NCERT Exemplar Class 12 Chemistry Solutions Chapter 7 p block elements

Question 1. On addition of conc. $\text {H}_{2}\text {SO}_{4}$ to a chloride salt, colourless fumes are evolved but in case of an iodide salt, violet fumes come out. This is because
(i) $\text {H}_{2}\text {SO}_{4}$ reduces $\text {HI to I}_{2}$
(ii) $\text {HI}$ is of violet colour
(iii) $\text {HI}$ gets oxidised to $\text {I}_{2}$
(iv) $\text {HI}$ changes to $\text {HIO}_{3}$
Answer:

The answer is the option (iii). When concentrated $\text {H}_{2}\text {SO}_{4}$ is added to iodine salt, HI, which is a strong reducing agent, is formed. During the reaction, it is oxidized to $\text {I}_{2}$ , which is violet in colour.
$\text {2NaCl + H}_{2}\text {SO}_{4}\rightarrow \text {Na}_{2}\text {SO}_{4}+\text {2HCl}$
In case of iodine the halogen acid $\text {HI}$ obtained reduces $\text {H}_{2}\text {SO}_{4}$ to $\text {SO}_{2}$ and itself is oxidised to free iodine.
$\text {2NaI + H}_{2}\text {SO}_{4}\rightarrow \text {Na}_{2}\text {SO}_{4}+ \text {2HI}\overset{\text {H}_{2}\text {SO}_{4}}{\longrightarrow} \text {2H}_{2}\text {O + SO}_{2}+\text {I}_{2}$

Question 2. In qualitative analysis when $\text {H}_{2}\text {S}$ is passed through an aqueous solution of salt acidified with dil.$\text {HCl}$, a black precipitate is obtained. On boiling the precipitate with dil. $\text {HNO}_{3}$, it forms a solution of blue colour. Addition of excess of aqueous solution of ammonia to this solution gives _________.
(i) a deep blue precipitate of $\text {Cu(OH)}_{2}$
(ii) a deep blue solution of $[Cu(NH_{3})_{4}]^{2+}$
(iii) a deep blue solution of $Cu(NO_{3})_{2}$
(iv) a deep blue solution of $Cu(OH)_{2}.Cu(NO_{3})_{2}$
Answer:

The answer is the option (ii). Black precipitate of copper sulphide is formed on passing $H_{2}S$ through acidified solution of salt. On boiling with dilute $HNO_{3}$, the precipitate gives blue colour of copper nitrate. On addition of excess aqueous solution of ammonia, deep blue coloured compound is found. The reactions involved are:

$Cu^{2+}+H_{2}S\rightarrow CuS (black)+2H^{+}$

$CuS+dil.HNO_{3}\rightarrow Cu\left ( NO_{3} \right )_{2}$

$Cu\left ( NO_{3} \right )_{2}+4NH_{3}\rightarrow \left [ Cu\left ( NH_{3} \right )_{4} \right ]^{2+}(Deep\; blue)$

Question 3. In a cyclotrimetaphosphoric acid molecule, how many single and double bonds are present?
(i) 3 double bonds; 9 single bonds
(ii) 6 double bonds; 6 single bonds
(iii) 3 double bonds; 12 single bonds
(iv) Zero double bonds; 12 single bonds

Answer:

The answer is the option (i) 3 double bonds; 9 single bonds
In Cyclotrimetaphosphoric acid, there are 3 double bonds and 9 single bonds as present

Question 4. Which of the following elements can be involved in $p\pi -d\pi$ bonding?
(i) Carbon
(ii) Nitrogen
(iii) Phosphorus
(iv) Boron
Answer:

The answer is the option (iii) Phosphorus
The involvement of phosphorous in pπ-dπ bonding is due to presence of vacant d orbitals. The elements C, N, and B do not have vacant d orbitals.

Question 5. Which of the following pairs of ions are isoelectronic and isostructural?
(i) $CO_{3}^{2-}, NO^{-}_{3}$
(ii) $ClO_{3}^{-} , CO_{3}^{2-}$
(iii) $SO_{3}^{2-},NO^{-}_{3}$
(iv) $ClO^{-}_{3} ,SO_{3}^{2-}$
Answer:

The answer is the option (i).

Isoelectronic means that two ions or molecules have the same number of electrons. Isostructural means they have the same shape or geometry. The number of electrons in both the isoelectronic species is 32 and they exhibit $sp^{2}$ hybridisation and thus trigonal planar structure.

Question 6. Affinity for hydrogen decreases in the group from fluorine to iodine. Which of the halogen acids should have the highest bond dissociation enthalpy?
$(i) HF$
$(ii) HCl$
$(iii) HBr$
$(iv) HI$
Answer:

The answer is the option (i).
On moving down the group, there is an increase in the atomic radii and a decrease in bond dissociation enthalpy. That is why; $HF$ should have the highest bond dissociation enthalpy.

Question 7. Bond dissociation enthalpy of $E-H \left ( E = element \right )$ bonds are given below. Which of the compounds will act as the strongest reducing agent?

(i) $NH_{3}$
(ii) $PH_{3}$
(iii) $AsH_{3}$
(iv) $SbH_{3}$

Answer:

The answer is the option (iv). The bond dissociation enthalpy decreases down the group. $SbH_{3}$ has the least bond enthalpy and thus acts as the strongest reducing agent.

Question 8. On heating with concentrated $\text {NaOH}$ solution in an inert atmosphere of $\text {CO}_{2}$, white phosphorus gives a gas. Which of the following statement is incorrect about the gas?
(i) It is highly poisonous and has smelled like rotten fish.
(ii) It’s a solution in water decomposes in the presence of light.
(iii) It is more basic than $\text {NH}_{3}$
(iv) It is less basic than $\text {NH}_{3}$
Answer:

The answer is the option (c)

$P_{4}+3NaOH+3H_{2}O\rightarrow PH_{3} \left ( Phosphine \right )+3NaH_{2}PO_{2}$
$PH_{3}$ is less basic than $NH_{3}$

Question 9. Which of the following acids forms three series of salts?
(i) $H_{3}PO_{2}$
(ii) $H_{3}BO_{3}$
(iii) $H_{3}PO_{4}$
(iv) $H_{3}PO_{3}$
Answer:

The answer is the option (iii) $H_{3}PO_{4}$
There are 3 ionisable H-atoms in $H_{3}PO_{4}$ as it has 3- $OH$ groups.

Thus, it forms three series of salts are as follows: .
$NaH_{2}PO_{4}, NaHPO_{4}\; and \; Na_{3}PO_{4}$.

Question 10. Strong reducing behaviour of $H_{3}PO_{2}$ is due to
(i) The low oxidation state of phosphorus
(ii) Presence of two $-OH$ groups and one $P-H$ bond
(iii) Presence of one $-OH$ group and two $P-H$ bonds
(iv) High electron gain enthalpy of phosphorus

Answer:

The answer is the option (iii). In the monobasic acid $H_{3}PO_{2}$, there are two $P-H$ bonds that impart reducing character to the acid.

Question 11. On heating lead, nitrate forms oxides of nitrogen and lead. The oxides formed are ______.
(i) $N_{2}O,PbO$
(ii) $NO_{2},PbO$
(iii) $NO,PbO$
(iv) $NO,PbO_{2}$
Answer:

The answer is the option (ii) $2Pb \left ( NO_{3} \right )_{2}\rightarrow 2PbO+4NO_{2}+O_{2}$

Question 12. Which of the following elements does not show allotropy?
(i) Nitrogen
(ii) Bismuth
(iii) Antimony
(iv) Arsenic
Answer:

The answer is the option (i). Nitrogen has weak N – N single bond because of high inter-electronic repulsion of non-bonding electrons. This is why, nitrogen does not exhibit allotropy.

Question 13. Maximum covalency of nitrogen is ______________.
(i) 3
(ii) 5
(iii) 4
(iv) 6

Answer:

The answer is the option (iii). Maximum covalency of nitrogen is four as it cannot extend its valency beyond four $\left [ NH_{4},R_{4}N \right ]$ due to absence of vacant d orbitals.

Question 14. Which of the following statements is wrong?
(i) Single $N-N$ bond is stronger than the single $P-P$ bond.
(ii) $PH_{3}$ can act as a ligand in the formation of a coordination compound with transition elements.
(iii) $NO_{2}$ is paramagnetic.
(iv) Covalency of nitrogen in $N_{2}O_{5}$ is four
Answer:

The answer is the option (i).Single $P-P$ bond is stronger than single $N-N$ bond, however, naturally occurring nitrogen forms $p(pi)-p(pi)$ multiple bonds and becomes much more stable than phosphorous.

Question 15. A brown ring is formed in the ring test for $NO_{3}^{-}$ ion. It is due to the formation of
(i) $\left [ Fe\left ( H_{2}O \right )_{5} \left ( NO \right ) \right ]^{2+}$
(ii) $FeSO_{4}.NO_{2}$
(iii) $\left [Fe\left ( H_{2}O \right )_{4} \left ( NO \right )_{2} \right]^{2+}$
(iv) $FeSO_{4}.HNO_{3}$

Answer:

The answer is the option (i). On addition of freshly prepared solution of $FeSO_{4}$ in a $NO_{3}^{-}$ ion containing solution, a brown coloured complex is formed. This is also known as brown ring test of nitrate.

$NO_{3}^{-}+3Fe^{2+}+4H^{+}\rightarrow NO+3Fe^{3+}+2H_{2}O$

$\left [ Fe\left ( H_{2}O \right )_{6} \right ]^{2+}+NO\rightarrow \left [ Fe\left ( H_{2}O \right )_{5}\left ( NO \right ) \right ]^{2+}\left ( Brown\; ring \right )+H_{2}O$

Hence, 2 moles of $NO$ are produced from 2 moles of ammonia.

Question 16. Elements of group-15 form compounds in +5 oxidation state. However, bismuth forms only one well characterised compound in +5 oxidation state. The compound is
(i) $Bi_{2}O_{5}$
(ii) $BiF_{5}$
(iii) $BiCl_{5}$
(iv) $Bi_{2}S_{5}$
Answer:

The answer is the option (ii). Due to inert pair effect, bismuth generally exhibits +3 oxidation state. However, due to the small size and high electronegativity of the fluoride ion it forms the compound $BiF_{3}$.

Question 17. On heating ammonium dichromate and barium azide separately we get
(i) $N_{2}$ in both cases
(ii) $N_{2}$ with ammonium dichromate and $NO$ with barium azide
(iii) $N_{2}O$ with ammonium dichromate and $N_{2}$ with barium azide
(iv) $N_{2}O$ with ammonium dichromate and $NO_{2}$ with barium azide
Answer:

The answer is the option (i)
$\left ( NH_{4} \right )_{2}Cr_{2}O_{7}\rightarrow N_{2}+Cr_{2}O_{3}+4H_{2}O$
$Ba \left ( N_{3} \right )_{2}\rightarrow 3N_{2}+Ba$

Question 18. In the preparation of $HNO_{3}$, we get $NO$ gas by catalytic oxidation of ammonia. The moles of $NO$ produced by the oxidation of two moles of $NH_{3}$ will be ______.
(i) 2
(ii) 3
(iii) 4
(iv) 6
Answer:

The answer is the option (i). In the process of preparation of nitric acid, catalytic oxidation of 2 moles of ammonia produces 2 moles of $NO.$
$4NH_{3}+5O_{2}\overset{\Delta }{\rightarrow}4NO \left ( g \right )+6H_{2}O \left ( l \right )$
Hence, 2 moles of ammonia will produce 2 moles of $NO$.

Question 19. The oxidation state of a central atom in the anion of compound $NaH_{2}PO_{2}$ will be ______.
(i) +3
(ii) +5
(iii) +1
(iv) –3
Answer:

The answer is the option (c)

$NaH_{2}PO_{2} :$

$\left ( +1 \right )+\left ( +1 \times 2 \right )+x+\left ( -2 \times 2 \right )=0$
$x=+1$

Question 20. Which of the following is not tetrahedral in shape ?
(i) $NH^{+}_{4}$
(ii) $SiCl_{4}$
(iii) $SF_{4}$
(iv) $SO_{4}^{2-}$
Answer:

The answer is the option (iii). $SF_{4}$ is not tetrahedral in shape. It is a see saw structure.

Question 21. Which of the following are peroxoacids of sulphur?
(i) $H_{2}SO_{5}\; and \; H_{2}S_{2}O_{8}$
(ii) $H_{2}SO_{5}\; and \; H_{2}S_{2}O_{7}$
(iii) $H_{2}S_{2}O_{7} \; and \; H_{2}S_{2}O_{8}$
(iv) $H_{2}S_{2}O_{6} \; and \; H_{2}S_{2}O_{7}$
Answer:

The answer is the option (i).Peroxoacids of sulphur should contain one $-O-O-$ bond as shown below

Question 22. Hot conc. $H_{2}SO_{4}$ acts as a moderately strong oxidising agent. It oxidises both metals and nonmetals. Which of the following element is oxidised by conc. $H_{2}SO_{4}$ into two gaseous products?
(i) $Cu$
(ii) $S$
(iii) $C$
(iv) $Zn$
Answer:

The answer is the option (c)

$\left [ H_{2}SO_{4}\rightarrow H_{2}O+SO_{2}+O \right ]\times 2$
$C+2O\rightarrow CO_{2}$
$C+2H_{2}SO_{4}\rightarrow CO_{2}+2SO_{2}+2H_{2}O$
$CO_{2}$ and $SO_{2}$ are two gaseous products formed by oxidation of $C$ by $H_{2}SO_{4}$

Question 23. A black compound of manganese reacts with a halogen acid to give greenish yellow gas. When an excess of this gas reacts with $NH_{3}$ an unstable trihalide is formed. In this process the oxidation state of nitrogen changes from _________.
(i) – 3 to +3
(ii) – 3 to 0
(iii) – 3 to +5
(iv) 0 to – 3
Answer:

The answer is the option (i).

$MnO_{2}+4HCl\rightarrow MnCl_{2}+2H_{2}O+Cl$ (greenish yellow gas)

on reaction of excess of $Cl_{2}$ with $NH_{3}$ we obtained $NCl_{3}$ and $HCl.$

$NH_{3}+3Cl_{2}\rightarrow NCl_{3}+3HCl$
O.S. (-3) O.S.(+3)

Question 24. In the preparation of compounds of $Xe$, Bartlett had taken $O_{2}^+Pt F^{-}_{6}$ as a base compound. This is because
(i) both $O_{2}$ and $Xe$ has the same size.
(ii) both $O_{2}$ and $Xe$ has the same electron gain enthalpy.
(iii) both $O_{2}$ and $Xe$ has almost same ionisation enthalpy.
(iv) both $Xe$ and $O_{2}$ are gases
Answer:

The answer is the option (iii). The ionization enthalpy of $O_{2}$ is nearly equal to that of $Xe$, because of which Barlett chose $O_{2}$ compounds as base.

Question 25. In solid state $PCl_{5}$ is a _________.
(i) covalent solid
(ii) octahedral structure
(iii) ionic solid with $\left [ PCl_{6} \right ]^{+}$ octahedral and $\left [ PCl_{4} \right ]^{-}$ tetrahedra
(iv) ionic solid with $\left [ PCl_{4} \right ]^{+}$tetrahedral and $\left [ PCl_{6} \right ]^{-}$ octahedral
Answer:

The answer is the option (iv). Structure of $PCl_{5}$ in solid state

Question 26.

Reduction potentials of some ions are given below. Arrange them in decreasing order of oxidising power.

(i) $ClO_{4}^{-} > IO_{4}^{-} > BrO_{4}^{-}$
(ii) $IO_{4}^{-} > BrO_{4}^{-}>ClO_{4}^{-}$
(iii) $BrO_{4}^{-} > IO_{4}^{-} >ClO_{4}^{-}$
(iv) $BrO_{4}^{-} > ClO_{4}^{-} >IO_{4}^{-}$

Answer:

The answer is the option (iii).A higher reduction potential means that there is greater tendency to get reduced. Therefore, the order of oxidizing power is as follows
$BrO_{4}^{-} > IO_{4}^{-} >ClO_{4}^{-}$

Question 27. Which of the following is isoelectronic pair?
(i) $ICl_{2},ClO_{2}$
(ii) $BrO_{2}^{-}, BrF_{2}^{+}$
(iii) $ClO_{2}, BrF$
(iv) $CN^{-}, O_{3}$
Answer:

The answer is the option (ii). Isoelectronic pair have same number of electrons
$(i)ICl_{2}=53+2\times 17$

$=87$
$ClO_{2}=17+16$

$=33$
$(ii)BrO_{2}^{-}=35+2\times 8+1$

$=52$
$BrF_{2}^{+}=35+9\times2-1 $

$= 52$
$(iii)\; ClO_{2}=17+16$

$=33$
$BrF=35+9$

$=44$
$(iv)CN^{-}=6+7+1$

$=14$
$O_{3}=8\times 3$

$=24$
Hence, option b is correct.

NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: MCQ (Type 2)

All the MCQ (type 2) questions with solutions are given below:

Question 28. If chlorine gas is passed through a hot $NaOH$ solution, two changes are observed in the oxidation number of chlorine during the reaction. These are ________ and _________.
(i) 0 to +5
(ii) 0 to +3
(iii) 0 to –1
(iv) 0 to +1
Answer:

The answer is the option (i) and (iii). The oxidation state changes from 0 to +5 and 0 to -1 respectively.

Question 29. Which of the following options are not in accordance with the property mentioned against them?
(i) $F_{2}> Cl_{2}> Br_{2}> I_{2}$ Oxidising power.
(ii) $MI > MBr > MCl > MF$ Ionic character of metal halide.
(iii) $F_{2}> Cl_{2}> Br_{2}> I_{2}$ Bond dissociation enthalpy.
(iv) $HI < HBr < HCl < HF$ Hydrogen-halogen bond strength.
Answer:

The answer is the option (ii, iii) The correct order of bond dissociation enthalpy is $Cl_{2}>Br_{2}>F_{2}>I_{2}$
There is an increase in the ionic character of metal halides with the electronegativity of halogen
$MI > MBr > MCI > MF$

Question 30. Which of the following is correct for P₄ the molecule of white phosphorus?
(i) It has 6 lone pairs of electrons.
(ii) It has six P–P single bonds.
(iii) It has three P–P single bonds.
(iv) It has four lone pairs of electrons.
Answer:

The answer is the option (ii, iv) Structure of $P_{4}$ molecule can be represented as

There are 4 lone pairs of electrons over each P-atom. It has six P – P single bond.

Question 31. Which of the following statements are correct?
(i) Among halogens, radius ratio between iodine and fluorine is maximum.
(ii) Leaving $F-F$ bond, all halogens have weaker $X-X$ bond than $X-X'$ bond in interhalogens.
(iii) Among interhalogen compounds, the maximum number of atoms are present in iodine fluoride.
(iv) Interhalogen compounds are more reactive than halogen compounds.
Answer:

The answer is the option (i, iii, iv)
Iodine is the largest halogen, while fluorine is the smallest. $\frac{R_{Iodine}}{R_{fluorine}}$ is thus the maximum. A-B bond in interhalogen compound is weaker than A-A bond in halogen compounds; this also means that interhalogen compounds are more reactive.

Question 32. Which of the following statements are correct for $SO_{2}$ gas?
(i) It acts as a bleaching agent in moist conditions.
(ii) Its molecule has linear geometry.
(iii) Its dilute solution is used as a disinfectant.
(iv) It can be prepared by the reaction of dilute $H_{2}SO_{4}$ with metal sulphide.
Answer:

The answer is the option (i, iii). $SO_{2}$ is used as a bleaching agent in presence of moisture because of the reducing nature of $SO_{2}$
$SO_{2}+2H_{2}O\rightarrow H_{2}SO_{4}+2H$
It is also used as a preservative, disinfectant and anti-chlor.

Question 33. Which of the following statements are correct?
(i) All the three N—O bond lengths in $HNO_{3}$ are equal.
(ii) All P—Cl bond lengths in $PCl_{5}$ molecule in the gaseous state are equal.
(iii) $P_{4}$ molecule in white phosphorus have angular strain therefore white phosphorus is very reactive.
(iv) $PCl$ is ionic in the solid-state in which cation is tetrahedral and the anion is octahedral.
Answer:

The answer is the option (iii, iv).
(i) All the N – O bond in $HNO_{3}$ are not of the same bond length.
(ii) In gaseous state, all P – Cl bond lengths in $PCl_{5}$ molecule are not equal. Axial bond is longer than equatorial bond.
(iii) White phosphorous is more reactive than the other solid phases because of angular strain in the $P_{4}$ molecule under normal conditions.
(iv) In Solid state $PCl_{5}$ exists as ionic solid in which cation is tetrahedral and anion is octahedral.
$Cation-\left [ PCl_{4} \right ]^{+}\; \; \; \; \; \; Anion-\left [ PCl_{6} \right ]^{-}$

Question 34. Which of the following orders are correct as per the properties mentioned against each?
(i) $As_{2}O_{3}<SiO_{2}<P_{2}O_{3}<SO_{2}$ Acid strength.
(ii) $AsH_{3}<PH_{3}<NH_{3}$ Enthalpy of vaporization.
(iii) $S < O < Cl < F$ More negative electron gain enthalpy.
(iv) $H_{2}O >H_{2}S>H_{2}Se>H_{2}Te$ Thermal stability.

Answer:

The answer is the option (i, iv) There is an increase in the acidic strength of the group as we move up the group or left to right along a period. As we move down the group, thermal stability of hydrides decreases.

Question 35. Which of the following statements are correct?
(i) S–S bond is present in $H_{2}S_{2}O_{6}$
(ii) In peroxosulphuric acid $\left ( H_{2}SO_{5} \right )$ sulphur is in +6 oxidation state.
(iii) Iron powder along with $Al_{2}O_{3}$ and $K_{2}O$ is used as a catalyst in the preparation of $NH_{3}$ by Haber’s process.
(iv) Change in enthalpy is positive for the preparation of $SO_{3}$ by catalytic oxidation of $SO_{2}$
Answer:

The answer is the option (i, ii)
In $H_{2}SO_{5}$, there is a peroxo-linkage.

$O$ in peroxide linkage has oxidation state - 1

Question 36. In which of the following reactions conc. $H_{2}SO_{4}$ is used as an oxidising reagent?
(i) $CaF_{2}+ H_{2}SO_{4} \rightarrow CaSO_{4}+ 2HF$
(ii) $2HI + H_{2}SO_{4} \rightarrow I_{2} + SO_{2}+ 2H_{2}O$
(iii) $Cu + H_{2}SO_{4} \rightarrow CuSO_{4} + SO_{2}+ 2H_{2}O$
(iv) $NaCl + H_{2}SO_{4} \rightarrow NaHSO_{4} + HCl$
Answer:

The answer is the option (ii, iii). Amongst the given options, (ii). and (iii). represent oxidizing behaviour of $H_{2}SO_{4}$. The oxidizing agent reduces itself in the process. In option b, it oxidises HI and gets reduced to $SO_{2}$ .Here,

Question 37. Which of the following statements are true?
(i) The only type of interactions between particles of noble gases is due to weak dispersion forces.
(ii) Ionisation enthalpy of molecular oxygen is very close to that of xenon.
(iii) Hydrolysis of XeF₆ is a redox reaction.
(iv) Xenon fluorides are not reactive.
Answer:

The answer is the option (i, ii). Weak dispersion forces are responsible for attraction in noble gases. Ionization enthalpy of molecular oxygen and xenon are very similar. Also, Xenon fluorides are reactive in nature.

NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: Short Answer Type

All the short-answer type questions with solutions are given below:

Question 38. In the preparation of $H_{2}SO_{4}$ by Contact Process, why is $SO_{3}$ not absorbed directly in water to form $H_{2}SO_{4}$ ?
Answer:

Formation of acid fog takes place which is difficult to condense.

Question 39. Write a balanced chemical equation for the reaction showing catalytic oxidation of $NH_{3}$ by atmospheric oxygen.
Answer:

$4NH_{3}+5O_{2}\left ( from\; air \right )\overset{Pt\; gauge}{\rightarrow}4NO+6H_{2}O$

Question 40. Write the structure of pyrophosphoric acid.

Answer:

$H_{4}P_{2}O_{7}$ is pyrophosphoric acid. Its structure is

Question 41. $PH_{3}$ forms bubbles when passed slowly in water but $NH_{3}$ dissolves. Explain why?
Answer:

$NH_{3}$ dissolves in water as it forms hydrogen bonds with water but $PH_{3}$ cannot form hydrogen bonds with water so it escapes as gas.

Question 42. In $PCl_{5}$, phosphorus is in $sp^{3}d$ hybridised state, but all its five bonds are not equivalent. Justify your answer with reason.
Answer:

$PCl_{5}$ has a trigonal bipyramidal structure in liquid and gaseous phases.

The 3 equatorial P–Cl bonds are equivalent whereas 2 axial bonds are longer than equatorial bonds. This is because axial bond pairs suffer more repulsion compared to the equatorial bond pairs.

Question 43. Why is nitric oxide paramagnetic in gaseous state but the solid obtained on cooling it is diamagnetic ?
Answer:

In the gaseous state, nitric oxide $\left ( NO \right )$ exists as a monomer having one unpaired electron as it is paramagnetic in nature.

However, in solid state it dimerises to $N_{2}O_{2}$ which is diamagnetic in nature.

Question 44. Give reason to explain why $ClF_{3}$ exists but $FCl_{3}$ does not exist.
Answer:

Though both $Cl \; and \; F$ are halogens, $F$ has a configuration $2s^{2}2p^{5}$ and doesn’t have a d-orbital. Chlorine on the other hand can undergo $sp^{3}d$-hybridisation, where three orbitals will have unpaired electrons as shown below. Three fluorine atoms with a single unpaired electron each can pair up with the $sp^{3}d$-hybridised chlorine atom.

Question 45. Out of $H_{2}O$ and $H_{2}S$, which one has a higher bond angle and why?
Answer:

Due to higher electronegativity of Oxygen than Sulphur, O-H bond will have a higher electron density closer to the Oxygen atom. The two O-H bonds will repel each other more strongly than S-H bond leading to a larger bond angle in $H_{2}O$.

Question 46. $SF_{6}$ is known but $SCl_{6}$ is not. Why?
Answer:

There are two reasons as to why $SF_{6}$ exists in nature, but $SCl_{6}$ doesn’t:

• Fluorine is smaller in size than Chlorine and is easily accommodated around Sulphur atom.

• Due to higher electronegativity possessed by fluorine, it can unpair the Sulphur atom to the highest degree (where it attains the oxidation state of +6).

Question 47. On reaction with $Cl_{2}$, phosphorus forms two types of halides ‘A’ and ‘B’. Halide A is a yellowish-white powder, but halide ‘B’ is a colourless oily liquid. Identify A and B and write the formulas of their hydrolysis products.
Answer:

On passing dry chlorine over white phosphorus, first colourless oily phosphorus trichloride is formed. $PCl_{3}$ again reacts with $Cl_{2}$ and forms yellowish white powder like phosphorus pentachloride $PCl_{5}$

Question 48. In the ring test of $NO^{-}_{3}$ ion, $Fe^{2+}$ ion reduces nitrate ion to nitric oxide, which combines with $Fe^{2+}(aq)$ ion to form the brown complex. Write the reactions involved in the formation of a brown ring.
Answer:

$NO_{3}^{-}+3Fe^{2+}+4H^{+}\rightarrow NO+3Fe^{3+}+2H_{2}O$

$\left [ Fe\left ( H_{2}O \right )_{6} \right ]SO_{4}+NO\rightarrow \left [ Fe\left ( H_{2}O \right )_{5}NO \right ]SO_{4}\; \left ( Brown \; Complex \right )+H_{2}O$

Question 49. Explain why the stability of oxoacids of chlorine increases in the order given below:
$HClO< HClO_{2}< HClO_{3}< HClO_{4}$
Answer:

As number of oxygen atoms around chlorine increases, the electron density moves away from chlorine. It is so because, oxygen atoms are more electronegative than chlorine atoms.
The stability of ions will also increase with a higher number of Oxygen atoms:
$ClO^{-}<ClO_{2}^{-}<ClO_{3}^{-}<ClO_{4}^{-}$
Thus due to increase in stability of conjugate base, acidic strength of corresponding acid increases in the following order
$HClO< HClO_{2}< HClO_{3}< HClO_{4}$

Question 50. Explain why ozone is thermodynamically less stable than oxygen.
Answer:

Ozone is thermodynamically unstable with respect to oxygen because its decomposition results in liberation of heat $\left ( \Delta H=-ve \right )$ and increase in entropy $\left ( \Delta H=+ve \right )$
$2O_{3}\rightleftharpoons 3O_{2}$
These 2 factors reinforce each other resulting in large negative Gibbs free energy change $\left ( \Delta G=-ve \right )$ for its conversion into oxygen.
Therefore, the high concentration of ozone can result into dangerous explosion

Question 51. $P_{4}O_{6}$ reacts with water according to equation $P_{4}O_{6}+6H_{2}O\rightarrow 4H_{3}PO_{3}.$
Calculate the volume of 0.1 M $NaOH$ solution required to neutralise the acid formed by dissolving 1.1 g of $P_{4}O_{6}$ in $H_{2}O$.
Answer:

$P_{4}O_{6}+6H_{2}O\rightarrow 4H_{3}PO_{3}$

$4H_{3}PO_{3}+8NaOH\rightarrow 4Na_{2}HPO_{3}+8H_{2}O$

($H_{3}PO_{3}$ can be neutralised with $NaOH$ )

$P_{4}O_{6}(1 \; mol)+8NaOH (8\; mol)\rightarrow 4Na_{2}HPO_{3}+2H_{2}O$
Moles of $P_{4}O_{6} = \frac{1.1}{220}$

$=0.005\; mol$

Acid formed by 1 mol of $P_{4}O_{6}$ Require 8 mol $NaOH$

Acid formed by 0.005 mol of $P_{4}O_{6}$ required $=8\times 0.005$

$=0.04\; mol$

0.1 M $NaOH$ means 0.1 mol of $NaOH$ is present in 1000 mL solution

0.4 M of $NaOH$ is present in solution $=\frac{1000}{0.1}\times 0.04$

$=400mL$

Question 52. White phosphorus reacts with chlorine and the product hydrolyses in the presence of water. Calculate the mass of $HCl$ obtained by the hydrolysis of the product formed by the reaction of 62 g of white phosphorus with chlorine in the presence of water.
Answer:

Phosphorus reacts with $Cl_{2}$ as :

$P_{4}+6Cl_{2}\rightarrow 4PCl_{3}$

$PCl_{3}+3H_{2}O\rightarrow H_{3}PO_{3}+3HCl] \times 4 \left ( PCl_{3}\; is \; hydrolysed \right )$

$P_{4}+6Cl_{2}+12H_{2}O\rightarrow 4H_{3}PO_{3}+12HCl$

Moles of white $P=\frac{62}{124}$

$=0.5\; mol$

1 mol of white $P_{4}$ produce $HCl=12\; mol$

0.5 mol of white $P_{4}$ will produce $HCl=12\; \times 0.5$

$=6\; mol$

Mass of $HCl=6\times 36.5$

$=219.0\; g$

Question 53. Name three oxoacids of nitrogen. Write the disproportionation reaction of that oxoacid of nitrogen in which nitrogen is in +3 oxidation state.

Answer:

The 3 oxyacids of nitrogen are:

1. Nitric acid $\left ( HNO_{3} \right )$

2. Nitrous acid $\left ( HNO_{2} \right )$

3. Hyponitrous acid $\left ( H_{2}N_{2}O_{2} \right )$

The disproportionation reaction wherein the nitrogen is in +3 oxidation state is:
$\\3HNO_{2}\rightarrow HNO_{3}+2NO+H_{2}O\\$

$+3\; \; \; \; \; \; \; \; \; \; \; \; +5\; \; \; \; \; \; \; \; \; +2$
Oxidation state changes from +3 to +5 and +2

Question 54. Nitric acid forms an oxide of nitrogen on reaction with $P_{4}O_{10}$. Write the reaction involved. Also, write the resonating structures of the oxide of nitrogen formed.

Answer:

$P_{4}O_{10}+4NHO_{3}\rightarrow 4HPO_{3}+2N_{2}O_{5}$

Reactivity: White phosphorous are more reactive compared to the Red phosphorous as it undergoes more angular strain since the bonds are angled at 60°.

Question 55. Phosphorus has three allotropic forms — (i) white phosphorus (ii) red phosphorus and (iii) black phosphorus. Write the difference between white and red phosphorus on the basis of their structure and reactivity.
Answer:

White phosphorous exists as a single molecule where the phosphorous atoms are $sp^{3}$-hybridised and has a tetrahedral structure with 6 P-P bonds. It is highly reactive due to the strained bond angles (60^o).

Red Phosphorous on the other hands also has a tetrahedral structure, but each tetrahedron is linked to two others as shown below. The bond angle is much more relaxed, leading to a lower reactivity than White phosphorus.

Question 56. Give an example to show the effect of concentration of nitric acid on the formation of oxidation product.

Answer:

We get different oxidation products on reaction of dilute and concentrated nitric acid with copper metals.

For example,

zinc with very dilute

$HNO_{3}$ ( 6%) forms $NH_{4}NO_{3},$

with dilute

$HNO_{3}$ ( 20%) forms $N_{2}O$

and with concentration.

$HNO_{3}$ ( 70%) forms $NO_{2}.$

(Very dilute) $HNO_{3}:4Zn+10HNO_{3}\rightarrow 4Zn(NO_{3})_{2}+NH_{4}NO_{3}+3H_{2}O$

(Dilute) $HNO_{3}:4Zn+10HNO_{3}\rightarrow 4Zn(NO_{3})_{2}+N_{2}O+5H_{2}O$

(Concentrated) $HNO_{3}:Zn+HNO_{3}\rightarrow Zn(NO_{3})_{2}+2NO_{2}+2H_{2}O$

Question 57. $PCl_{5}$ reacts with finely divided silver on heating and a white silver salt is obtained, which dissolves on adding excess aqueous $NH_{3}$ solution. Write the reactions involved to explain what happens.

Answer:

$PCl_{5}$ reacts with finely divided silver on heating to form white silver salt $(AgCl)$. This dissolves on addition of excess aqueous ammonia to form soluble complex.
$PCl_{5}+2Ag\rightarrow 2AgCl+PCl_{3}$
$AgCl+2NH _3(aq)\rightarrow [Ag(NH_3 )_ 2]+Cl ^{-}$

Question 58. Phosphorus forms a number of oxoacids. Out of these oxoacidsphosphinic acid has strong reducing property. Write its structure and also write a reaction showing its reducing behaviour.
Answer:

One of the oxoacids of phosphorous that has strong reducing property is phosphinic acid or hypophosphorus acid $\left ( H_{3}PO_{2} \right )$. This is due to possession of P – H bonds.

Its reducing properties are :

$4AgNO_{3}+H_{3}PO_{2}+2H_{2}O\rightarrow 4Ag+H_{3}PO_{4}+4HNO_{3}$

$2HgCl_{2}+H_{3}PO_{2}+2H_{2}O\rightarrow 2Hg+H_{3}PO_{4}+4HCl$

$H_{3}PO_{2}+2H_{2}O+2Cl_{2}\rightarrow H_{3}PO_{4}+4HCl$

NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: Matching Type

All the matching type questions with solutions are given below:

Question 59. Match the compounds given in Column I with the hybridization and shape given in Column II and mark the correct option.

(a) A (1) B(3) C(4) D(2)
(b) A (1) B(2) C(4) D(3)
(c) A (4) B(3) C(1) D(2)
(d) A (4) B(1) C(2) D(3)

Answer:

(a) $(A\rightarrow 1),(B\rightarrow 3),(C\rightarrow 4),(D\rightarrow 2)$

Question 60. Match the formulas of oxides given in Column I with the type of oxide given in Column II and mark the correct option.

(a) A (1) B(2) C(3) D(4)
(b) A (4) B(1) C(2) D(3)
(c) A (3) B(2) C(4) D(1)
(d) A (4) B(3) C(1) D(2)

Answer:

(b). (A $\rightarrow$ 4), (B $\rightarrow$ 1), (C $\rightarrow$ 2), (D $\rightarrow$ 3)

Question 61. Match the items of Column I and Column II and mark the correct option.

(a) A (4) B(3) C(1) D(2)
(b) A (3) B(4) C(1) D(2)
(c) A (4) B(1) C(2) D(3)
(d) A (2) B(1) C(3) D(4)

Answer:

(a). (A $\rightarrow$ 4), (B $\rightarrow$ 3), (C $\rightarrow$ 1), (D $\rightarrow$ 2)

Question 62. Match the species given in Column I with the shape given in Column II and mark the correct option.

(a) A (3) B(2) C(1) D(4)
(b) A (3) B(4) C(2) D(1)
(c) A (1) B(2) C(3) D(4)
(d) A (1) B(4) C(3) D(2)

Answer:

(b) (A $\rightarrow$3), (B $\rightarrow$ 4), (C $\rightarrow$ 2), (D $\rightarrow$ 1)

Question 63. Match the items of Column I and Column II and mark the correct option.

(a) A (1) B(4) C(2) D(3)
(b) A (1) B(2) C(3) D(4)
(c) A (2) B(1) C(4) D(3)
(d) A (1) B(3) C(2) D(4)

Answer:

(c). (A $\rightarrow$ 2), (B $\rightarrow$ 1), (C $\rightarrow$ 4), (D $\rightarrow$ 3)

1. Oxidation state of $\text {Xe}$ is unaffected by partial hydrolysis of $\text {XeF}_{6}$

2. He is utilized in the modern driving apparatus.

3. Air is used for filling electrical bulbs to provide inert atmosphere.

4. Xe has 4 bond pair and 2 lone pair and a hybridisation state of $\text {sp}^{3}\text {d}^{2}$ in $\text {XeF}_{4}$

NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: Assertion and Reason Type

All the assertion-reason type questions with solutions are given below:

Question 64. In the following question, a statement of Assertion (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion (I): $N_{2}$ is less reactive than $P_{4}$
Reason (R): Nitrogen has more electron gain enthalpy than phosphorus.
(i) Both Assertion and Reason are correct statements and Reason is the correct explanation of Assertion.
(ii) Both Assertion and Reason are correct statements but Reason is not the correct explanation of Assertion.
(iii) Assertion is correct but the reason is wrong.
(iv) Assertion is wrong but the Reason is correct.
(v) Both Assertion and Reason are wrong.
Answer:

The answer is the option (iii). Nitrogen has a higher electron gain enthalpy than Phosphorous. N-N atoms in Nitrogen are bonded by triple bond, whereas Phosphorous atoms are bonded by single P-P bond, due to which Nitrogen has a high bond dissociation energy and is less reactive.

Question 65. In the following question, a statement of Assertion (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion (I): $HNO_{3}$ makes iron passive.
Reason (R): $HNO_{3}$ forms a protective layer of ferric nitrate on the surface of iron.
(i) Both Assertion and Reason are correct statements and Reason is the correct explanation of Assertion.
(ii) Both Assertion and Reason are correct statements but Reason is not the correct explanation of Assertion.
(iii) Assertion is correct but Beason is wrong.
(iv) Assertion is wrong but Reason is correct.
(v) Both Assertion and Reason are wrong.
Answer:

The answer is the option (iii). Iron attains passivity due to formation of a thin layer of oxide.

Question 66. In the following question, a statement of Assertion (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion (I): $HI$ cannot be prepared by the reaction of $KI$ with concentrated $H_{2}SO_{4}$.
Reason (R): $HI$ has lowest H – X bond strength among halogen acids.
(i) Both Assertion and Reason are correct statements and Reason is the correct explanation of Assertion.
(ii) Both Assertion and Reason are correct statements but Reason is not the correct explanation of Assertion.
(iii) Assertion is correct but Beason is wrong.
(iv) Assertion is wrong but Reason is correct.
(v) Both Assertion and Reason are wrong.
Answer:

The answer is the option (ii). Though correct, both the statements are independent of each other.

Question 67. In the following question, a statement of Assertion (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion (i): Both rhombic and monoclinic sulphur exist as $S_{8}$ but oxygen exists as $O_{2}$.
Reason (R): Oxygen forms $p\pi -p\pi$ multiple bond due to small size and small bond length but $p\pi -p\pi$ bonding is not possible in sulphur.
(i) Both Assertion and Reason are correct statements and Reason is the correct explanation of Assertion.
(ii) Both Assertion and Reason are correct statements but Reason is not the correct explanation of Assertion.
(iii) Assertion is correct but Beason is wrong.
(iv) Assertion is wrong but Reason is correct.
(v) Both Assertion and Reason are wrong.
Answer:

The answer is the option (i). Sulphur cannot form $p\pi -p\pi$ multiple bonds like oxygen, so it exists as $S_{8}$.

Question 68. In the following question, a statement of Assertion (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion (I): $NaCl$ reacts with concentrated $H_{2}SO_{4}$ to give colourless fumes with pungent smell. But on adding $MnO_{2}$ the fumes become greenish yellow.
Reason (R): $MnO_{2}$ oxidises $HCl$ to chlorine gas which is greenish yellow.
(i) Both Assertion and Reason are correct statements and Reason is the correct explanation of Assertion.
(ii) Both Assertion and Reason are correct statements but Reason is not the correct explanation of Assertion.
(iii) Assertion is correct but Beason is wrong.
(iv) Assertion is wrong but Reason is correct.
(v) Both Assertion and Reason are wrong.
Answer:

The answer is the option (i). $MnO_{2}$ is a strong oxidising agent and it oxidises colourless fumes of $HCl$ to greenish yellow $Cl_{2}$

Question 69. In the following question, a statement of Assertion (i). followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion (i): $SF_{6}$ cannot be hydrolysed but $SF_{4}$ can be.
Reason (R): Six $F$ atoms in $SF_{6}$ prevent the attack of $H_{2}O$ on sulphur atom of $SF_{6}$ .
(i) Both Assertion and Reason are correct statements and Reason is the correct explanation of Assertion.
(ii) Both Assertion and Reason are correct statements but Reason is not the correct explanation of Assertion.
(iii) Assertion is correct but Beason is wrong.
(iv) Assertion is wrong but Reason is correct.
(v) Both Assertion and Reason are wrong.
Answer:

The answer is the option (i). $H_{2}O$ can’t react with $SF_{6}$ because the fluorine atoms don’t allow $H_{2}O$ to attack $SF_{6}$, and thus, is sterically protected.

NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: Long Answer Type

All the long-answer type questions with solutions are given below:

Question 70. An amorphous solid “A” burns in air to form a gas “B” which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified aqueous $KMnO_{4}$ solution and reduces $Fe^{3+}$ to $Fe^{+2}$. Identify the solid “A” and the gas “B” and write the reactions involved.

Answer:

A is sulphur $(S_{8})$ while B is sulphur dioxide $(SO_{2})$.
$S_{8}+8O_{2}\rightarrow 8SO_{2}$

$2ZnS+3O_{2}\rightarrow 2ZnO+2SO_{2}$

$2KMnO_{4}+3H_{2}SO_{4}\rightarrow K_{2}SO_{4}+2MnSO_{4}+3H_{2}O+5\left [ O \right ]$

$\left [ SO_{2}+\left [ O \right ]+H_{2}O\rightarrow H_{2}SO_{4} \right ]\times 5$

$2KMnO_{4}+3H_{2}SO_{4}\rightarrow K_{2}SO_{4}+2MnSO_{4}+2H_{2}SO_{4}$

$SO_{2}+2H_{2}O\rightarrow H_{2}SO_{4}+2\left [ H \right ]$

$Fe_{2}\left ( SO_{4} \right )_{3}+2\left [ H \right ]\rightarrow 2FeSO_{4}+H_{2}SO_{4}$

$Fe_{2}\left ( SO_{4} \right )_{3}+SO_{2}+2H_{2}O\rightarrow 2FeSO_{4}+2H_{2}SO_{4}$

Question 71. On heating lead (II) nitrate gives a brown gas “A”. The gas “A” on cooling changes to colourless solid “B”. Solid “B” on heating with NO changes to a blue solid C. Identify A, B and C and also write reactions involved and draw the structures of B and C.
Answer:

The gas A is $NO_{2}$ . The reactions are given as under:
$2Pb\left ( NO_{3} \right )_{2}\overset{Heat}{\rightarrow}2PbO+4NO_{2}+O_{2}$

$2NO_{2}\overset{Cooling}{\rightarrow}N_{2}O_{4}$

$2NO+N_{2}O_{4}\overset{Heat}{\rightarrow}2N_{2}O_{3}$

Question 72. On heating compound (A) gives a gas (B) which is a constituent of air. This gas when treated with 3 mol of hydrogen $(H_{2})$ in the presence of a catalyst gives another gas (C) which is basic in nature. Gas $C$ on further oxidation in moist condition gives a compound (D) which is a part of acid rain. Identify compounds (A) to (D) and also give necessary equations of all the steps involved.
Answer:

$(A)=NH_{4}NO_{2},\; \; (B)=N_{2},\; \; \; \; (C)=NH_{3},\; \; \; \; \; (D)=HNO_{3}$
$NH_{4}NO_{2}\overset{Heat}{\rightarrow}N_{2}+2H_{2}O$

$N_{2}+3H_{2}\overset{Catalyst}{\rightarrow}2NH_{3}$

$4NH_{3}+5O_{2}\rightarrow 4NO+6H_{2}O$

$4NO+O_{2}\rightarrow 2NO_{2}$

$3NO_{2}+H_{2}O\rightarrow 2HNO_{3}+NO$