NCERT Exemplar Class 12 Chemistry Solutions Chapter 7 p block elements

NCERT Exemplar Class 12 Chemistry Solutions Chapter 7: MCQ (Type 1)

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.
${\text{2NaCl + H}}_2{\text{SO}}_4\to {\text{Na}}_2{\text{SO}}_4+\text{2HCl}$
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.
${\text{2NaI + H}}_2{\text{SO}}_4\to {\text{Na}}_2{\text{SO}}_4+\text{2HI}\stackrel{{\text{H}}_2{\text{SO}}_4}{\to }{\text{2H}}_2{\text{O + SO}}_2+{\text{I}}_2$

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 $[Cu(N{H}_3{)}_4{]}^{2+}$
(iii) a deep blue solution of $Cu(N{O}_3{)}_2$
(iv) a deep blue solution of $Cu(OH{)}_2.Cu(N{O}_3{)}_2$
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 $HN{O}_3$, 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:
$C{u}^{2+}+H_{2}S\to CuS(black)+2H^{+}$
$CuS+dil.HN{O}_3\to Cu{\left(N{O}_3\right)}_2$
$Cu{\left(N{O}_3\right)}_2+4N{H}_3\to {\left[Cu{\left(N{H}_3\right)}_4\right]}^{2+}(Deepblue)$

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) $C{O}_3^{2-},N{O}_3^{-}$
(ii) $Cl{O}_3^{-},C{O}_3^{2-}$
(iii) $S{O}_3^{2-},N{O}_3^{-}$
(iv) $Cl{O}_3^{-},S{O}_3^{2-}$
Answer:

The answer is the option (i). The number of electrons in both the isoelectronic species is 32 and they exhibit $s{p}^2$ 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 $E-H(E=element)$ bonds are given below. Which of the compounds will act as the strongest reducing agent?

(i) $N{H}_3$
(ii) $P{H}_3$
(iii) $As{H}_3$
(iv) $Sb{H}_3$

Answer:

The answer is the option (iv). The bond dissociation enthalpy decreases down the group. $Sb{H}_3$ 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) $P_4+3NaOH+3H_{2}O\to P{H}_3\left(Phosphine\right)+3Na{H}_{2}P{O}_2$
$P{H}_3$ is less basic than $N{H}_3$

Question:9

Which of the following acids forms three series of salts?
(i) $H_{3}P{O}_2$
(ii) $H_{3}B{O}_3$
(iii) $H_{3}P{O}_4$
(iv) $H_{3}P{O}_3$
Answer:

The answer is the option (iii) $H_{3}P{O}_4$
There are 3 ionisable H-atoms in $H_{3}P{O}_4$ as it has 3- $OH$ groups. Thus, it forms three series of salts are as follows: .
$Na{H}_{2}P{O}_4,NaHP{O}_{4}andN{a}_{3}P{O}_4$.

Question:10

Strong reducing behaviour of $H_{3}P{O}_2$ 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 $H_{3}P{O}_2$, 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) $N{O}_2,PbO$
(iii) $NO,PbO$
(iv) $NO,Pb{O}_2$
Answer:

The answer is the option (ii) $2Pb{\left(N{O}_3\right)}_2\to 2PbO+4N{O}_2+O_2$

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 $[N{H}_4,R_{4}N]$ 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) $P{H}_3$ can act as a ligand in the formation of a coordination compound with transition elements.
(iii) $N{O}_2$ 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 $N{O}_3^{-}$ ion. It is due to the formation of
(i) ${\left[Fe{\left(H_{2}O\right)}_5\left(NO\right)\right]}^{2+}$
(ii) $FeS{O}_4.N{O}_2$
(iii) ${\left[Fe{\left(H_{2}O\right)}_4{\left(NO\right)}_2\right]}^{2+}$
(iv) $FeS{O}_4.HN{O}_3$

Answer:

The answer is the option (i). On addition of freshly prepared solution of $FeS{O}_4$ in a $N{O}_3^{-}$ ion containing solution, a brown coloured complex is formed. This is also known as brown ring test of nitrate.
$N{O}_3^{-}+3F{e}^{2+}+4H^{+}\to NO+3F{e}^{3+}+2H_{2}O$
${\left[Fe{\left(H_{2}O\right)}_6\right]}^{2+}+NO\to {\left[Fe{\left(H_{2}O\right)}_5\left(NO\right)\right]}^{2+}\left(Brownring\right)+H_{2}O$
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) $B{i}_2{O}_5$
(ii) $Bi{F}_5$
(iii) $BiC{l}_5$
(iv) $B{i}_2{S}_5$
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 $Bi{F}_3$.

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 $N{O}_2$ with barium azide
Answer:

The answer is the option (i)
${\left(N{H}_4\right)}_{2}C{r}_2{O}_7\to N_2+C{r}_2{O}_3+4H_{2}O$
$Ba{\left(N_3\right)}_2\to 3N_2+Ba$

Question:18

In the preparation of $HN{O}_3$, we get $NO$ gas by catalytic oxidation of ammonia. The moles of $NO$ produced by the oxidation of two moles of $N{H}_3$ 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 $NO.$
$4N{H}_3+5O_2\stackrel{\mathrm{\Delta }}{\to }4NO\left(g\right)+6H_{2}O\left(l\right)$
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 $Na{H}_{2}P{O}_2$ will be ______.
(i) +3
(ii) +5
(iii) +1
(iv) –3
Answer:

The answer is the option (c) $Na{H}_{2}P{O}_2:(+1)+(+1\times 2)+x+(-2\times 2)=0$
$x=+1$

Question:20

Which of the following is not tetrahedral in shape ?
(i) $N{H}_4^{+}$
(ii) $SiC{l}_4$
(iii) $S{F}_4$
(iv) $S{O}_4^{2-}$
Answer:

The answer is the option (iii). $S{F}_4$ is not tetrahedral in shape. It is a see saw structure.

Question:21

Which of the following are peroxoacids of sulphur?
(i) $H_{2}S{O}_{5}and{H}_2{S}_2{O}_8$
(ii) $H_{2}S{O}_{5}and{H}_2{S}_2{O}_7$
(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.$H_{2}S{O}_4$ acts as a moderately strong oxidising agent. It oxidises both metals and nonmetals. Which of the following element is oxidised by conc. $H_{2}S{O}_4$ into two gaseous products?
(i) $Cu$
(ii) $S$
(iii) $C$
(iv) $Zn$
Answer:

The answer is the option (c) $[H_{2}S{O}_4\to H_{2}O+S{O}_2+O]\times 2$
$C+2O\to C{O}_2$
$C+2H_{2}S{O}_4\to C{O}_2+2S{O}_2+2H_{2}O$
$C{O}_2$ and $S{O}_2$ are two gaseous products formed by oxidation of $C$ by $H_{2}S{O}_4$

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 $N{H}_3$ 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). $Mn{O}_2+4HCl\to MnC{l}_2+2H_{2}O+Cl$ (greenish yellow gas)
on reaction of excess of $C{l}_2$ with $N{H}_3$ we obtained $NC{l}_3$ and $HCl.$
$N{H}_3+3C{l}_2\to NC{l}_3+3HCl$
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 $PC{l}_5$ is a _________.
(i) covalent solid
(ii) octahedral structure
(iii) ionic solid with ${\left[PC{l}_6\right]}^{+}$ octahedral and ${\left[PC{l}_4\right]}^{-}$ tetrahedra
(iv) ionic solid with ${\left[PC{l}_4\right]}^{+}$tetrahedral and ${\left[PC{l}_6\right]}^{-}$ octahedral
Answer:

The answer is the option (iv). Structure of $PC{l}_5$ in solid state

Question:26

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

(i) $Cl{O}_4^{-}>I{O}_4^{-}>Br{O}_4^{-}$
(ii) $I{O}_4^{-}>Br{O}_4^{-}>Cl{O}_4^{-}$
(iii) $Br{O}_4^{-}>I{O}_4^{-}>Cl{O}_4^{-}$
(iv) $Br{O}_4^{-}>Cl{O}_4^{-}>I{O}_4^{-}$

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
$Br{O}_4^{-}>I{O}_4^{-}>Cl{O}_4^{-}$

Question:27

Which of the following is isoelectronic pair?
(i) $IC{l}_2,Cl{O}_2$
(ii) $Br{O}_2^{-},Br{F}_2^{+}$
(iii) $Cl{O}_2,BrF$
(iv) $C{N}^{-},O_3$
Answer:

The answer is the option (ii). Isoelectronic pair have same number of electrons
$(i)IC{l}_2=53+2\times 17=87$
$Cl{O}_2=17+16=33$
$(ii)Br{O}_2^{-}=35+2\times 8+1=52$
$Br{F}_2^{+}=35+9\times 2-1=52$
$(iii)Cl{O}_2=17+16=33$
$BrF=35+9=44$
$(iv)C{N}^{-}=6+7+1=14$
$O_3=8\times 3=24$
Hence, option b is correct.

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

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) $F_2>C{l}_2>B{r}_2>I_2$ Oxidising power.
(ii) $MI>MBr>MCl>MF$ Ionic character of metal halide.
(iii) $F_2>C{l}_2>B{r}_2>I_2$ 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 $C{l}_2>B{r}_2>F_2>I_2$
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 $X-X^{\prime }$ 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 $S{O}_2$ 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 $H_{2}S{O}_4$ with metal sulphide.
Answer:

The answer is the option (i, iii). $S{O}_2$ is used as a bleaching agent in presence of moisture because of the reducing nature of $S{O}_2$
$S{O}_2+2H_{2}O\to H_{2}S{O}_4+2H$
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 $HN{O}_3$ are equal.
(ii) All P—Cl bond lengths in $PC{l}_5$ 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 $HN{O}_3$ are not of the same bond length.
(ii) In gaseous state, all P – Cl bond lengths in $PC{l}_5$ 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 $PC{l}_5$ exists as ionic solid in which cation is tetrahedral and anion is octahedral.
$Cation-{\left[PC{l}_4\right]}^{+}Anion-{\left[PC{l}_6\right]}^{-}$

Question:34

Which of the following orders are correct as per the properties mentioned against each?
(i) $A{s}_2{O}_3<Si{O}_2<P_2{O}_3<S{O}_2$ Acid strength.
(ii) $As{H}_3<P{H}_3<N{H}_3$ 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 $\left(H_{2}S{O}_5\right)$ sulphur is in +6 oxidation state.
(iii) Iron powder along with $A{l}_2{O}_3$ and $K_{2}O$ is used as a catalyst in the preparation of $N{H}_3$ by Haber’s process.
(iv) Change in enthalpy is positive for the preparation of $S{O}_3$ by catalytic oxidation of $S{O}_2$
Answer:

The answer is the option (i, ii)
In $H_{2}S{O}_5$, there is a peroxo-linkage.

$O$ in peroxide linkage has oxidation state - 1

Question:36

In which of the following reactions conc. $H_{2}S{O}_4$ is used as an oxidising reagent?
(i) $Ca{F}_2+H_{2}S{O}_4\to CaS{O}_4+2HF$
(ii) $2HI+H_{2}S{O}_4\to I_2+S{O}_2+2H_{2}O$
(iii) $Cu+H_{2}S{O}_4\to CuS{O}_4+S{O}_2+2H_{2}O$
(iv) $NaCl+H_{2}S{O}_4\to NaHS{O}_4+HCl$
Answer:

The answer is the option (ii, iii). Amongst the given options, (ii). and (iii). represent oxidizing behaviour of $H_{2}S{O}_4$. The oxidizing agent reduces itself in the process. In option b, it oxidises HI and gets reduced to $S{O}_2$ .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 Class 12 Chemistry Solutions Chapter 7: Short Answer Type

Question:38

In the preparation of $H_{2}S{O}_4$ by Contact Process, why is $S{O}_3$ not absorbed directly in water to form $H_{2}S{O}_4$ ?
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 $N{H}_3$ by atmospheric oxygen.
Answer:

$4N{H}_3+5O_2\left(fromair\right)\stackrel{Ptgauge}{\to }4NO+6H_{2}O$

Question:40

Write the structure of pyrophosphoric acid.
Answer:

$H_4{P}_2{O}_7$ is pyrophosphoric acid. Its structure is

Question:41

$P{H}_3$ forms bubbles when passed slowly in water but $N{H}_3$ dissolves. Explain why?
Answer:

$N{H}_3$ dissolves in water as it forms hydrogen bonds with water but $P{H}_3$ cannot form hydrogen bonds with water so it escapes as gas.

Question:42

In $PC{l}_5$, phosphorus is in $s{p}^{3}d$ hybridised state, but all its five bonds are not equivalent. Justify your answer with reason.
Answer:

$PC{l}_5$ 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 $Cl{F}_3$ exists but $FC{l}_3$ does not exist.
Answer:

Though both $ClandF$ are halogens,$F$ has a configuration $2s^{2}2p^5$ and doesn’t have a d-orbital. Chlorine on the other hand can undergo $s{p}^{3}d$-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 $s{p}^{3}d$-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

$S{F}_6$ is known but $SC{l}_6$ is not. Why?
Answer:

There are two reasons as to why $S{F}_6$ exists in nature, but $SC{l}_6$ 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 $C{l}_2$, 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. $PC{l}_3$ again reacts with $C{l}_2$ and forms yellowish white powder like phosphorus pentachloride $PC{l}_5$

Question:48

In the ring test of $N{O}_3^{-}$ ion, $F{e}^{2+}$ ion reduces nitrate ion to nitric oxide, which combines with $F{e}^{2+}(aq)$ ion to form the brown complex. Write the reactions involved in the formation of a brown ring.
Answer:

$N{O}_3^{-}+3F{e}^{2+}+4H^{+}\to NO+3F{e}^{3+}+2H_{2}O$
$\left[Fe{\left(H_{2}O\right)}_6\right]S{O}_4+NO\to \left[Fe{\left(H_{2}O\right)}_{5}NO\right]S{O}_4\left(BrownComplex\right)+H_{2}O$

Question:49

Explain why the stability of oxoacids of chlorine increases in the order given below:
$HClO<HCl{O}_2<HCl{O}_3<HCl{O}_4$
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:
$Cl{O}^{-}<Cl{O}_2^{-}<Cl{O}_3^{-}<Cl{O}_4^{-}$
Thus due to increase in stability of conjugate base, acidic strength of corresponding acid increases in the following order
$HClO<HCl{O}_2<HCl{O}_3<HCl{O}_4$

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 $(\mathrm{\Delta }H=-ve)$ and increase in entropy $(\mathrm{\Delta }H=+ve)$
$2O_3\rightleftharpoons 3O_2$
These 2 factors reinforce each other resulting in large negative Gibbs free energy change $(\mathrm{\Delta }G=-ve)$ 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 $P_4{O}_6+6H_{2}O\to 4H_{3}P{O}_3.$
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:

$P_4{O}_6+6H_{2}O\to 4H_{3}P{O}_3$
$4H_{3}P{O}_3+8NaOH\to 4N{a}_{2}HP{O}_3+8H_{2}O$ ($H_{3}P{O}_3$ can be neutralised with $NaOH$ )
$P_4{O}_6(1mol)+8NaOH(8mol)\to 4N{a}_{2}HP{O}_3+2H_{2}O$
Moles of $P_4{O}_6=\frac{1.1}{220}=0.005mol$
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 $=8\times 0.005=0.04mol$
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 $C{l}_2$ as : $P_4+6C{l}_2\to 4PC{l}_3;$
$PC{l}_3+3H_{2}O\to H_{3}P{O}_3+3HCl]\times 4\left(PC{l}_{3}ishydrolysed\right)$
$P_4+6C{l}_2+12H_{2}O\to 4H_{3}P{O}_3+12HCl$
Moles of white $P=\frac{62}{124}=0.5mol$
1 mol of white $P_4$ produce $HCl=12mol$
0.5 mol of white $P_4$ will produce $HCl=12\times 0.5=6mol$
Mass of $HCl=6\times 36.5=219.0g$

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 $\left(HN{O}_3\right)$

2. Nitrous acid $\left(HN{O}_2\right)$

3. Hyponitrous acid $\left(H_2{N}_2{O}_2\right)$

The disproportionation reaction wherein the nitrogen is in +3 oxidation state is:
$$\begin{gathered} 3HN{O}_2\to HN{O}_3+2NO+H_{2}O \\ +3+5+2 \end{gathered}$$
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:

$P_4{O}_{10}+4NH{O}_3\to 4HP{O}_3+2N_2{O}_5$

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 $s{p}^3$-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 phosphorous.

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 $HN{O}_3\left(6{}^o{/}_o\right)$ forms $N{H}_{4}N{O}_3,$ with dilute $HN{O}_3\left(20{}^o{/}_o\right)$ forms $N_{2}O$ and with cone. $HN{O}_3\left(70{}^o{/}_o\right)$ forms $N{O}_2.$
(Very dilute) $HN{O}_3:4Zn+10HN{O}_3\to 4Zn(N{O}_3{)}_2+N{H}_{4}N{O}_3+3H_{2}O$
(Dilute) $HN{O}_3:4Zn+10HN{O}_3\to 4Zn(N{O}_3{)}_2+N_{2}O+5H_{2}O$
(Concentrated) $HN{O}_3:Zn+HN{O}_3\to Zn(N{O}_3{)}_2+2N{O}_2+2H_{2}O$

Question:57

$PC{l}_5$ reacts with finely divided silver on heating and a white silver salt is obtained, which dissolves on adding excess aqueous $N{H}_3$ solution. Write the reactions involved to explain what happens.

Answer:

$PC{l}_5$ 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.
$PC{l}_5+2Ag\to 2AgCl+PC{l}_3$
$AgCl+2N{H}_3(aq)\to [Ag(N{H}_3{)}_2]+C{l}^{-}$

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 $\left(H_{3}P{O}_2\right)$. This is due to possession of P – H bonds.

Its reducing properties are :
$4AgN{O}_3+H_{3}P{O}_2+2H_{2}O\to 4Ag+H_{3}P{O}_4+4HN{O}_3$
$2HgC{l}_2+H_{3}P{O}_2+2H_{2}O\to 2Hg+H_{3}P{O}_4+4HCl$
$H_{3}P{O}_2+2H_{2}O+2C{l}_2\to H_{3}P{O}_4+4HCl$

NCERT Exemplar Class 12 Chemistry Solutions Chapter 7: Matching Type

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