NCERT Exemplar Class 12 Chemistry Solutions Chapter 7 p block elements

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NCERT Exemplar Class 12 Chemistry Solutions Chapter 7 p block elements

Q: What are the important topics covered in the NCERT Exemplar Class 12 Chemistry Solutions Chapter 7 ?

There are different topics covered in the Chapter 7 of NCERT Solutions for Class 12 Chemistry are Group 15 Elements, Nitric Acid, Phosphine, Dioxygen, Ozone, Sulphuric Acid, and different topics are all important topics that are covered in the chapter 7.

Q: What are the benefits of using the NCERT Solutions for Class 12 Chemistry Chapter 7?

The benefits of using the NCERT Solutions for Class 12 Chemistry chapter 7 is the solutions created are completely based on the latest syllabus and guidelines.

Q: How to download these NCERT Exemplar Class 12 Chemistry Solutions Chapter 7?

The students can download the solutions in the PDF format and they can download it from a single link.

Q: What one can learn from this NCERT Exemplar Class 12 Chemistry Solutions Chapter 7?

The students learn about the different elements and the use of the group in the P-block elements.

Edited By Sumit Saini | Updated on Sep 16, 2022 05:37 PM IST | #CBSE Class 12th

NCERT exemplar Class 12 Chemistry solutions chapter 7 is one of the most important chapters from the examination point of view. The experts in this field who have immense and thorough knowledge in this field have prepared the NCERT exemplar Class 12 Chemistry chapter 7 solutions to help the students with all the required information. To answer the questions, the experts have used simple language so that the students will understand it easily. The students can refer to NCERT exemplar Class 12 Chemistry chapter 7 solutions to get the best learning experience. NCERT exemplar Class 12 Chemistry solutions chapter 6 pdf download is useful to access it offline.

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This Story also Contains

NCERT Exemplar Class 12 Chemistry Solutions Chapter 7: MCQ (Type 1)
NCERT Exemplar Class 12 Chemistry Solutions Chapter 7: MCQ (Type 2)
NCERT Exemplar Class 12 Chemistry Solutions Chapter 7: Short Answer Type
NCERT Exemplar Class 12 Chemistry Solutions Chapter 7: Matching Type
NCERT Exemplar Class 12 Chemistry Solutions Chapter 7: Assertion and Reason Type
NCERT Exemplar Class 12 Chemistry Solutions Chapter 7: Long Answer Type
Introduction of NCERT Exemplar Class 12 Chemistry Solutions Chapter 7 p-Block Elements
Major Subtopics in NCERT Exemplar Class 12 Chemistry Solutions Chapter 7 p-Block Elements
NCERT Exemplar Class 12 Chemistry Solutions Chapter 7 p-Block Elements - Learning Outcome
NCERT Exemplar Class 12 Chemistry Solutions
Important Topics in NCERT Exemplar Class 12 Chemistry Solutions Chapter 7

Also, check - NCERT Solutions for Class 12, other subjects

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.
$\text {2NaCl + H}_{2}\text {SO}_{4}\rightarrow \text {Na}_{2}\text {SO}_{4}+\text {2HCl}$
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.
$\text {2NaI + H}_{2}\text {SO}_{4}\rightarrow \text {Na}_{2}\text {SO}_{4}+ \text {2HI}\overset{\text {H}_{2}\text {SO}_{4}}{\rightarrow} \text {2H}_{2}\text {O + SO}_{2}+\text {I}_{2}$

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). 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?

Compound
$\Delta _{diss} \left ( E-H \right )/kJ\; mol^{-1}$

$NH_{3}$
389

$PH_{3}$
322

$AsH_{3}$
297

$SbH_{3}$

255

(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.

Ion
Reduction potential $E^{\ominus }/V$

$ClO_{4}^{-}$
$E^{\ominus }=1.19V$

$IO_{4}^{-}$
$E^{\ominus }=1.65V$

$BrO_{4}^{-}$
$E^{\ominus }=1.74V$

(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 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) $F_{2}> Cl_{2}> Br_{2}> I_{2}$ 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 Class 12 Chemistry Solutions Chapter 7: Short Answer Type

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 $PCl_{5}$

(i)	(ii)
$PCl_{5}$	$PCl_{3}$
Yellowish	Colorless Oily
White Powder	Liquid

$PCl_{3}$ reacts violently with water forming phosphosphorous acid $\left ( H_{3}PO_{3} \right )$
$PCl_{3}+3H_{2}O\rightarrow H_{3}PO_{3}+3HCl$

$PCl_{5}$ undergoes a violent hydrolysis
$PCl_{5}+H_{2}O\rightarrow POCl_{3}+2HCl$
$PCl_{5}+4H_{2}O\rightarrow H_{3}PO_{4}+5HCl$

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:

Nitric acid $\left ( HNO_{3} \right )$
Nitrous acid $\left ( HNO_{2} \right )$
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 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 $HNO_{3}\left ( 6 \; ^{o}/_{o} \right )$ forms $NH_{4}NO_{3},$ with dilute $HNO_{3}\left ( 20 \; ^{o}/_{o} \right )$ forms $N_{2}O$ and with cone. $HNO_{3}\left ( 70 \; ^{o}/_{o} \right )$ 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 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.

Column I		Column II
(A)	$\text {XeF}_{6}$	(1)	$\text {sp}^{3}\text {d}^{2}$ - distorted octahedral
(B)	$\text {XeO}_{3}$	(2)	$\text {sp}^{3}\text {d}^{2}$ -square planar
(C)	$\text {XeOF}_{4}$	(3)	$\text {sp}^{3}$ - pyramidal
(D)	$\text {XeF}_{4}$	(4)	$\text {sp}^{3}\text {d}^{2}$ -square pyramidal

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

Compound	Hybridisation
(i).	(1)	$(\text {sp}^{3}\text {d}^{2})$ - distorted octahedral
(ii).	(2)	$(\text {sp}^{3})$ - pyramidal
(iii).	(3)	$(\text {sp}^{3}\text {d}^{2})$ - square pyramidal
(iv).	(4)	$(\text {sp}^{3}\text {d}^{2})$ - square planar

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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.

Column I		Column II
(A)	$Pb_{3}O_{4}$	(1)	Neutral Oxide
(B)	$N_{2}O$	(2)	Acidic Oxide
(C)	$Mn_{2}O_{7}$	(3)	Basic Oxide
(D)	$Bi_{2}O_{3}$	(4)	Mixed Oxide

(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)

Formulas		Type of Oxide
(i)	$Pb_{3}O_{4}$ $(PbO.Pb_{2}O_{3})$	(4)	Mixed Oxide
(ii)	$N_{2}O$	(1)	Neutral Oxide
(iii)	$Mn_{2}O_{7}$	(2)	Acidic Oxide
(iv)	$Bi_{2}O_{3}$	(3)	Basic Oxide

Question:61

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

Column I		Column II
(A)	$\text {H}_{2}\text {SO}_{4}$	(1)	Highest electron gain enthalpy
(B)	$\text {CCl}_{3}\text {NO}_{2}$	(2)	Chalcogen
(C)	$\text {Cl}_{2}$	(3)	Tear gas
(D)	$\text {Sulphur}$	(4)	storage batteries

(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.

Column I		Column II
(A)	$\text {SF}_{4}$	(1)	Tetrahedral
(B)	$\text {BrF}_{3}$	(2)	Pyramidal
(C)	$\text {BrO}_{3}^{-}$	(3)	Sea-Saw shaped
(D)	$\text {NH}_{4}^{+}$	(4)	Bent-T-shaped

(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)

Species	Shape	Structure
$\text {SF}_{4}$	Sea-Saw shaped
$\text {BrF}_{3}$	Bent-T-shaped
$\text {BrO}_{3}^{-}$	Pyramidal
$\text {NH}_{4}^{+}$	Tetrahedral

Question:63

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

Column I		Column II
(A)	Its partial hydrolysis does not change oxidation state of central atom	(1)	$\text {He}$
(B)	It is used in modern diving apparatus	(2)	$\text {XeF}_{6}$
(C)	It is used to provide inert atmosphere for filling electrical bulbs	(3)	$\text {XeF}_{4}$
(D)	Its central atom is in $\text {sp}^{3}\text {d}^{2}$ hybridisation	(4)	$\text {Ar}$

(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)

Oxidation state of $\text {Xe}$ is unaffected by partial hydrolysis of $\text {XeF}_{6}$
He is utilized in the modern driving apparatus.
Air is used for filling electrical bulbs to provide inert atmosphere.
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 Class 12 Chemistry Solutions Chapter 7: Assertion and Reason Type

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 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). 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 Class 12 Chemistry Solutions Chapter 7: Long Answer Type

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:

$(A)=NH_{4}NO_{2},\; \; (B)=N_{2},\; \; \; \; (C)=NH_{3},\; \; \; \; \; (D)=HNO_{3}$
$NH_{4}NO_{2}\overset{Heat}{\rightarrow}N_{2}+2H_{2}O$
$N_{2}+3H_{2}\overset{Catalyst}{\rightarrow}2NH_{3}$
$4NH_{3}+5O_{2}\rightarrow 4NO+6H_{2}O$
$4NO+O_{2}\rightarrow 2NO_{2}$
$3NO_{2}+H_{2}O\rightarrow 2HNO_{3}+NO$

Introduction of NCERT Exemplar Class 12 Chemistry Solutions Chapter 7 p-Block Elements

The chemistry of different inorganic ring systems of the p-block elements has a long and a history that dates back to the early 19th century. There are different elements of different Groups that Group 13, 14, 15, 16, 17 and 18 are called P block elements. The elements are often represented by the outer electronic configuration that are general ns2np1-6. P block elements often exist in all the three different physical states and also metals, nonmetals or metalloids.

As per NCERT exemplar Class 12 Chemistry solutions chapter 7, P-block elements refer to the elements that exist on the right side of the periodic table. They also involve carbon, boron, oxygen, nitrogen, and different noble gases. This chapter goes around p-block elements.

Major Subtopics in NCERT Exemplar Class 12 Chemistry Solutions Chapter 7 p-Block Elements

Occurrence
Electronic Configuration
Atomic and Ionic Radii
Ionisation Enthalpy
Electronegativity
Physical Properties
Chemical Properties

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● Group 15 Elements

● Dinitrogen

● Ammonia

● Oxides of Nitrogen

● Nitric Acid

● Phosphorus — Allotropic Forms

● Phosphine

● Phosphorus Halides

● Oxoacids of Phosphorus

● Group 16 Elements

● Dioxygen

● Simple Oxides

● Ozone

● Sulphur — Allotropic Forms

● Sulphur Dioxide

● Oxoacids of Sulphur

● Sulphuric Acid

● Group 17 Elements

● Chlorine

● Hydrogen Chloride

● Oxoacids of Halogens

● Interhalogen Compounds

● Group 18 Elements

NCERT Exemplar Class 12 Chemistry Solutions Chapter 7 p-Block Elements - Learning Outcome

In Class 12 NCERT exemplar Chemistry solutions chapter 7 the students get to learn about the different topics and subtopics that are covered in this chapter that one needs to know and understand. The p-block elements are found on the right side of the periodic table. The p-block is the region of the periodic table that includes columns 3A to column 8A and does not include helium. Every topic in this NCERT Exemplar Class 12 Chemistry Chapter 7 Solutions are detailed and accurately explained.

P block elements are the elements in which the last electron enters any of the three p-orbitals of their respective shells. P block elements are shiny and usually a good conductor of electricity. The uses of p block elements are mentioned here. Overall, NCERT exemplar solutions for Class 12 Chemistry chapter 7 along with giving the required information is also interesting and for all those students who want to learn about the different p block elements and work on it.

NCERT Exemplar Class 12 Chemistry Solutions

Chapter 1 Solid State

Chapter 2 Solutions

Chapter 3 Electrochemistry

Chapter 4 Chemical Kinetics

Chapter 5 Surface Chemistry

Chapter 6 General Principles and Processes of Isolation of Elements

Chapter 8 The d & f Block Elements

Chapter 9 Coordination Compounds

Chapter 10 Haloalkanes and Haloarenes

Chapter 11 Alcohols, Phenols, and Ethers

Chapter 12 Aldehydes, Ketones, and Carboxylic Acids

Chapter 13 Amines

Chapter 14 Biomolecules

Chapter 15 Polymers

Chapter 16 Chemistry in Everyday Life

Important Topics in NCERT Exemplar Class 12 Chemistry Solutions Chapter 7

- The P-block elements and its working and all the different internal and external factors are given in this chapter. All the required information regarding the P-block elements are given in the NCERT exemplar Class 12 Chemistry solutions chapter 7. The elements included in the chapter are equally important for the students to refer to.

- NCERT exemplar Class 12 Chemistry chapter 7 solutions provides the required information about the elements and the use of p-block elements.

- The chapter also provides the required information about the different groups in the p-block elements. The chapter also includes different diagrams for a better understanding for the students.

NCERT Exemplar Class 12 Solutions

NCERT Exemplar Class 12 Mathematics Solutions

NCERT Exemplar Class 12 Physics Solutions

NCERT Exemplar Class 12 Biology Solutions

Check NCERT Solutions for Class 12 Chemistry

Chapter 1	The Solid State
Chapter 2	Solutions
Chapter 3	Electrochemistry
Chapter 4	Chemical Kinetics
Chapter 5	Surface chemistry
Chapter 6	General Principles and Processes of isolation of elements
Chapter 7	The P-block elements
Chapter 8	The d and f block elements
Chapter 9	Coordination compounds
Chapter 10	Haloalkanes and Haloarenes
Chapter 11	Alcohols, Phenols, and Ethers
Chapter 12	Aldehydes, Ketones and Carboxylic Acids
Chapter 13	Amines
Chapter 14	Biomolecules
Chapter 15	Polymers
Chapter 16	Chemistry in Everyday life

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Frequently Asked Question (FAQs)

1. What are the important topics covered in the NCERT Exemplar Class 12 Chemistry Solutions Chapter 7 ?

There are different topics covered in the Chapter 7 of NCERT Solutions for Class 12 Chemistry are Group 15 Elements, Nitric Acid, Phosphine, Dioxygen, Ozone, Sulphuric Acid, and different topics are all important topics that are covered in the chapter 7.

2. What are the benefits of using the NCERT Solutions for Class 12 Chemistry Chapter 7?

The benefits of using the NCERT Solutions for Class 12 Chemistry chapter 7 is the solutions created are completely based on the latest syllabus and guidelines.

3. How to download these NCERT Exemplar Class 12 Chemistry Solutions Chapter 7?

The students can download the solutions in the PDF format and they can download it from a single link.

4. What one can learn from this NCERT Exemplar Class 12 Chemistry Solutions Chapter 7?

The students learn about the different elements and the use of the group in the P-block elements.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

I am interested in graphic designing and coding. how do I make a career .which stream should I go.what engineering subject to take

Hello aspirant,

The purpose of graphic design extends beyond the brand's look. Nevertheless, by conveying what the brand stands for, it significantly aids in the development of a sense of understanding between a company and its audience. The future in the field of graphic designing is very promising.

There are various courses available for graphic designing. To know more information about these courses and much more details, you can visit our website by clicking on the link given below.

https://www.careers360.com/courses/graphic-designing-course

Thank you

Hope this information helps you.

Read Complete Answer

Sajal Trivedi 29 January,2024

i couldn't clear 1st compartment (which held in August 2022) cbse class 12th so can I give improvement in all subjects which to be held on March 2023. tell me a solution i need more than 75% in class 12th

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

Read Complete Answer

Upasana Barman 24 August,2022

i was not able to clear 1st compartment and now I have to give 2nd compartment so can I give improvement exam next year? plz tell me very much tensed(cbse class 12th)

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.

As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.

Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.

Believe in Yourself! You can make anything happen

All the very best.

Read Complete Answer

Mayankjha.student2007 23 August,2022

if i am giving improvement in one subject this year, will i be able to give improvement in more subjects next year?

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects and we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

Read Complete Answer

Nuthalapati Vyshnavi 21 August,2022

I got an RT for 2 subjects, so do I have to give exams for all the subjects in the coming year or just the 2. pls help me

Hi,

You just need to give the exams for the concerned two subjects in which you have got RT. There is no need to give exam for all of your subjects, you can just fill the form for the two subjects only.

Read Complete Answer

Muskan Dhalwal 17 August,2022

View All

Popular CBSE Class 12th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms^-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8×10⁷J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms⁻² :

Option 1)

2.45×10⁻³ kg

Option 2)

6.45×10⁻³ kg

Option 3)

9.89×10⁻³ kg

Option 4)

12.89×10⁻³ kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

A particle is projected at 60⁰ to the horizontal with a kinetic energy $K$ . The kinetic energy at the highest point

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

In the reaction,

$2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, $Mg_{3}(PO_{4})_{2}$ will contain 0.25 mole of oxygen atoms?

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol^-1) is

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t²) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m² , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

Data Administrator

Database professionals use software to store and organise data such as financial information, and customer shipping records. Individuals who opt for a career as data administrators ensure that data is available for users and secured from unauthorised sales. DB administrators may work in various types of industries. It may involve computer systems design, service firms, insurance companies, banks and hospitals.

4 Jobs Available

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The field of biomedical engineering opens up a universe of expert chances. An Individual in the biomedical engineering career path work in the field of engineering as well as medicine, in order to find out solutions to common problems of the two fields. The biomedical engineering job opportunities are to collaborate with doctors and researchers to develop medical systems, equipment, or devices that can solve clinical problems. Here we will be discussing jobs after biomedical engineering, how to get a job in biomedical engineering, biomedical engineering scope, and salary.

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A career as ethical hacker involves various challenges and provides lucrative opportunities in the digital era where every giant business and startup owns its cyberspace on the world wide web. Individuals in the ethical hacker career path try to find the vulnerabilities in the cyber system to get its authority. If he or she succeeds in it then he or she gets its illegal authority. Individuals in the ethical hacker career path then steal information or delete the file that could affect the business, functioning, or services of the organization.

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GIS officer work on various GIS software to conduct a study and gather spatial and non-spatial information. GIS experts update the GIS data and maintain it. The databases include aerial or satellite imagery, latitudinal and longitudinal coordinates, and manually digitized images of maps. In a career as GIS expert, one is responsible for creating online and mobile maps.

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The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.

3 Jobs Available

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If you are intrigued by the programming world and are interested in developing communications networks then a career as database architect may be a good option for you. Data architect roles and responsibilities include building design models for data communication networks. Wide Area Networks (WANs), local area networks (LANs), and intranets are included in the database networks. It is expected that database architects will have in-depth knowledge of a company's business to develop a network to fulfil the requirements of the organisation. Stay tuned as we look at the larger picture and give you more information on what is db architecture, why you should pursue database architecture, what to expect from such a degree and what your job opportunities will be after graduation. Here, we will be discussing how to become a data architect. Students can visit NIT Trichy, IIT Kharagpur, JMI New Delhi.

3 Jobs Available

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Remote sensing scientists use remote sensing technology to support scientists in fields such as community planning, flight planning or the management of natural resources. Analysing data collected from aircraft, satellites or ground-based platforms using statistical analysis software, image analysis software or Geographic Information Systems (GIS) is a significant part of their work. Do you want to learn how to become remote sensing technician? There's no need to be concerned; we've devised a simple remote sensing technician career path for you. Scroll through the pages and read.

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2 Jobs Available

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A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

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An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems.