrm{PH}_3$
Hint: $\mathrm{PH}_3$ is hydrides of phosphorous and others are oxo acids of phosphorous.
Question 5.
(a) $\mathrm{He}$
(b) $\mathrm{Ne}$
(c) $\mathrm{Ar}$
(d) $\mathrm{Xe}$
Answer:
(d) $\mathrm{Xe}$
Hint: Xe forms several chemical compounds than others.
VI. Find out the correct pair.
Question 1.

(a) Helium - filament bulbs
(b) Krypton - prevent bonds
(c) Xenon - Lasers
(d) Radon - flash bulbs
Answer:
(c) Xenon-Lasers
Question 2.
(a) $\mathrm{Ra}$ - gamma rays
(b) $\mathrm{Xe}$ - cancer growth
(c) $\mathrm{Ne}$ - balloons
(d) $\mathrm{Kr}$ - advertisement bulb
Answer:
(a) $\mathrm{Ra}$ - gamma rays

Question 3.
(a) $\mathrm{ClF}_3$ - Linear
(b) $\mathrm{BrF}_5$ - $\mathrm{T}$ Shaped
(c) $\mathrm{IF}_4$ - square pyrimidal
(d) $\mathrm{BrF}_5$ - square pyrimidal
Answer:
(d) $\mathrm{BrF}_5-$ square pyrimidal
Question 4.
(a) $\mathrm{XeOF}_2-\mathrm{sp}^3$
(b) $\mathrm{XeF}_6-\mathrm{sp}^3 \mathrm{~d}^3$
(c) $\mathrm{XeF}_4-\mathrm{sp}^3$
(d) $\mathrm{XeOF}_4-\mathrm{sp}^3 \mathrm{~d}$
Answer:
(b) $\mathrm{XeF}_6-\mathrm{sp}^3 \mathrm{~d}^3$
Question 5.
(a) $\mathrm{OF}_2=-1$
(b) $\mathrm{Cl}_4 \mathrm{O}_4=-1$
(c) $\mathrm{I}_2 \mathrm{O}_4=-1$
(d) $\mathrm{I}_2 \mathrm{O}_9=-1$
Answer:
(a) $\mathrm{OF}_2=-1$
Question 6.
(a) Royal water - Dissolving gold
(b) Chlorine - Entraction of glue from bone
(c) $\mathrm{HCl}-$ Extraction of gold
(d) Chlorine - Temporary bleaching
Answer:
(a) Royal water - Dissolving gold

Question 7.
(a) $\mathrm{O}-2 \mathrm{~s}^2 2 \mathrm{p}^3$
(b) $S-3 s^2 3 p^4$
(c) $\mathrm{Se}-4 \mathrm{~d}^{10} 5 \mathrm{~s}^2 5 \mathrm{p}^4$
(d) $\mathrm{Te}-3 \mathrm{~d}^{10} 4 \mathrm{~s}^2 4 \mathrm{p}^4$
Answer:
(b) $S-3 s^2 3 p^4$
VI.Find out the incorrect pair.
Question 1.

(a) Nitrogen gas - inert
(b) Ammonia - pungent smelling gas
(c) Nitric acid-oxidizing agent
(d) Phosphine - rotten egg smell
Answer:
(d) Phosphine - rotten egg smell
Question 2.
(a) Liquid nitrogen - biological preservation
(b) Nitric acid - photography
(c) white phosphorous - yellow phosphorous
(d) phosphorous - welding
Answer:
(d) phosphorous - welding
Question 3.
(a) $\mathrm{N}_2 \mathrm{O}=+1$
(b) $\mathrm{N}_2 \mathrm{O}=+2$
(c) $\mathrm{N}_2 \mathrm{O}_3=+5$
(d) $\mathrm{NO}_2=+4$
Answer:
(c) $\mathrm{N}_2 \mathrm{O}_3=+5$

Question 4.
(a) Hvponitrous acid $-\mathrm{N}_2 \mathrm{O}$
(b) Nitrous acid $-\mathrm{HNO}_2$
(c) pernitric acid $-\mathrm{HNO}_4$
(d) pernitrous acid - HOONO
Answer:
(a) Hvponitrous acid $-\mathrm{N}_2 \mathrm{O}$
Question 5.
(a) $\mathrm{PH}_3$ - Holme's signal
(b) $\mathrm{O}_2$ - welding
(c) $\mathrm{H}_2 \mathrm{SO}_4-$ Disinfecting crops
(d) $\mathrm{SO}_3$ - Bleaching hair
Answer:
(c) $\mathrm{H}_2 \mathrm{SO}_4-$ Disinfecting crops
Question 6.
(a) $\mathrm{H}_2 \mathrm{SO}_4$ drying agent
(b) Chlorine - Deacon's process
(c) $\mathrm{HCI}$ - purification of bone black
(d) Helium - flash bulbs
Answer:
(d) Helium - flash bulbs
Question 7.
(a) $\mathrm{ICl}$ - Linear
(b) $\mathrm{CIF}_3-\mathrm{T}$ shape
(c) $\mathrm{IF}_5$ - pentagonal bipyramidal
(d) $\mathrm{IF}_7-$ pentagonal bipyramidal
Answer:
(c) $\mathrm{IF}_5$ - pentagonal bipyramidal
Question 8.
(a) $\mathrm{HOCl}=+2$
(b) $\mathrm{HOCl}=+3$
(e) $\mathrm{HOCI}=+5$
(d) $\mathrm{HOCl}_4=+7$
Answer:
(a) $\mathrm{HOCl}=+2$

Question 9.
(a) $\mathrm{XeF}_4-\mathrm{sp}^3 \mathrm{~d}^3$
(b) $\mathrm{XeF}_6-\mathrm{sp}^3 \mathrm{~d}^3$
(e) $\mathrm{XeOF}_4-\mathrm{sp}^3 \mathrm{~d}^2$
(d) $\mathrm{XeO}_3-\mathrm{sp}^3 \mathrm{~d}$
Answer:
(d) $\mathrm{XeO}_3-\mathrm{sp}^3 \mathrm{~d}$
Question 10 .
(a) $\mathrm{He}$ - cryogenics
(b) $\mathrm{Ne}$ - advertisement
(c) $\mathrm{Kr}$ - flurescent bulbs
(d) $\mathrm{Ra}$ - Lasers.
Answer:
(d) Ra - Lasers.
2 Mark Questions and Answers
Question 1.

What are pnictogens?
Answer:
The group - 15 elements like nitrogen, phosphorous Arsenic, Antimony and Bismuth are collectively called as pnictogens. Their general outer electronic configuration is $\mathrm{ns}^2 \mathrm{np}^3$.
Question 2.
What happen when sodium azide undergoes thermal decomposition?
Answer:
Pure nitrogen gas can be obtained by the thermal decomposition of sodium azide about $575 \mathrm{~K}$ $2 \mathrm{NaN}_3 \stackrel{575 K}{\longrightarrow} 2 \mathrm{Na}+3 \mathrm{~N}_2$
Question 3.
How will you prepare ammonia from nitrogen? and mention the name of process?
Answer:
Nitrogen directly reacts with hydrogen gives ammonia. This reaction is favoured by high pressures and at optimum temperature in the presence of iron catalyst,
$\mathrm{N}_2+3 \mathrm{H}_3 \rightleftharpoons 2 \mathrm{NH}_3$
This process is called as Haber's process.

Question 4.
Why nitrogen gas is chemically inert?
Answer:
The chemically inert character of nitrogen is largely due to high bonding energy of the molecules $225 \mathrm{cal}$ $\mathrm{mol}^{-1}$. Interestingly the triply bonded species is notable for its less reactivity in comparison with other isoelectronic triply bonded systems such as $-\mathrm{C} \equiv \mathrm{C}-\mathrm{C} \equiv \mathrm{Q}, \mathrm{X}-\mathrm{C} \equiv \mathrm{N}$, etc.
Question 5.
Mention the uses of nitrogen?
Answer:
1. Nitrogen is used for the manufacture of ammonia, nitric acid and calcium cyanamide etc.
2. Liquid nitrogen is used for producing low temperature required in cryosurgery, and so in biological preservation
Question 6.
Explain the action of heat on ammonia.
Answer:
Above $500^{\circ} \mathrm{C}$ ammonia decomposes into its elements. The decomposition may be accelerated by metallic catalysts like Nickel, Iron. Almost complete dissociation occurs on continuous sparking.


Question 7.
Prove ammonia act as a reducing agent?
Answer:
Ammonia reduces the metal oxides to metal when passed over heated metallic oxide.
$3 \mathrm{PbO}+2 \mathrm{NH}_3 \rightarrow 3 \mathrm{~Pb}+\mathrm{N}_2+3 \mathrm{H}_2 \mathrm{O}$
Question 8.
What happen when copper sulphate reacts with ammonia?
Answer:
When excess ammonia is added to aqueous solution copper sulphate a deep blue colour Complex $\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+}$ is formed.

Question 9.
Pure nitric acid is colourless, on standing it becomes yellow. Justify your answer.
Answer:
Nitric acid decomposes on exposure to sunlight or on being heated, into nitrogen dioxide, water and oxygen.
$
4 \mathrm{HNO}_4 \rightarrow 4 \mathrm{NO}_2+2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2
$
Due to this reaction pure acid or its concentrated solution becomes yellow on standing.
Question 10.
Write the products formed in the reaction of nitric acid with dilute and concentrated with magnesium.

Answer:
1. Magnesium with cone. $\mathrm{HNO}_3$ :
$
4 \mathrm{Mg}+10 \mathrm{HNO}_3 \rightarrow 4 \mathrm{Mg}\left(\mathrm{NO}_3\right)_2+\mathrm{NH}_4 \mathrm{NO}_3+3 \mathrm{H}_2 \mathrm{O}
$
Magnesium reacts with concentraated nitric acid to gives both magnesium nitrate and ammonium nitrate.
2. Magnesium with dil $\mathrm{HNO}_3$ :
Magnesium reacts with dilute nitric acid to gives both magnesium nitrate and nitrous oxide.
$
4 \mathrm{Mg}+10 \mathrm{HNO}_3 \rightarrow 4 \mathrm{Mg}\left(\mathrm{NO}_3\right)_2+\mathrm{N}_2 \mathrm{O}+5 \mathrm{H}_2 \mathrm{O}
$
Question 11.
Give the uses of nitric acid.
Answer:
1. Nitric acid is used as a oxidising agent and in the preparation of aquaregia.
2. 4. Salts of nitric acid are used in photography $\left(\mathrm{AgNO}_3\right)$ and gunpowder for firearms. $\left(\mathrm{NaNO}_3\right)$.
Question 12.
How will you prepare nitric oxide from sodium nitrite?
Answer:
When sodium nitrite reacts with ferrous sulphate in the presence of sulphuric acid to gives nitric oxide.
$
2 \mathrm{NaNO}_2+2 \mathrm{FeSO}_4+3 \mathrm{H}_2 \mathrm{~S} \rightarrow \mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3+2 \mathrm{NaHSO}_4+2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{NO}
$
Question 13.
How will you prepare nitrogen pentoxide?
Answer:
Nitric acid reacts with phosphorous pentaoxide to give nitrogen pentoxide.
$
2 \mathrm{HNO}_3+\mathrm{P}_2 \mathrm{O}_5 \rightarrow \mathrm{N}_2 \mathrm{O}_5+2 \mathrm{HPO}_3
$

Question 14.
Draw the structure of
1. $\mathrm{N}_2 \mathrm{O}$
2. $\mathrm{N}_2 \mathrm{O}_3$
Answer:
1. $\mathrm{N}_2 \mathrm{O}$ (Nitrous oxide)

2. $\mathrm{N}_2 \mathrm{O}_3$ (Nitrogen sesquoxide)


Question 15.
Draw- the structure of
1. $\mathrm{N}_2 \mathrm{O}_4$
2. $\mathrm{N}_2 \mathrm{O}_3$
Answer:
1. $\mathrm{N}_2 \mathrm{O}_4$ (Nitrogen tetraoxide)

2. $\mathrm{N}_2 \mathrm{O}_5$


Question 16.
Draw the structure of

1. Hyponitrous acid
2. Hydronitrous acid.
Answer:
(a) Hypon trous acid $-\mathrm{H}_2 \mathrm{~N}_2 \mathrm{O}_2$
$
\mathrm{HO}-\mathrm{N}=\mathrm{H}-\mathrm{OH}
$
(b) Hydronitrous acid:

Question 17.
Mention the allotropic forms of phosphorous?
Answer:
The most common allotropic forms of phosphorous
1. white phosphorous
2. Red phosphorous
3. Black phosphorous
Question 18.
Why white phosphorous is also known as yellow phosphorous?
Answer:
The freshly prepared white phosphorus is colourless but becomes pale yellow due to formation of a layer of red phosphorus upon standing. Hence it is also known as yellow phosphorus.
Question 19.
What is phosphorescence?
Answer:
White (yellow) phosphorous glows in the dark due to oxidation which is called phosphorescence.
Question 20.
Why white phosphorous undergoes spontaneous combustion in air?
Answer:
White phosphorous ignition temperature is very low and hence it undergoes spontaneous combustion in air at room temperature and during in the combusition process, white phosphorous produces $\mathrm{P}_2 \mathrm{O}_5$.
Question 21.
Draw the structure of
1. white phosphorous
2. red phosphorous

Answer:
1. White phosphorous

2. Red phosphorous


Question 22.
How will you convert red phosphorous into $\mathrm{P}_2 \mathrm{O}_3$ and $\mathrm{P}_2 \mathrm{Q}_5$ ?
Answer:
Red phosphorus reacts with oxygen on heating to give phosphorus trioxide or phosphorus pentoxide. $\mathrm{P}_4+3 \mathrm{O}_2 \triangle \mathrm{P}_4 \mathrm{O}_6$ (phosphorous trioxide)
$\mathrm{P}_4+5 \mathrm{O}_2 \stackrel{\triangle}{\rightarrow} \mathrm{P}_4 \mathrm{Q}_{10}$ (phosphorous pentaoxide)
Question 23.
How will you prepare orthophosphoric acid from phosphorous?
Answer:
When phosphorous is treated with cone nitric acid it is oxidised to orthophosphoric acid. This reaction is catalysed by iodine crystals.

Question 24.
Mention the uses of phosphorous?
Answer:
1. The red phosphorus is used in the match boxes.
2. It is also used for the production of certain alloys such as phosphor bronze.

Question 25 .
Show that phosphine is weakly basic?
Answer:
Phosphine is weakly basic and forms phosphonium salts with halogen acids.
$\mathrm{PH}_3+\mathrm{HI} \rightarrow \mathrm{PH}_4 \mathrm{I}$
$
\mathrm{PH}_4 \mathrm{I}+\mathrm{H}_2 \mathrm{O} \underset{\rightarrow}{\rightarrow} \mathrm{PH}_3+\mathrm{H}_3 \mathrm{O}^{-}+\mathrm{I}^{-}
$
Question 26.
Draw the structure of $\mathrm{PCl}_3$
Answer:

Question 27.
How will you prepare $\mathrm{PCl}_3$ from white phosphorous?
1. When a slow stream of chlorine is passed over white phosphorous, $\mathrm{PCl}_3$ is formed
2. It can also be obtained by treating white phosphorous with thionyl chloride.
Question 28.
What happen when $\mathrm{PCl}_3$ is treated with cold water?
Answer:
When $\mathrm{PCl}_3$ is hydrolysed with cold water it gives phosphorous acid.


Question 29.
How will you prepare $\mathrm{PCl}_3$ ?
Answer:
When $\mathrm{PCl}_3$ is treated with excess chlorine, phosphorous pentachloride is obtained.
$
\mathrm{PCl}_3+\mathrm{Cl}_2 \rightarrow \mathrm{PCl}_5
$
Question 30.
Mention the uses of $\mathrm{PCl}_3$ and $\mathrm{PCl}_5$ ?
answer:
1. Phosphorus trichloride is used as a chlorinating agent and for the preparation of $\mathrm{H}_2 \mathrm{PO}_3$.
2. Phosphorous pentachloride is a chlorinating agent and is useful for replacing hydroxyl groups by chlorine atom.
Question 31.
Draw the structure of $\mathrm{H}_3 \mathrm{PO}_2$ and $\mathrm{H}_3 \mathrm{PO}_3$ ?
Answer:

Question 32.
Draw the structure of $\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_6$ and $\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_3$ ?
Answwer:
(i) $\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_6$
(ii) $\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_3$
(i) $\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_6$
(ii) $\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_7$

Question 33.
Give the method to prepare hypophosphorous acid and pyrophosphoric acid?
Answer:
1. Hypophosphorous acid - Phosphorous reacts with water to give hypophosphorous acid.


2. Pyrophosphoric acid - Phosphorous acid is heated they produce pyrophosphoric acid.


Question 34.
Complete the reactions
1. $\mathrm{HgO} \triangle$ ?
2. $\mathrm{BaO}_2 \stackrel{\triangle}{\rightarrow}$ ?
Answer:

1. 
2. $2 \mathrm{BaO}_2 \stackrel{\triangle}{\rightarrow} 2 \mathrm{BaO}+\mathrm{O}_2$

Question 35 .
Give and explain the reaction used to estimation of ozone.
Answer:
Ozone oxidises potassium iodide to iodine.
$
\mathrm{O}_3+2 \mathrm{KI}+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{KOH}+\mathrm{O}_2+\mathrm{I}_2
$
This reaction is quantitative and can be used for estimation of ozone.
Question 36 .
Write a short notes on Rhombic sulphur?
Answer:
Rhombic sulphur also known as a sulphur, is the only thermodynamically stable allotropic form at ordinary temperature and pressure. The crystals have a characteristic yellow colour and composed of $\mathrm{S}_8$ molecules. When heated slowly above $96^{\circ} \mathrm{C}$, it converts into monoclinic sulphur. Upon cooling below $96^{\circ} \mathrm{C}$ the $\beta$ form converts back to a form.
Question 37.
Write a notes on monoclinic sulphur?
Answer:
Monoclinic sulphur also contains $\mathrm{S}_8$ molecules in addition to small amount of $\mathrm{S}_6$ molecules. It exists as a long needle like prism and is also called as prismatic sulphur. It is stable between $96^{\circ}-119^{\circ} \mathrm{C}$ and slowly changes into rhombic sulphur.
Question 38.
What are $\lambda$ - sulphur?
Answer:
Sulphur also exists in liquid and gaseous states. At around $140^{\circ} \mathrm{C}$ the monoclinic sulphur melts to form mobile pale yellow liquid called $\lambda$ sulphur.

Question 39.
How will you prepare sulphurdioxide by laboratory method?
Answer:
Sulphur dioxide is prepared in the laboratory treating a metal or metal sulphite with sulphuric acid.
$\mathrm{Cu}+2 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CuSO}_4+\mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O}$
$
\mathrm{SO}_3^{-}+2 \mathrm{H}^{+} \rightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{SO}_2
$
Question 40.
Complete the reactions,
Answer:
1. $\mathrm{Zns}+\mathrm{O}_2 \stackrel{\Delta}{\rightarrow}$ ?

2. $\mathrm{FeS}_2+\mathrm{O}_2 \stackrel{\Delta}{\rightarrow}$ ?
Answer:

1. 
$
\begin{aligned}
& \text { 2. } 4 \mathrm{FeS}_2+11 \mathrm{O}_2 \underset{\rightarrow}{\rightarrow} 2 \mathrm{Fe}_2 \mathrm{O}_3+8 \mathrm{SO}_2 \\
&
\end{aligned}
$
Question 41.
What happen when sulphurdioxdie reacts with sodium hydroxide and sodium carbonate?
Answer:
Sulphur dioxide reacts with sodium hydroxide and sodium carbonate to form sodium bisulphite and sodium sulphite respectively.

Question 42.
Mention the uses of sulphurdioxide?
Answer:
1. Sulphur dioxide is used in bleaching hair, silk, wool etc...
2. It can be used for disinfecting crops and plants in agriculture.
Question 43.
Why sulphuric acid is high boiling and viscous liquid ?
Answer:
Sulphuric acid is high boiling and viscous liquid this is due to the association of molecules together through hydrogen bonding.
Question 44.
Explain the reaction between benzene and sulphuric acid.
Answer:
Sulphuric acid reacts with benzene to give benzene sulphuric acid.
$\mathrm{C}_6 \mathrm{H}_6+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{SO}_3 \mathrm{H}+\mathrm{H}_2 \mathrm{O}$
(benzene) (Benzene sulphonic acia)
Question 45 .
Give the uses of sulphuric acid?
Answer:
1. Sulphuric acid is used in the manufacture of fertilisers, ammonium sulphate and super phosphates and other chemicals such as hydrochloric acid, nitric acid etc...
2. It is used as a drying agent and also used in the preparation of pigments, explosives etc..
Question 46.
Draw the structure of $\mathrm{H}_2 \mathrm{SO}_3$ and $\mathrm{H}_2 \mathrm{SO}_4$ ?
Answer:
1. $\mathrm{H}_2 \mathrm{SO}_3$

2. $\mathrm{H}_2 \mathrm{SO}_4$


Question 48.
Draw the structure of
1. Pyrosulphuric acid
2. Peroromonosulphuric acid?
Answer:
1. Pyrosulphuric acid $\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7$
2. Peroromonosulphuric acid $\mathrm{H}_2 \mathrm{SO}_5$

Question 49.
What happen when chlorine reacts with trupentine?
Answer:
When chlorine burnt with turpentine it forms carbon and hydrochloric acid.


Question 50.
How will you prepare hydrochloric acid by laboratory method?
Answer:
It is prepared by the action of sodium chloride and concentrated sulphuric acid

- $\mathrm{NaCl}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{NaHSO}_4+\mathrm{HCl}$
- $\mathrm{NaHSO}_4+\mathrm{NaCl} \rightarrow \mathrm{Na}_2 \mathrm{SO}_4+\mathrm{HCl}$
Dry hydrochloric acid is obtained by passing the gas through conc. sulphuric acid
Question 51.
Mention the uses of hydrochloric acid?
Answer:
1. Hydrochloric acid is used for the manufacture of chlorine, ammonium chloride, glucose from com starch etc.,
2. It is used in the extraction of glue from bone and also for purification of bone black.
Question 52.
Complet the following reaction?
(i) $\mathrm{BrF}_5 \stackrel{\mathrm{OH}^{\ominus}}{\longrightarrow}$ ?
(ii) $\mathrm{ICl} \stackrel{\mathrm{OH}^{\Theta}}{\longrightarrow}$ ?

Answer:

(i) 

(ii) 

Question 53.
Why noble gases have the largest ionisation energy?
Answer:
Noble gases have the largest ionisation energy compared to any other elements in a given row as they h completely filled orbital $\left(\mathrm{ns}^2 n \mathrm{p}^6\right)$ in their outer most shell. They are extremely stable and have a small tendency to gain or lose electrons. Hence noble gases have the largest ionisation energy.
Question 54
Mention the uses of Neon?
Answer:
Neon is used in advertisement as neon sign and the brilliant red glow is caused by passing electric curre through neon gas under low pressure.
Question 55.
Give the uses of Krypton?
Answer:
1. Krypton is used in fluorescent bulbs, flash bulbs etc...
2. Lamps filed with krypton are used in airports as approaching lights as they can penetrate through dense fog.
Question 56.
Mention the application of Xenon?
Answer:
1. Xenon is used in fluorescent bulbs, flash bulbs and lasers.
2. Xenon emits an intense light in discharge tubes instantly. Due to this it is used in high speed electronic flash bulbs used by photographers.
Question 57.
Give the uses of Radon?
Answer:
1. Radon is radioactive and used as a source of gamma rays
2. Radon gas is sealed as small capsules and implanted in the body to destroy malignant i.e. cancer growth.

Question 58.
Why are pentahalides more covalent than trihalides?
Answer:
Since elements in the +5 oxidation state have less tendency to lose electrons than in the +3 oxidation state, therefore elements in the +5 oxidation state are more covalent than in the +3 oxidation state. In other words, pentahalides are more covalent than trihalides.
59. Why is $\mathrm{N}_2$ less reactive at room temperature?
Answer:
Due to the presence of triple bond between the two nitrogen atoms, the bond dissociation energy of $\mathrm{N}_2$ $\left(941.4 \mathrm{~kJ} \mathrm{~mol}^{-}\right)$is very high. Therefore, $\mathrm{N}_2$ is less reactive at room temperature.
Question 60.
Why is $\mathrm{ICl}$ more reactive than $\mathrm{I}_2$ ?
Answer:
$\mathrm{ICl}$ bond is weaker than $\mathrm{I}$ - I bond. Therefore $\mathrm{ICl}$ can break easily to form halogen atoms which readily brings about the reaction.
Question 61.
Why is helium used in diving apparatus?
Answer:
It is used as a diluent for oxygen in modern diving apparatus because of its very low solubility in blood.
Question 62 .
Give the disproportionation reaction of $\mathrm{H}_3 \mathrm{PO}_3$ ?
Answer:
$\mathrm{H}_3 \mathrm{PO}_3$ on heating undergoes self-oxidation reduction as:

Question 63.
Why is red Phosphorus less reactive than white Phosphorus?
Answer:
White Phosphorus is more reactive than red Phosphorus under normal conditions because of angular strain in the $\mathrm{P}_4$ molecule where the angles are only $60^{\circ}$.
Question 64.
Nitrogen does not form any pentahalide like Phosphorus. Why?
Answer:
Nitrogen does not form pentahalide due to non-availability of the d-orbitals in its valence shell.
Question 65.
Give reason for the following. Among the noble gases only Xenon is well known to form chemical compounds?
Answer:
$\mathrm{Xe}$ is largest in size and has the highest polarising power.
Question 66.
Why is hydrogen sulphide, with greater molar mass a gas, while water a liquid at room temperature?
Answer:
$\mathrm{H}_2 \mathrm{O}$ molecules are associated with intermolecular $\mathrm{H}$-bonding, $\mathrm{H}_2 \mathrm{~S}$ is not because oxygen is more electronegative and smaller in size than sulphur. That is why $\mathrm{H}_2 \mathrm{O}$ is a liquid and $\mathrm{H}_2 \mathrm{~S}$ is a gas.
Question 67.
Noble gases are chemically inert. Give reasons.
Answer:
Noble gases are chemically inert because they have their octet complete except Helium, i.e. they have a stable electronic configurations.
Question 68.
Why do noble gases exist as monoatomic?
Answer:
Noble gases have stable electronic configuration, that is why they have no tendency to lose or gain electrons. Therefore they do not form covalent bond.
Question 69.
Nitrogen exists as diatomic molecule and Phosphrus as $\mathrm{P}^4$. Why?
Answer:
Nitrogen has a triple bond between its two atoms because of its small size and high electro negativity.
Phosphorus $\mathrm{P}^4$ has single bond, that is why it is tetra-atomic.

Question 70.
Why is $\mathrm{H}_2 \mathrm{~S}$ more acidic than water?
Answer:
It is because bond dissociation energy of $\mathrm{S}-\mathrm{H}$ bond is less than $\mathrm{O}-\mathrm{H}$ bond due to longer bond length.
3 Marks Questions and Answers
Question 1.

Complete the following reactions.
(i) $6 \mathrm{Li}+\mathrm{N}_2 \rightarrow$ ?
(ii) $3 \mathrm{Ca}+\mathrm{N}_2 \rightarrow$ ?
(iii) $2 \mathrm{~B}+\mathrm{N}_2 \rightarrow$ ?
Answer:
(i) $6 \mathrm{Li}+\mathrm{N}_2 \longrightarrow 2 \mathrm{Li}_3 \mathrm{~N}$ (Lithium nitride)
(ii) $3 \mathrm{Ca}+\mathrm{N}_2 \stackrel{\text { Red hot }}{\longrightarrow} \mathrm{Ca}_3 \mathrm{~N}_2$ (Calcium Nitride)
(iii) $2 \mathrm{~B}+\mathrm{N}_2 \stackrel{\text { Bright red hot }}{\longrightarrow} 2 \mathrm{BN}$ (Boron nitride)
Qustion 2.
How is ammonia prepared?
Answer:
1. Ammonia is formed by the hydrolysis of urea.
$
\mathrm{NH}_2 \mathrm{CONH}_2+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NH}_3+\mathrm{CO}_2
$
2. Ammonia is prepared in the laboratory by heating an ammonium salt with a base.
$
\begin{aligned}
& 2 \mathrm{NH}_4^{+}+\mathrm{OH}^{-} \rightarrow 2 \mathrm{NH}_3+\mathrm{H}_2 \mathrm{O} \\
& 2 \mathrm{NH}_4 \mathrm{Cl}+\mathrm{CaO} \rightarrow \mathrm{CaCl}_2+2 \mathrm{NH}_3+\mathrm{H}_2 \mathrm{O}
\end{aligned}
$
3. It can also be prepared by heating a metal nitrides such as magnesium nitride with water.
$
\mathrm{Mg}_3 \mathrm{~N}_2+6 \mathrm{H}_2 \mathrm{O} \rightarrow 3 \mathrm{Mg}(\mathrm{OH})_2+2 \mathrm{NH}_3
$
Question 3.
Explain the structure of ammonia.
Answer:
Ammonia molecule is pyramidal in shape $\mathrm{N}-\mathrm{H}$ bond distance is $1.016 \mathrm{~A}$ and $\mathrm{H}-\mathrm{H}$ bond distance is 1.645 A with a bond angle $107^{\circ}$. The structure of ammonia may be regarded as a tetrahedral with one lone pair of electrons in one tetrahedral position hence it has a pyramidal shape as shown in the figure.

Question 4.
How will you prepare nitric acid?
Answer:
Nitric acid is prepared by heating equal amounts of potassium or sodium nitrate with concentrated sulphuric acid.
$
\mathrm{KNO}_3+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{KHSO}_4+\mathrm{HNO}_3
$
The temperature is kept as low as possible to avoid decomposition of nitric acid. The acid condenses to a fuming liquids which is coloured brown by the presence of a little nitrogen dioxide which is formed due to the decomposition of nitric acid.
$
4 \mathrm{HNO}_3 \rightarrow 4 \mathrm{NO}_2+2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2
$
Question 5.
Discuss the Commercial method to prepare Nitric acid.
(Or)
How will you prepare nitric acid by Ostwald's process?
Answer:
Nitric acid prepared in large scales using Ostwald's process. In this method ammonia from Haber's process is mixed about 10 times of air. This mixture is preheated and passed into the catalyst chamber where they come in contact with platinum gauze. The temperature rises to about $1275 \mathrm{~K}$ and the metallic gauze brings about the rapid catalytic oxidation of ammonia resulting in the formation of $\mathrm{NO}$, which then oxidised to nitrogen dioxide.
$
\begin{aligned}
& 4 \mathrm{NH}_3+5 \mathrm{O}_2 \rightarrow 4 \mathrm{NO}+6 \mathrm{H}_3 \mathrm{O}+120 \mathrm{~kJ} \\
& 2 \mathrm{NO}+\mathrm{O}_2 \rightarrow 2 \mathrm{NO}_3
\end{aligned}
$
The nitrogen dioxide produced is passed through a series of adsorption towers. It reacts with water to give nitric acid. Nitric acid formed is bleached by blowing air.
$
6 \mathrm{NO}_2+3 \mathrm{H}_2 \mathrm{O} \rightarrow 4 \mathrm{HNO}_3+2 \mathrm{NO}+\mathrm{H}_2 \mathrm{O}
$
Question 6.
Draw the structure of the following compounds.
(a) Nitrous acid
(b) Nitric acid
(c) Pernitric acid
Answer:

Question 7.
Identify $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ from the following reactions.
1. $\mathrm{P}_4+\mathrm{Mg} \rightarrow \mathrm{A}$
2. $\mathrm{P}_4+\mathrm{Ca} \rightarrow \mathrm{B}$
3. $\mathrm{P}_4+\mathrm{Na} \rightarrow \mathrm{C}$
Answer:
1. $\mathrm{P}_4+6 \mathrm{Mg} \rightarrow 2 \mathrm{Mg}_3 \mathrm{P}_2$ (Magnesium phosphide)
2. $\mathrm{P}_4+6 \mathrm{Ca} \rightarrow 2 \mathrm{Ca}_3 \mathrm{P}_2$ (Calcium phosphide)
3. $\mathrm{P}_4+12 \mathrm{Na} \rightarrow 2 \mathrm{Na}_3 \mathrm{P}$ (Sodium phosphide)
Question 8.
How will you prepare phosphine and explain the purification of phosphine?
Answer:
Phosphine is prepared by action of sodium hydroxide with white phosphorous in an inert atmosphere of carbon dioxide or hydrogen.


Phosphine is freed from phosphine dihydride $\left(\mathrm{P}_2 \mathrm{H}_4\right)$ by passing through a freezing mixture. The dihydride condenses while phosphine does not. Phosphine can also prepared by the hydrolysis of metallic phosphides with water or dilute mineral acids.
$
\begin{aligned}
\mathrm{Ca}_3 \mathrm{P}_2+6 \mathrm{H}_2 \mathrm{O} \longrightarrow & 2 \mathrm{PH}_3 \uparrow+3 \mathrm{Ca}(\mathrm{OH})_2 \\
& \text { Phosphine } \\
\mathrm{AlP}+3 \mathrm{HCl} \longrightarrow & \mathrm{PH}_3 \uparrow+\mathrm{AlCl}_3 \\
& \text { Phosphine }
\end{aligned}
$
Phosphine is prepared in pure form by heating phosphorous acid.


Phosphorous acid Ortho phosphoric acid Phosphine
A pure sample of phosphine is prepared by heating phosphonium iodide with caustic soda solution.

Question 9.
What happens when $\mathrm{PH}_3$ reacts with oxygen or air?
Answer:
When phosphine air or oxygen it bums to give meta phosphoric acid.


Question 10.
Explain the structure of phosphine.
Answer:
In phosphine, phosphorus shows $\mathrm{sp}_3$ hybridisation. Three orbitals are occupied by bond pair and fourth comer is occupied by lone pair of electrons. Hence, bond angle is reduced to $94^{\circ}$. Phosphine has a pyramidal shape.

Question 11.
Discuss the uses of phosphine.
Answer:
Phosphine is used for producing smoke screen as it gives large smoke. In a ship, a pierced container with a mixture of calcium carbide and calcium phosphide, liberates phosphine and acetylene when thrown into sea. The liberated phosphine catches fire and ignites acetylene. These burning gases serves as a signal to the approaching ships. This is known as Holmes signal.
Question 12.
How does $\mathrm{PCl}_3$ reacts with the following reagents?
1. $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$
2. $\mathrm{C}_2 \mathrm{H}_5 \mathrm{COOH}$
Answer:
1. $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+\mathrm{PCl}_3 \rightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl}+\mathrm{H}_3 \mathrm{PO}_3$
2. $\mathrm{C}_2 \mathrm{H}_5 \mathrm{COOH}+\mathrm{PCl}_3 \rightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{COCl}+\mathrm{H}_3 \mathrm{PO}_3$
Question 13.
Explain the reaction between $\mathrm{PCl}_5$ and water.
Answer:
Phosphorous pentachloride reacts with water to give phosphoryl chloride and orthophosphoric acid.

Question 14.
Explain the structure of phosphorous trioxide $\left(\mathrm{P}_2 \mathrm{O}_3\right)$.
Answer:
In phosphorous trioxide four phosphorous atoms lie at the comers of a tetrahedron and six oxygen atoms along the edges. The $\mathrm{P}-\mathrm{O}$ bond distance is $165.6 \mathrm{pm}$ which is shorter than the single bond distance of $\mathrm{P}$ $\mathrm{O}(184 \mathrm{pm})$ due to $\mathrm{p} \pi-\mathrm{d} \pi$ bonding and results in considerable double bond character.


Question 15.
Discuss the structure of phosphorous pentaoxide $\left(\mathrm{P}_2 \mathrm{O}_3\right)$.
Answer:
In $\mathrm{P}_4 \mathrm{O}_{10}$ each $\mathrm{P}$ atoms form three bonds to oxygen atom and also an additional coordinate bond with an oxygen atom. Terminal coordinate $\mathrm{P}-\mathrm{O}$ bond length is $143 \mathrm{pm}$, which is less than the expected single bond distance. This may be due to lateral overlap of filled p orbitals of an oxygen atom with empty d-orbital on phosphorous.


Question 16.

Mention the uses of oxygen.
Answer:
1. Oxygen is one of the essential component for the survival of living organisms.
2. It is used in welding (oxyacetylene welding)
3. Liquid oxygen is used as fuel in rockets etc.
Question 17.
Give and explain reducing behaviour of sulphur dioxide.
Answer:
As Sulphur dioxide can readily be oxidised, it acts as a reducing agent. It reduces chlorine into hydrochloric acid.
$
\mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2 \rightarrow \mathrm{H}_2 \mathrm{SO}_4+2 \mathrm{HCl}
$
It also reduces potassium permanganate and dichromate to $\mathrm{Mn}^{2+}$ and $\mathrm{Cr}^{3+}$ respectively.
- $2 \mathrm{KMnO}_4+5 \mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4+2 \mathrm{H}_2 \mathrm{SO}_4$
- $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+3 \mathrm{SO}_2+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{K}_2 \mathrm{SO}_4+\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3+\mathrm{H}_2 \mathrm{O}$
Question 18 .
Why bleaching action of sulphur dioxide is temporary?
Answer:
In presence of water, sulphur dioxide bleaches coloured wool, silk, sponges and straw into colourless due to its reducing property.

However, the bleached product (colourless) is allowed to stand in air, it is reoxidised by atmospheric oxygen to its original colour. Hence bleaching action of sulphur dioxide is temporary.
Question 19.
Explain the structure of sulphur dioxide.
Answer:
In sulphur dioxide, sulphur atom undergoes $\mathrm{sp}^2$ hybridisation. A double bond arises between $\mathrm{S}$ and $\mathrm{O}$ is due to $\mathrm{p} \pi-\mathrm{d} \pi$ overlapping.


Question 20 .
How will you manufacture sulphuric acid by contact process?
Answer:
Manufacture of sulphuric acid by contact process involves the following steps:
1. Initially sulphur dioxide is produced by burning sulphur or iron pyrites in oxygen/air.
$
\begin{aligned}
& \mathrm{S}+\mathrm{O}_2 \rightarrow \mathrm{SO}_2 \\
& 4 \mathrm{FeS}_2+11 \mathrm{O}_2 \rightarrow 2 \mathrm{Fe}_2 \mathrm{O}_3+8 \mathrm{SO}_2
\end{aligned}
$
2. Sulphur dioxide formed is oxidised to sulphur trioxide by air in the presence of a catalyst such as $\mathrm{V}_2 \mathrm{O}_5$ or platinised asbestos.
3. The sulphur trioxide is absorbed in concentrated sulphuric acid and produces oleum $\left(\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7\right)$. The oleum is converted into sulphuric acid by diluting it with water.

$
\mathrm{SO}_3+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7 \stackrel{\mathrm{H}_2 \mathrm{O}}{\longrightarrow} 2 \mathrm{H}_2 \mathrm{SO}_4
$
To maximise the yield the plant is operated at 2 bar pressure and $720 \mathrm{~K}$. The sulphuric acid obtained in this process is over $96 \%$ pure.
Question 21.
Prove that $\mathrm{H}_2 \mathrm{SO}_4$ is a strong dibasic acid.
Answer:
Sulphuric acid forms two types of salts namely sulphates and bisulphates.

Question 22 .
Draw the structure of
(a) Marshall's acid
(b) polythionic acid
(c) Dithionic acid
Answer:

Question 23.
What happens when chlorine reacts with ammonia?
Answer:
1. If chlorine reacts with excess ammonia, it gives nitrogen and ammonium chloride. $8 \mathrm{NH}_3+3 \mathrm{Cl}_2 \rightarrow \mathrm{N}_2+6 \mathrm{NH}_4 \mathrm{Cl}$
2. If ammonia reacts with excess chlorine, it gives nitrogen trichloride and ammonium chloride. $4 \mathrm{NH}_2+3 \mathrm{Cl}_2 \rightarrow \mathrm{NCl}_2+3 \mathrm{NH}_4 \mathrm{Cl}$
Question 24.
Bleaching action of chlorine is permanent. Justify this statement.
Answer:
Chlorine is a strong oxidising and bleaching agent because of the nascent oxygen.


Colouring matter + Nascent oxygen $\rightarrow$ Colourless oxidation product
Therefore, bleaching of chlorine is permanent. It oxidises ferrous salts to ferric, sulphites to sulphates and
hydrogen sulphide to sulphur.

Question 25.
Give the uses of chlorine.
Answer:
1. Purification of drinking water
2. Bleaching of cotton textiles, paper and rayon
3. It is used in extraction of gold and platinum
Question 26.
What is Royal water?
Answer:
When three parts of concentrated hydrochloric acid and one part of concentrated nitric acid are mixed, aquaregia is obtained. This is also known as Royal water. This is used for dissolving gold, platinum etc.
$
\begin{aligned}
& \mathrm{Au}+4 \mathrm{H}^{+}+4 \mathrm{NO}_3^{-}+4 \mathrm{Cl}^{-} \rightarrow \mathrm{AuCl}_4^{-}+\mathrm{NO}+2 \mathrm{H}_2 \mathrm{O} \\
& 3 \mathrm{Pt}+16 \mathrm{H}^{+}+4 \mathrm{NO}_3^{-}+18 \mathrm{Cl}^{-} \rightarrow 3 \mathrm{PtCl}_3^{2-}+\mathrm{NO}+8 \mathrm{H}_2 \mathrm{O}
\end{aligned}
$
Question 27.
Why HF is not stored in glass bottles?
Answer:
HF rapidly reacts with silica and glass to form soluble salt and that process leads to break the glass bottles. Hence $\mathrm{HF}$ is not stored in glass bottles.
$
\begin{aligned}
& \mathrm{SiO}_2+4 \mathrm{HF} \rightarrow \mathrm{SiF}_4+2 \mathrm{H}_2 \mathrm{O} \\
& \mathrm{Na}_2 \mathrm{SiO}_3+6 \mathrm{HF} \rightarrow \mathrm{Na}_2 \mathrm{SiF}_6+3 \mathrm{H}_2 \mathrm{O}
\end{aligned}
$
Question 28 .
Give the properties of inter halogen compounds.
Answer:
Properties of inter halogen compounds:
1. The central atom will be the larger one.
2. It can be formed only between two halogen and not more than two halogens.
3. Fluorine can't act as a central metal atom being the smallest one.
4. Due to high electronegativity with small size fluorine helps the central atom to attain high coordination number
5. They can undergo the auto ionization.
6. They are strong oxidizing agents.
Question 29.
How will you prepare Xenon fluoride?
Answer:
Xenon fluorides are prepare by direct reaction of xenon and fluorine under different conditions as shown below.
$
\begin{gathered}
\mathrm{Xe}+\mathrm{F}_2 \frac{\mathrm{Ni}}{400^{\circ} \mathrm{C}} \mathrm{XeF}_2 \\
\mathrm{Xe}+2 \mathrm{~F}_2 \frac{\mathrm{Ni} / \text { acetone }}{400^{\circ} \mathrm{C}} \mathrm{XeF}_4 \\
\mathrm{Xe}+3 \mathrm{~F}_2 \frac{\mathrm{Ni} / 200 \mathrm{~atm}}{400^{\circ} \mathrm{C}} \mathrm{XeF} 6
\end{gathered}
$
Question 30.
Complete the following reaction.
(i) $\mathrm{XeOF}_4+\mathrm{SiO}_2 \rightarrow \mathrm{A}+\mathrm{SiF}_6$
(ii) $\mathrm{A}+\mathrm{SiO}_2 \rightarrow \mathrm{B}+\mathrm{SiF}_6$
Answer:
(i) 
(ii)  
Question 31.
What happens when $\mathrm{XeF}_6$ reacts $2.5 \mathrm{M}$ solution of $\mathrm{NaOH}$ ?
Answer:
When $\mathrm{XeF}_6$ reacts with $2.5 \mathrm{M}$ of $\mathrm{NaOH}$, sodium per xenate is obtained.

Question 32.
Give two examples to show the anomalous behaviour of fluorine.
Answer:
The anomalous behaviour of fluorine is due to its:
1. Small size
2. highest electronegativity
3. low F-F bond dissociation enethalpy
4. non-availability of d-orbitals in its valence shell.
The two examples are:
1. Due to non-availability of d-orbitals in its valence shell, fluorine cannot expand its octet, therefore, shows only -1 oxidation state while all other halogens due to the presence of d-orbitals shows positive oxidation states of $+1,+3,+5$ and +7 besides oxidation state of -1 .
2. Due to its small size, the three lone pair of electrons on each $F$ atom in $F-F$ molecule, repel the bond pair. As a result, $\mathrm{F}-\mathrm{F}$ bond dissociation energy is lower than that of $\mathrm{Cl}-\mathrm{Cl}$ bond.

Question 33.
Why does the reactivity of nitrogen differ from Phosphorus?
Answer:
Nitrogen exists as a diatomic molecule $(\mathrm{N} \equiv \mathrm{N})$. Due to the presence of a triple bond between the two $\mathrm{N}-$ atoms the bond dissociation energy is large $\left(941.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$. As a result nitrogen is said to be chemically
inert in its elemental state. In contrast, $\mathrm{P}-\mathrm{P}$ single bond is much weaker $\left(213 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$ than $\mathrm{N} \equiv \mathrm{N}$ triple bond Therefore, phosphorus is much more reactive than nitrogen.
Question 34
Why does $\mathrm{NH}_3$ form hydrogen bond but $\mathrm{PH}_3$ does not?
Answer:
The electronegativity of $\mathrm{N}(3.0)$ is much higher than that of $\mathrm{FI}$ (2.1). As a result, $\mathrm{N}-\mathrm{H}$ bond is quite polar and hence $\mathrm{NH}_3$ undergoes intermolecular $\mathrm{H}$-bonding. In contrast, both $\mathrm{P}$ and $\mathrm{H}$ have an electronegativity of 2.1. Therefore $\mathrm{P}-\mathrm{H}$ bond is non-polar and hence $\mathrm{PH}_3$ does not undergo $\mathrm{H}-$ bonding.
Question 35.
Can $\mathrm{PCl}_5$ act as an oxidising as well as a reducing agent? Justify.
Answer:
Oxidation state of $\mathrm{P}$ in $\mathrm{PCl}_5$ is +5 . Since $\mathrm{P}$ has five valence electrons in its valence shell, therefore it cannot increase its oxidation state beyond +5 by donating its electrons, therefore $\mathrm{PCl}_5$ cannot act as a reducing agent. However, it can decrease its oxidation number from +5 to +3 or some lower value, therefore $\mathrm{PCl}_5$ acts as an oxidising agent. For example, it oxidises $\mathrm{Ag}$ to $\mathrm{AgCl}$.
5 Mark Questions and Answers
Question 1.

How does ammonia react with
1. Excess $\mathrm{Cl}_2$
2. $\mathrm{Na}$
3. $\mathrm{CuSO}_4$
4. $\mathrm{O}_2 / \Delta$
Answer:
1. Reaction with Excess $\mathrm{Cl}_2$

2. Reaction with $\mathrm{Na}$ :

3. Reaction with $\mathrm{CuSO}_4$

4. Reaction with $\mathrm{O}_2$

Question 2.
Explain the reaction of metals with nitric acid.
Answer:
The reactions of metals with nitric acid are explained in 3 steps as follows:
Primary reaction:
Metal nitrate is formed with the release of nascent hydrogen
$\mathrm{M}+\mathrm{HNO}_3 \rightarrow \mathrm{MNO}_3+(\mathrm{H})$
Secondary reaction:
Nascent hydrogen produces the reduction products of nitric acid.

Tertiary reaction:
The secondary products either decompose or react to give final products

For examples:
Copper reacts with nitric acid in the following manner
$
\begin{aligned}
& 3 \mathrm{Cu}+6 \mathrm{HNO}_3 \rightarrow 3 \mathrm{Cu}\left(\mathrm{NO}_3\right)_2+6(\mathrm{H}) \\
& 6(\mathrm{H})+3 \mathrm{HNO}_3 \rightarrow 3 \mathrm{HNO}_2+3 \mathrm{H}_2 \mathrm{O} \\
& 3 \mathrm{HNO}_2 \rightarrow \mathrm{HNO}_3+2 \mathrm{NO}+\mathrm{H}_2 \mathrm{O}
\end{aligned}
$
Overall reaction
$
3 \mathrm{Cu}+8 \mathrm{HNO}_3 \rightarrow 3 \mathrm{CU}\left(\mathrm{NO}_3\right)_2+2 \mathrm{NO}+4 \mathrm{H}_2 \mathrm{O}
$
The concentrated acid has a tendency to form nitrogen dioxide
$
\mathrm{Cu}+4 \mathrm{HNO}_3 \rightarrow 3 \mathrm{CU}\left(\mathrm{NO}_3\right)_2+2 \mathrm{NO}_2+2 \mathrm{H}_2 \mathrm{O}
$
Question 3.
How will you prepare ozone by laboratory method? Explain the structure of ozone.
Answer:
In the laboratory ozone is prepared by passing electrical discharge through oxygen. At a potential of 20,000
$\mathrm{V}$ about $10 \%$ of oxygen is converted into ozone it gives a mixture known as ozonised oxygen. Pure ozone is obtained as a pale blue gas by the fractional distillation of liquefied ozonised oxygen.

The ozone molecule have a bent shape and symmetrical with delocalised bonding between the oxygen atoms.
Question 4.
$A$ is a king of acid. A reacts with $\mathrm{HBr}$ to give $\mathrm{B}$ and Bromine. A reacts with $\mathrm{Na}_2 \mathrm{CO}_3$ to give $\mathrm{C}$ and carbon dioxide. Identify A, B and C. Give the reaction.
Answer:
1. King of acid is sulphuric acid (A).
2. Sulphuric acid (A) reacts with $\mathrm{HBr}$ to give $\mathrm{SO}_2$ (B) and Bromine.


3. Sulphuric acid (A) reacts with $\mathrm{Na}_2 \mathrm{CO}_3$ to give sodium sulphate $(\mathrm{C})$ and $\mathrm{CO}_2$.


Question 5.
How does Sulphuric acid react with the following:
(a) $\mathrm{Al}$
(b) $\mathrm{KNO}_3$
(c) $\mathrm{NaBr}$
(d) $\mathrm{C}_6 \mathrm{H}_6$

Answer:

(a) 

(b) 

(c) 

(d) 

Question 6.
1. Explain the test for sulphate (or) sulphuric acid.
2. What happens when sulphuric acid reacts with oxalic acid?
Answer:
1. Dilute solution of sulphuric acid or aqueous solution of sulphates gives white precipitate with barium chloride solution. It can also be detected using lead acetate solution. Here a white precipitate of lead sulphate is obtained.


2. Sulphuric acid reacts with oxalic acid to give $\mathrm{CO}$ and $\mathrm{CO}_2$


Question 7.
Discuss the manufacture of chlorine.
Answer:
Electrolytic process:
When a solution of brine $(\mathrm{NaCl})$ is electrolysed, $\mathrm{Na}+$ and $\mathrm{CP}$ ions are formed. $\mathrm{Na}^{+}$ion reacts with $\mathrm{OH}^{-}$ions of water and forms sodium hydroxide. Hydrogen and chlorine are liberated as gases.
$
\begin{array}{lll}
\mathrm{NaCl} \longrightarrow \mathrm{Na}^{+}+\mathrm{Cl}^{-} & \text {At the cathode, } & \text { At the cathode, } \\
\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{H}^{+}+\mathrm{OH}^{-} & \mathrm{H}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{H} & \mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}+\mathrm{e}^{-} \\
\mathrm{Na}^{+}+\mathrm{OH}^{-} \longrightarrow \mathrm{NaOH} & \mathrm{H}+\mathrm{H} \longrightarrow \mathrm{H}_2 & \mathrm{Cl}+\mathrm{Cl} \longrightarrow \mathrm{Cl}_2
\end{array}
$
Deacons process:
In this process a mixture of air and hydrochloric acid is passed up a chamber containing a number of shelves, pumice stones soaked in cuprous chloride are placed. Hot gases at about $723 \mathrm{~K}$ are passed through
a jacket that surrounds the chamber.
$
4 \mathrm{HCl}+\mathrm{O}_2 \underset{\mathrm{Cu}_2 \mathrm{Cl}_2}{\stackrel{400^{\circ} \mathrm{C}}{\longrightarrow}} 2 \mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2 \uparrow
$

The chlorine obtained by this method is dilute and is employed for the manufacture of bleaching powder. The catalysed reaction is given below,

Question 8.
1. How is bleaching powder prepared?
2. What happens when benzene reacts with chlorine?
Answer:
1. Bleaching powder is prepared by passing chlorine gas through dry slaked lime (Calcium hydroxide).

2. Benzene reacts with chlorine in the presence of ferric chloride to give chlorobenzene.


Question 9.
Cone. $\mathrm{H}_2 \mathrm{SO}_4$ is added followed by heating to each of the following test tubes labelled (I) to (IV). Identify in which of the above test tube the following change will be observed. Support your answer with the help of chemical equation:
(a) formation of black substance
(b) evolution of brown gas
(c) evolution of colourless gas
(d) formation of a brown substance which on dilution becomes blue.
(e) disappearance of yellow powder along with evolution of colourless gas.

Answer:

(a) 

(b) 

(c) 

(d) 

(e) 

 

Question 10.
Give reasons for each of the following:
1. Bleaching of flowers by $\mathrm{Cl}_2$ is permanent while by $\mathrm{SO}_2$ is temporary.
2. Molten aluminium bromide is a poor conductor of electricity.
3. Nitric oxide becomes brown when released in air.
4. $\mathrm{PCl}_5$ is ionic in nature in the solid state.
5. Ammonia is a good complexing agent.
Answer:
1. $\mathrm{Cl}_2$ bleaches by oxidation, while $\mathrm{SO}_2$ does it by reduction. The reduced product gets oxidise again and the colour is regained back.
2. Aluminium bromide exists as a dimer, $\mathrm{Al}_2 \mathrm{Br}_6$. In this structure, each aluminium atom forms one coordinate bond by accepting a lone pair of electrons from the bromine atom of another aluminium bromide molecule and thus complete the octet of electrons. Due to lack of free electrons, molten aluminium bromide is a poor conductor of electricity.
3. Nitric oxide reacts with air and oxidised into $\mathrm{NO}_2$ which is brown in colour.
$
2 \mathrm{NO}+\mathrm{O}_2 \rightarrow 2 \mathrm{NO}_2
$
4. In solid state $\mathrm{PCl}_5$ exists as $\left[\mathrm{PCl}_4\right]^{+}\left[\mathrm{PCl}_6\right]^{-}$and hence it is ionic in nature. Due to its ionic nature, it conducts current on fusion.
5. $\mathrm{N}$ atom in ammonia has lone pair of electrons which can coordinate with other atoms or cations required for the stability of electron pair.
Question 11.
How will you prepare the following compounds.
(a) Hyponitrous acid
(b) Nitrous acid
(c) Pernitrous acid
(d) Pernitric acid
Answer:
(a) Hyponitrous acid: $\mathrm{Ag}_2 \mathrm{~N}_2 \mathrm{O}_2+2 \mathrm{HCl} \longrightarrow 2 \mathrm{AgCl}+\mathrm{H}_2 \mathrm{~N}_2 \mathrm{O}_2$
(b) Nitrous acid $: \mathrm{Ba}\left(\mathrm{NO}_2\right)_2+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow 2 \mathrm{HNO}_2+\mathrm{BaSO}_4$
(c) Pernitrous acid : $\mathrm{H}_2 \mathrm{O}_2+\mathrm{ON}(\mathrm{OH}) \longrightarrow \mathrm{ON}(\mathrm{OOH})+\mathrm{H}_2 \mathrm{O}$

(d) Pernitric acid : $\mathrm{H}_2 \mathrm{O}_2+\mathrm{N}_2 \mathrm{O}_5 \longrightarrow \mathrm{NO}_2 \mathrm{OOH}+\mathrm{HNO}_3$
Common Errors And Its Rectificaations:
Common Errors:
1. Oxidation number rules may be confused.
2. Oxidation number of oxygen may get confused.
3. Molecular formula and compound name may get confused.
Rectifications:
1. Oxidation number of fluorine is always -1 .
2. Always oxygen is -2 but in peroxide it is -1 . When it comes as a first element its oxidation number is positive.
3. Student should memorise the formula using short cut method, e.g. Phosphorous acid $-\mathrm{H}_3 \mathrm{PO}_3$ Phosphoric acid $-\mathrm{H}_3 \mathrm{PO}_4$ e.g. Phosphorous acid $-\mathrm{H}_3 \mathrm{PO}_3$ By pronunciation, ' $\mathrm{O}$ ' indicates oxygen is more.

-->
$ C while hydrogen fluoride and hydrogen chloride are stable at this temperature.
Reason (R) - Thermal stability of hydrogen halides decreases from fluoride to iodide.
(a) $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ explains $\mathrm{A}$
(b) A and R are correct but doesn't explains A
(c) $\mathrm{A}$ is correct but $\mathrm{R}$ is wrong
(d) $A$ is wrong but $R$ is correct
Answer:
(a) $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ explains $\mathrm{A}$
Question 4.
Assertion (A) - The bleaching of chlorine is temporary.
Reason (R) - Chlorine oxidises ferrous salts to ferric salts.
(a) $A$ and $R$ are correct and $R$ explains $A$
(b) $A$ and $R$ are correct but doesn't explains $A$
(c) $A$ is correct but $R$ is wrong
(d) $\mathrm{A}$ is wrong but $\mathrm{R}$ is correct
Answer:
(d) $\mathrm{A}$ is wrong but $\mathrm{R}$ is correct
Question 5.
Assertion (A) - Sulphuric acid is highly reactive.
Reason (R) - Sulphuric acid can act as strong acid and an oxidising agent.
(a) $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ explains $\mathrm{A}$
(b) $A$ and $R$ are correct but doesn't explains $A$
(c) $A$ is correct but $R$ is wrong
(d) $\mathrm{A}$ is wrong but $\mathrm{R}$ is correct
Answer:
(a) $\mathrm{A}$ and $\mathrm{R}$ are correct but doesn't explains $\mathrm{A}$
Question 6.
Assertion (A) - Sulphuric acid is a high boiling point and viscous liquid.
Reason $(\mathrm{R})$ - This is due to the association of molecules together through hydrogen bonding.
(a) $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ explains $\mathrm{A}$
(b) $A$ and $R$ are correct but doesn't explains $A$
(c) $\mathrm{A}$ is correct but $\mathrm{R}$ is wrong
(d) $\mathrm{A}$ is wrong but $R$ is correct
Answer:
(a) $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ explains $\mathrm{A}$

Question 7.
Assertion (A) - Monoclinic sulphur is less stable than rhomobic sulphur.
Reason (R) - Monoclinic sulphur is stable between $96^{\circ} \mathrm{C}-119^{\circ} \mathrm{C}$ and slowly changes into rhombic sulphur.
(a) $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ explains $\mathrm{A}$
(b) $\mathrm{A}$ and $\mathrm{R}$ are correct but doesn't explains $\mathrm{A}$
(c) $\mathrm{A}$ is correct but $\mathrm{R}$ is wrong
(d) $A$ is wrong but $R$ is correct
Answer:
(a) $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ explains $\mathrm{A}$
Question 8.
Assertion (A) - Nitrogen gas is chemically inert.
Reason (R) - Nitrogen has low bonding energy.
(a) $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ explains $\mathrm{A}$
(b) A and $\mathrm{R}$ are correct but doesn't explains $\mathrm{A}$
(c) $\mathrm{A}$ is correct but $\mathrm{R}$ is wrong
(d) $A$ is wrong but $R$ is correct
Answer:
(c) $\mathrm{A}$ is correct but $\mathrm{R}$ is wrong
V. Find the odd one out
Question 1.

(a) $\mathrm{NO}$
(b) $\mathrm{HNO}_3$
(C) $\mathrm{NO}_2$
(d) $\mathrm{N}_2 \mathrm{O}$
Answer:
(b) $\mathrm{HNO}_3$
Hint: $\mathrm{HNO}_3$ is acid and other are oxides.
Question 2.
(a) Nitrous acid
(b) Nitric acid
(c) Hyponitrous acid
(d) Pemitrous acid
Answer:
(d) Pemitrous acid
Hint: Pemitrous acid contains peroxide linkage others doesn't have peroxide linkage.
Question 3.
(a) White phosphorous
(b) Red phosphorous
(c) phosphorous pentaoxide
(d) black phosphorous
Answer:

(c) phosphorous pentaoxide
Hint: $\mathrm{P}_2 \mathrm{O}_5$ is a compound of phosphorous and others are allotropic form of phosphorous.
Question 4.
(a) $\mathrm{PH}_3$
(b) $\mathrm{HPO}_3$
(c) $\mathrm{H}_3 \mathrm{PO}_3$
(d) $\mathrm{H}_3 \mathrm{PO}_4$
Answer:
(a) $\mathrm{PH}_3$
Hint: $\mathrm{PH}_3$ is hydrides of phosphorous and others are oxo acids of phosphorous.
Question 5.
(a) $\mathrm{He}$
(b) $\mathrm{Ne}$
(c) $\mathrm{Ar}$
(d) $\mathrm{Xe}$
Answer:
(d) $\mathrm{Xe}$
Hint: Xe forms several chemical compounds than others.
VI. Find out the correct pair.
Question 1.

(a) Helium - filament bulbs
(b) Krypton - prevent bonds
(c) Xenon - Lasers
(d) Radon - flash bulbs
Answer:
(c) Xenon-Lasers
Question 2.
(a) $\mathrm{Ra}$ - gamma rays
(b) $\mathrm{Xe}$ - cancer growth
(c) $\mathrm{Ne}$ - balloons
(d) $\mathrm{Kr}$ - advertisement bulb
Answer:
(a) $\mathrm{Ra}$ - gamma rays

Question 3.
(a) $\mathrm{ClF}_3$ - Linear
(b) $\mathrm{BrF}_5$ - $\mathrm{T}$ Shaped
(c) $\mathrm{IF}_4$ - square pyrimidal
(d) $\mathrm{BrF}_5$ - square pyrimidal
Answer:
(d) $\mathrm{BrF}_5-$ square pyrimidal
Question 4.
(a) $\mathrm{XeOF}_2-\mathrm{sp}^3$
(b) $\mathrm{XeF}_6-\mathrm{sp}^3 \mathrm{~d}^3$
(c) $\mathrm{XeF}_4-\mathrm{sp}^3$
(d) $\mathrm{XeOF}_4-\mathrm{sp}^3 \mathrm{~d}$
Answer:
(b) $\mathrm{XeF}_6-\mathrm{sp}^3 \mathrm{~d}^3$
Question 5.
(a) $\mathrm{OF}_2=-1$
(b) $\mathrm{Cl}_4 \mathrm{O}_4=-1$
(c) $\mathrm{I}_2 \mathrm{O}_4=-1$
(d) $\mathrm{I}_2 \mathrm{O}_9=-1$
Answer:
(a) $\mathrm{OF}_2=-1$
Question 6.
(a) Royal water - Dissolving gold
(b) Chlorine - Entraction of glue from bone
(c) $\mathrm{HCl}-$ Extraction of gold
(d) Chlorine - Temporary bleaching
Answer:
(a) Royal water - Dissolving gold

Question 7.
(a) $\mathrm{O}-2 \mathrm{~s}^2 2 \mathrm{p}^3$
(b) $S-3 s^2 3 p^4$
(c) $\mathrm{Se}-4 \mathrm{~d}^{10} 5 \mathrm{~s}^2 5 \mathrm{p}^4$
(d) $\mathrm{Te}-3 \mathrm{~d}^{10} 4 \mathrm{~s}^2 4 \mathrm{p}^4$
Answer:
(b) $S-3 s^2 3 p^4$
VI.Find out the incorrect pair.
Question 1.

(a) Nitrogen gas - inert
(b) Ammonia - pungent smelling gas
(c) Nitric acid-oxidizing agent
(d) Phosphine - rotten egg smell
Answer:
(d) Phosphine - rotten egg smell
Question 2.
(a) Liquid nitrogen - biological preservation
(b) Nitric acid - photography
(c) white phosphorous - yellow phosphorous
(d) phosphorous - welding
Answer:
(d) phosphorous - welding
Question 3.
(a) $\mathrm{N}_2 \mathrm{O}=+1$
(b) $\mathrm{N}_2 \mathrm{O}=+2$
(c) $\mathrm{N}_2 \mathrm{O}_3=+5$
(d) $\mathrm{NO}_2=+4$
Answer:
(c) $\mathrm{N}_2 \mathrm{O}_3=+5$

Question 4.
(a) Hvponitrous acid $-\mathrm{N}_2 \mathrm{O}$
(b) Nitrous acid $-\mathrm{HNO}_2$
(c) pernitric acid $-\mathrm{HNO}_4$
(d) pernitrous acid - HOONO
Answer:
(a) Hvponitrous acid $-\mathrm{N}_2 \mathrm{O}$
Question 5.
(a) $\mathrm{PH}_3$ - Holme's signal
(b) $\mathrm{O}_2$ - welding
(c) $\mathrm{H}_2 \mathrm{SO}_4-$ Disinfecting crops
(d) $\mathrm{SO}_3$ - Bleaching hair
Answer:
(c) $\mathrm{H}_2 \mathrm{SO}_4-$ Disinfecting crops
Question 6.
(a) $\mathrm{H}_2 \mathrm{SO}_4$ drying agent
(b) Chlorine - Deacon's process
(c) $\mathrm{HCI}$ - purification of bone black
(d) Helium - flash bulbs
Answer:
(d) Helium - flash bulbs
Question 7.
(a) $\mathrm{ICl}$ - Linear
(b) $\mathrm{CIF}_3-\mathrm{T}$ shape
(c) $\mathrm{IF}_5$ - pentagonal bipyramidal
(d) $\mathrm{IF}_7-$ pentagonal bipyramidal
Answer:
(c) $\mathrm{IF}_5$ - pentagonal bipyramidal
Question 8.
(a) $\mathrm{HOCl}=+2$
(b) $\mathrm{HOCl}=+3$
(e) $\mathrm{HOCI}=+5$
(d) $\mathrm{HOCl}_4=+7$
Answer:
(a) $\mathrm{HOCl}=+2$

Question 9.
(a) $\mathrm{XeF}_4-\mathrm{sp}^3 \mathrm{~d}^3$
(b) $\mathrm{XeF}_6-\mathrm{sp}^3 \mathrm{~d}^3$
(e) $\mathrm{XeOF}_4-\mathrm{sp}^3 \mathrm{~d}^2$
(d) $\mathrm{XeO}_3-\mathrm{sp}^3 \mathrm{~d}$
Answer:
(d) $\mathrm{XeO}_3-\mathrm{sp}^3 \mathrm{~d}$
Question 10 .
(a) $\mathrm{He}$ - cryogenics
(b) $\mathrm{Ne}$ - advertisement
(c) $\mathrm{Kr}$ - flurescent bulbs
(d) $\mathrm{Ra}$ - Lasers.
Answer:
(d) Ra - Lasers.
2 Mark Questions and Answers
Question 1.

What are pnictogens?
Answer:
The group - 15 elements like nitrogen, phosphorous Arsenic, Antimony and Bismuth are collectively called as pnictogens. Their general outer electronic configuration is $\mathrm{ns}^2 \mathrm{np}^3$.
Question 2.
What happen when sodium azide undergoes thermal decomposition?
Answer:
Pure nitrogen gas can be obtained by the thermal decomposition of sodium azide about $575 \mathrm{~K}$ $2 \mathrm{NaN}_3 \stackrel{575 K}{\longrightarrow} 2 \mathrm{Na}+3 \mathrm{~N}_2$
Question 3.
How will you prepare ammonia from nitrogen? and mention the name of process?
Answer:
Nitrogen directly reacts with hydrogen gives ammonia. This reaction is favoured by high pressures and at optimum temperature in the presence of iron catalyst,
$\mathrm{N}_2+3 \mathrm{H}_3 \rightleftharpoons 2 \mathrm{NH}_3$
This process is called as Haber's process.

Question 4.
Why nitrogen gas is chemically inert?
Answer:
The chemically inert character of nitrogen is largely due to high bonding energy of the molecules $225 \mathrm{cal}$ $\mathrm{mol}^{-1}$. Interestingly the triply bonded species is notable for its less reactivity in comparison with other isoelectronic triply bonded systems such as $-\mathrm{C} \equiv \mathrm{C}-\mathrm{C} \equiv \mathrm{Q}, \mathrm{X}-\mathrm{C} \equiv \mathrm{N}$, etc.
Question 5.
Mention the uses of nitrogen?
Answer:
1. Nitrogen is used for the manufacture of ammonia, nitric acid and calcium cyanamide etc.
2. Liquid nitrogen is used for producing low temperature required in cryosurgery, and so in biological preservation
Question 6.
Explain the action of heat on ammonia.
Answer:
Above $500^{\circ} \mathrm{C}$ ammonia decomposes into its elements. The decomposition may be accelerated by metallic catalysts like Nickel, Iron. Almost complete dissociation occurs on continuous sparking.


Question 7.
Prove ammonia act as a reducing agent?
Answer:
Ammonia reduces the metal oxides to metal when passed over heated metallic oxide.
$3 \mathrm{PbO}+2 \mathrm{NH}_3 \rightarrow 3 \mathrm{~Pb}+\mathrm{N}_2+3 \mathrm{H}_2 \mathrm{O}$
Question 8.
What happen when copper sulphate reacts with ammonia?
Answer:
When excess ammonia is added to aqueous solution copper sulphate a deep blue colour Complex $\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+}$ is formed.

Question 9.
Pure nitric acid is colourless, on standing it becomes yellow. Justify your answer.
Answer:
Nitric acid decomposes on exposure to sunlight or on being heated, into nitrogen dioxide, water and oxygen.
$
4 \mathrm{HNO}_4 \rightarrow 4 \mathrm{NO}_2+2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2
$
Due to this reaction pure acid or its concentrated solution becomes yellow on standing.
Question 10.
Write the products formed in the reaction of nitric acid with dilute and concentrated with magnesium.

Answer:
1. Magnesium with cone. $\mathrm{HNO}_3$ :
$
4 \mathrm{Mg}+10 \mathrm{HNO}_3 \rightarrow 4 \mathrm{Mg}\left(\mathrm{NO}_3\right)_2+\mathrm{NH}_4 \mathrm{NO}_3+3 \mathrm{H}_2 \mathrm{O}
$
Magnesium reacts with concentraated nitric acid to gives both magnesium nitrate and ammonium nitrate.
2. Magnesium with dil $\mathrm{HNO}_3$ :
Magnesium reacts with dilute nitric acid to gives both magnesium nitrate and nitrous oxide.
$
4 \mathrm{Mg}+10 \mathrm{HNO}_3 \rightarrow 4 \mathrm{Mg}\left(\mathrm{NO}_3\right)_2+\mathrm{N}_2 \mathrm{O}+5 \mathrm{H}_2 \mathrm{O}
$
Question 11.
Give the uses of nitric acid.
Answer:
1. Nitric acid is used as a oxidising agent and in the preparation of aquaregia.
2. 4. Salts of nitric acid are used in photography $\left(\mathrm{AgNO}_3\right)$ and gunpowder for firearms. $\left(\mathrm{NaNO}_3\right)$.
Question 12.
How will you prepare nitric oxide from sodium nitrite?
Answer:
When sodium nitrite reacts with ferrous sulphate in the presence of sulphuric acid to gives nitric oxide.
$
2 \mathrm{NaNO}_2+2 \mathrm{FeSO}_4+3 \mathrm{H}_2 \mathrm{~S} \rightarrow \mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3+2 \mathrm{NaHSO}_4+2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{NO}
$
Question 13.
How will you prepare nitrogen pentoxide?
Answer:
Nitric acid reacts with phosphorous pentaoxide to give nitrogen pentoxide.
$
2 \mathrm{HNO}_3+\mathrm{P}_2 \mathrm{O}_5 \rightarrow \mathrm{N}_2 \mathrm{O}_5+2 \mathrm{HPO}_3
$

Question 14.
Draw the structure of
1. $\mathrm{N}_2 \mathrm{O}$
2. $\mathrm{N}_2 \mathrm{O}_3$
Answer:
1. $\mathrm{N}_2 \mathrm{O}$ (Nitrous oxide)

2. $\mathrm{N}_2 \mathrm{O}_3$ (Nitrogen sesquoxide)


Question 15.
Draw- the structure of
1. $\mathrm{N}_2 \mathrm{O}_4$
2. $\mathrm{N}_2 \mathrm{O}_3$
Answer:
1. $\mathrm{N}_2 \mathrm{O}_4$ (Nitrogen tetraoxide)

2. $\mathrm{N}_2 \mathrm{O}_5$


Question 16.
Draw the structure of

1. Hyponitrous acid
2. Hydronitrous acid.
Answer:
(a) Hypon trous acid $-\mathrm{H}_2 \mathrm{~N}_2 \mathrm{O}_2$
$
\mathrm{HO}-\mathrm{N}=\mathrm{H}-\mathrm{OH}
$
(b) Hydronitrous acid:

Question 17.
Mention the allotropic forms of phosphorous?
Answer:
The most common allotropic forms of phosphorous
1. white phosphorous
2. Red phosphorous
3. Black phosphorous
Question 18.
Why white phosphorous is also known as yellow phosphorous?
Answer:
The freshly prepared white phosphorus is colourless but becomes pale yellow due to formation of a layer of red phosphorus upon standing. Hence it is also known as yellow phosphorus.
Question 19.
What is phosphorescence?
Answer:
White (yellow) phosphorous glows in the dark due to oxidation which is called phosphorescence.
Question 20.
Why white phosphorous undergoes spontaneous combustion in air?
Answer:
White phosphorous ignition temperature is very low and hence it undergoes spontaneous combustion in air at room temperature and during in the combusition process, white phosphorous produces $\mathrm{P}_2 \mathrm{O}_5$.
Question 21.
Draw the structure of
1. white phosphorous
2. red phosphorous

Answer:
1. White phosphorous

2. Red phosphorous


Question 22.
How will you convert red phosphorous into $\mathrm{P}_2 \mathrm{O}_3$ and $\mathrm{P}_2 \mathrm{Q}_5$ ?
Answer:
Red phosphorus reacts with oxygen on heating to give phosphorus trioxide or phosphorus pentoxide. $\mathrm{P}_4+3 \mathrm{O}_2 \triangle \mathrm{P}_4 \mathrm{O}_6$ (phosphorous trioxide)
$\mathrm{P}_4+5 \mathrm{O}_2 \stackrel{\triangle}{\rightarrow} \mathrm{P}_4 \mathrm{Q}_{10}$ (phosphorous pentaoxide)
Question 23.
How will you prepare orthophosphoric acid from phosphorous?
Answer:
When phosphorous is treated with cone nitric acid it is oxidised to orthophosphoric acid. This reaction is catalysed by iodine crystals.

Question 24.
Mention the uses of phosphorous?
Answer:
1. The red phosphorus is used in the match boxes.
2. It is also used for the production of certain alloys such as phosphor bronze.

Question 25 .
Show that phosphine is weakly basic?
Answer:
Phosphine is weakly basic and forms phosphonium salts with halogen acids.
$\mathrm{PH}_3+\mathrm{HI} \rightarrow \mathrm{PH}_4 \mathrm{I}$
$
\mathrm{PH}_4 \mathrm{I}+\mathrm{H}_2 \mathrm{O} \underset{\rightarrow}{\rightarrow} \mathrm{PH}_3+\mathrm{H}_3 \mathrm{O}^{-}+\mathrm{I}^{-}
$
Question 26.
Draw the structure of $\mathrm{PCl}_3$
Answer:

Question 27.
How will you prepare $\mathrm{PCl}_3$ from white phosphorous?
1. When a slow stream of chlorine is passed over white phosphorous, $\mathrm{PCl}_3$ is formed
2. It can also be obtained by treating white phosphorous with thionyl chloride.
Question 28.
What happen when $\mathrm{PCl}_3$ is treated with cold water?
Answer:
When $\mathrm{PCl}_3$ is hydrolysed with cold water it gives phosphorous acid.


Question 29.
How will you prepare $\mathrm{PCl}_3$ ?
Answer:
When $\mathrm{PCl}_3$ is treated with excess chlorine, phosphorous pentachloride is obtained.
$
\mathrm{PCl}_3+\mathrm{Cl}_2 \rightarrow \mathrm{PCl}_5
$
Question 30.
Mention the uses of $\mathrm{PCl}_3$ and $\mathrm{PCl}_5$ ?
answer:
1. Phosphorus trichloride is used as a chlorinating agent and for the preparation of $\mathrm{H}_2 \mathrm{PO}_3$.
2. Phosphorous pentachloride is a chlorinating agent and is useful for replacing hydroxyl groups by chlorine atom.
Question 31.
Draw the structure of $\mathrm{H}_3 \mathrm{PO}_2$ and $\mathrm{H}_3 \mathrm{PO}_3$ ?
Answer:

Question 32.
Draw the structure of $\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_6$ and $\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_3$ ?
Answwer:
(i) $\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_6$
(ii) $\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_3$
(i) $\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_6$
(ii) $\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_7$

Question 33.
Give the method to prepare hypophosphorous acid and pyrophosphoric acid?
Answer:
1. Hypophosphorous acid - Phosphorous reacts with water to give hypophosphorous acid.


2. Pyrophosphoric acid - Phosphorous acid is heated they produce pyrophosphoric acid.


Question 34.
Complete the reactions
1. $\mathrm{HgO} \triangle$ ?
2. $\mathrm{BaO}_2 \stackrel{\triangle}{\rightarrow}$ ?
Answer:

1. 
2. $2 \mathrm{BaO}_2 \stackrel{\triangle}{\rightarrow} 2 \mathrm{BaO}+\mathrm{O}_2$

Question 35 .
Give and explain the reaction used to estimation of ozone.
Answer:
Ozone oxidises potassium iodide to iodine.
$
\mathrm{O}_3+2 \mathrm{KI}+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{KOH}+\mathrm{O}_2+\mathrm{I}_2
$
This reaction is quantitative and can be used for estimation of ozone.
Question 36 .
Write a short notes on Rhombic sulphur?
Answer:
Rhombic sulphur also known as a sulphur, is the only thermodynamically stable allotropic form at ordinary temperature and pressure. The crystals have a characteristic yellow colour and composed of $\mathrm{S}_8$ molecules. When heated slowly above $96^{\circ} \mathrm{C}$, it converts into monoclinic sulphur. Upon cooling below $96^{\circ} \mathrm{C}$ the $\beta$ form converts back to a form.
Question 37.
Write a notes on monoclinic sulphur?
Answer:
Monoclinic sulphur also contains $\mathrm{S}_8$ molecules in addition to small amount of $\mathrm{S}_6$ molecules. It exists as a long needle like prism and is also called as prismatic sulphur. It is stable between $96^{\circ}-119^{\circ} \mathrm{C}$ and slowly changes into rhombic sulphur.
Question 38.
What are $\lambda$ - sulphur?
Answer:
Sulphur also exists in liquid and gaseous states. At around $140^{\circ} \mathrm{C}$ the monoclinic sulphur melts to form mobile pale yellow liquid called $\lambda$ sulphur.

Question 39.
How will you prepare sulphurdioxide by laboratory method?
Answer:
Sulphur dioxide is prepared in the laboratory treating a metal or metal sulphite with sulphuric acid.
$\mathrm{Cu}+2 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CuSO}_4+\mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O}$
$
\mathrm{SO}_3^{-}+2 \mathrm{H}^{+} \rightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{SO}_2
$
Question 40.
Complete the reactions,
Answer:
1. $\mathrm{Zns}+\mathrm{O}_2 \stackrel{\Delta}{\rightarrow}$ ?

2. $\mathrm{FeS}_2+\mathrm{O}_2 \stackrel{\Delta}{\rightarrow}$ ?
Answer:

1. 
$
\begin{aligned}
& \text { 2. } 4 \mathrm{FeS}_2+11 \mathrm{O}_2 \underset{\rightarrow}{\rightarrow} 2 \mathrm{Fe}_2 \mathrm{O}_3+8 \mathrm{SO}_2 \\
&
\end{aligned}
$
Question 41.
What happen when sulphurdioxdie reacts with sodium hydroxide and sodium carbonate?
Answer:
Sulphur dioxide reacts with sodium hydroxide and sodium carbonate to form sodium bisulphite and sodium sulphite respectively.

Question 42.
Mention the uses of sulphurdioxide?
Answer:
1. Sulphur dioxide is used in bleaching hair, silk, wool etc...
2. It can be used for disinfecting crops and plants in agriculture.
Question 43.
Why sulphuric acid is high boiling and viscous liquid ?
Answer:
Sulphuric acid is high boiling and viscous liquid this is due to the association of molecules together through hydrogen bonding.
Question 44.
Explain the reaction between benzene and sulphuric acid.
Answer:
Sulphuric acid reacts with benzene to give benzene sulphuric acid.
$\mathrm{C}_6 \mathrm{H}_6+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{SO}_3 \mathrm{H}+\mathrm{H}_2 \mathrm{O}$
(benzene) (Benzene sulphonic acia)
Question 45 .
Give the uses of sulphuric acid?
Answer:
1. Sulphuric acid is used in the manufacture of fertilisers, ammonium sulphate and super phosphates and other chemicals such as hydrochloric acid, nitric acid etc...
2. It is used as a drying agent and also used in the preparation of pigments, explosives etc..
Question 46.
Draw the structure of $\mathrm{H}_2 \mathrm{SO}_3$ and $\mathrm{H}_2 \mathrm{SO}_4$ ?
Answer:
1. $\mathrm{H}_2 \mathrm{SO}_3$

2. $\mathrm{H}_2 \mathrm{SO}_4$


Question 48.
Draw the structure of
1. Pyrosulphuric acid
2. Peroromonosulphuric acid?
Answer:
1. Pyrosulphuric acid $\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7$
2. Peroromonosulphuric acid $\mathrm{H}_2 \mathrm{SO}_5$

Question 49.
What happen when chlorine reacts with trupentine?
Answer:
When chlorine burnt with turpentine it forms carbon and hydrochloric acid.


Question 50.
How will you prepare hydrochloric acid by laboratory method?
Answer:
It is prepared by the action of sodium chloride and concentrated sulphuric acid

- $\mathrm{NaCl}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{NaHSO}_4+\mathrm{HCl}$
- $\mathrm{NaHSO}_4+\mathrm{NaCl} \rightarrow \mathrm{Na}_2 \mathrm{SO}_4+\mathrm{HCl}$
Dry hydrochloric acid is obtained by passing the gas through conc. sulphuric acid
Question 51.
Mention the uses of hydrochloric acid?
Answer:
1. Hydrochloric acid is used for the manufacture of chlorine, ammonium chloride, glucose from com starch etc.,
2. It is used in the extraction of glue from bone and also for purification of bone black.
Question 52.
Complet the following reaction?
(i) $\mathrm{BrF}_5 \stackrel{\mathrm{OH}^{\ominus}}{\longrightarrow}$ ?
(ii) $\mathrm{ICl} \stackrel{\mathrm{OH}^{\Theta}}{\longrightarrow}$ ?

Answer:

(i) 

(ii) 

Question 53.
Why noble gases have the largest ionisation energy?
Answer:
Noble gases have the largest ionisation energy compared to any other elements in a given row as they h completely filled orbital $\left(\mathrm{ns}^2 n \mathrm{p}^6\right)$ in their outer most shell. They are extremely stable and have a small tendency to gain or lose electrons. Hence noble gases have the largest ionisation energy.
Question 54
Mention the uses of Neon?
Answer:
Neon is used in advertisement as neon sign and the brilliant red glow is caused by passing electric curre through neon gas under low pressure.
Question 55.
Give the uses of Krypton?
Answer:
1. Krypton is used in fluorescent bulbs, flash bulbs etc...
2. Lamps filed with krypton are used in airports as approaching lights as they can penetrate through dense fog.
Question 56.
Mention the application of Xenon?
Answer:
1. Xenon is used in fluorescent bulbs, flash bulbs and lasers.
2. Xenon emits an intense light in discharge tubes instantly. Due to this it is used in high speed electronic flash bulbs used by photographers.
Question 57.
Give the uses of Radon?
Answer:
1. Radon is radioactive and used as a source of gamma rays
2. Radon gas is sealed as small capsules and implanted in the body to destroy malignant i.e. cancer growth.

Question 58.
Why are pentahalides more covalent than trihalides?
Answer:
Since elements in the +5 oxidation state have less tendency to lose electrons than in the +3 oxidation state, therefore elements in the +5 oxidation state are more covalent than in the +3 oxidation state. In other words, pentahalides are more covalent than trihalides.
59. Why is $\mathrm{N}_2$ less reactive at room temperature?
Answer:
Due to the presence of triple bond between the two nitrogen atoms, the bond dissociation energy of $\mathrm{N}_2$ $\left(941.4 \mathrm{~kJ} \mathrm{~mol}^{-}\right)$is very high. Therefore, $\mathrm{N}_2$ is less reactive at room temperature.
Question 60.
Why is $\mathrm{ICl}$ more reactive than $\mathrm{I}_2$ ?
Answer:
$\mathrm{ICl}$ bond is weaker than $\mathrm{I}$ - I bond. Therefore $\mathrm{ICl}$ can break easily to form halogen atoms which readily brings about the reaction.
Question 61.
Why is helium used in diving apparatus?
Answer:
It is used as a diluent for oxygen in modern diving apparatus because of its very low solubility in blood.
Question 62 .
Give the disproportionation reaction of $\mathrm{H}_3 \mathrm{PO}_3$ ?
Answer:
$\mathrm{H}_3 \mathrm{PO}_3$ on heating undergoes self-oxidation reduction as:

Question 63.
Why is red Phosphorus less reactive than white Phosphorus?
Answer:
White Phosphorus is more reactive than red Phosphorus under normal conditions because of angular strain in the $\mathrm{P}_4$ molecule where the angles are only $60^{\circ}$.
Question 64.
Nitrogen does not form any pentahalide like Phosphorus. Why?
Answer:
Nitrogen does not form pentahalide due to non-availability of the d-orbitals in its valence shell.
Question 65.
Give reason for the following. Among the noble gases only Xenon is well known to form chemical compounds?
Answer:
$\mathrm{Xe}$ is largest in size and has the highest polarising power.
Question 66.
Why is hydrogen sulphide, with greater molar mass a gas, while water a liquid at room temperature?
Answer:
$\mathrm{H}_2 \mathrm{O}$ molecules are associated with intermolecular $\mathrm{H}$-bonding, $\mathrm{H}_2 \mathrm{~S}$ is not because oxygen is more electronegative and smaller in size than sulphur. That is why $\mathrm{H}_2 \mathrm{O}$ is a liquid and $\mathrm{H}_2 \mathrm{~S}$ is a gas.
Question 67.
Noble gases are chemically inert. Give reasons.
Answer:
Noble gases are chemically inert because they have their octet complete except Helium, i.e. they have a stable electronic configurations.
Question 68.
Why do noble gases exist as monoatomic?
Answer:
Noble gases have stable electronic configuration, that is why they have no tendency to lose or gain electrons. Therefore they do not form covalent bond.
Question 69.
Nitrogen exists as diatomic molecule and Phosphrus as $\mathrm{P}^4$. Why?
Answer:
Nitrogen has a triple bond between its two atoms because of its small size and high electro negativity.
Phosphorus $\mathrm{P}^4$ has single bond, that is why it is tetra-atomic.

Question 70.
Why is $\mathrm{H}_2 \mathrm{~S}$ more acidic than water?
Answer:
It is because bond dissociation energy of $\mathrm{S}-\mathrm{H}$ bond is less than $\mathrm{O}-\mathrm{H}$ bond due to longer bond length.
3 Marks Questions and Answers
Question 1.

Complete the following reactions.
(i) $6 \mathrm{Li}+\mathrm{N}_2 \rightarrow$ ?
(ii) $3 \mathrm{Ca}+\mathrm{N}_2 \rightarrow$ ?
(iii) $2 \mathrm{~B}+\mathrm{N}_2 \rightarrow$ ?
Answer:
(i) $6 \mathrm{Li}+\mathrm{N}_2 \longrightarrow 2 \mathrm{Li}_3 \mathrm{~N}$ (Lithium nitride)
(ii) $3 \mathrm{Ca}+\mathrm{N}_2 \stackrel{\text { Red hot }}{\longrightarrow} \mathrm{Ca}_3 \mathrm{~N}_2$ (Calcium Nitride)
(iii) $2 \mathrm{~B}+\mathrm{N}_2 \stackrel{\text { Bright red hot }}{\longrightarrow} 2 \mathrm{BN}$ (Boron nitride)
Qustion 2.
How is ammonia prepared?
Answer:
1. Ammonia is formed by the hydrolysis of urea.
$
\mathrm{NH}_2 \mathrm{CONH}_2+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NH}_3+\mathrm{CO}_2
$
2. Ammonia is prepared in the laboratory by heating an ammonium salt with a base.
$
\begin{aligned}
& 2 \mathrm{NH}_4^{+}+\mathrm{OH}^{-} \rightarrow 2 \mathrm{NH}_3+\mathrm{H}_2 \mathrm{O} \\
& 2 \mathrm{NH}_4 \mathrm{Cl}+\mathrm{CaO} \rightarrow \mathrm{CaCl}_2+2 \mathrm{NH}_3+\mathrm{H}_2 \mathrm{O}
\end{aligned}
$
3. It can also be prepared by heating a metal nitrides such as magnesium nitride with water.
$
\mathrm{Mg}_3 \mathrm{~N}_2+6 \mathrm{H}_2 \mathrm{O} \rightarrow 3 \mathrm{Mg}(\mathrm{OH})_2+2 \mathrm{NH}_3
$
Question 3.
Explain the structure of ammonia.
Answer:
Ammonia molecule is pyramidal in shape $\mathrm{N}-\mathrm{H}$ bond distance is $1.016 \mathrm{~A}$ and $\mathrm{H}-\mathrm{H}$ bond distance is 1.645 A with a bond angle $107^{\circ}$. The structure of ammonia may be regarded as a tetrahedral with one lone pair of electrons in one tetrahedral position hence it has a pyramidal shape as shown in the figure.

Question 4.
How will you prepare nitric acid?
Answer:
Nitric acid is prepared by heating equal amounts of potassium or sodium nitrate with concentrated sulphuric acid.
$
\mathrm{KNO}_3+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{KHSO}_4+\mathrm{HNO}_3
$
The temperature is kept as low as possible to avoid decomposition of nitric acid. The acid condenses to a fuming liquids which is coloured brown by the presence of a little nitrogen dioxide which is formed due to the decomposition of nitric acid.
$
4 \mathrm{HNO}_3 \rightarrow 4 \mathrm{NO}_2+2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2
$
Question 5.
Discuss the Commercial method to prepare Nitric acid.
(Or)
How will you prepare nitric acid by Ostwald's process?
Answer:
Nitric acid prepared in large scales using Ostwald's process. In this method ammonia from Haber's process is mixed about 10 times of air. This mixture is preheated and passed into the catalyst chamber where they come in contact with platinum gauze. The temperature rises to about $1275 \mathrm{~K}$ and the metallic gauze brings about the rapid catalytic oxidation of ammonia resulting in the formation of $\mathrm{NO}$, which then oxidised to nitrogen dioxide.
$
\begin{aligned}
& 4 \mathrm{NH}_3+5 \mathrm{O}_2 \rightarrow 4 \mathrm{NO}+6 \mathrm{H}_3 \mathrm{O}+120 \mathrm{~kJ} \\
& 2 \mathrm{NO}+\mathrm{O}_2 \rightarrow 2 \mathrm{NO}_3
\end{aligned}
$
The nitrogen dioxide produced is passed through a series of adsorption towers. It reacts with water to give nitric acid. Nitric acid formed is bleached by blowing air.
$
6 \mathrm{NO}_2+3 \mathrm{H}_2 \mathrm{O} \rightarrow 4 \mathrm{HNO}_3+2 \mathrm{NO}+\mathrm{H}_2 \mathrm{O}
$
Question 6.
Draw the structure of the following compounds.
(a) Nitrous acid
(b) Nitric acid
(c) Pernitric acid
Answer:

Question 7.
Identify $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ from the following reactions.
1. $\mathrm{P}_4+\mathrm{Mg} \rightarrow \mathrm{A}$
2. $\mathrm{P}_4+\mathrm{Ca} \rightarrow \mathrm{B}$
3. $\mathrm{P}_4+\mathrm{Na} \rightarrow \mathrm{C}$
Answer:
1. $\mathrm{P}_4+6 \mathrm{Mg} \rightarrow 2 \mathrm{Mg}_3 \mathrm{P}_2$ (Magnesium phosphide)
2. $\mathrm{P}_4+6 \mathrm{Ca} \rightarrow 2 \mathrm{Ca}_3 \mathrm{P}_2$ (Calcium phosphide)
3. $\mathrm{P}_4+12 \mathrm{Na} \rightarrow 2 \mathrm{Na}_3 \mathrm{P}$ (Sodium phosphide)
Question 8.
How will you prepare phosphine and explain the purification of phosphine?
Answer:
Phosphine is prepared by action of sodium hydroxide with white phosphorous in an inert atmosphere of carbon dioxide or hydrogen.


Phosphine is freed from phosphine dihydride $\left(\mathrm{P}_2 \mathrm{H}_4\right)$ by passing through a freezing mixture. The dihydride condenses while phosphine does not. Phosphine can also prepared by the hydrolysis of metallic phosphides with water or dilute mineral acids.
$
\begin{aligned}
\mathrm{Ca}_3 \mathrm{P}_2+6 \mathrm{H}_2 \mathrm{O} \longrightarrow & 2 \mathrm{PH}_3 \uparrow+3 \mathrm{Ca}(\mathrm{OH})_2 \\
& \text { Phosphine } \\
\mathrm{AlP}+3 \mathrm{HCl} \longrightarrow & \mathrm{PH}_3 \uparrow+\mathrm{AlCl}_3 \\
& \text { Phosphine }
\end{aligned}
$
Phosphine is prepared in pure form by heating phosphorous acid.


Phosphorous acid Ortho phosphoric acid Phosphine
A pure sample of phosphine is prepared by heating phosphonium iodide with caustic soda solution.

Question 9.
What happens when $\mathrm{PH}_3$ reacts with oxygen or air?
Answer:
When phosphine air or oxygen it bums to give meta phosphoric acid.


Question 10.
Explain the structure of phosphine.
Answer:
In phosphine, phosphorus shows $\mathrm{sp}_3$ hybridisation. Three orbitals are occupied by bond pair and fourth comer is occupied by lone pair of electrons. Hence, bond angle is reduced to $94^{\circ}$. Phosphine has a pyramidal shape.

Question 11.
Discuss the uses of phosphine.
Answer:
Phosphine is used for producing smoke screen as it gives large smoke. In a ship, a pierced container with a mixture of calcium carbide and calcium phosphide, liberates phosphine and acetylene when thrown into sea. The liberated phosphine catches fire and ignites acetylene. These burning gases serves as a signal to the approaching ships. This is known as Holmes signal.
Question 12.
How does $\mathrm{PCl}_3$ reacts with the following reagents?
1. $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$
2. $\mathrm{C}_2 \mathrm{H}_5 \mathrm{COOH}$
Answer:
1. $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+\mathrm{PCl}_3 \rightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl}+\mathrm{H}_3 \mathrm{PO}_3$
2. $\mathrm{C}_2 \mathrm{H}_5 \mathrm{COOH}+\mathrm{PCl}_3 \rightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{COCl}+\mathrm{H}_3 \mathrm{PO}_3$
Question 13.
Explain the reaction between $\mathrm{PCl}_5$ and water.
Answer:
Phosphorous pentachloride reacts with water to give phosphoryl chloride and orthophosphoric acid.

Question 14.
Explain the structure of phosphorous trioxide $\left(\mathrm{P}_2 \mathrm{O}_3\right)$.
Answer:
In phosphorous trioxide four phosphorous atoms lie at the comers of a tetrahedron and six oxygen atoms along the edges. The $\mathrm{P}-\mathrm{O}$ bond distance is $165.6 \mathrm{pm}$ which is shorter than the single bond distance of $\mathrm{P}$ $\mathrm{O}(184 \mathrm{pm})$ due to $\mathrm{p} \pi-\mathrm{d} \pi$ bonding and results in considerable double bond character.


Question 15.
Discuss the structure of phosphorous pentaoxide $\left(\mathrm{P}_2 \mathrm{O}_3\right)$.
Answer:
In $\mathrm{P}_4 \mathrm{O}_{10}$ each $\mathrm{P}$ atoms form three bonds to oxygen atom and also an additional coordinate bond with an oxygen atom. Terminal coordinate $\mathrm{P}-\mathrm{O}$ bond length is $143 \mathrm{pm}$, which is less than the expected single bond distance. This may be due to lateral overlap of filled p orbitals of an oxygen atom with empty d-orbital on phosphorous.


Question 16.

Mention the uses of oxygen.
Answer:
1. Oxygen is one of the essential component for the survival of living organisms.
2. It is used in welding (oxyacetylene welding)
3. Liquid oxygen is used as fuel in rockets etc.
Question 17.
Give and explain reducing behaviour of sulphur dioxide.
Answer:
As Sulphur dioxide can readily be oxidised, it acts as a reducing agent. It reduces chlorine into hydrochloric acid.
$
\mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2 \rightarrow \mathrm{H}_2 \mathrm{SO}_4+2 \mathrm{HCl}
$
It also reduces potassium permanganate and dichromate to $\mathrm{Mn}^{2+}$ and $\mathrm{Cr}^{3+}$ respectively.
- $2 \mathrm{KMnO}_4+5 \mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4+2 \mathrm{H}_2 \mathrm{SO}_4$
- $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+3 \mathrm{SO}_2+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{K}_2 \mathrm{SO}_4+\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3+\mathrm{H}_2 \mathrm{O}$
Question 18 .
Why bleaching action of sulphur dioxide is temporary?
Answer:
In presence of water, sulphur dioxide bleaches coloured wool, silk, sponges and straw into colourless due to its reducing property.

However, the bleached product (colourless) is allowed to stand in air, it is reoxidised by atmospheric oxygen to its original colour. Hence bleaching action of sulphur dioxide is temporary.
Question 19.
Explain the structure of sulphur dioxide.
Answer:
In sulphur dioxide, sulphur atom undergoes $\mathrm{sp}^2$ hybridisation. A double bond arises between $\mathrm{S}$ and $\mathrm{O}$ is due to $\mathrm{p} \pi-\mathrm{d} \pi$ overlapping.


Question 20 .
How will you manufacture sulphuric acid by contact process?
Answer:
Manufacture of sulphuric acid by contact process involves the following steps:
1. Initially sulphur dioxide is produced by burning sulphur or iron pyrites in oxygen/air.
$
\begin{aligned}
& \mathrm{S}+\mathrm{O}_2 \rightarrow \mathrm{SO}_2 \\
& 4 \mathrm{FeS}_2+11 \mathrm{O}_2 \rightarrow 2 \mathrm{Fe}_2 \mathrm{O}_3+8 \mathrm{SO}_2
\end{aligned}
$
2. Sulphur dioxide formed is oxidised to sulphur trioxide by air in the presence of a catalyst such as $\mathrm{V}_2 \mathrm{O}_5$ or platinised asbestos.
3. The sulphur trioxide is absorbed in concentrated sulphuric acid and produces oleum $\left(\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7\right)$. The oleum is converted into sulphuric acid by diluting it with water.

$
\mathrm{SO}_3+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7 \stackrel{\mathrm{H}_2 \mathrm{O}}{\longrightarrow} 2 \mathrm{H}_2 \mathrm{SO}_4
$
To maximise the yield the plant is operated at 2 bar pressure and $720 \mathrm{~K}$. The sulphuric acid obtained in this process is over $96 \%$ pure.
Question 21.
Prove that $\mathrm{H}_2 \mathrm{SO}_4$ is a strong dibasic acid.
Answer:
Sulphuric acid forms two types of salts namely sulphates and bisulphates.

Question 22 .
Draw the structure of
(a) Marshall's acid
(b) polythionic acid
(c) Dithionic acid
Answer:

Question 23.
What happens when chlorine reacts with ammonia?
Answer:
1. If chlorine reacts with excess ammonia, it gives nitrogen and ammonium chloride. $8 \mathrm{NH}_3+3 \mathrm{Cl}_2 \rightarrow \mathrm{N}_2+6 \mathrm{NH}_4 \mathrm{Cl}$
2. If ammonia reacts with excess chlorine, it gives nitrogen trichloride and ammonium chloride. $4 \mathrm{NH}_2+3 \mathrm{Cl}_2 \rightarrow \mathrm{NCl}_2+3 \mathrm{NH}_4 \mathrm{Cl}$
Question 24.
Bleaching action of chlorine is permanent. Justify this statement.
Answer:
Chlorine is a strong oxidising and bleaching agent because of the nascent oxygen.


Colouring matter + Nascent oxygen $\rightarrow$ Colourless oxidation product
Therefore, bleaching of chlorine is permanent. It oxidises ferrous salts to ferric, sulphites to sulphates and
hydrogen sulphide to sulphur.

Question 25.
Give the uses of chlorine.
Answer:
1. Purification of drinking water
2. Bleaching of cotton textiles, paper and rayon
3. It is used in extraction of gold and platinum
Question 26.
What is Royal water?
Answer:
When three parts of concentrated hydrochloric acid and one part of concentrated nitric acid are mixed, aquaregia is obtained. This is also known as Royal water. This is used for dissolving gold, platinum etc.
$
\begin{aligned}
& \mathrm{Au}+4 \mathrm{H}^{+}+4 \mathrm{NO}_3^{-}+4 \mathrm{Cl}^{-} \rightarrow \mathrm{AuCl}_4^{-}+\mathrm{NO}+2 \mathrm{H}_2 \mathrm{O} \\
& 3 \mathrm{Pt}+16 \mathrm{H}^{+}+4 \mathrm{NO}_3^{-}+18 \mathrm{Cl}^{-} \rightarrow 3 \mathrm{PtCl}_3^{2-}+\mathrm{NO}+8 \mathrm{H}_2 \mathrm{O}
\end{aligned}
$
Question 27.
Why HF is not stored in glass bottles?
Answer:
HF rapidly reacts with silica and glass to form soluble salt and that process leads to break the glass bottles. Hence $\mathrm{HF}$ is not stored in glass bottles.
$
\begin{aligned}
& \mathrm{SiO}_2+4 \mathrm{HF} \rightarrow \mathrm{SiF}_4+2 \mathrm{H}_2 \mathrm{O} \\
& \mathrm{Na}_2 \mathrm{SiO}_3+6 \mathrm{HF} \rightarrow \mathrm{Na}_2 \mathrm{SiF}_6+3 \mathrm{H}_2 \mathrm{O}
\end{aligned}
$
Question 28 .
Give the properties of inter halogen compounds.
Answer:
Properties of inter halogen compounds:
1. The central atom will be the larger one.
2. It can be formed only between two halogen and not more than two halogens.
3. Fluorine can't act as a central metal atom being the smallest one.
4. Due to high electronegativity with small size fluorine helps the central atom to attain high coordination number
5. They can undergo the auto ionization.
6. They are strong oxidizing agents.
Question 29.
How will you prepare Xenon fluoride?
Answer:
Xenon fluorides are prepare by direct reaction of xenon and fluorine under different conditions as shown below.
$
\begin{gathered}
\mathrm{Xe}+\mathrm{F}_2 \frac{\mathrm{Ni}}{400^{\circ} \mathrm{C}} \mathrm{XeF}_2 \\
\mathrm{Xe}+2 \mathrm{~F}_2 \frac{\mathrm{Ni} / \text { acetone }}{400^{\circ} \mathrm{C}} \mathrm{XeF}_4 \\
\mathrm{Xe}+3 \mathrm{~F}_2 \frac{\mathrm{Ni} / 200 \mathrm{~atm}}{400^{\circ} \mathrm{C}} \mathrm{XeF} 6
\end{gathered}
$
Question 30.
Complete the following reaction.
(i) $\mathrm{XeOF}_4+\mathrm{SiO}_2 \rightarrow \mathrm{A}+\mathrm{SiF}_6$
(ii) $\mathrm{A}+\mathrm{SiO}_2 \rightarrow \mathrm{B}+\mathrm{SiF}_6$
Answer:
(i) 
(ii)  
Question 31.
What happens when $\mathrm{XeF}_6$ reacts $2.5 \mathrm{M}$ solution of $\mathrm{NaOH}$ ?
Answer:
When $\mathrm{XeF}_6$ reacts with $2.5 \mathrm{M}$ of $\mathrm{NaOH}$, sodium per xenate is obtained.

Question 32.
Give two examples to show the anomalous behaviour of fluorine.
Answer:
The anomalous behaviour of fluorine is due to its:
1. Small size
2. highest electronegativity
3. low F-F bond dissociation enethalpy
4. non-availability of d-orbitals in its valence shell.
The two examples are:
1. Due to non-availability of d-orbitals in its valence shell, fluorine cannot expand its octet, therefore, shows only -1 oxidation state while all other halogens due to the presence of d-orbitals shows positive oxidation states of $+1,+3,+5$ and +7 besides oxidation state of -1 .
2. Due to its small size, the three lone pair of electrons on each $F$ atom in $F-F$ molecule, repel the bond pair. As a result, $\mathrm{F}-\mathrm{F}$ bond dissociation energy is lower than that of $\mathrm{Cl}-\mathrm{Cl}$ bond.

Question 33.
Why does the reactivity of nitrogen differ from Phosphorus?
Answer:
Nitrogen exists as a diatomic molecule $(\mathrm{N} \equiv \mathrm{N})$. Due to the presence of a triple bond between the two $\mathrm{N}-$ atoms the bond dissociation energy is large $\left(941.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$. As a result nitrogen is said to be chemically
inert in its elemental state. In contrast, $\mathrm{P}-\mathrm{P}$ single bond is much weaker $\left(213 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$ than $\mathrm{N} \equiv \mathrm{N}$ triple bond Therefore, phosphorus is much more reactive than nitrogen.
Question 34
Why does $\mathrm{NH}_3$ form hydrogen bond but $\mathrm{PH}_3$ does not?
Answer:
The electronegativity of $\mathrm{N}(3.0)$ is much higher than that of $\mathrm{FI}$ (2.1). As a result, $\mathrm{N}-\mathrm{H}$ bond is quite polar and hence $\mathrm{NH}_3$ undergoes intermolecular $\mathrm{H}$-bonding. In contrast, both $\mathrm{P}$ and $\mathrm{H}$ have an electronegativity of 2.1. Therefore $\mathrm{P}-\mathrm{H}$ bond is non-polar and hence $\mathrm{PH}_3$ does not undergo $\mathrm{H}-$ bonding.
Question 35.
Can $\mathrm{PCl}_5$ act as an oxidising as well as a reducing agent? Justify.
Answer:
Oxidation state of $\mathrm{P}$ in $\mathrm{PCl}_5$ is +5 . Since $\mathrm{P}$ has five valence electrons in its valence shell, therefore it cannot increase its oxidation state beyond +5 by donating its electrons, therefore $\mathrm{PCl}_5$ cannot act as a reducing agent. However, it can decrease its oxidation number from +5 to +3 or some lower value, therefore $\mathrm{PCl}_5$ acts as an oxidising agent. For example, it oxidises $\mathrm{Ag}$ to $\mathrm{AgCl}$.
5 Mark Questions and Answers
Question 1.

How does ammonia react with
1. Excess $\mathrm{Cl}_2$
2. $\mathrm{Na}$
3. $\mathrm{CuSO}_4$
4. $\mathrm{O}_2 / \Delta$
Answer:
1. Reaction with Excess $\mathrm{Cl}_2$

2. Reaction with $\mathrm{Na}$ :

3. Reaction with $\mathrm{CuSO}_4$

4. Reaction with $\mathrm{O}_2$

Question 2.
Explain the reaction of metals with nitric acid.
Answer:
The reactions of metals with nitric acid are explained in 3 steps as follows:
Primary reaction:
Metal nitrate is formed with the release of nascent hydrogen
$\mathrm{M}+\mathrm{HNO}_3 \rightarrow \mathrm{MNO}_3+(\mathrm{H})$
Secondary reaction:
Nascent hydrogen produces the reduction products of nitric acid.

Tertiary reaction:
The secondary products either decompose or react to give final products

For examples:
Copper reacts with nitric acid in the following manner
$
\begin{aligned}
& 3 \mathrm{Cu}+6 \mathrm{HNO}_3 \rightarrow 3 \mathrm{Cu}\left(\mathrm{NO}_3\right)_2+6(\mathrm{H}) \\
& 6(\mathrm{H})+3 \mathrm{HNO}_3 \rightarrow 3 \mathrm{HNO}_2+3 \mathrm{H}_2 \mathrm{O} \\
& 3 \mathrm{HNO}_2 \rightarrow \mathrm{HNO}_3+2 \mathrm{NO}+\mathrm{H}_2 \mathrm{O}
\end{aligned}
$
Overall reaction
$
3 \mathrm{Cu}+8 \mathrm{HNO}_3 \rightarrow 3 \mathrm{CU}\left(\mathrm{NO}_3\right)_2+2 \mathrm{NO}+4 \mathrm{H}_2 \mathrm{O}
$
The concentrated acid has a tendency to form nitrogen dioxide
$
\mathrm{Cu}+4 \mathrm{HNO}_3 \rightarrow 3 \mathrm{CU}\left(\mathrm{NO}_3\right)_2+2 \mathrm{NO}_2+2 \mathrm{H}_2 \mathrm{O}
$
Question 3.
How will you prepare ozone by laboratory method? Explain the structure of ozone.
Answer:
In the laboratory ozone is prepared by passing electrical discharge through oxygen. At a potential of 20,000
$\mathrm{V}$ about $10 \%$ of oxygen is converted into ozone it gives a mixture known as ozonised oxygen. Pure ozone is obtained as a pale blue gas by the fractional distillation of liquefied ozonised oxygen.

The ozone molecule have a bent shape and symmetrical with delocalised bonding between the oxygen atoms.
Question 4.
$A$ is a king of acid. A reacts with $\mathrm{HBr}$ to give $\mathrm{B}$ and Bromine. A reacts with $\mathrm{Na}_2 \mathrm{CO}_3$ to give $\mathrm{C}$ and carbon dioxide. Identify A, B and C. Give the reaction.
Answer:
1. King of acid is sulphuric acid (A).
2. Sulphuric acid (A) reacts with $\mathrm{HBr}$ to give $\mathrm{SO}_2$ (B) and Bromine.


3. Sulphuric acid (A) reacts with $\mathrm{Na}_2 \mathrm{CO}_3$ to give sodium sulphate $(\mathrm{C})$ and $\mathrm{CO}_2$.


Question 5.
How does Sulphuric acid react with the following:
(a) $\mathrm{Al}$
(b) $\mathrm{KNO}_3$
(c) $\mathrm{NaBr}$
(d) $\mathrm{C}_6 \mathrm{H}_6$

Answer:

(a) 

(b) 

(c) 

(d) 

Question 6.
1. Explain the test for sulphate (or) sulphuric acid.
2. What happens when sulphuric acid reacts with oxalic acid?
Answer:
1. Dilute solution of sulphuric acid or aqueous solution of sulphates gives white precipitate with barium chloride solution. It can also be detected using lead acetate solution. Here a white precipitate of lead sulphate is obtained.


2. Sulphuric acid reacts with oxalic acid to give $\mathrm{CO}$ and $\mathrm{CO}_2$


Question 7.
Discuss the manufacture of chlorine.
Answer:
Electrolytic process:
When a solution of brine $(\mathrm{NaCl})$ is electrolysed, $\mathrm{Na}+$ and $\mathrm{CP}$ ions are formed. $\mathrm{Na}^{+}$ion reacts with $\mathrm{OH}^{-}$ions of water and forms sodium hydroxide. Hydrogen and chlorine are liberated as gases.
$
\begin{array}{lll}
\mathrm{NaCl} \longrightarrow \mathrm{Na}^{+}+\mathrm{Cl}^{-} & \text {At the cathode, } & \text { At the cathode, } \\
\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{H}^{+}+\mathrm{OH}^{-} & \mathrm{H}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{H} & \mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}+\mathrm{e}^{-} \\
\mathrm{Na}^{+}+\mathrm{OH}^{-} \longrightarrow \mathrm{NaOH} & \mathrm{H}+\mathrm{H} \longrightarrow \mathrm{H}_2 & \mathrm{Cl}+\mathrm{Cl} \longrightarrow \mathrm{Cl}_2
\end{array}
$
Deacons process:
In this process a mixture of air and hydrochloric acid is passed up a chamber containing a number of shelves, pumice stones soaked in cuprous chloride are placed. Hot gases at about $723 \mathrm{~K}$ are passed through
a jacket that surrounds the chamber.
$
4 \mathrm{HCl}+\mathrm{O}_2 \underset{\mathrm{Cu}_2 \mathrm{Cl}_2}{\stackrel{400^{\circ} \mathrm{C}}{\longrightarrow}} 2 \mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2 \uparrow
$

The chlorine obtained by this method is dilute and is employed for the manufacture of bleaching powder. The catalysed reaction is given below,

Question 8.
1. How is bleaching powder prepared?
2. What happens when benzene reacts with chlorine?
Answer:
1. Bleaching powder is prepared by passing chlorine gas through dry slaked lime (Calcium hydroxide).

2. Benzene reacts with chlorine in the presence of ferric chloride to give chlorobenzene.


Question 9.
Cone. $\mathrm{H}_2 \mathrm{SO}_4$ is added followed by heating to each of the following test tubes labelled (I) to (IV). Identify in which of the above test tube the following change will be observed. Support your answer with the help of chemical equation:
(a) formation of black substance
(b) evolution of brown gas
(c) evolution of colourless gas
(d) formation of a brown substance which on dilution becomes blue.
(e) disappearance of yellow powder along with evolution of colourless gas.

Answer:

(a) 

(b) 

(c) 

(d) 

(e) 

 

Question 10.
Give reasons for each of the following:
1. Bleaching of flowers by $\mathrm{Cl}_2$ is permanent while by $\mathrm{SO}_2$ is temporary.
2. Molten aluminium bromide is a poor conductor of electricity.
3. Nitric oxide becomes brown when released in air.
4. $\mathrm{PCl}_5$ is ionic in nature in the solid state.
5. Ammonia is a good complexing agent.
Answer:
1. $\mathrm{Cl}_2$ bleaches by oxidation, while $\mathrm{SO}_2$ does it by reduction. The reduced product gets oxidise again and the colour is regained back.
2. Aluminium bromide exists as a dimer, $\mathrm{Al}_2 \mathrm{Br}_6$. In this structure, each aluminium atom forms one coordinate bond by accepting a lone pair of electrons from the bromine atom of another aluminium bromide molecule and thus complete the octet of electrons. Due to lack of free electrons, molten aluminium bromide is a poor conductor of electricity.
3. Nitric oxide reacts with air and oxidised into $\mathrm{NO}_2$ which is brown in colour.
$
2 \mathrm{NO}+\mathrm{O}_2 \rightarrow 2 \mathrm{NO}_2
$
4. In solid state $\mathrm{PCl}_5$ exists as $\left[\mathrm{PCl}_4\right]^{+}\left[\mathrm{PCl}_6\right]^{-}$and hence it is ionic in nature. Due to its ionic nature, it conducts current on fusion.
5. $\mathrm{N}$ atom in ammonia has lone pair of electrons which can coordinate with other atoms or cations required for the stability of electron pair.
Question 11.
How will you prepare the following compounds.
(a) Hyponitrous acid
(b) Nitrous acid
(c) Pernitrous acid
(d) Pernitric acid
Answer:
(a) Hyponitrous acid: $\mathrm{Ag}_2 \mathrm{~N}_2 \mathrm{O}_2+2 \mathrm{HCl} \longrightarrow 2 \mathrm{AgCl}+\mathrm{H}_2 \mathrm{~N}_2 \mathrm{O}_2$
(b) Nitrous acid $: \mathrm{Ba}\left(\mathrm{NO}_2\right)_2+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow 2 \mathrm{HNO}_2+\mathrm{BaSO}_4$
(c) Pernitrous acid : $\mathrm{H}_2 \mathrm{O}_2+\mathrm{ON}(\mathrm{OH}) \longrightarrow \mathrm{ON}(\mathrm{OOH})+\mathrm{H}_2 \mathrm{O}$

(d) Pernitric acid : $\mathrm{H}_2 \mathrm{O}_2+\mathrm{N}_2 \mathrm{O}_5 \longrightarrow \mathrm{NO}_2 \mathrm{OOH}+\mathrm{HNO}_3$
Common Errors And Its Rectificaations:
Common Errors:
1. Oxidation number rules may be confused.
2. Oxidation number of oxygen may get confused.
3. Molecular formula and compound name may get confused.
Rectifications:
1. Oxidation number of fluorine is always -1 .
2. Always oxygen is -2 but in peroxide it is -1 . When it comes as a first element its oxidation number is positive.
3. Student should memorise the formula using short cut method, e.g. Phosphorous acid $-\mathrm{H}_3 \mathrm{PO}_3$ Phosphoric acid $-\mathrm{H}_3 \mathrm{PO}_4$ e.g. Phosphorous acid $-\mathrm{H}_3 \mathrm{PO}_3$ By pronunciation, ' $\mathrm{O}$ ' indicates oxygen is more.

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