NCERT Solutions for Class 12 Chemistry Chapter 7 The p-block elements

Q: What is the weightage of NCERT Class 12 Chemistry chapter 7 in JEE Mains?

This chapter holds Weightage of around 5-6 marks. Other than NCERT questions students can refer to the NCERT exemplar questions and JEE Main previous year papers.

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Around 8 marks questions can be expected for the CBSE board exam. Follow NCERT syllabus for a good score.

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For complete solutions : https://school.careers360.com/ncert/ncert-solutions-class-12-chemistry

Q: What are the important topics of this chapter?

Structures Preparation of components by commercial and laboratory preparation of N2,P4010,P4O5,O2,etc . Difference and general properties like atomic radii, melting point increase or decrease or just like questions are coming most .

NCERT Solutions for Class 12 Chemistry Chapter 7 The p-block elements

Edited By Sumit Saini | Updated on Aug 16, 2022 12:51 PM IST | #CBSE Class 12th

NCERT S olutions for Class 12 Chemistry Chapter 7 The P-block Elements - In Class 11th, you must have learnt that the p-block elements are placed from 13 to 18 groups of the periodic table but in NCERT syllabus of Class 11 you have studied only two groups 13 and 14, so in NCERT solutions for Class 12 Chemistry chapter 7 the p-block elements, you are going to study and get questions from groups from 15 to 18 of the periodic elements and their answers.

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If you are appearing for boards exam then NCERT solutions for p block elements Class 12 Chemistry chapter 7 can be of a great help because this chapter carries a huge weightage of 8 marks out of 70 marks and are also important for competitive exams like JEE, SRMJEEE, VITEEE, BITSAT, etc. The step-by-step NCERT solutions are arranged in a sequential manner which are prepared by subject experts. By referring to the NCERT solutions for class 12 , students can understand all the important concepts and practice questions well enough before their examination.

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Find NCERT Solutions for Class 12 Chemistry Chapter 7 The p-block elements below:

Solutions to In Text Questions Ex 7.1 to 7.34

Question 7.1 Why are pentahalides more covalent than trihalides ?

Answer :

Pentahalides are more covalent than trihalides. This is due to the fact that in pentahalides +5 oxidation state exists while in the case of trihalides +3 oxidation state exists. So, Higher the +ve O.S of the central atom more will be the polarising power and more will be the covalent character in the bond between the central atom and a halogen atom. Since elements in +5 oxidation state will have more polarising power than in +3 oxidation state, the covalent character of bonds is more in pentahalides.

Question 7.2 Why is $BiH_{3}$ the strongest reducing agent amongst all the hydrides of Group 15 elements ?

Answer :

We see that the stability of hydrides becomes lesser as we go from $NH_3$ to $BiH_3$ .this can be seen from dissociation enthalpy of their bond. due to that, the reducing character of the hydrides will be more. Ammonia is only a very mild reducing agent while $BiH_3$ is the strongest reducing agent amongst all of the hydrides.

Question 7.3 Why is $N_{2}$ less reactive at room temperature?

Answer :

$N_{2}$ reacts poorly at room temperature. high bond enthalpy of N≡N bond is the reason behind this. Reactivity, however, increases rapidly with increase in temperature.

Question 7.4 Mention the conditions required to maximise the yield of ammonia.

Answer :

Ammonia is produced by Haber's process-

$N_{2}+3H_{2} \overset{700k}{\leftrightharpoons} 2NH_{3}\:\:\:\:\: \Delta H^{o}=-46.1KJmol^{-1}$

The maximum yield conditions for the production of ammonia are-

pressure = $200 \:atm \:or\:200\times 10^5 Pa$ ,
Temperature of around 700 K
catalyst = Iron oxide
Promotor= small amounts of $K_2O$ and $Al_2O_3$ to increase the rate of attainment of equilibrium.

Question 7.5 How does ammonia react with a solution of $Cu^{2+}$ ?

Answer :

Ammonia is a Lewis base due to The presence of a lone pair of electrons on the nitrogen atom of the ammonia molecule. It forms a linkage with metal ions by donating the electron pair.

$Cu^{2+} (aq)(blue) +4NH_3\rightarrow [Cu(NH_3)_4]^{2+}(aq)(deepblue)$

Question 7.6 What is the covalence of nitrogen in $N_{2}O_{5}$ ?

Answer :

We can see From the structure of $N_2O_5$ that covalence of nitrogen is four.

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Question 7.7 Bond angle in $PH_{4}^{+}$ is higher than that in $PH_{3}$ . Why?

Answer :

As we can see Both are $sp^3$ hybridised. In $PH_4^+$ all of the 4 orbitals are bonded whereas in $PH_3$ there is a lone pair of electrons on P.this lone pair is responsible for lone pair-bond pair repulsion in $PH_3$ , which results in reducing the bond angle to less than 109° 28′.

Question 7.7 What is formed when $PH_{3}$ reacts with an acid?

Answer :

$PH_3$ reacts with acids like $HI$ to form $PH_4I$ which shows that it is basic in nature. Because of lone pair on phosphorus atom, $PH_3$ is acting as a Lewis base in the above reaction

$PH_3 + HI\rightarrow PH_4I$

$PH_3 + Acid\rightarrow salt$

Question 7.8 What happens when white phosphorus is heated with concentrated $NaOH$ solution in an inert atmosphere of $CO_{2}$ ?

Answer :

When white phosphorus is heated with the concentrated $NaOH$ solution in an inert atmosphere of $CO_{2}$ we see that phosphine and sodium hypophosphite is formed.

$P_{4}+\:3NaOH\:+H_{2}O\rightarrow 3NaH_{2}PO_{2}\:+ \:PH_{3}$

Question 7.9 What happens when $PCI_{5}$ is heated?

Answer :

When we heat $PCI_{5}$ , it sublimes but decomposes on stronger heating and phosphorus trichloride is formed.

$PCI_{5}$ + Heat $\rightarrow$ $PCI_{3}$ + $Cl_{2}$

Question 7.10 Write a balanced equation for the reaction of $PCI_{5}$ with heavy water.

Answer :

$\\Step\: I\: : \:PCl_{5}\:\:+\:\:\:D_{2}O\rightarrow POCl_{3}+2DCl\\Step\:II:POCl_{3}+3D_{2}O\rightarrow D_{3}PO_{4}+3DCl$

$Overall \:reaction\:PCl_5 + 4D_2O \rightarrow D_{3}PO_4 + 5DCl$

Question 7.11 What is the basicity of $H_{3}PO_{4}$ ?

Answer :

There are three P–OH bonds present in the molecule of $H_{3}PO_{4}$ . Hence, its basicity is three.

$H_{3}PO_{4}\overset{H_{2}O}{\rightleftharpoons } 3H^{+}\:+\:PO_{4}^{3-}$

Question 7.12 What happens when $H_{3}PO_{3}$ is heated?

Answer :

$H_3PO_4$ when heated, it will disproportionates and give orthophosphoric acid (or phosphoric acid) and phosphine.

$4H_3PO_3 \rightarrow 3H_3PO_4 + PH_3$

Question 7.13 List the important sources of sulphur.

Answer :

The presence of sulphur in the earth’s crust is only about 0.03-0.1%.
mixed sulphur exists primarily as sulphates such as gypsum $CaSO_4.2H_2O$ , Epsom salt $MgSO_4.7H_2O$ , baryte $BaSO_4$ .
another source is by sulphides such as galena $PbS$ , zinc blende $ZnS$ , copper pyrites $CuFeS_2$ . some sulphur also occurs as hydrogen sulphide in volcanoes. eggs, proteins, garlic, onion, mustard, hair and wool also contain sulphur.

Question 7.14 Write the order of thermal stability of the hydrides of Group 16 elements.

Answer :

Hydrides of group 16 elements become less thermally stable as we go down the group, i.e., $H_2O$ > $H_2S$ > $H_2Se$ > $H_2Te$ > $H_2Po$ . This is due to M-H bond dissociation energy decreases down the group as we increase in the size of the atom.

Question 7.15 Why is $H_{2}O$ a liquid and $H_{2}S$ a gas ?

Answer :

$H_{2}O$ has oxygen as the centre atom. oxygen is small in size as well as high electronegative when we compare it with sulpher.molecules of water are highly associated through hydrogen bonding which is not present in sulpher.molecules of $H_{2}S$ are connected to each other through weak van der Wal force only. Hence a $H_{2}O$ liquid and $H_{2}S$ a gas.

Question 7.16 Which of the following does not react with oxygen directly? Zn, Ti, Pt, Fe

Answer :

Since Platinum(Pt) is a noble metal .it will not react with oxygen directly

Question 7.17 Complete the following reactions:

(i) $C_{2}H_{4}+O_{2}\rightarrow$

Answer :

The reaction is:

$C_2H_4 + 3O_2\rightarrow 2CO_2+2H_2O$

Question 7.17 Complete the following reactions:

(ii) $4AI+3 O_{2}\rightarrow$

Answer :

The complete reaction is:

$4AI+3 O_{2}\rightarrow$ $2Al_2O_3$

Question 7.18 Why does $O_{3}$ act as a powerful oxidising agent?

Answer :

$O_{3}$ act as a powerful oxidising agent .This is because of the ease with which it frees atoms of nascent oxygen.i.e.

$O_3\rightarrow O_2+O$

Question 7.19 How is $O_{3}$ estimated quantitatively?

Answer :

A quantitative method for estimating O3 gas is:

When we reacts $O_3$ with an excess of potassium iodide solution which is buffered with a borate buffer (pH 9.2), iodine is released which can be then titrated against a standard solution of sodium thiosulphate.

$2I^-+H_2O+O_3\rightarrow 2OH^-+I_2+O_2$

we use starch as an indicator when $I_2$ liberated is titrated against a standard solution of sodium thiosulphate.

Question 7.20 What happens when sulphur dioxide is passed through an aqueous solution of Fe(III) salt?

Answer :

When we pass sulphur dioxide through an aqueous solution of Fe(III) salt, it converts iron(III) ions to iron(II) ions.

$2Fe^{3+}+SO_2+2H_20\rightarrow 2Fe^{2+}+SO_4^{2-}+4H^+$

Question 7.21 Comment on the nature of two $S-O$ bonds formed in $SO_{2}$ molecule. Are the two $S-O$ bonds in this molecule equal ?

Answer :

The two $S-O$ bonds in $SO_{2}$ molecule are covalent and have equal strength because of having resonating structures.

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Question 7.22 How is the presence of $SO_{2}$ detected ?

Answer :

Presence of sulphur dioxide is measured by the following reaction. it decolourises acidified potassium permanganate(VII) solution.

$5SO_2+2MnO_4^-+2H_2O\rightarrow 5SO_4^{2-}+4H^++2Mn^{2+}$ .

it decolourises acidified potassium permanganate(VII) solution.

Hence This can be used to detect the presence of $SO_{2}$ .

Question 7.23 Mention three areas in which $H_{2}SO_{4}$ plays an important role.

Answer :

Manufacture of fertilisers (e.g., ammonium sulphate, superphosphate) from $H_2SO_4$ .
Use is petroleum refining
Manufacture of pigments, paints and dyestuff intermediates and detergent industry.

Question 7.24 Write the conditions to maximise the yield of $H_{2}SO_{4}$ by Contact process.

Answer :

Contact process which we use to create sulphuric acid is exothermic, reversible and the forward reaction which leads to a decrease in volume. hence, low temperature and high pressure are the optimum conditions for maximum yield. But if the temperature will be very low then the rate of reaction will become slow. Also, the presence of catalyst $V_2O_5$ fastens the reaction.

Question 7.25 Why is $K_{a_{2}}<<K_{a}_{_{1}}$ for $H_{2}SO_{4}$ in water ?

Answer :

$H_2SO_4$ is a very strong acid in water mostly due to its first ionisation to $H_3O^+$ and $HSO_4^-$ .The ionisation of $HSO_4^-$ to $H_3O^+$ and $SO_4^{2-}$ is minuscule. That is the reason why Ka₂ << Ka₁.

Question 7.26 Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of $F_{2}$ and $CI_{2}$ .

Answer :

$F_2$ is much more powerful in oxidising, than $Cl_2$ .The reason being, hydration enthalpy of F– ions (515 kJ mol^–1) is much higher than that of Cl– ion (381 kJ mol^–1). the dissociation energy of bond F-F is less than Cl-Cl bond but The former factor more than compensate the less negative electron gain enthalpy of F2. Hence it is a much stronger oxidising agent.

Question 7.27 Give two examples to show the anomalous behaviour of fluorine.

Answer :

We see anomalous behaviour of fluorine and this is because of its small size, highest electronegativity, very low F-F bond dissociation enthalpy, and non-availability of d orbitals in the valence shell.

1. ionisation enthalpy, electronegativity, and electrode potentials are all higher for fluorine than expected from the trends set by other halogens

2.ionic and covalent radii, melting point and boiling point., enthalpy for bond dissociation and electron gain enthalpy are very much lower than expected

3. Fluorine shows only an oxidation state of –1 due to non-availability of d-orbitals in its valence shell.

Question 7.28 Sea is the greatest source of some halogens. Comment.

Answer :

The water of the sea contains bromides, chlorides, and iodides of sodium, magnesium, potassium and calcium, but mainly solution of sodium chloride (2.5% by mass). The deposits of dried up seas have these compounds in it, e.g., sodium chloride and carnallite, KCl.MgCl2 .6H2O. iodine is also formed in Certain forms of marine life in their systems; many seaweeds, for example, contain up to 0.5% of iodine. Chile saltpeter contains up to 0.2% of sodium iodate. That's why the sea is the greatest source of halogens.

Question 7.29 Give the reason for bleaching action of $CI_{2}$ .

Answer :

Chlorine is a powerful bleaching agent and its bleaching action happens due to oxidation.

$Cl_{2}\:+\:H_{2}O\rightarrow 2HCl + \left [ O \right ]$

Chlorine + Water $\rightarrow$ Hydrochloric acid + nascent Oxygen

Coloured substance + nascent Oxygen $\left [ O \right ]$ $\rightarrow$ Colourless substance

Question 7.30 Name two poisonous gases which can be prepared from chlorine gas.

Answer :

The poisonous gases which we can be prepared from chlorine are

phosgene ( $COCl_2$ )
mustard gas $ClCH_2CH_2SCH_2CH_2CL$ .

Question 7.31 Why is ICl more reactive than $I_{2}$ ?

Answer :

ICl is more reactive than $I_{2}$ . this is because interhalogen compounds are more reactive than halogens (except fluorine). This is because X–X′ bond in interhalogens is weaker than X–X bond in halogens except F–F bond.

Question 7.32 Why is helium used in diving apparatus?

Answer :

Helium is used in diving apparatus because it is very low soluble in blood.

Question 7.33 Balance the following equation:

$XeF_{6}+H_{2}O\rightarrow XeO_{2}F_{2}+HF$

Answer :

The balanced reaction is:

$XeF_{6}+2H_{2}O\rightarrow XeO_{2}F_{2}+4HF$

Question 7.34 Why has it been difficult to study the chemistry of radon?

Answer :

It has been difficult to study the chemistry of radon because radon is a radioactive element and it has a short half-life.

NCERT Solutions for Class 12 Chemistry Chapter 7 The p-block elements- Exercise Questions

Question 7.1 Discuss the general characteristics of Group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionisation enthalpy and electronegativity

Answer :

Since all the elements in group 15 have 5 valence electrons, Electronic configuration of group 15 element is ns2np3 where n= 2 to 6. All element requires three more electrons to complete their octets. However, gaining electrons is very difficult as the nucleus will have to attract three more electrons. This can take place only with nitrogen as it is the smallest in size and the distance between the nucleus and the valence shell is relatively small. The rest elements of this group show a formal oxidation state of -3 in their covalent compounds. N and P also show -1 and -2 oxidation states In addition to the -3 state. every element which is present in this group shows +3 and +5 oxidation states. whereas, the stability of the +5 oxidation state decreases as we go down a group, whereas the stability of +3 oxidation state increases. This happens because of the inert pair effect.

First ionization decreases on moving down a group. This is because of increasing atomic sizes. As we move down a group, electronegativity decreases, due to an increase in size. As we go down in the group, the atomic size increases. This increase in the atomic size is credited to an increase in the number of shells.

Question 7.2 Why does the reactivity of nitrogen differ from phosphorus?

Answer :

Nitrogen is a diatomic molecule $N\equiv N$ . The two atoms of nitrogen form a triple bond which makes it highly stable. The triple bond present is very strong and difficult to break due to the small size of the nitrogen atom, this is not the case in phosphorus atom and phosphorus exists in a tetra-atomic molecule. Since $P-P$ single bond (213KJ/mol) is weaker than $N\equiv N$ triple bond (941KJ/mol) hence they both react differently..

Question 7.3 Discuss the trends in chemical reactivity of group 15 elements.

Answer :

The element of group 15 :

React with hydrogen in order to form hydrides of type $EH_3$ , where E = N, P, As, Sb, or Bi.

React with oxygen in order to form two types of oxides: $E_2O_3\: and\: E_2O_5$ where E = N, P, As, Sb, or Bi.

React with halogens in order to form two series of salts: $EX_3$ and $EX_5$ . Except $NBr_{3},\:NI_{3} \:and \:NX_{5}$ because it lacks the d -orbital.

React with metals for forming binary compounds in which metals exhibit -3 oxidation states.

Question 7.4 Why does $NH_{3}$ form hydrogen bond but $PH_{3}$ does not?

Answer :

$NH_{3}$ form hydrogen bond but $PH_{3}$ does not because Nitrogen has the massive attraction of the electron to the nucleus due to its higher electronegativity in comparison to the phosphorus. hence H-bonding in PH3 is very less as compared to NH3.

Note: Conditions for the formation of H-bond are-

high electronegativity
small size

Question 7.5 How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions involved.

Answer :

We prepare nitrogen by the following method,

when An aqueous solution of ammonium chloride is reacted with sodium nitrite.

$NH_4Cl(aq)+NaNO_2(aq)\rightarrow N_2(g)+2H_2O(l)+NaCl$

here, NO and HNO ₃ are produced in small amounts. These are counted in impurities that we can remove by passing nitrogen gas through aqueous sulphuric acid, containing potassium dichromate.

Question 7.6 How is ammonia manufactured industrially?

Answer :

Ammonia is produced by Haber's process-

$N_{2}+3H_{2} \overset{700k}{\leftrightharpoons} 2NH_{3}\:\:\:\:\: \Delta H^{o}=-46.1KJmol^{-1}$

According to Le-Chatelier's principle, High pressure would favour the production. The maximum yield conditions for the production of ammonia are-

pressure = $200 \:atm \:or\:200\times 10^5 Pa$ ,
Temperature of around 700 K
catalyst = Iron oxide
Promotor= small amounts of $K_2O$ and $Al_2O_3$ to increase the rate of attainment of equilibrium.

Question 7.7 Illustrate how copper metal can give different products on reaction with $HNO_{3}$ .

Answer :

Concentrated nitric acid has a strong oxidizing property. It is used for oxidizing most metals. The concentration of the acid and temperature decides the products of oxidation.

(i) Cu reacts with dilute $HNO_3$

$3Cu + 8HNO_3(dilute)\rightarrow 3Cu(NO_3)_2+2NO +4H_2O$

(i) Cu reacts with conc. $HNO_3$

$Cu + 4HNO_3(conc)\rightarrow Cu(NO_3)_2+2NO_{2} +2H_2O$

Question 7.8 Give the resonating structures of $NO_{2}$ and $N_{2}O_{5}$ .

Answer :

Resonance structure of $NO_{2}$ and $N_{2}O_{5}$ are

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Question 7.9 The HNH angle value is higher than HPH, HAsH and HSbH angles. Why?

Answer :

The angle value of HNH is higher than HPH, HAsH and HSbH angles. This is due to the higher electronegativity of the electron. Since nitrogen is highly electronegative, there is high electron density around the atom of nitrogen. This causes greater repulsion between the electron pairs which are around nitrogen, resulting in maximum bond angle.

Question 7.10 Why does $R_{3}P=0$ exist but $R_{3}N=0$ does not $(R = alkyl\: group)$ ?

Answer :

$N$ does not have any $d$ -orbitals but phosphorus( $P$ ) does. This is the restriction which comes in nitrogen( $N$ ) to expand its coordination number beyond four. Hence, $R_3N=O$ does not exist.

Question 7.11 Explain why $NH_{3}$ is basic while $BiH_{3}$ is only feebly basic.

Answer :

Nitrogen has a small size because of which the lone pair of electrons are concentrated in a small region. As we go down a group, the size of the central atom increases and the charge gets distributed over a large area which results in decreasing the electron density. Hence, the electron donating capacity(Basicity) of group 15 element hydrides decreases on moving down the group. And that's why electron releasing tendency(basicity) of $BiH_{3}$ is less than ammonia.

Question 7.12 Nitrogen exists as diatomic molecule and phosphorus as $P_{4}$ . Why?

Answer :

The nitrogen atom has small size and high electronegativity due to this nitrogen form $p\pi -p\pi$ multiple bonds with itself and with other elements which have small size and high electronegativity (e.g., C, O). The elements which are heavier of this group do not form $p\pi -p\pi$ bonds because their atomic orbitals are so large and diffuse that they cannot have effective overlapping. Thus, nitrogen exists as a diatomic molecule with a triple bond (one s and two p) between the two atoms. On the contrary, phosphorus has less the tendency to form pπ-pπ bonds and hence it exists in the form $P_{4}$ .

Question 7.13 Write main differences between the properties of white phosphorus and red phosphorus.

Answer :

White phosphorus	Red phosphorus
It is a translucent white waxy solid	It is crystalline solid.
It is insoluble in water but soluble in carbon disulphide	It is insoluble in water as well as in carbon disulphide
poisonous	non-poisonous
It consists of discrete tetrahedral P4 molecule	red phosphorus is polymeric, consisting of chains of P4 tetrahedra linked together

Question 7.14 Why does nitrogen show catenation properties less than phosphorus?

Answer :

The single $N-N$ bond in nitrogen is weaker than the single $P-P$ bond because of high interelectronic repulsion of the non-bonding electrons in $N_{2}$ , due to the small bond length. Therefor, the catenation tendency is weaker in nitrogen.

Question 7.15 Give the disproportionation reaction of $H_{3}PO_{3}$ .

Answer :

When we heat, orthophosphorus acid (H ₃ PO ₃ ) disproportionates into orthophosphoric acid (H ₃ PO ₄ ) and phosphine (PH ₃ ).

$\dpi{100} \overset{\:\:\:+3}{4H_3PO_3}\rightarrow \overset{\:\:\:-5}{3H_3PO_4}+\overset{-3}{\:\:\:\:PH_3}$

Question 7.16 Can $PCI_{5}$ act as an oxidising as well as a reducing agent? Justify.

Answer :

No $PCI_{5}$ can not act as reducing agent but it can act as an oxidising. In $PCI_{5}$ , phosphorus have its highest oxidation state (+5) which cannot be increased further but it can decrease its oxidation state and act as an oxidizing agent. For example-

$Sn \:+\:\overset{+5}{\:\:2PCl_{5}}\rightarrow SnCl_{4}+\:\overset{+3}{\:2PCl_{3}}$

Question 7.17 Justify the placement of $O$ , $S$ , $Se$ , $Te$ and $Po$ in the same group of the periodic table in terms of electronic configuration, oxidation state and hydride formation.

Answer :

Electronic Configuration-
$O$ , $S$ , $Se$ , $Te$ and $Po$ , all have six valance electron each. The general electronic configuration of these elements is $ns^2,\ np^4$ , where $n$ varies from 2 to 6.
Oxidation state-
As all of these elements have six valence electrons, they should display an oxidation state of -2. The stability of the -2 oxidation state decreases on moving down a group due to a decrease in the electronegativity of the elements. The heavier elements show +2, +4 and +6 oxidation state due to availability of $d$ -orbitals. It also exhibits the oxidation state of -1 ( $H_{2}O_{2}$ ), zero ( $O_{2}$ ), and +2 ( $OF_{2}$ )
Hydrides-
They all form hydrides of formula $H_{2}E$ , where $E = O, S, Se, Te, Po.$ Oxygen and sulphur also form hydrides of type $H_{2}E_{2}$ . These hydrides are volatile in nature.

Question 7.18 Why is dioxygen a gas but sulphur a solid?

Answer :

Oxygen is smaller in size as compared to the sulphur. Thus it can effectively form $p\pi-p\pi$ bond and form $O_{2}(O=O)$ molecules. The intermolecular forces in oxygen are weak van der Wall's, which cause it to exist as gas.whereas sulphur exists as a puckered structure held together by strong covalent bonds. Hence, it is solid.

Question 7.19 Knowing the electron gain enthalpy values for $O\rightarrow O^{-}$ and $O\rightarrow O^{2-}$ as $-141$ and $702 kJ mol^{-1}$ respectively, how can you account for the formation of a large number of oxides having $O^{2-}$ species and not $O^{-}$ ?

Answer :

Lattice energy directly depends on the charge carried by an ion. More the lattice energy, more stable the compound will be. When metal and oxygen combine, the lattice energy of the oxide, which involves $O^{2-}$ ion is much more than the oxide which involves $O^{-}$ ion. Ionic compound stability depends on the lattice energy of the compound. Thus the oxides of $O^{2-}$ is more stable than oxides having $O^{-}$ .

Question 7.20 Which aerosols deplete ozone?

Answer :

Freons which are used in aerosol sprays and as refrigerants is accountable for the depletion of the ozone layer, freons are also called chlorofluorocarbons.\

Question 7.21 Describe the manufacture of $H_{2}SO_{4}$ by contact process?

Answer :

Sulphuric acid is manufactured by the Contact Process that involves three steps:

(i) burning of sulphide ores or sulphur in the air to generate. $SO_2$

(ii) conversion of $SO_2$ to $SO_3$ by the reaction with oxygen in the presence of a catalyst ( $V_2O_5$ ), and

(iii) absorption of $SO_3$ in $H_2SO_4$ to give Oleum ( $H_2S_2O_7$ )

Diluting the oleum with water gives $H_2SO_4$ of the desired concentration.

$2SO_2+O_2\rightarrow 2SO_3$

$SO_3+H_2SO_4\rightarrow H_2S_2O_7(olium)$

Question 7.22 How is $SO_{2}$ an air pollutant?

Answer:

Sulphur dioxide $SO_2$ is considered an air pollutant because it readily undergoes oxidation in the atmosphere to form Sulphur trioxide $SO_3$ which then reacts with water vapour to form sulphuric acid $H_2SO_4$ . Which comes down in the form of acid rain. acid rain causes deforestation which is also not good for the environment.
Even in low concentration of $SO_2$ causes the irritation in the eyes, respiratory problem and due to this it affects the larynx to cause breathlessness.
Harmful for plants also, long exposure to $SO_2$ can reduce the colour of the leaves. It is because the formation of chlorophyll is affected by sulphur dioxide.

Question 7.23 Why are halogens strong oxidising agents?

Answer :

Halogens have 7 electrons in their valance shell and they need only one more electron to complete their octet and to attain the stable noble gas configuration. So they have a high tendency to gain an electron. Also, halogens are highly electronegative with low dissociation energies and high negative electron gain enthalpies which just increase the tendency to gain an electron. Hence they are strong oxidising agent.

Question 7.24 Explain why fluorine forms only one oxoacid, $HOF$ .

Answer :

In fluorine d-orbitals are absent and also it has very high electronegativity and small size.o it shows only +1 oxidation state in oxo-acid, but not + 3, + 5 or + 7. Hence It forms only one oxoacid $HOF$ and doesn't form oxoacid having other oxidation states than +1 like $HOFO,HOFO_2 \:and \:HOFO_3.$

Question 7.25 Explain why inspite of nearly the same electronegativity, nitrogen forms hydrogen bonding while chlorine does not.

Answer :

Inspite of nearly the same electronegativity, nitrogen forms hydrogen bonding while chlorine does not, the reason behind this is the small size of nitrogen atom as compared to the chlorine atom. The small size makes electron density per volume higher.

Question 7.26 Write two uses of $CIO_{2}$ .

Answer :

Uses of $CIO_{2}$ .

$ClO_2$ is used as a bleaching agent for paper pulp and in textiles
$ClO_2$ is used as a germicide in water treatment.

Question 7.27 Why are halogens coloured?

Answer :

All halogens are coloured due to the absorption of radiations which comes under visible region, which results in the excitation of outer electrons to higher energy level. The different quanta of radiation absorb by different halogens and they display different colours, for example is, $F_2$ , has yellow, $Cl_2$ , greenish yellow, $Br_2$ , red and $I_2$ , violet colour.

Question 7.28 Write the reactions of $F_{2}$ and $CI_{2}$ with water.

Answer :

$Cl_2$ with water

$Cl_2+H_2O\rightarrow HCl+HOCl$

$F_2$ with water

$2F_2+2H_2O\rightarrow 4H^++4F^-+O_2+4HF$

Question 7.29 How can you prepare $CI_{2}$ from $HCL$ and $HCL$ from $CI_{2}$ ? Write reactions only.

Answer :

Chlorine has a great affinity for hydrogen. Chlorine reacts with hydrogen-containing compounds to form HCl.

$H_2+Cl_2\rightarrow 2HCl$

$H_2S+Cl_2\rightarrow 2HCl+S$

$C_{10}H{16}+8Cl_2\rightarrow 16HCl+10C$

HCL to Chlorine

$4HCL+O_2\rightarrow 2Cl_2+2H_2O$

Question 7.30 What inspired $N$ . Bartlett for carrying out reaction between $Xe$ and $PtF_{6}$ ?

Answer :

Initially, he prepared a red compound of formula $O_2^+ PtF_6^-$ with the help of oxygen and $PtF_{6}$ . Later, he realised that the first ionisation enthalpy of molecular oxygen (1175 kJ/mol) and that of xenon (1170 kJ/mol) are almost identical and then he tried to prepare the same type of compound with Xe and $PtF_{6}$ . He was successful in preparing another red colour compound. $Xe^+$ $PtF_6^-$

Question 7.31 What are the oxidation states of phosphorus in the following:

$H_{3}PO_{_{3}}$

Answer :

It is known that the oxidation state of H = 1 and O is -2.
Let the oxidation state of $P$ be x
$\\3 + x + 3(-2) = 0\\ 3 + x - 6 = 0\\ x - 3 = 0\\ x = 3$

hence oxidation state of $H_{3}PO_{_{3}}$ is 3

Question 7.31 What are the oxidation states of phosphorus in the following:

$PCl_{3}$

Answer:

It is known that the oxidation state of chlorine is -1
let oxidation state $P$ be $x$
$\\x + 3(-1) = 0\\ x - 3 = 0\\ x = 3$

hence oxidation state phosphorus in $PCl_{3}$ is +3

Question 7.31 What are the oxidation states of phosphorus in the following:

$Ca_{3}P_{2}$

Answer :

We know that the oxidation state of calcium is +2
let oxidation state $P$ be $x$
$\\3(+2) + 2(x) = 0\\ 6 + 2x = 0\\ 2x = -6\\ x = -6 / 2\\ x = -3$

Hence the oxidation state of the phosphorus is -3

Question 7.31 What are the oxidation states of phosphorus in the following:

$Na_{3}PO_{4}$

Answer :

we know the oxidation state of sodium( $Na$ ) is +1 and oxygen( $O_{2}$ ) is -2
Let Oxidation state = x

$\\3(+1) + x + 4(-2) = 0\\ 3 + x - 8 = 0\\ x - 5 = 0\\ x = 5$

Thus the oxidation state of phosphorus in $Na_{3}PO_{4}$ is +5

Question 7.31 What are the oxidation states of phosphorus in the following:

$POF_{3}$

Answer :

It is known that the oxidation state of the oxygen and fluorine are -2 and -1 respectively
Let oxidation state be x
$\\x + (-2) + 3(-1) = 0\\ x - 2 - 3 = 0\\ x - 5 = 0 \\ x = 5$
So, the oxidation state of the phosphorus in $POF_{3}$ is +5

Question 7.32 Write balanced equations for the following:

(i) $NaCl$ is heated with sulphuric acid in the presence of $MnO_{2}$ .

Answer :

Nacl is heated with sulphuric acid in presence of Kmno4

$4NaCl + MnO_2+4H_2SO_4\rightarrow MnCl_2+4NaHSO_4+2H_2O+Cl_2$

Question 7.32 Write balanced equations for the following:

(ii) Chlorine gas is passed into a solution of $NaI$ in water.

Answer :

Chlorine gas is passed into a solution of water

$Cl_2+2NaI\rightarrow 2NaCl+I_2$

Question 7.33 How are xenon fluorides $XeF_{2}$ , $XeF_{4}$ and $XeF_{6}$ obtained?

Answer :

Under different concentration of Xenon, it forms , XeF2 , XeF4 and XeF6 by the direct reaction.

$F_2+Xe(in-excess)\rightarrow XeF_2;under-673k,1bar$

$2F_2+Xe(1:5ratio)\rightarrow XeF_4;under-873k,7bar$

$3F_2+Xe(1:20ratio)\rightarrow XeF_6;under-573k,65bar$

XeF6 can also be made by the interacting $XeF_4$ and $O_2F_2$ at 143K.

$XeF_4+O_2F_2\rightarrow XeF_6+O_2$

Question 7.34 With what neutral molecule is $ClO^{-}$ isoelectronic? Is that molecule a Lewis base?

Answer :

Total electrons in $ClO^- = 17 + 8 + 1 = 26$

$ClO^-$ is isoelectronic with two neutral molecules. And these two are $ClF$ and $OF_2$

In $ClF= 17 + 9 = 26$

In $OF_2=8+ \:9\times 2=26$

both species also contain 26 electrons.

Question 7.35 How are $XeO_{3}$ and $XeOF_{4}$ prepared?

Answer :

When we do Hydrolysis of $XeF_4$ and $XeF_6$ with water we get $XeOF_3$

$6XeF_4+12H_2O\rightarrow 4Xe+2XeO_3+24HF+3O_2$

$XeF_6+3H_2O\rightarrow XeO_3+6HF$

And Partial hydrolysis of $XeF_6$ gives us, $XeOF_4$

$XeF_6+H_2O\rightarrow XeOF_4+2HF$

Question 7.36 Arrange the following in the order of property indicated for each set:

(i) $F_{2}$ , $CI_{2}$ , $Br_{2}$ , $I_{2}$ - increasing bond dissociation enthalpy.

Answer :

Bond dissociation energy usually decreases as we move down in a group, Bond dissociation energy usually decreases as the atomic size increases. whereas, the bond dissociation energy of $F_2$ is lower than that of $Cl_2$ and $Br_2$ . This is due to the small atomic size of fluorine. hence,

Question 7.36 Arrange the following in the order of property indicated for each set:

(ii) $HF$ , $HCl$ , $HBr$ , $HI$ - increasing acid strength.

Answer :

The dissociation energy of bond of H-X molecules where X = F, Cl, Br, I, decreases as we increase the atomic size.

HI is the strongest acid Since H-I bond is the weakest

$HF<HCl<HBr<HI$

Question 7.36 Arrange the following in the order of property indicated for each set:

(iii) $NH_{3}$ , $PH_{3}$ , $AsH_{3}$ , $SbH_{3}$ , $BiH_{3}$ – increasing base strength.

Answer :

As we move from nitrogen to bismuth, the size of the atom increases and the electron density on the atom decreases. hence, the basic strength will decrease.

$BiH_3<SbH_3<AsH_3<PH_3<NH_3$

Question 7.37 Which one of the following does not exist?

$(i)\:XeOF_{4}$

$(ii)\:NeF_{2}$

$(iii)\:XeF_{2}$

$(iv)\:XeF_{6}$

Answer :

$NeF_{2}$ does not exist because neon has very high ionization enthalpy. But ionization enthalpy of xenon is low.

Question 7.37 Which one of the following does not exist?

$NeF_{2}$

Answer :

NeF2 Does not Exists.

Question 7.37 Which one of the following does not exist?

$XeF_{2}$

Answer :

XeF2 Exists.

Question 7.37 Which one of the following does not exist?

$XeF_{6}$

Answer :

XeF6 Exists.

Question 7.38 Give the formula and describe the structure of a noble gas species which is isostructural with:

$ICI_{4}^{-}$

Answer :

XeF ₄ has square planar geometry and is isoelectronic with ICl ^- _{4 .}

1649395779836

Question 7.38 Give the formula and describe the structure of a noble gas species which is isostructural with:

$IBr_{2}^{-}$

Answer :

XeF ₂ has a linear structure and is isoelectronic to IBr ^- ₂ and

1649395817482

Question 7.38 Give the formula and describe the structure of a noble gas species which is isostructural with:

$BrO_{3}^{-}$

Answer :

XeO ₃ has a pyramidal molecular structure and is isostructural to BrO ^- ₃ and.

1649395849919

Question 7.39 Why do noble gases have comparatively large atomic sizes?

Answer :

The atomic radius of an element corresponds to the covalent radius. but noble gases do not form any molecule, so for them, the radius is Vander walls radius. Vander wall radius is larger than the covalent radius.

Question 7.40 List the uses of neon and argon gases.

Answer :

We use Neon in discharge tubes and fluorescent bulbs for advertisement display purposes. Neon bulbs are used in botanical gardens and in green houses. We use Argon mainly to provide an inert atmosphere during metallurgical processes involving high temperature(arc welding of metals or alloys) and for filling electric bulbs. We use It in the laboratory too for handling substances that are air-sensitive

NCERT Solutions Class 12 Chemistry

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

NCERT Solutions for Class 12 Subject wise

NCERT Solutions for class 12 biology

NCERT solutions for class 12 maths

NCERT solutions for class 12 chemistry

NCERT Solutions for class 12 physics

Features of P block elements Class 12 NCERT Solutions

P block elements Class 12 chapter 7 the p-block elements, there are 34 intext questions and 40 questions are given in the exercise. P block elements Class 12 , you will learn the preparation, properties, and uses of dinitrogen, phosphorous, dioxygen, ozone, simple oxides, chlorine and hydrochloric acid and also study the uses of noble gases.

The properties of p-block elements in comparison to others are greatly influenced by atomic sizes, ionisation enthalpy, electron gain enthalpy, and electronegativity. Also, this group has all the three types of elements, metals, non-metals, and metalloids. As this is one of the most important chapter of inorganic chemistry hence it is necessary for you to clear your doubts before moving further to other chapters. The NCERT solutions for Class 12 Chemistry chapter 7 The p-block elements will help you to score well in your exams.

After completing NCERT textbook Class 12 Chemistry chapter 7 the p-block elements students will be able to explain the general trends in the chemistry of elements of groups 15, 16, 17 and 18 and also able to explain the importance of these elements and their compounds in our daily life. The p-block elements are given in the following table-

13	14	15	16	17	18
					He
B	C	N	O	F	Ne
Al	Si	P	S	Cl	Ar
Ga	Ge	As	Se	Br	Kr
In	Sn	Sb	Te	I	Xe
Ti	Pb	Bi	Po	At	Rn

Benefits of NCERT solutions for class 12 chemistry chapter 7 The p-block elements

The step-by-step answers given in the NCERT solutions for Class 12 Chemistry chapter 7 The p-block elements will help you in understanding chapter easily.
Revision will be easier because with the help of solutions, you will always remember the concepts and can get very good marks in your class.
Homework will not be a problem now, all you need to do is check the detailed NCERT solutions for Class 12 chemistry a nd you are good to go.

If you have a doubt or question that is not available here or in any of the chapters, contact us. You will get all the answers that will help you score well in your exams.

Also Check NCERT Books and NCERT Syllabus here:

Frequently Asked Questions (FAQs)

1. What is the weightage of NCERT Class 12 Chemistry chapter 7 in JEE Mains?

This chapter holds Weightage of around 5-6 marks. Other than NCERT questions students can refer to the NCERT exemplar questions and JEE Main previous year papers.

2. What is the weightage of NCERT Class 12 Chemistry chapter 7 in CBSE board exam ?

Around 8 marks questions can be expected for the CBSE board exam. Follow NCERT syllabus for a good score.

3. Where can I find complete solutions of NCERT class 12 Chemistry?

For complete solutions : https://school.careers360.com/ncert/ncert-solutions-class-12-chemistry

4. What are the important topics of this chapter?

Structures
Preparation of components by commercial and laboratory preparation of N2,P4010,P4O5,O2,etc .
Difference and general properties like atomic radii, melting point increase or decrease or just like questions are coming most .

Also Read

NCERT Class 12 Maths Notes - Download Chapter wise Free PDFs

NCERT Class 12 Physics Chapter 9 Notes Ray Optics and Optical Instruments - Download PDF

NCERT Class 12 Physics Chapter 8 Notes Electromagnetic Waves - Download PDF

NCERT Class 12 Physics Chapter 7 Notes Alternating Current - Download PDF

NCERT Class 12 Physics Chapter 6 Notes Electromagnetic Induction - Download PDF

NCERT Class 12 Physics Chapter 5 Notes Magnetism and Matter - Download PDF

NCERT Books for Class 12 - Download All Subjects PDFs Here

NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12 - Integrals

NCERT Solutions for Exercise 4.1 Class 11 Maths Chapter 4 - Principle of Mathematical Induction

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry

NCERT Exemplar Class 12 Physics Solutions Chapter 8 Electromagnetic Waves

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer

Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

Read Complete Answer

Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

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Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024

Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

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kumardurgesh1802 10 July,2024

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Central Board of Secondary Education Class 12th Examination

JEE Test Series

Applications Open

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

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) $20,000 \, \, J - 50,000 \, \, J$

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

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 .

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

13	14	15	16	17	18
					He
B	C	N	O	F	Ne
Al	Si	P	S	Cl	Ar
Ga	Ge	As	Se	Br	Kr
In	Sn	Sb	Te	I	Xe
Ti	Pb	Bi	Po	At	Rn

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.

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

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

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

13	14	15	16	17	18
					He
B	C	N	O	F	Ne
Al	Si	P	S	Cl	Ar
Ga	Ge	As	Se	Br	Kr
In	Sn	Sb	Te	I	Xe
Ti	Pb	Bi	Po	At	Rn

NCERT Solutions for Class 12 Chemistry Chapter 7 The p-block elements

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13	14	15	16	17	18
					He
B	C	N	O	F	Ne
Al	Si	P	S	Cl	Ar
Ga	Ge	As	Se	Br	Kr
In	Sn	Sb	Te	I	Xe
Ti	Pb	Bi	Po	At	Rn