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

NCERT solutions 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 class 11 you have studied only two groups 13 and 14, so in solutions of NCERT 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. 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 CBSE NCERT solutions for class 12 chemistry chapter 7 The p-block elements will help you to score well in your exams.

If you are appearing for boards exam then NCERT solutions for class 12 chemistry chapter 7 The p-Block elements 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. In chapter 7 the p-block elements, there are 34 intext questions and 40 questions are given in the exercise. The step-by-step NCERT solutions are arranged in a sequential manner which are prepared by subject experts. In this chapter, 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.

Intext Questions

Exercise Questions

After completing 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

## Question 7.1     Why are pentahalides more covalent than trihalides ?

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.

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.

$N_{2}$  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.

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-

1. pressure = $200 \:atm \:or\:200\times 10^5 Pa$,

2. Temperature of around 700 K

3. catalyst = Iron oxide

4. Promotor= small amounts of $K_2O$ and $Al_2O_3$ to increase the rate of attainment of equilibrium.

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.

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

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

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′.

$PH_3$ 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

$PH_3 + HI\rightarrow PH_4I$

$PH_3 + Acid\rightarrow salt$

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.

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

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}$

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

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

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

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

$H_3PO_4$ 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.

• 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.

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.

$H_{2}O$  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.

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

Question 7.17     Complete the following reactions:

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

The reaction is:

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

Question 7.17     Complete the following reactions:

(ii)  $4AI+3 O_{2}\rightarrow$

The complete reaction is:

$4AI+3 O_{2}\rightarrow$ $2Al_2O_3$

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

$O_3\rightarrow O_2+O$

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.

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

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

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

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 $SO_{2}$.

1. Manufacture of fertilisers (e.g., ammonium sulphate, superphosphate) from $H_2SO_4$.

2. Use is petroleum refining

3. Manufacture of pigments, paints and dyestuff intermediates and detergent industry.

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.

$H_2SO_4$ 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.

$F_2$ 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.

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.

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.

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$\dpi{80} \left [ O \right ]$  $\rightarrow$Colourless substance

The poisonous gases which we can be prepared from chlorine are

1. phosgene ($COCl_2$

2. mustard gas $ClCH_2CH_2SCH_2CH_2CL$.

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?

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

Question 7.33     Balance the following equation:

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

The balanced reaction is:

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

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

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.

Nitrogen is a diatomic molecule$\dpi{80} N\equiv N$. 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 $\dpi{80} P-P$ single bond (213KJ/mol) is weaker than $\dpi{80} N\equiv N$ triple bond (941KJ/mol) hence they both react differently..

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: $\dpi{100} E_2O_3\: and\: E_2O_5$  where E = N, P, As, Sb, or Bi.

React with halogens in order to form two series of salts: $\dpi{100} EX_3$ and $\dpi{100} EX_5$. Except $\dpi{100} NBr_{3},\:NI_{3} \:and \:NX_{5}$ because it lacks the d-orbital.

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

$NH_{3}$  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

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 HNO3 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?

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-

1. pressure = $200 \:atm \:or\:200\times 10^5 Pa$,

2. Temperature of around 700 K

3. catalyst = Iron oxide

4. Promotor= small amounts of $K_2O$ and $Al_2O_3$ to increase the rate of attainment of equilibrium.

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 $\dpi{80} HNO_3$

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

(i) Cu reacts with conc. $\dpi{80} HNO_3$

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

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

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.

$N$ 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.

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 $\dpi{100} BiH_{3}$ is less than ammonia.

The nitrogen atom has small size and high electronegativity due to this nitrogen form $\dpi{100} p\pi -p\pi$ 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 $\dpi{100} p\pi -p\pi$ 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 $P_{4}$.

 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

The single $\dpi{80} N-N$ bond in nitrogen is weaker than the single $\dpi{80} P-P$ bond because of high interelectronic repulsion of the non-bonding electrons in $\dpi{80} N_{2}$, due to the small bond length. Therefor,  the catenation tendency is weaker in nitrogen.

When we heat, orthophosphorus acid (H3PO3) disproportionates into orthophosphoric acid (H3PO4) and phosphine (PH3).

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

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-

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

• 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?

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.

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?

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

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.

$2SO_2+O_2\rightarrow 2SO_3$

$SO_3+H_2SO_4\rightarrow H_2S_2O_7(olium)$

•  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?

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.

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$HOFO,HOFO_2 \:and \:HOFO_3.$

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.

Uses of $\dpi{80} CIO_{2}$.

1.  $\dpi{80} ClO_2$  is used as a bleaching agent for paper pulp and in textiles

2. $\dpi{80} ClO_2$  is used as a germicide in water treatment.

Question 7.27   Why are halogens coloured?

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.

$Cl_2$ with water

$Cl_2+H_2O\rightarrow HCl+HOCl$

$F_2$ with water

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

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

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

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^-$

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

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

$PCl_{3}$

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

$Ca_{3}P_{2}$

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

$Na_{3}PO_{4}$

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

$POF_{3}$

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  $MnO_{2}$.

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.

Chlorine gas is passed into a solution of water

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

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.

$XeF_4+O_2F_2\rightarrow XeF_6+O_2$

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.

• 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, $XeOF_4$

$XeF_6+H_2O\rightarrow XeOF_4+2HF$

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

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,

$I_2< F_2

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

Question 7.37    Which one of the following does not exist?

$(i)\:XeOF_{4}$

$(ii)\:NeF_{2}$

$(iii)\:XeF_{2}$

$(iv)\:XeF_{6}$

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

$ICI_{4}^{-}$

XeF4  has square planar geometry and is isoelectronic with  ICl-4 .

$IBr_{2}^{-}$

XeF2 has a linear structure and is isoelectronic to  IBr-2 and

$BrO_{3}^{-}$

XeOhas a pyramidal molecular structure and is isostructural to BrO-and.

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.

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 CBSE NCERT solutions for class 12 chapter 1 The Solid State Chapter 2 NCERT solutions for class 12 chemistry chapter 2 Solutions Chapter 3 Solutions of NCERT class 12 chemistry chapter 3 Electrochemistry Chapter 4 CBSE NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics Chapter 5 Solutions of NCERT 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 CBSE NCERT solutions for class 12 chemistry chapter 7 The P-block elements Chapter 8 Solutions of NCERT 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 Solutions of NCERT class 12 chemistry chapter 10 Haloalkanes and Haloarenes Chapter 11 CBSE NCERT solutions for class 12 chemistry Alcohols, Phenols, and Ethers Chapter 12 Solutions of NCERT class 12 chemistry chapter 12 Aldehydes, Ketones and Carboxylic Acids Chapter 13 NCERT solutions for class 12 chemistry chapter 13 Amines Chapter 14 CBSE NCERT solutions for class 12 chemistry chapter 14 Biomolecules Chapter 15 Solutions of NCERT 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

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