Text Book Back Questions and Answers - Chapter 3 - p-Block Elements – II - 12th Chemistry Guide Samacheer Kalvi Solutions

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p-Block Elements - II
TextBook Evalution
I. Choose the correct answer.
Question 1.
In which of the following, $\mathrm{NH}_3$ is not used?
(a) Nessler's reagent
(b) Reagent for the analysis of IV group basic radical
(c) Reagent for the analysis of III group basic radical
(d) Tollen's reagent
Answer:
(a) Nessler's reagent
Question 2.
Which is time regarding nitrogen?
(a) least electronegative element
(b) has low ionisation enthalpy than oxygen
(c) d-orbitals available
(d) ability to form $\mathrm{p} \pi-\mathrm{p} \pi$ bonds with itself
Answer:
(d) ability to form $\mathrm{p} \pi-\mathrm{p} \pi$ bonds with itself
Question 3.
An element belongs to group 15 and 3 rd period of the periodic table, its electronic configuration would be
(a) $1 s^2 2 s^2 2 p^4$
(b) $1 s^2 2 s^2 2 p^3$
(c) $1 s^2 2 s^2 2 p^6 3 s^2 3 p^2$
(d) $1 s^2 2 s^2 2 p^6 3 s^2 3 p^3$
Answer:
(d) $1 s^2 2 s^2 2 p^6 3 s^2 3 p^3$

Question 4.
Solid (A) reacts with strong aqueous $\mathrm{NaOH}$ liberating a foul smelling gas(B) which spontaneously bum in air giving smoky rings. $A$ and $B$ are respectively
(a) $\mathrm{P}_4$ (red) and $\mathrm{PH}_3$
(b) $\mathrm{P}_4$ (white) and $\mathrm{PH}_3$
(c) $\mathrm{S}_8$ and $\mathrm{H}_2 \mathrm{~S}$
(d) $\mathrm{P}_4$ (white) and $\mathrm{H}_2 \mathrm{~S}$
Answer:
(b) $\mathrm{P}_4$ (white) and $\mathrm{PH}_3$
Question 5.
In the brown ring test, brown colour of the ring is due to
(a) a mixture of $\mathrm{NO}$ and $\mathrm{NO}_2$
(b) Nitroso ferrous sulphate
(c) Ferrous nitrate
(d) Ferric nitrate
Answer:
(b) Nitroso ferrous sulphate
Question 6.
On hydrolysis, $\mathrm{PCl}_3$ gives
(a) $\mathrm{H}_3 \mathrm{PO}_3$
(b) $\mathrm{PH}_3$
(c) $\mathrm{H}_3 \mathrm{PO}_4$
(d) POOL
Answer:
(a) $\mathrm{H}_3 \mathrm{PO}_3$
Question 7.
$\mathrm{P}_4 \mathrm{O}_6$ reacts with cold water to give
(a) $\mathrm{H}_3 \mathrm{PO}_3$
(b) $\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_7$
(c) $\mathrm{HPO}_3$
(d) $\mathrm{H}_3 \mathrm{PO}_4$
Answer:
(a) $\mathrm{H}_3 \mathrm{PO}_3$
Question 8.
The basicity of pyrophosphorous acid $\left(\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_5\right)$ is
(a) 4
(b) 2
(c) 3
(d) 5

Answer:

(b) 2

Question 9.
The molarity of given orthophosphoric acid solution is $2 \mathrm{M}$. its normality is
(a) $6 \mathrm{~N}$
(b) $4 \mathrm{~N}$
(c) $2 \mathrm{~N}$
(d) none of these
Answer:
(a) $6 \mathrm{~N}$
Question 10.
Assertion - bond dissociation energy of fluorine is greater than chlorine gas Reason - chlorine has more electronic repulsion than fluorine
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer:
(d) Both assertion and reason are false. The converse is true.
Question 11.
Among the following, which is the strongest oxidizing agent?
(a) $\mathrm{Cl}_2$
(b) $\mathrm{F}_2$
(c) $\mathrm{Br}_2$
(d) $\mathrm{I}_2$
Answer:
(b) $\mathrm{F}_2$
Question 12.
The correct order of the thermal stability of hydrogen halide is
(a) $\mathrm{HI}>\mathrm{HBr}>\mathrm{HCl}>\mathrm{HF}$
(b) $\mathrm{HF}>\mathrm{HCl}>\mathrm{HBr}>\mathrm{HI}$
(c) $\mathrm{HCl}>\mathrm{HF}>\mathrm{HBr}>\mathrm{HI}$
(d) $\mathrm{HI}>\mathrm{HCl}>\mathrm{HF}>\mathrm{HBr}$
Answer:
(b) $\mathrm{HF}>\mathrm{HCl}>\mathrm{HBr}>\mathrm{HI}$
Question 13.
Which one of the following compounds is not formed?
(a) $\mathrm{XeOF}_4$
(b) $\mathrm{XeO}_3$
(c) $\mathrm{XeF}_2$
(d) $\mathrm{NeF}_2$
Answer:
(d) $\mathrm{NeF}_2$

Question 14.
Most easily liquefiable gas is
(a) $\mathrm{Ar}$
(b) $\mathrm{Ne}$
(c) $\mathrm{He}$
(d) $\mathrm{Kr}$
Answer:
(c) $\mathrm{He}$
Question 15.
$\mathrm{XeF}_6$ on complete hydrolysis produces
(a) $\mathrm{XeOF}_4$
(b) $\mathrm{XeO}_2 \mathrm{~F}_4$
(c) $\mathrm{XeO}_3$
(d) $\mathrm{XeO}_2$
Answer:
(c) $\mathrm{XeO}_3$
Question 16.
On oxidation with iodine, sulphite ion is transformed to
(a) $\mathrm{S}_4 O_6^{2-}$
(b) $\mathrm{S}_2 O_6^{2-}$
(c) $\mathrm{SO}_4^{2-}$
(d) $\mathrm{SO}_3^{2-}$
Answer:
(c) $\mathrm{SO}_4^{2-}$
Question 17.
Which of the following is strongest acid among all?
(a) $\mathrm{HI}$
(b) $\mathrm{HF}$
(c) $\mathrm{HBr}$
(d) $\mathrm{HCl}$
Answer:
(a) $\mathrm{HI}$
Question 18.
Which one of the following orders is correct for the bond dissociation enthalpy of halogen molecules?
(a) $\mathrm{Br}_2>\mathrm{I}_2>\mathrm{F}_2>\mathrm{Cl}_2$

(b) $\mathrm{F}_2>\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2$
(c) $\mathrm{I}_2>\mathrm{Br}_2>\mathrm{Cl}_2>\mathrm{F}_2$
(d) $\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{F}_2>\mathrm{I}_2$
Answer:
(d) $\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{F}_2>\mathrm{I}_2$
Question 19.
Among the following the correct order of acidity is
(a) $\mathrm{HClO}_2<\mathrm{HCIO}<\mathrm{HClO}_3<\mathrm{HClO}_4$
(b) $\mathrm{HClO}_4<\mathrm{HClO}_2<\mathrm{HCIO}<\mathrm{HClO}_3$
(c) $\mathrm{HClO}_3<\mathrm{HClO}_4<\mathrm{HClO}_2<\mathrm{HCIO}$
(d) $\mathrm{HCIO}<\mathrm{HClO}_2<\mathrm{HClO}_3<\mathrm{HClO}_4$
Answer:
(d) $\mathrm{HCIO}<\mathrm{HClO}_2<\mathrm{HClO}_3<\mathrm{HClO}_4$
Question 20.
When copper is heated with cone $\mathrm{HNO}_3$ it produces
(a) $\mathrm{CU}\left(\mathrm{NO}_3\right)_2, \mathrm{NO}$ and $\mathrm{NO}_2$
(b) $\mathrm{Cu}\left(\mathrm{NO}_3\right)_2$ and $\mathrm{N}_2 \mathrm{O}$
(c) $\mathrm{CU}\left(\mathrm{NO}_3\right)_2$ and $\mathrm{NO}_2$
(d) $\mathrm{Cu}\left(\mathrm{NO}_3\right)_2$ and $\mathrm{NO}$
Answer:
(c) $\mathrm{CU}\left(\mathrm{NO}_3\right)_2$ and $\mathrm{NO}_2$
II. Answer the following questions:
Question 1.
What is inert pair effect?
Answer:
In p-block elements, as we go down the group, two electrons present in the valence s-orbital become inert and are not available for bonding (only p-orbital involves chemical bonding). This is called inert pair effect.
Question 2.
Chalcogens belongs to p-block. Give reason.
Answer:
Chalcogens are ore forming elements. Most of the ores are oxides and sulphides, therefore oxygen, sulphur and other group 16 elements are called Chalcogens. In $\mathrm{O}, \mathrm{S}, \mathrm{Se}, \mathrm{Te}$ and Po last electron enters to p-orbital.

Therefore Chalcogens belongs to p-block.
Question 3.
Explain why fluorine always exhibit an oxidation state of -1 ?
Answer:
Fluorine the most electronegative element than other halogens and cannot exhibit any positive oxidation state. Fluorine does not have d-orbital while other halogens have d-orbitals. Therefore fluorine always exhibit an oxidation state of- 1 and others in halogen family shows $+1,+3,+5$ and +7 oxidation states.
Question 4.
Give the oxidation state of halogen in the following.
1. $\mathrm{OF}_2$
2. $\mathrm{O}_2 \mathrm{~F}_2$
3. $\mathrm{Cl}_2 \mathrm{O}_3$
4. $\mathrm{I}_2 \mathrm{O}_4$
Answer:
1. $\mathrm{OF}_2$
$
\begin{aligned}
& +2+2(\mathrm{x})=0 \\
& +2=-2 \mathrm{x} \\
& 2 \mathrm{x}=-2 \\
& \mathrm{x}=-1
\end{aligned}
$
$
\begin{aligned}
& \text { 2. } \mathrm{O}_2 \mathrm{~F}_2 \\
& 2(+1)+2 \mathrm{x}=0 \\
& 2 \mathrm{x}=-2 \\
& \mathrm{x}=-1
\end{aligned}
$
3. $\mathrm{Cl}_2 \mathrm{O}_3$
$
\begin{aligned}
& 2(x)+3(-2)=0 \\
& 2 x=+6 \\
& x=+3
\end{aligned}
$
4. $\mathrm{I}_2 \mathrm{O}_4$
$
\begin{aligned}
& 2(x)+4(-2)=0 \\
& 2 x=+8 \\
& x=+4
\end{aligned}
$

Question 5 .
What are interhalogen compounds? Give examples.
Answer:
Each halogen combines with other halogens to form a series of compounds called interhalogen compounds. For example, Fluorine reacts readily with oxygen and forms difluorine oxide $\left(\mathrm{F}_2 \mathrm{O}\right)$ and difluorine dioxide $\left(\mathrm{F}_2 \mathrm{O}_2\right)$
Question 6.
Why fluorine is more reactive than other halogens?
Answer:
Fluorine is the most reactive element among halogen. This is due to the minimum value of $F-F$ bond dissociation energy. Hence fluorine is more reactive than other halogens.
Question 7.
Give the uses of helium.
Answer:
1. Helium and oxygen mixture is used by divers in place of air oxygen mixture. This prevents the painful dangerous condition called bends.
2. Helium is used to provide inert atmosphere in electric arc welding of metals
3. Helium has lowest boiling point hence used in cryogenics (low temperature science).
4. It is much less denser than air and hence used for filling air balloons
Question 8.
What is the hybridisation of iodine in $\mathrm{IF}_7$ ? Give its structure.
Answer:
Hybridisation of iodine in $\mathrm{IF}_7$ is $\mathrm{sp}^3 \mathrm{~d}^3$ Structure of $\mathrm{IF}_7$ is pentagonal bipyramidal.

Question 9.
Give the balanced equation for the reaction between chlorine with cold $\mathrm{NaOH}$ and hot $\mathrm{NaOH}$.
Answer:
1. Reaction between chlorine with cold $\mathrm{NaOH}$ :
$
\begin{aligned}
& \mathrm{Cl}_2+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{HCl}+\mathrm{HOCl} \\
& \mathrm{HCl}+\mathrm{NaOH} \rightarrow \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O} \\
& \mathrm{HOCl}+\mathrm{NaOH} \rightarrow \mathrm{NaOCl}+\mathrm{H}_2 \mathrm{O}
\end{aligned}
$
Overall reaction

Chlorine reacts with cold $\mathrm{NaOH}$ to give sodium chloride and sodium hypochlorite.
2. Reaction between chlorine with hot $\mathrm{NaOH}$ :
$
\begin{aligned}
& \mathrm{Cl}_2+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{HCl}+\mathrm{HOCl} \\
& \mathrm{HCl}+\mathrm{NaOH} \rightarrow \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O} \\
& \mathrm{HOCl}+\mathrm{NaOH} \rightarrow \mathrm{NaOCl}+\mathrm{H}_2 \mathrm{O} \\
& 3 \mathrm{NaOCl} \rightarrow \mathrm{NaClO}_3+2 \mathrm{NaCl}
\end{aligned}
$
Overall reaction

Chlorine reacts with hot $\mathrm{NaOH}$ to give sodium chlorate and sodium chloride.
Question 10.
How will you prepare chlorine in the laboratory?
Answer:
1. Chlorine is prepared by the action of cone, sulphuric acid on chlorides in presence of manganese dioxide.
$
4 \mathrm{NaCl}+\mathrm{MnO}_2+4 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{Cl}_2+\mathrm{MnCl}_2+4 \mathrm{NaHSO}_4+2 \mathrm{H}_2 \mathrm{O}
$
2. It can also be prepared by oxidising hydrochloric acid using various oxidising agents such as manganese dioxide, lead dioxide, potassium permanganate or dichromate.
$
\begin{aligned}
& \mathrm{PbO}_2+4 \mathrm{HCl} \rightarrow \mathrm{PbCl}_2+2 \mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2 \\
& \mathrm{MnO}_2+4 \mathrm{HCl} \rightarrow \mathrm{MnCl}_2+2 \mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2 \\
& 2 \mathrm{KMnO}_4+16 \mathrm{HCl} \rightarrow 2 \mathrm{KCl}+2 \mathrm{MnCl}+8 \mathrm{H}_2 \mathrm{O}+5 \mathrm{Cl}_2 \\
& \mathrm{~K}_2 \mathrm{Cr}_2 \mathrm{O}_7+14 \mathrm{HCl} \rightarrow 2 \mathrm{KCl}+2 \mathrm{CrCl}_3+7 \mathrm{H}_2 \mathrm{O}+3 \mathrm{Cl}_2
\end{aligned}
$

3. When bleaching powder is treated with mineral acids chlorine is liberated
$
\begin{aligned}
& \mathrm{CaOCl}_2+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_2+\mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2 \\
& \mathrm{CaOCl}_2+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CaSO}_4+\mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2 \\
&
\end{aligned}
$
Question 11.
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 12.
Give a reason to support that sulphuric acid is a dehydrating agent.
Answer:
Sulphuric acid is highly soluble in water and has strong affinity towards water and hence it can be used as a dehydrating agent. When dissolved in water it forms mono $\left(\mathrm{H}_2 \mathrm{SO}_4 \cdot \mathrm{H}_2 \mathrm{O}\right)$ and di $\left(\mathrm{H}_2 \mathrm{SO}_4 \cdot 2 \mathrm{H}_2 \mathrm{O}\right)$
hydrates and the reaction is exothermic. The dehydration property can also be illustrated by its reaction with organic compounds such as sugar, oxalic acid and formic acid.
$
\begin{aligned}
& \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow 12 \mathrm{C}+\mathrm{H}_2 \mathrm{SO}_4 \cdot 11 \mathrm{H}_2 \mathrm{O} \\
& \quad \text { (Sucrose) } \\
& \mathrm{HCOOH}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{CO}+\mathrm{H}_2 \mathrm{SO}_4 \cdot \mathrm{H}_2 \mathrm{O} \\
& \quad \text { (Formic acid) } \\
& (\mathrm{COOH})_2+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{CO}+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{SO}_4 \cdot \mathrm{H}_2 \mathrm{O}
\end{aligned}
$
(Oxalic acid)
Question 13.
Write the reason for the anamolous behaviour of Nitrogen.
Answer:
1. Due to its small size, high electro negativity, high ionisation enthalpy and absence of d-orbitals.
2. $N$, has a unique ability to form $\mathrm{p} \pi-\mathrm{p} \pi$ multiple bond whereas the heavier members of this group (15) do not form $\mathrm{p} \pi-\mathrm{p} \pi$ bond, because their atomic orbitals are so large and diffused that they cannot have effective overlapping.
3. Nitrogen exists a diatomic molecule with triple bond between the two atoms whereas other elements form single bond in the elemental state.
4. N cannot form $\mathrm{d} \pi-\mathrm{p} \pi$ bond due to the absence of $\mathrm{d}$ - orbitals whereas other elements can.

Question 14.
Write the molecular formula and structural formula for the following molecules.
(a) Nitric acid
(b) dinitrogen pentoxide
(c) phosphoric acid
(d) phosphine
Answer:

Question 15.
Give the uses of argon.
Answer:
Argon prevents the oxidation of hot filament and prolongs the life in filament bulbs.
Question 16.
Write the valence shell electronic configuration of group-15 elements.
Answer:
General electronic configuration of group 15 elements are $\mathrm{ns}^2 \mathrm{np}^3$.
1. Nitrogen $-[\mathrm{He}] 2 \mathrm{~s}^2 2 \mathrm{p}^3$
2. Phosphorous - $[\mathrm{Ne}] 3 \mathrm{~s}^2 3 \mathrm{p}^3$
3. Arsenic - [Ar] $3 \mathrm{~d}^{10} 4 \mathrm{~s}^2 4 \mathrm{p}^3$
4. Antimony $-[\mathrm{Kr}] 4 \mathrm{~d}^{10} 5 \mathrm{~s}^2 5 \mathrm{p}^3$
5. Bismuth - [Ne] $4 \mathrm{f}^{14} 5 \mathrm{~s}^{10} 6 \mathrm{~s}^2 6 \mathrm{p}^3$
Question 17.
Give two equations to illustrate the chemical behaviour of phosphine.
Answer:
1. Phosphine reacts with halogens to give phosphorous penta halides.
$
\mathrm{PH}_3+4 \mathrm{Cl}_2 \rightarrow \mathrm{PCl}_5+3 \mathrm{HCl}
$
2. Phosphine forms coordination compound with lewis acids such as boron trichloride.

Coordination compound
3. Phosphine precipitates some metal from their salt solutions.
$
3 \mathrm{AgNO}_3+\mathrm{PH}_3 \rightarrow \mathrm{Ag}_3 \mathrm{P}+3 \mathrm{HNO}_3
$
Question 18.
Give a reaction between nitric acid and a basic oxide.
Answer:
Nitric acid reacts with bases and basic oxides to form salts and water.
1. $\mathrm{ZnO}+2 \mathrm{HNO}_3 \rightarrow \mathrm{Zn}\left(\mathrm{NO}_3\right)_2+\mathrm{H}_2 \mathrm{O}$
2. $3 \mathrm{FeO}+10 \mathrm{HNO}_3 \rightarrow 3 \mathrm{Fe}\left(\mathrm{NO}_3\right)_3+\mathrm{NO}+5 \mathrm{H}_2 \mathrm{O}$

Question 19.
What happens when $\mathrm{PCl}_5$ is heated?
Answer:
On heating phosphorous pentachloride, it decomposes into phosphorus trichloride and chlorine. $\mathrm{PCl}_5 \underset{\rightarrow}{\rightarrow} \mathrm{PCl}_3+\mathrm{Cl}_2$
Question 20.
Suggest a reason why HF is a weak acid, whereas binary acids of the all other halogens are strong acids.
Answer:
The hydrogen halides are extremely soluble in water due to the ionisation.
$
\mathrm{X}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{X}^{-}
$
$(\mathrm{X}=\mathrm{F}, \mathrm{Cl}, \mathrm{Br}$ or $\mathrm{I})$
Solutions of hydrogen halides are therefore acidic and known as hydrohalic acids. Hydrochloric, hydrobromic and hydroiodic acids are almost completely ionised and are therefore strong acids but $\mathrm{HF}$ is a weak acid. For $\mathrm{HF}$,
- $\mathrm{HF}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{F}^{-}$
- $\mathrm{HF}+\mathrm{F}^{-} \rightarrow \mathrm{HF}_2^{-}$
At high concentration, the equilibrium involves the removal of fluoride ions is important. Since it affects the dissociation of hydrogen fluoride, therefore it is a weak acid.
Question 21 .
Deduce the oxidation number of oxygen in hypofluorous acid - HOF.
Answer:
In case of $\mathrm{O}-\mathrm{F}$ bond is $\mathrm{HOF}$, fluorine is most electronegative element. So its oxidation number is -1 . Thereby oxidation number of $\mathrm{O}$ is +1 . Similarly in case of $\mathrm{O}-\mathrm{H}$ bond is $\mathrm{HOF}$. $\mathrm{O}$ is highly electronegative than $\mathrm{H}$. So its oxidation number is -1 and oxidation number of $\mathrm{H}$ is +1 . So, Net oxidation of oxygen is $-1+$ $1=0$.

Question 22.
What type of hybridisation occur in
1. $\mathrm{BrF}_5$
2. $\mathrm{BrF}_3$
Answer:
1. $\mathrm{BrF}_5$
$\mathrm{BrF}_5$ is a $\mathrm{AX}_5$ type. Therefore is has $\mathrm{sp}^3 \mathrm{~d}^2$ hybridisation. Hence, $\mathrm{BrF}_5$ molecule has square pyramidal shape.
2. $\mathrm{BrF}_3$
$\mathrm{BrF}_3$ is a $\mathrm{AX}_3$ type. Therefore it has $\mathrm{sp}^3 \mathrm{~d}$ hybridisation. Hence, $\mathrm{BrF}_3$ molecule has T-shape.
Question 23.
Complete the following reactions.
Answer:

1. $\mathrm{NaCl}+\mathrm{MnO}_2+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow$
$
4 \mathrm{NaCl}+\mathrm{MnO}_2+4 \mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{Cl}_2+\mathrm{MnCl}_2+4 \mathrm{NaHSO}_4+2 \mathrm{H}_2 \mathrm{O}
$
2. $\mathrm{NaNO}_2+\mathrm{HCl} \longrightarrow$
$
\mathrm{NaNO}_2+\mathrm{HCl} \longrightarrow \mathrm{NaCl}+\mathrm{HNO}_2
$
3. $\mathrm{IO}_3^{-}+\mathrm{I}^{-}+\mathrm{H}^{+} \longrightarrow$
$
\mathrm{IO}_3^{-}+5 \mathrm{I}^{-}+6 \mathrm{H}^{+} \longrightarrow 3 \mathrm{I}_2+3 \mathrm{H}_2 \mathrm{O}
$
4. $\begin{aligned} & \mathrm{I}_2+\mathrm{S}_2 \mathrm{O}_3{ }^{2-} \longrightarrow \\ & \mathrm{I}_2+2 \mathrm{~S}_2 \mathrm{O}_3{ }^{2-} \longrightarrow \mathrm{S}_4 \mathrm{O}_6{ }^{2-}+2 \mathrm{I}^{-}\end{aligned}$
5. $\mathrm{P}_4+\mathrm{NaOH}+\mathrm{H}_2 \mathrm{O} \longrightarrow$
$
\mathrm{P}_4+3 \mathrm{NaOH}+3 \mathrm{H}_2 \mathrm{O} \longrightarrow 3 \mathrm{NaH}_2 \mathrm{PO}_2+\mathrm{PH}_3 \uparrow
$
6. $\mathrm{AgNO}_3+\mathrm{PH}_3 \longrightarrow$
$
3 \mathrm{AgNO}_3+\mathrm{PH}_3 \longrightarrow \mathrm{Ag}_3 \mathrm{P}+3 \mathrm{HNO}_3
$
7. $\mathrm{Mg}+\mathrm{HNO}_3 \longrightarrow$
$
4 \mathrm{Mg}+10 \mathrm{HNO}_3 \longrightarrow 4 \mathrm{Mg}\left(\mathrm{NO}_3\right)_2+\mathrm{NH}_4 \mathrm{NO}_3+3 \mathrm{H}_2 \mathrm{O}
$
8. $\mathrm{KClO}_3 \stackrel{\Delta}{\longrightarrow}$
$
2 \mathrm{KClO}_3 \underset{\mathrm{MnO}_2}{\stackrel{\Delta}{\longrightarrow}} 2 \mathrm{KCl}+3 \mathrm{O}_2
$

9. $\mathrm{Cu}+$ Hot Conc. $\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow$
$\mathrm{Cu}+$ Hot Conc. $2 \mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{CuSO}_4+2 \mathrm{H}_2 \mathrm{O}+\mathrm{SO}_2 \uparrow$
10. $\mathrm{Sb}+\mathrm{Cl}_2 \longrightarrow$
$2 \mathrm{Sb}+3 \mathrm{Cl}_2 \longrightarrow 2 \mathrm{SbCl}_3$
11. $\mathrm{HBr}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow$
$2 \mathrm{HBr}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow 2 \mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O}+\mathrm{Br}_2$
12. $\mathrm{XeF}_6+\mathrm{H}_2 \mathrm{O} \longrightarrow$
$\mathrm{XeF}_6+3 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{XeO}_3+6 \mathrm{HF}$
13. $\mathrm{XeO}_6^{4-}+\mathrm{Mn}^{2+}+\mathrm{H}^{+} \longrightarrow$
$5 \mathrm{XeO}_6^{4-}+2 \mathrm{Mn}^{2+}+14 \mathrm{H}^{-} \longrightarrow 2 \mathrm{MnO}_4^{-}+5 \mathrm{XeO}_3+7 \mathrm{H}_2 \mathrm{O}$
14. $\mathrm{XeOF}_4+\mathrm{SiO}_2 \longrightarrow$
$2 \mathrm{XeOF}_4+\mathrm{SiO}_2 \longrightarrow 2 \mathrm{XeO}_2 \mathrm{~F}_2+\mathrm{SiF}_4$

Question 1.
Write the products formed in the reaction of nitric acid (both dilute and concentrated) with zinc.

Answer: