CHAPTER – 7 p-Block Elements

Anomalous properties : The first member of each group i.e second period element differ considerably from the other family members in terms of properties due to small size ,high charge/ radius ratio , high electronegativityand absence of vacant d-orbital .

Anomalous properties of nitrogen : Nitrogen differs from the rest of the members of this group due to its smaller size, high electronegativity, high ionisation enthalpy and non-availability of d orbitals.

Absence or Presence of Vacant d- orbital : Another factor which affects the chemistry of nitrogen is the absence of d orbitals in its valence shell. Besides restricting its covalency to four, nitrogen cannot form d–pbond as the heavier elements can e.g., R3P = O or R3P = CH2 (R = alkyl group). Phosphorus and arsenic can form d–pbond also with transition metals when their compounds like P(C2H5)3 and As(C6H5)3 act as ligands

Nitrogen is restricted to a maximum covalency of 4 since only four (one s and three p) orbitals are available for bonding.

The heavier elements have vacant d orbitals in the outermost shell which can be used for bonding (covalency) and hence, expand their covalence as in PF6ˉ.

Multiple Bond formation ability ( PΠ --- PΠ ) with atoms of Small size and high electronegativity /Heavier elements don’t form PΠ --- PΠ bonds because their atomic orbitals are too large and diffuse to have effective overlapping .( the combined effect of size and availability of d orbitals considerably influence the ability of these elements to form pi –bonds/ lighter elements form PΠ --- PΠ bonds, heavier elements form d Π --- PΠ or d Π --- d Π bonds

Nitrogen has unique ability to form p-pmultiple bonds with itself and with other elements having small size and high electronegativity (e.g., C, O). Heavier elements of this group do not form p-pmultiple bonds as 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.

Consequently, its bond enthalpy (941.4 kJ mol–1) is very high. On the contrary, phosphorus, arsenic and antimony form single bonds as P–P, As–As and Sb–Sb while bismuth forms metallic bonds in elemental state.

However, the single N–N bond is weaker than the single P–P bond because of high interelectronic repulsion of the non-bonding electrons, owing to the small bond length. As a result the catenation tendency is weaker in nitrogen.

Anomalous behaviour of oxygen : Due to its small size and high electronegativity. small size and high electronegativity is the

presence of strong hydrogen bonding in H2O which is not found in H2S. BDE :-Their acidic character increases from H2O to H2Te. The increase in acidic character can

be explained in terms of decrease in bond (H–E) dissociation enthalpy down the group. Owing to the decrease in bond (H–E) dissociation enthalpy down the group, the thermal stability of hydrides also decreases from H2O to H2Po.

Anomalous behaviour of fluorine Like other elements of p-block present in second period of the periodic table, fluorine is

anomalous in many properties. For example, ionization enthalpy, electronegativity, enthalpy of bond dissociation and electrode potentials are all higher for fluorine than expected from the trends set by other halogens.

Also, ionic and covalent radii, m.p. and b.p. and electron gain enthalpy are quite lower than expected.

The anomalous behaviour of fluorine is due to its small size, highest electronegativity .low F-F bond dissociation enthalpy , and non availability of d orbitals in valence shell.

Most of the reactions of fluorine are exothermic (due to the small and strong bond formed by it with other elements). E.g NF3 exothermic compound BUT NCl3 endothermic compound .

It forms only one oxoacid while other halogens form a number of oxoacids. Hydrogen fluoride is a liquid (b.p. 293 K) due to strong hydrogen bonding. Other hydrogen halides are gases.

The acidic strength of these acids varies in the order: HF <HCl<HBr< HI. The stability of these halides decreases down the group due to decrease in bond (H–X) dissociation enthalpy in the order: H–F > H–Cl> H–Br > H–I.

The ionic character of the halides decreases in the order MF >MCl>MBr> MI where M is a monovalent metal. If a metal exhibits more than one oxidation state, the halides in higher oxidation state will be more covalent than the one in lower oxidation state.For e.g., SnCl4, PbCl4, SbCl5 and UF6 are more covalent than SnCl2, PbCl2, SbCl3 and UF4 respectively.

Hydrides Of group-15 , 16 and 17

GROUP-15 GROUP-16 GROUP-17NH3 H2O HFPH3 H2S HClAsH3 H2Se HBrSbH3 H2Te HIBiH3

(1) Bond Dissociation Enthalpy (BDE) : Smaller the size of the central atom higher the BDE e.g N—H > P—H > As –H > SB—H > Bi—H (a)Thermal Stability :[ Smaller the size of the central atom Higher the Bond Dissociation Enthalpy , higher the thermal Stability of hydride molecule ] (i) H2Te < H2Se < H2S <H2O (ii) BiH3 < SbH3 < AsH3< PH3< NH3 (iii) HI <HBr<HCl< HF (b) Acidic Strength :[ Larger the size of the central atom Lower the bond diossociation Enthalpy Higher the acidic strength Higher the tendency to make the H+ ion available (easy to break the bond ) ] (i) HF <HClHBr< HI (ii) H2O< H2S < H2Se < H2Te (iii) BiH3 < SbH3 < AsH3< PH3< NH3(c ) Reducing Strength :[ Larger the size of the central atom Lower the bond diossociation Enthalpy Higher the reducing strength Higher the tendency to make the H-atom available for reduction (easy to break the bond ) ](i) HF <HClHBr< HI (ii) H2O< H2S < H2Se < H2Te (iii) NH3 < PH3< AsH3<SbH3 <BiH3

(2) Boiling Point : It depends on the intermolecular force of attraction ( both Van der wall’s force and H- bond ) As per the size and molar mass , down the group size and mass of hydride increases and hence Van der wall’s force of attraction is more and Boiling point . BUT due to nH- bonding some anomaly do observed . In some cases H- Bonding is stronger than even Van der waal’’s force of Larger molecule . [ H- bonding : NH3 H2O HF ] (i) HClHBr< HI<HF (ii) H2S< H2Se < H2Te <H2O (iii) PH3< AsH3<NH3< SbH3<BiH3 ( ** Higher the volatility lesser boiling point : so the order for volatility is reverse of boiling point ) (3 ) Bond angle : [Smaller the size and more electro-negativity Larger the bond angle] (i)H2Te < H2Se < H2S < H2O (ii) BiH3 < SbH3 < AsH3< PH3< NH3

(sp3hybridisation in NH3 and only s-p bonding between hydrogen and other elements of the group )

(4) Basic Strength ( of Gr-15 elements) : Smaller the size of Gr-15 elements , larger the electron density and hence more electron donating capacity -> more basic strength ) BiH3< SbH3< AsH3< PH3< NH3

Miscellaneous concepts :-

# GROUP-15 Inert pair effect (Lower O.S gets stabilized If you go down the group-13,14,15 ):

(Tl,Pb,Bi)/the energy required to unpair the ns2 e- is not compensated by the energy realesed in the forming the two additional bonds

The only well characterized Bi (V) compound is BiF5. The stability of +5 oxidation state decreases and that of +3 stateincreases (due toinert pair effect) down the group

*Basicity also decreases in the order NH3 > PH3> AsH3> SbH3> BiH3. (small size of Nitrogen in ammonia has larger electron density and can have better electron donating capacity)

Their valence shell electronic configuration is ns2np1–6 (except He which has 1s2 configuration). The properties of p-block elements like that of others are greatly influenced by atomic sizes, ionisation enthalpy, electron gain enthalpy and electronegativity.

The absence of d-orbitals in second period and presence of d -and/or f -orbitals in heavier elements (starting from third period onwards) have significant effects on the properties of elements .

The stability of hydrides decreases from NH3 to BiH3 which can be observed from their bond dissociation enthalpy.

The reducing character of the hydrides increases. Ammonia is only a mild reducing agent while BiH3 is the strongest reducing agent amongst all the hydrides.(due to weak bond dissociation enthalpy of Bi—H , Bi being larger in size)

E2O3 and E2O5. The oxide in the higher oxidation state of the element is more acidic than that of lower oxidation state. Their acidic character decreases down the group. The oxides of the type E2O3 of nitrogen and phosphorus are purely acidic, that of arsenic and antimony amphoteric and those of bismuth is predominantly basic.

Nitrogen does not form pentahalide due to non-availability of the d orbitals in its valence shell. Pentahalides are more covalent than trihalides (Fajan’s Rule)

Dinitrogen is rather inert at room temperature because of the high bond enthalpy of N = N bond. Reactivity, however, increases rapidly with rise in temperature.

The optimum conditions for the production of ammonia are a pressure of 200 × 10 5 Pa (about 200 atm), a temperature of ~ 700 K and the use of a catalyst such as iron oxide with small amounts of K2O and Al2O3 to increase the rate of attainment of equilibrium

The presence of a lone pair of electrons on the nitrogen atom of the ammonia molecule makes it a Lewis base. It donates the electron pair and forms linkage with metal ions and the formation of such complex compounds finds applications in detection of metal ions such as Cu2+, Ag+:

Cu2+ (aq) + 4 NH3(aq) [Cu(NH3)4]2+(aq)

(blue) (deep blue) * In gaseous and liquid phases, PCl5 has a trigonalbipyramidal structure. The three equatorial P–Cl bonds are equivalent, while the two axial bonds are longer than equatorial bonds. This is due to the fact that the axial bond pairs suffer more repulsion as compared to equatorial bond pairs. That is why PCl5 is reactive.* The acids which contain P–H bond have strong reducing properties. Thus, hypophosphorous acid is a good reducing agent as it contains two P–H bonds and reduces * These P–H bonds are not ionisable to give H+ and do not play any role in basicity. Only those H atoms which are attached with oxygen in P–OH form are ionisable and cause the basicity. Thus, H3PO3 and H3PO4 are dibasic and tribasic, respectively as the structure of H3PO3 has two P–OH bonds and H3PO4 three. Brown Ring Test: The familiar brown ring test for nitrates depends on the ability of Fe2+ to reduce nitrates to nitric oxide, which reacts with Fe2+ to form a brown coloured complex. The test is usually carried out by adding dilute ferrous sulphate solution to an aqueous solution containing nitrate ion, and then carefully adding concentrated sulphuric acid along the sides of the test tube.

A brown ring at the interface between the solution and sulphuric acid layers indicate the presence of nitrate ion in solution.

NO3+ 3Fe2+ + 4H+ NO + 3Fe3+ + 2H2O [Fe (H2O)6 ]2+ + NO [Fe (H2O)5 (NO)]2+ + H2O

(brown) GROUP-16 (Chalcogens)

* The elements of group-16 have lower ionisation enthalpy values compared to those of Group15 in the corresponding periods. This is due to the fact that Group 15 elements have extra stable half-filled p orbitals electronic configurations .Catenation properties:(higher sigma bond strength-higher self linking property/also based electronegativity) Sigma Bond Strength : O—O < S—S (small size 2p orbital , inter electronic -lone pair repulsion of Oxygen weakens the bond ) ( Catenation properties)

* Oxides can be simple (e.g., MgO, Al2O3 ) or mixed (Pb3O4, Fe3O4). Simple oxides can be classified on the basis of their acidic, basic or amphoteric character. Basic Oxide- An oxide that combines with water to give an base is termed basic oxide e.g Metal oxide(CaO, BaO, Na2O, etc.)

Acidic oxide-An oxide that combines with water to give an acid is termed acidic oxide (e.g., SO2, Cl2O7, CO2, N2O5 ). As a general rule, only non-metal oxides are acidic but oxides of some metals in high oxidation state also have acidic character (e.g., Mn2O7, CrO3, V2O5)Amphoteric oxides :- They show characteristics of both acidic as well as basic oxides. Al2O3, PbO, ZnONeutral oxides:- neither acidic nor basic e.g CO, NO and N2O.

Passive oxide layer – BeO , Al2 O3 , PbO , PbO2presence of unpaired electron., odd electron , paramagnetism oxidizing and dehydrating action of conc. H2SO4

* Ozone Ozone is thermodynamically unstable with respect to oxygen since its decomposition into oxygen results in the liberation of heat (H is negative) and an increase in entropy (S is positive). These two effects reinforce each other, resulting in large negative Gibbs energy change (G) for its conversion into oxygen. It is not really surprising, therefore, high concentrations of ozone can be dangerously explosive. =>Due to the ease with which it liberates atoms of nascent oxygen (O3O2 + O), it acts as a powerful oxidising agent. For e.g., it oxidizes lead sulphide to lead sulphate and iodide ions to iodine.

* Ozone layer depletion

NO combine very rapidly with ozone and there is, thus, the possibility that nitrogen oxides emitted from the exhaust systems of supersonic jet aeroplanes might be slowly depleting the concentration of the ozone layer in the upper atmosphere.

NO + O3 NO2 + O2

GROUP-17: HALOGENS

Halogens have maximum negative electron gain enthalpy in the corresponding periods. This is due to the fact that the atoms of these elements have only one electron less than stable noble gas configurations.Electron gain enthalpy of the elements of the group becomes less negative down the group. *However, the negative electron gain enthalpy of fluorine is less than that of chlorine. It is due to small size of fluorine atom. As a result, there are strong interelectronic repulsions in the relatively small 2p orbitals of fluorine and thus, the incoming electron does not experience much attraction. * Fluorine is the most electronegative element in the periodic table.* All halogens are coloured. This is due to absorption of radiations in visible region which results in the excitation of outer electrons to higher energy level. By absorbing different quanta of radiation, they display different colours. For example, F2, ------ yellow Cl2 --------greenish yellow , Br2,------------ red and I2------------violet colour.* The smaller enthalpy of dissociation of F2 compared to that of Cl2 whereas X-X bond dissociation enthalpies from chlorine onwards show the expected trend: Cl – Cl> Br – Br > I – I. A reason for this anomaly is the relatively large electron-electron repulsion among the lone pairs in F2 molecule where they are much closer to each other than in case of Cl2.* The ready acceptance of an electron is the reason for the strong oxidising nature of halogens. F2 is the strongest oxidising halogen and it oxidises other halide ions in solution or even in the solid phase. * Fluorine oxidises water to oxygen whereas chlorine and bromine react with water to form corresponding hydrohalic and hypohalous acids. The reaction of iodine with water is nonspontaneous

Fajan’sRule :- Explains about covalent character in ionic bond . e.g -- BeCl2 is more covalent than CaCl2Be has higher charge by radius ratio has higher polarizing power and hence higher covalent character [ Condition for higher covalent character : -- FOR CATION :- Higher polarizing power of cation , FOR ANION :- larger the size -- higher polarizability.) e.g SnCl4 is more covalent than SnCl2 .

Bleaching action of Chlorine :-Chlorine water on standing loses its yellow colour due to the formation of HCl and HOCl. Hypochlorous acid (HOCl) so formed, gives nascent oxygen which is responsible for oxidising and bleaching properties of chlorine .

Bleaching action of Chlorine is permanent due its oxidizing action(atmosphere is an oxidizing character) but of SO2 it is temporary due to its reducing action(atmospheric O2 oxidizes back to its previous state).

Aqua regia :When three parts of concentrated HCl and one part of concentrated HNO3 are mixed, aqua regia is formed which is used for dissolving noble metals, e.g., gold, platinum

Oxo acids of halogensDue to high electronegativity and small size, fluorine forms only one oxoacid, HOF known as fluoric (I) acid or hypofluorous acid. The other halogens form several oxoacids. Most of them cannot be isolated in pure state. They are stable only in aqueous solutions or in the form of their salts 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.

Group-18 : Noble gases and its compounds

* Due to stable electronic configuration these gases exhibit very high ionisation enthalpy. However, it decreases down the group with increase in atomic size. *Since noble gases have stable electronic configurations, they have no tendency to accept the electron and therefore, have large positive values of electron gain enthalpy.

* Only Xe forms large no .of compounds because of comparatively low I.E .Xe forms compounds only with Oxygen and Fluorine . O and F being the small size and highly electronegative form strong bond with Xe and large amount of energy is released which can compensate the energy required for ionization in Xe.*They are sparingly soluble in water, very low melting and boiling points because the only type of interatomic interaction in these elements is weak dispersion forces. Xe is more soluble than He,Ne,Ar,Kr in water. Being larger size of Xe ,Itschage cloud easily gets distorted or polarized and have more London Force . *Neil Bartlett , First, he prepared a red compound which is formulated as O2+PtF6ˉ. He, then realised that the first ionization enthalpy of molecular oxygen (1175 kJmol–1) was almost identical with that of xenon (1170 kJ mol–1). He made efforts to prepare same type of compound with Xe and was successful in preparing another red colour compound Xe+PtF6 ˉ by mixing PtF6 and xenon . LOGIC—If O2 reacts with PtF6 then why not Xe with PtF6 ?

CHECK-UP LIST:- ARRANGEMENT IN terms of properties mentioned.

1# I—I <F—F< Br—Br <Cl—Cl (Bond Dissociation Enthalpy)—Inter-electronic repulsion 2# H—F < H—Cl< H—Br < H—I (Acid strength)—Lower BDE of HI,large size of I 3# M—I < M—Br < M—Cl< M—F ( Ionic Character)—Fajan’s Rule—Lager polarizability of Iˉ4# BiH3< SbH3< AsH3< PH3< NH3 ( Base Strength) – small size of N – High electron density in Ammonia5# H2O < H2S < H2Se < H2Te ( Acid Strength and Reducing Character) --- BDE6# H2Te < H2Se < H2S < H2O ( Thermal Stability) ---BDE7# H2O < H2Te < H2Se < H2S (Volatility)--- H-Bond and Vander waal’s force 8# H2S <H2Se < H2Te < H2O ( Boiling Point) -- H-Bond and Vander waal’s force9# PH3< AsH3< NH3< SbH3< BiH3 (Boiling Point) -- H-Bond and Vander waal’s force 10# I2< Br2< Cl2< F2 (Oxidizing Ability) --- Electron gain enthalpy, hydration enthalpy, dissociation enthalpy .11# H2Te < H2Se < H2S < H2O ( Bond Angle )----- Size of central atom , electronegativity, repulsion of bond pairs. 12# HClO4< HClO3< HClO2<HClO ( Oxidizing Power) 13# HOCl<HOClO< HOClO2< HOClO3 ( Acid Strength) – Stability of its conjugate base , charge dispersal , Oxidation states. 14 # HOI <HOBr<HOCl ( Acid Strength) ---Stronger the O—X bond – Weaker the O—H bond – More the acidic character.15# Cl2O <ClO2< Cl2O6< Cl2O7 ( Acid Strength)-- Higher oxidation states, covalent character 16# ClO4ˉ < BrO4ˉ > IO4ˉ or BrO4ˉ > IO4ˉ > ClO4ˉ (Oxidizing Power)17# Ga2O3< GeO2< As2O3< ClO2 (Acidic Character) 18# BF3< BCl3< BBr3 (Acidic Character)—Effective 2p—2p overlap in BF3 reduces the electron deficiency of B , make it less acidic.19# Pb<Sn = Ge< Si << C ( Catenation Property) – Sigma bond strength20# C++< Si++<Sn++<Pb++ ( Stability) ---- Inert pair effect21# N2O < NO < N2O3< N2O4< N2O5 (Acidic Character) – Higher oxidation states, covalent character 23# HCl< H2SO4 < HNO3 (Oxidizing Acid) 24# (i) N < P > As >Sb (ii) O < S > Se >Te ( catenation property) --- ( P , N ) ; ( O ; S )25# (a) HNO2< HNO3 (b) H2SO3< H2SO4 (Acidic character)26# Iˉ < Brˉ <Clˉ < Fˉ (Stability) 27# Reactivity---ClF3> BrF5> IF7>ClF7> BrF3>IF5>BrF> IF3>IF28# Stability—ClO4ˉ > ClO3ˉ > ClO2ˉ >ClOˉ

CHECK LIST :- SHAPE AND STRUCTURE BASED ON VSEPR THEORY

SN

HYBRIDISATION

NO OF ELECTRON PAIR

NO OF LP AND BP

Geometry Shape Important Examples (CBSE)(ISOSTRUCTURAL)

Additional e.g

1 sp 2 e- pair 2 BP linear Linear NO2 +

2 sp2 3 electron pair

3 BP Trigonalplanar

Trigonalplannar BF3 , N(SiH3)3 , NO3

‾ , CO32‾ ,

2 BP +1 LP

Bent NO2 , NO2ֿ

3 sp3 4 electron pair

4 BP tetrahedral tetrahedral ClO4‾

(perchlorate), BrO4

‾ (perbromate) H3PO4 , H3PO3 , H3PO2 , SO4

2-

Ni(CO)4 , MnO4

‾ (manganate) , CrO4

2 -

(chromate)[NiCl4]2‾

3 BP + 1 LP

Trigonal pyramidal

BrO3‾ , XeO3 ,

PCl3 ,N(CH3)3

NH3,PH3, AsH3, SbH3, BiH3

2 BP + 2 LP

Bent or v-shape

H2O , H2S , XeO2 , OCl2 , OF2

1 BP +3 LP

Linear OCl-

Sp2d 4 electron pair

4 BP Square Plannar

Square Plannar

[PtCl4]2‾ , [Ni(CN)4]2‾

4 sp3d 5electron pair

5 BP Trigonalbipyramidal

Trigonalbipyramidal

PCl5 (gas and Liq)

Fe(CO)5

4 BP + 1 LP

See-saw SF4 , XeO2F2

3 BP + 2 LP

T-shape ClF3 , BrF3, XeOF2

2 BP + 3 LP

Lnear XeF2 , IBr2‾ ,

I3‾

5 sp3d2 6 electron pair

6 BP octahedral octahedral SF6 [Fe(CN)6]3‾

[Co(NH)3]3+ , [CrF6]3‾

5BP + 1 LP

Square Pyramidal

BrF5, XeOF4 , IF5

4 BP + 2 LP

Square Plannar

XeF4 , ICl4‾ ,

6 sp3d3 7 electron pair

7 BP Pentagonal bipyramidal

Pentagonal bipyramidal

IF7

6 BP + 1 LP

Distorted Octahedral

XeF6 (gas) , IF6

‾

Tetrahedral and octahedral

PCl5 ( solid )- [PCl4]+[ PCl6 ] ֿ

O3 (bent str.) XeF6 ( solid phase )

[XeF5]+

[ XeF7]¯

Structures of Oxoacids of N ,P ,S , Cl

STRUCTURES – OXIDES , OXOANIONS, OXOACIDS OF SULPHUR , SULPHUR COMPOUNDS , ALLOTROPES OF –SH2SO4 H2S2O3(thiosulphuric

acid )H2S2O7 (Oleum)Pyrosulphuric acid

H2S2O8(Peroxodisulphuric acid or Marshall’s acid )

H2SO3(sulphurous acid )

H2SO5(peroxomonosulphuric acid or Caro’s acid )

H2S2O6 (dithionic acid )

S6 , S8 ring(rhombic sulphur )

SO3 2‾ (sulphite )

STRUCTURES – OXIDES , OXOACIDS OF PHOSPHOROUS , PHOSPHOROUS COMPOUNDS , ALLOTROPES OF –PH3PO4(ortho-phosphoric acid )

H3PO3(phosphonic acid or phosphorous acid )

H3PO2(phosphinic acid or Hypophosphorous acid )

H4P2O7(pyrophosphoric acid )

(HPO3)n (polymetaphosphoric acid )

(HPO3)3 - cyclotrimetaphosphoric acid

White –P , Red-P P4O6 , P4O10

PF3 , PF5 , NH3,PH3, AsH3, SbH3, BiH3

HNO3 N2O5 , N2O4 N2O3 , NO2

STRUCTURES – OXIDES , OXOANIONS , OXOACIDS OF CHLORINE , CHLORINE COMPOUNDS HClO4 (perchloric acid )

HClO3(chloric acid ) HClO2 HClO(hypochlorous acid )

ClO4‾

(perchlorate), BrO4

‾ (perbromate)

Cl2O7 OCl2 , ClO2

HOTSI# IDENTIFY THE FOLLOWING:1 # An element ‘A’ exists as a yellow solid in standard state. It forms a volatile hydride ‘B’ which is a foul smelling gas and is extensively used in qualitative analysis of salts. When treated with oxygen, ‘B’ forms an oxide ‘C’ which is colourless, pungent smelling gas. This gas when passed through acidified KMnO4solution,decolourizes it. ‘C’ gets oxidized to another oxide ‘D’ in the presence of a Heterogeneous catalyst.Identify A, B, C, D and also give the chemical equation of reaction of ‘C’ with acidified KMnO4 solution and for conversion of ‘C’ to ‘D’.2# When conc. sulphuric acid was added to an unknown salt present in a test tube, a brown gas (A) was evolved. This gas intensified when copper turnings were also added into this tube. On cooling, the gas ‘A’ changed into a colourless gas ‘B’. (a) Identify the gases A and B. (b) Write the equations for the reactions involved.3# A translucent white waxy solid ‘A’ on heating in an inert atmosphere is converted in to its allotropic form (B). Allotrope ‘A’ on reaction with very dilute aqueous KOH librates a highly

poisonous gas ‘C’ having rotten fish smell. With excess of chlorine ‘A’ forms ‘D’ which hydrolysis to compound ‘E’. Identify compounds ‘A’ to ‘E’ .4# Concentrated sulphuric acid is added followed by heating to each of the following test tubes labelled (i) to (v)

Identify in which of the above test tube the following change will be observed.Support your answer with the help of a chemical equation.(a) Formation of black substance(b) Evolution of brown gas(c) Evolution of colourless gas(d) Formation of brown substance which on dilution becomes blue

5 # A gas “X” is soluble in water . Its aq. Solution turns red litmus blue with excess of aq. CuSO4 solution it give s deep blue colour with FeCl3 solution a brownish ppt. soluble in HNO3 is obtained. Identify gas”X” and write reactions for changes observed .6# Element “A” burns in N2 to give a compound “B” which is ionic in nature .” B” on reaction with water gives ‘C” and “D”.The solution of “C” becomes milky on passing CO2 through it. Identify the compound A ,B ,C ,D . 7# An orange solid “A” on heating gives a green residue”B”, water vapours and a colourless gas “C” . The gas “C” in dry condition is passed over heated Mg to give solid “D” which further reacts with water to form “E” which gives dense white fumes with HCl . Identify the compound A ,B ,C ,D and E and write the reactions involved .8# A colourless inorganic salt “A” decomposes at 2500C to form “B” and “C” leaving no residue . The oxide “C” is liquid at room temperature and is neutral towards litmus while “B” is neutral oxide and is used as anaesthetic in minor dental surgery . When Phosphorous burns in excess of –B produces compound which is a strong dehydrating agent which when treated with HNO3 produces acid “D” and a gas “E” Identify A” to “E” and write the reactions.

I# ANSWERS OF IDENTIFY THE FOLLOWING:

ANS-1#S8[A] +8H2 + heat 8H2S[B] 8H2S + 12O2 8SO2 [C] ;5SO2 + 2KMnO4(pink)/H+ +2H2O K2SO4 +2MnSO4(Colourless solution)2SO2 +O2 + Pt/heat 2SO3 (D)ANS-2# 2NaNO3 +H2SO4 +heatNa2SO4 + 2HNO34HNO3+ heat4NO2+2H2O+O2Cu+4HNO3Cu(NO3)2 +2NO2 +2H2O2NO2[A] ═(cool)=> N2O4[B]ANS-3#P4(white Phosphorous)[A] +very dil KOHPH3[C]P4(white Phosphorous)[A] +heatRed- P(red Phosphorous)[B] P4(white Phosphorous)[A] +Cl2(excess)PCl5[D]PCl5[D] +H2O(excess)H3PO4[E]ANS-4#(a) Test tube(i): C12H22O11+Conc.H2SO411H2O +12C (black subs.)(b) Test tube(ii):2NaBr +3H2SO42NaHSO4 +SO2+Br2(brown gas)(c) Test tube(v): KCl +H2SO4 +heat KHSO4+HCl(colourless gas)(d) Test tube(iii): Cu +2 H2SO4CuO(brown subs) On dilution: CuO(brown subs) +2 dil.H2SO4 CuSO4(blue substance)(e) Test Tube(iv) : S(yellow powder) + conc.H2SO4 ----SO2 +H2O

ANS-5:NH3(g) +H2O NH4OH(aq) [basic and turn red litmus blue] 3 NH4OH(aq) + CuSO4[Cu(NH3)4]SO4 +4H2O

(deep blue) 3 NH4OH(aq) +FeCl3Fe(OH)3 +3 NH4Cl Fe(OH)3 + 3HNO3 Fe(NO3)3( soluble) +3H2O ANS-6:3Ca + N2Ca3N2 [B] [A]Ca3N2 + 6 H2O3 Ca(OH)2 [C] + 2NH3 [D]Ca(OH)2 +CO2 CaCO3 + H2OANS-7: (NH4)2Cr2O7 +heat N2 + Cr2O3+ H2O (Orange solid-A) [C] [green solid B ] N2 +3Mg +heat Mg3N2 [D] Mg3N2 [D] +6H2O 3 Mg(OH)2 +2NH3 [E]NH3 [E] +HCl NH4Cl (dense white fumes)ANS-8:NH4NO3[A]+heat at 2500C N2O(g) [B] + 2 H2O [C] 10 N2O +P4 10 N2 + P4O1010 P4O10 is a dehydrating agent P4O10 + 4 HNO3 4 HPO3 [D] + 2 N2O5 (g)

Explain Why ? / Give reason ? / Account for the following ? (HOTS)QUESTIONS ANSWERS1# HBr and HI can’t be prepared by treating metal bromides or iodides with conc. H2SO4 .

ANS-1# Because Br- and I - will get oxidized by conc. H2SO4 to respective Br2 and I2

2# Why NO2 readily forms a dimmer, whereas ClO2 does not though both are odd electron molecule.

ANS-2# odd electron molecule often get dimerise in order to pair the electrons but ClO2 does not .Thisis thought to be because the odd electron is delocalized.

3# Write the balanced chemical equation for the reaction of Cl2 with hot and concentrated NaOH. Is this reaction a disproportionation reaction? Justify.

ANS-3# 6 NaOH( hot and conc.)+ 3Cl2 5 NaCl + NaClO3 + H2O .yes it a disproportionation reaction because the OS of Cl2 is Zero and is oxidized to +5 (in NaClO3) and reduced to -1 (in NaCl) . Due to simultaneous oxidation and reduction occurs of the same species i.eCl2 .

4# SCl6 is not known but SF6 is known .

ANS-4# SF6 is known because of small size and more electronegativity of Fluorine which can form strong bond and hence the energy released can compensate to promote electrons in vacant d-orbitals of S which is not possible in case of Chlorine . Due to large size of Cl which cannot be accommodated around Sulphur and stability will also be less due to inter electronic repulsion between lone pairs of Chlorine ( whose size is large as compared to F)

5# In trimethylamine, the nitrogen has a pyramidal geometry whereas in trisilylamine,it has a planar

ANS-5# Due to presence of Vacant d-orbitals in Si , it can form Pл-dл bond with nitrogen . so reduced to SP2hybridisation and becomes plannar with three sigma bonds .Which is not possible in case of trimethylamine as Carbon does have vacant d-orbital.

6# SF6 is resistant to hydrolysis whereas SF4 is readily hydrolysed.

ANS-6# SF6 is octahedral ( which is symmetrical) where six F-atoms surrounding”S” protect it from attack of H2O molecules. So ,SF6 is sterically protected then SF4 towards the even thermodynamically favorable hydrolysis reaction.Also F-atoms do not have Vacant d-orbitals to accommodate electrons from water molecules . This makes SF6 chemically inert. Due to this it suppresses internal charges hence used in high voltage generators(gaseous insulators, and switch gears.

7# NCl3 getshydrolysed to form NH3 and HOCl while PCl3 gets

ANS-7# TheCl of NCl3 has vacant d-orbital which can accommodate electrons from water molecules and get

hydrolysed to give H3PO3 and HCl . Explain

hydrolysed to produce HOCl and NH3 . But in case of PCl3 , it is the vacant d- orbital of P can accommodate electrons from water molecules and get hydrolysed to give H3PO3 and HCl . [P—O bond strength >Cl—O]

8# In the solid state , PCl5 is in ionic nature . Explain.

ANS-8# Due to unequal bond length of equatorial and axial bond that is why it is reactive . So, in solid state (in close range) , it can be stabilized by forming [PCl4]+ and [PCl6]- which are symmetrically stable tetrahedral and octahedral str. respectively .

9# I2 form I3− but F2 does not formF3− .

ANS-9# Due to presence of vacant d-orbital in Iodine it can accept the electrons of iodide ion .

10# NO is paramagnetic in gaseous state but diamagnetic in the solid and liquid state.

ANS-10# NO is an odd electron molecule (paramagnetic) . In solid and liquid state , dimer is formed and there is no unpaired electrons and act as diamagnetic

11#Yellow ppt. of Sulphur disappear when boiled with sodiumsulphite .

ANS-11# Na2SO3 + S ( yellow solid ) Na2 S2O3 ( sodium thiosulphate)

12# NF3 is more stable but NCl3 is less stable .

ANS-12# Because of small size and more electronegativity of Fluorine which can form strong bond with Nitrogen and becomes stable as compared to large size Chlorine

13# SH6 and PH5 are not formed but SF6 and PF5 are formed .

ANS-13# It is because of two reason why hydrogen cannot form SH6 and PH5 though SF6 and PF5 are formed Reason-1: Hydrogen is less electronegative element as compared to fluorine . S—H bond is weaker than S—H bond . Reason-2: The enthalpy of atomization of H—H is very high as compared to F—F . High enthalpy of dissociation can not be compensated by energy released during bond formation .

14# KHF2 is known but compound of formula KHCl2 or KHBr2 are not known .Why ?

ANS-14# KHF2 contains HF2− ion which is stable due to formation os H-bonding between H—F and F− . This is possible because of high electronegativity of F ( F—H …..H) − on the other hand , Cl and Br cannot form hydrogen bonds due to low electronegativity

15# Iodine is more soluble in KI solution than in water .

ANS-15# Due to formation of polyiodide ion .KI +I2 K+ [I3]− . Ion –dipole interaction between polyiodide ion and water responsible for solubility .

16# HF is stored in wax -coated bottle .

ANS-16# Due to formation of complex called flurosilicic acid , H2[SiF6]

17# HCl is not used to make the medium acidic in titrations involving KMnO4 .

ANS-17# Chloride gets oxidized to chlorine by conc. Sulphuric acid ( an oxidizing agent)

18 # Addition of Cl2 to KI solution gives it a brown colour but excess of Cl2 turns it colourless.

ANS-18# Cl2 + KI ( solution) KCl + I2 ( brown ) . Iodide reduces chlorine to chloride ion. . The I2formed will react with excess Cl2 in presence of water to form HIO3 which makes it colourless .

19# CN− ion is known but CP− is not ?

ANS-19# Nitrogen being small size can form stable and effective Pл-Pл bond ( in case of cyanide ion) but which is not possible in C≡P− . P is larger size which cannot form stable and effective Pл-Pл bond .

20# Comment on the nature of two S—O bonds formed in SO2 molecule. Are the two S—O bonds in this molecule.

ANS-20# S 3s2 3px23py13pz1 (Ground state) ; S 3s2 3px13py13pz1 3d1 (excited state) Sp2hybridisation in Sulphur . one half filled 3pz1 can form Pл-Pл bond with half-filled 2p orbital of one Oxygen and one half filled 3d1 of Sulphur with half-filled 2p orbital of another Oxygen can form Pл-dл bond . And due to resonance the bond lengths are equal.

21# Why anhydrous CaCl2 is not used to dry up Ammonia ?

ANS-21# Because it t forms CaCl2.8NH3 , addition compound .

22#What happens when Zinc is treated with Conc.H2SO4

ANS-22# Zn reduces Sulphate to SO2 and ultimately colloidal sulphur .As Conc. H2SO4 is an oxidizing agent .

23# Why conc.H2SO4 is not ANS-23# H2S + Conc.H2SO4 ( an oxidizing agent ) Sulphur

used for drying of H2S gas .24# When HCl reacts with finely powdered iron , it forms ferrous Chloride and not ferric chloride .

ANS-24# Its reaction with iron produces H2 . Fe + 2 HCl FeCl2 + Cl2 . Liberation of hydrogen creates a reducing environment and prevents the formation of ferric chloride .

25# Conc.HNO3 turns yellow on exposure to sunlight .Why ?

ANS-25# In presence of light nitric acid decomposes to form NO2 (brown) , O2 and water

PREVIOUS YEARS QUESTIONS – SHAPE AND STRUCTURES (P-BLOCK) – 2 MDraw the structure and write the shape of the following :-

XeF2 AI-08 ,F-10 , D-11,F-12 , AI,D -13,D,AI-14, AI-15

HClO4 AI-09 , F-10 , AI-12 ,AI,F-14

XeF4 D-08 ,F, D-09 , D,AI-10 ,D-11,D-13, D,AI-14

(HPO3)n

XeOF4 AI-09, F-10, D,AI-12 ,AI-13 , D,F-14

(HPO3)3 AI-11 ,D-13

XeO2F2 H4P2O7 F-13XeOF2 H2S2O7

(Oleum) AI-08 ,F,AI-09 ,D-12 ,AI-13

XeO3 F-13 , AI-14 H2S2O8 AI-09, F-11, D,F-12 , D, F-13 ,F-14XeF6 D-12 , F, AI-13 H2SO5ClF3 , AI-12 ,F-13 White –

P , D-10 D-14

BrF3, AI-08 ,F,D,AI-09 , D,F-10 , D,AI-11 ,F-12 ,D-13 ,D,AI-14

Red-P D-10 ,D,F-14

BrF5 PCl5 (gas)

F-13

IF7 PCl5 ( solid )

AI-09 ,F-13 ,AI-COMPTT-14

ICl4‾ H2SO4 D,AI-14

IF6‾ H3PO4

ClO4‾ H3PO3 F-11,12 ,AI-13

BrO4‾ H3PO2 AI-09 , D,AI-12 ,F-13

SF4 D-08 , AI-2015 NF3 F-11N2O5 AI-09 ,AI-12 , F,AI-13 ,D-14 H4P2O5 AI-10HClO3 AI-09 SO3

2‾ AI-09IBr2ֿ AI-comptt-14BrO3

‾ AI-comptt-14

MASTER THE REACTION – PREVIOUS YEAR BOARD QUESTIONS (P-Block)- 2M

SN Complete the following chemical reaction equations:

SN Complete the following chemical reaction equations:

P-BLOCK REACTION CARD-1 P-BLOCK REACTION CARD-61 Cu + HNO3 (CONC.) 26 P4 + Conc. HNO32 NaOH (hot and conc.) + Cl2 27 Sn + 2PCl5 (heating)3 XeF2 (s) + H2O(l) 28 AgCl(s) + NH3(aq)4 HgCl2 +2PH3 29 NH3 + Cl2 ( excess) 5 P4 + NaOH + H2O 30 Cu (s) + H 2SO 4 (conc.)

P-BLOCK REACTION CARD-2 P-BLOCK REACTION CARD-76 Cu + HNO3 (DIL.) 31 PCl5 + H2O (excess) 7 NaOH (cold and dil.) + Cl2 32 I¯ (aq ) +H2O(aq) + O3(g)8 XeF4(s) + H2O(l) 33 XeF4 + O2F29 CuSO4+2PH3 34 XeF6 (s) + H2O(l) 10 Ca3P2 + H2O 35 AgNO3 +H3PO2+ H2O

P-BLOCK REACTION CARD-3 P-BLOCK REACTION CARD-811 P4 + SO2Cl2 36 PCl3 + H2O 12 SO2 + MnO 4- + H2O 37 Br2 (g) + F2(g) (excess)

13 Xe (g) (in excess) + F2(g) , 673K 1Bar

38 SO3 + H 2SO 4 (conc.)

14 H3PO3 + (Heat) 39 XeF6 (s) + 2H2O(l) 15 XeF2 + PF5 40 CaF2 + H2SO4

P-BLOCK REACTION CARD-4 P-BLOCK REACTION CARD-916 P4 + SOCl2 41 KClO3 + Heat( in presence of MnO2)17 SO2 + Fe3+ + H2O 42 SiO2 + HF 18 Cl2 (g) + F2(g) (excess)(3000C) 43 XeF6 (s) + H2O(excess) 19 F2(g) + H2O 44 NaF + XeF620 PCl5 + Heat 45 (NH4)2 Cr2O7 + Heat

P-BLOCK REACTION CARD-5 P-BLOCK REACTION CARD-1021 I2 + HNO3 (conc.) 46 Ca(OH)2 + Cl2 (g) 22 Ag + 2PCl5 (heating) 47 Cl2 (g) + H2O 23 Cu (aq) + NH 3(excess (aq) 48 NH3 (excess) + Cl224 NH4Cl(aq) + NaNO2 (aq) 49 NaN3 + Heat25 C + H 2SO 4 (conc.) 50 Na2SO3 + Cl2 + H2O

51 Cu + Conc.H2SO452 Cl2 + F2(excess) 53 CaF2 +H2SO454 Ag+PCl5

**** Answer can be directly obtained from the reactions given below

R E A C T I O N S O F p-B L O C K E L E M E N T S 1(a)# Cu + 8 HNO3(Dil. ) Cu(NO3)2 + 2NO +4 H2O1(b)# Cu + 4 HNO3( conc.)Cu(NO3)2 + 2NO2 +2H2O

2(a)# 2NaOH (cold and dilute) + Cl2NaCl +NaOCl +H2O2(b)# 6NaOH (hot and conc.) + 3Cl2 5NaCl +NaClO3 +H2O

3(a)# 8NH3 (excess) + 3 Cl2 6NH4Cl + N2 3(b)# NH3 + Cl2 ( excess) NCl3 (explosive ) + 3 HCl

4(a)# PCl3 + H2O H3PO3 + 3 HCl4(b)# PCl5 + H2O POCl3 + 2 HCl └> + H2O H3PO4 + 3 HCl

5(a)# F2 + H2O HF + O25(b)# X2(g) + H2O(l) HX(aq) + HOX (aq) ] 1/3 ClF + H2O HF + H—O—Cl

X2(Hydrated) ] 2/3

6(a) # 2XeF2 (s) + 2H2O(l) 2Xe (g) + 4 HF(aq) + O2(g)6(b) # 6XeF4 + 12 H2O 4Xe + 2Xe03 + 24 HF + 3 O2[Hydrolysis of XeF4 and XeF6 with water gives XeO3.]

7(a)## C + 2H2SO4 (Conc.) CO2 + 2 SO2 +2H2O 7(b)# # Cu + 2H2SO4 (Conc.) CuSO4 + SO2 +2H2O

8(a)# # 4 H3PO3 3 H3PO4 + PH38(b)# (NH4)2Cr2O7 —(Heat ) - N2 + Cr2O3 +H2O8(c)# HNO2 --(Heat) HNO3 + H2O + 2NO

8(d)#PCl5 + Heat PCl3 + Cl2

9(a)# Cu + 2H2SO4 (Conc.) CuSO4 + SO2 +2H2O 9(b)# Cu + 4 HNO3( conc.)Cu(NO3)2 + 2NO2 +2H2O

10(a)#XeF6 + 3 H2O XeO3 + 6 HF10(b)#XeF6 + H2O XeOF4 + 2 HF [Partial hydrolysis of XeF6 gives oxyfluorides, XeOF4 and XeO2F2]10(c)#XeF6 + 2 H2O XeO2F2 + 4HF

11(a)#XeF2 + PF5[XeF]+ [PF6]– ; 11(b)#XeF4 + SbF5[XeF3]+ [SbF6]–

11(c)#XeF6 + MF M+ [XeF7]– (M = Na, K, Rb or Cs)

12(a)# Ca3P2 + 6H2O 3 Ca(OH)2 + 2 PH312(b)# P4 + 3NaOH + 3 H2O PH3 + 3 NaH2PO2

13(a)# 2 Fe3+ + SO2 + 2H2O 2 Fe2+ + SO42ˉ + 4 H+

13(b)# 5 SO2 + 2 MnO4 ˉ + 2 H2O 5 SO42ˉ + 4 H+ + 2 Mn2+

14(a)#P4 + 8 SOCl2 4 PCl3 +4 SO2 + 2 S2Cl214(c)#P4 + SO2Cl2 4 PCl5 + 10 SO2

15(a)# 3CuSO4 +2PH3 Cu3P2 + 3 H2SO415(b)# 3HgCl2 +2 PH3 Hg3P2 + 6 HCl

16(a)#Cu2+ (aq) + 4 NH3(aq) [Cu(NH3)4]2+(aq) (blue) (deep blue) 16(b)#AgCl (s) + 2 NH3(aq) [Ag(NH3)2] Cl (aq) {colourless soluble complex}

17(a)# NH4Cl(aq) + NaNO2 (aq) N2(g) +2H2O (l) + NaCl (aq) 17(b)# (NH4)2Cr2O7+Heat N2 + 4H2O + Cr2O317(c)# Ba(N3)2 Ba + N2

18(a)#PCl3 + H2O H3PO3 + 3 HCl18(b)# PCl5 + H2O POCl3 + 2 HCl └> + H2O H3PO4 + 3 HCl

19(a)#C2 H5 OH + PCl5C2 H 5Cl + POCl3 + HCl19(a)#CH 3COOH + PCl5CH3COCl + POCl3 +HCl

20(a)#2Ag (Finely divided )+ PCl5 2AgCl + PCl320(b)#Sn(Finely divided ) + 2PCl5 SnCl4 + 2PCl3

21(a)# NCl3 + H2O NH3 + 3 HOCl21(b)# PCl3 + H2O H3PO3 + 3 HCl

22(a)#2FeSO4 + H2SO4 + Cl2 → Fe2(SO4)3 + 2HCl22(b)#Na2SO3 + Cl2 + H2O → Na2SO4 + 2HCl22(c)#SO2 + 2H2O + Cl2 → H2SO4 + 2HCl

23(a)#Cl2+F2 [437K]→2ClF;

(equal volume)23(b)#Cl2 + 3F2 (excess) [573K ]→2ClF3 ;

24(a)#I2 +3Cl2 [437K]→2ICl3;24(b)# I2 +Cl2 (equimolar)→2ICl

25(a)#Br2 + 3F2 →2BrF3(Diluted with water)25(b)# Br2 + 5F2(excess) →2BrF5

26(a)# 2 Na2S2O3 + I2 Na2S4O6 + 2 NaI26(a)# Na2S2O3 + 4 Cl2 +5 H2O 2 NaHSO4 + 8 HCl

CONCEPT : ANOMALOUS PROPERTIES CHAP: P-BLOCK ELEMENTS CLASS-XII Give Reason for each of the following :-

CARD-1: [1×5=5]1 # NCl5 does not exist but NCl3 exits BUT both PCl3 & PCl5 exists .2# Why does R3P = O exist but R3N = O does not (R = alkyl group)?3# Nitrogen exists as diatomic molecule and phosphorus as P4. Why?4# Oxygen &nitrogen are gases BUT sulphur and phosphorus are found in solid state at room temp.5# BiCl3 is more stable than BiCl5.Explain

CARD-2: [1×5=5]

6 # Although electron gain enthalpy of fluorine is less negative as compared to chlorine, fluorine is a stronger oxidizing agent than chlorine. 7# Explain why fluorine forms only one oxoacid, HOF. 8# ClF3 exists but FCl3 does not Explain9# Why does nitrogen show catenation properties less than phosphorus? 10 # The electron gain enthalpy of Sulphur is more than Oxygen .

CARD-3: [1×5=5]

11# Are all the five bonds in PCl5 molecule equivalent? Justify your answer 12# The O—O bond energy is less than the S—S bond energy. ( sigma bond) OR Sulphur exhibits a stronger tendency for catenation as compared to oxygen.13 # NH3 is a good complexing agent but NF3 is not .14# On being slowly passed through water PH3 forms bubbles but NH3 dissolves . 15# Why does NH3 form hydrogen bond but PH3 does not?

CARD-4: [1×5=5]16# Why does NH3 act as a Lewis base ?OR NH3acts as ligand or good complexing agent OR , NH3 has higher H+ affinity than PH3.17# Why is H2O a liquid and H2S a gas ?18# SCl6 is not known but SF6 is known .19# SF6 exists but SH6 does not 20# SF6 is known but OF6 is not formed .Explain.

CARD-5: [1×5=5]21# I3ˉ is known but F3ˉ is not.22# HF is least volatile , whereas HCl is the most volatile.OR,HF has higher B.P than HCl OR, HF is liquid and HCl is gas23#Oxygen and fluorine both stabilize higher oxidation states of metals but oxygen exceeds fluorine in doing so .24 # Bismuth is a strong oxidizing agent in pentavalent state.25# PH3 has lower boiling point than NH3. Why?

CARD-6: [1×5=5]

26 # Explain why NH3 is basic while BiH3 is only feebly basic.27 # Why does the reactivity of nitrogen differ from phosphorus? 28# Why does white ppt. of AgCl dissolves in ammonia solution.29# Though nitrogen exhibits +5 oxidation state, it does not form pentahalide. Give reason.30#Why is N2 less reactive at room temperature?

CARD-7: [[1×5=5]

31# There is a large difference between the melting and boiling points of Oxygen and Sulphur .32# Fluorine exhibits only –1 oxidation state whereas other halogens exhibit + 1, + 3, + 5 and + 7 oxidation states also. Explain. 33# Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F2 and Cl2. 34# Fluorine never acts as a central atom in its compounds with other halogens .35 # In trimethylamine, the nitrogen has a pyramidal geometry whereas in trisilylamine,it has a planar II # CHOOSE THE CORRECT ANSWER FROM GIVEN OPTIONS .

1# Whose boiling point is more ? (H2O, H2S) 2# Which dissolves more in water ? (PH3 , NH3 )3# Which is more basic ?( NH3 , BiH3) 4# Which has more oxidizing ability ( Cl2 , F2) 5# Which has more bond dissociation enthalpy ? ( F—F , Cl—Cl) 6# Which one exists ? ( R3P=O , R3N=O )7# Which is more reactive ? ( Red— P , White – P ) 8# Which has more catenation properties ? ( Nitrogen or Phosphorus) 9# Which is possible ( ClF3 or FCl3) 10# Which is more reactive ( Nitrogen gas , Phosphorus) 11# Whose sigma bond strength is more ? ( O—O , S—S) 12# WhoseIonisation energy is more ? ( N ,O ) 13# Which one does not release white fumes of HClupon hydrolysis ( PCl5 , PCl3 , SiCl4 ,NCl5)14# Which is more stable ? (NF3 , NCl3) 15# Which has more negative gain enthalpy ?( F , Cl ) III # Arrange the Following in increasing order against the properties mentioned :-

1# Bond Dissociation Enthalpy:-

(a) Br—Br , I—I , Cl—Cl , F—F (b) H—I , H—F, H—Br,H—Cl

(c) O—H, H—Te, H—Se, H—S. (d) N—N, P—P, As—As

2# Catenation property:- (a) As , N, P , Sb(b) Se ,S , Te ,O (c) Si , Sn , C , Ge

3# Electron Gain Enthalpy :- (a) I , Br , Cl , F (b) N , O, P ,S (c) F, Cl , O , S

5 # Ionisation Enthalpy:- (a) O , N , F , C (b) Ar , Ne , He , Xe , Kr

6# Electronegativity:- (a) Cl ,F, Br, I (b) O , N , F , C

CONCEPTS :- HYDRIDES , OXIDES AND OXOACIDS CHAP: P-BLOCK ELEMENTS CLASS-XII

I# ( Fill in the blanks ) [1 ˟ 10=10]

1# Arrange In increasing order of basic strength -- NH3 , BiH3 , PH3 , AsH3 , SbH3 -------------------------------------2# Arrange In increasing order of acidic strength -- HF ,HCl ,HBr , HI . --------------------------------------------------3# The optimum conditions for the production of ammonia are ----------------------------------------------------------4# Which of the following does not have two different N—O bond length ? ( NO2 , N2O3 , N2O4 , N2O5) 5# The chemical compound responsible for Brown –Ring in nitrate test is --------------------------------------------6# Write the conditions to maximize the yield of sulphuric acid by Contact Process are ---------------------------7# The two areas in which H2SO4 plays an important role are 1.-------------------------- 2.----------------------------8# Out of HOF and HOCl , relatively stable oxo-acid is ------------------------------------ 9# HClO4 is more acidic than HOCl because ------------------------------------------------- 10# Give one chemical equation to show the dehydrating action of conc. H2SO4 .-----------------------------II # Arrange the Following in increasing order against the properties mentioned :-1# Bond Dissociation Enthalpy:- (a) Br—Br , I—I , Cl—Cl , F—F (b) H—I , H—F, H—Br,H—Cl (c) O—H, H—Te, H—Se, H—S. (d) N—N, P—P, As—As 2# Acid strength:- (a) H—I , H—F , H—Br , H—Cl (b) HF, CH4 , H2O , NH3

(c) H2O, H2Te , H2Se , H2S 3# Base Strength:- BiH3 , NH3 , AsH3 , SbH3 , PH34# Thermal Stability:- (a) H2O , H2Te , H2Se , H2S (b) PH3 , BiH3 , AsH3 , SbH3 , NH3 5# Bond Angle:- (a) H2Se , H2O, H2S ,H2Te (b) PH3 , BiH3 , AsH3 , SbH3 , NH3 6# Boiling Point :- (a) H2S , H2O , H2Te , H2Se (b) PH3 , BiH3 , AsH3 , SbH3 , NH3 7# Covalent Character :- (a) Cr2O3 , CrO, CrO3 (b) P2O5,Sb2O5, As2O5 (c) BeCl2, MgCl2 ,CaCl2, BaCl28# Acid Strength:- (a) HOClO2 , HOClO , HOCl ,HOClO39# Acidic Character -- (a) N2O, N2O5, N2O3 ,NO , N2O4 (b) ClO2 , Cl2O7 ,Cl2O , Cl2O6 (c) HNO2& HNO3 (d) H2SO3&H2SO4 (e) P2O5 ,SO3, N2O5 , CO2 , SiO2 (f) Al2O3 ,CaO, Cl2O7 ,SO310# Reducing properties: (a) H2O, H2Te , H2Se , H2S (b) H3PO4 , H3PO2 , H3PO3

III # Give reason / Account for the following : [1 ˟ 20=20]

1# In aqueous solution , HI is stronger acid than HCl .2# Hydrogen fluoride has a much higher boiling point than hydrogen Chloride .3# NH3 is a stronger base than PH3 .4# In the structure of HNO3 molecule , The N—O bond (121pm) is shorter than N—OH bond(140pm) 5# H3PO2 and H3PO3 act as good reducing agents while H3PO4 does not ?6# Iron dissolves in HCl to form FeCl2 and not FeCl3 .7# H2O is a liquid while ,inspite of higher molecular mass , H2S is gas .8# HBr and HI can’t be prepared by treating metal bromides or iodides with conc. H2SO4 .9#Draw the structure of SO2 molecule Comment on the nature of two S–O bonds formed in SO2 molecule. Are the two S–O bonds in this molecule equal ?10# Why BiH3 the strongest reducing agent among all the hydrides of group -15 elements ?

11# In solution of H2SO4 in water , the second dissociation constant Ka2 , is less than the first dissociation constant Ka112# H2O is a liquid while ,inspite of higher molecular mass , H2S is gas .13# In which one of the following structures , NO2+ and NO2 ˉ , the bond angle has higher value ?14# NH3 is a stronger base than PH3 .OR , ammonia has greater affinity for protons than phosphIne ?15# Why the bond angle of PH3 molecule is lesser than that in NH3molecule ?

CBSE-MOST WANTED

16# Describe the favourable conditions for the manufacture of (i) ammonia by Haber’s Process (ii) Sulphuric acid by Contact Process (2)17# Which is stronger acid in aqueous solution ,HCl or HI ? 18# ArrangeHClO3 , HClO2 , HClO ,HClO4 in order of increasing acid strength . Give reason for your answer (2m) 19# Although the H-bonding in hydrogen fluoride is much stronger than that in water , yet water has a much higher boiling point than hydrogen fluoride . Why ?20# Why do chlorine water on standing loses its yellow colour?

VALUE BASED QUESTION-CHAPTER-7 (p-BLOCK ELEMENTS)

Q. 1. Ammonium nitrate is used as a high nitrogen fertilizer in agriculture. It is also used in instant cold packs. It is mixed with fuel oil and used as explosive. Due to its various uses its production cannot be banned. (a) What are cold packs? (b) What steps Indian government has taken to avoid the misuse of ammonium nitrate by terrorists? (c) What values are needed by an individual while handling such chemicals? Ans: (a) They contain ammonium nitrate and water . Endothermic reaction gives cooling effect and relieves pain (b) Only authorized dealers can sell ammonium nitrate (c)Proper knowledge about the properties of the chemicals, misuse should be avoided, Chemical should be used only for the welfare of mankind

Q.2. In recent years, the excessive uses of some aerosol propellants and refrigerants have depleted the ozone layer. (a)Name two chemicals which deplete the ozone layer in the stratosphere. (b)What is the importance of ozone in the stratosphere? (c)Suggest some measures for minimizing global warming.Answera)NO, CFC’S (b)Prevents entry of harmful ultraviolet radiations in earth’s atmosphere. (c)Minimising the usage of CFC’S, fossil fuels etc.

Q.3. Student accidently spills concentrated H2SO4 on his hand. Before the teacher gets to know, his friend washed his hands with water and also with soap but the burning sensation on hand was still going on. The friend then rubs solid sodium bicarbonate on his hand and then washed with water; finally the burning sensation is relieved. (i) Mention the values shown by student’s friend. (ii) Can you recommend any other substance available in the laboratory which can be used instead of sodium bicarbonate?Answer:- (i)Concern and care for others and application of scientific knowledge (ii) Slaked limeQ.4.India's top court ruled that authorities must regulate the sale of acid used in a spate of attacks on women. An acid called "Tezaab", which is designed to clean rusted tools but is often used in the attacks, can currently be bought across the counter. But the judges said the buyer of such acids should in future have to provide a photo identity card to any retailer when they make a purchase. The retailers must register the name and address of the buyer. (a) Why are acid burns harmful? (b) What are the values associated with the above discussion? (c) What are responsibilities of the citizens in this regard? Answer:- (a) Stringent measures to prevent misuse of acid (b)They are highly corrosive. (c) Help in preventing such incidences and helping victims , bringing to the notice of authorities such incidences Q.5. In a paper industry hydrogen peroxide is used bleaching agent. Satish asked his friend about using chlorine as bleaching agent. His friend told that using hydrogen peroxide was better. (a) What is the disadvantage of using chlorine as bleaching agent? (b) Why is use of hydrogen peroxide better?Answer: (a) HCl is the bye product in case Cl is used as a bleaching agent (b) Water is the bye product.