GENERAL PRINCIPLES OF EXTRACTION
P- BLOCK ELEMENTS 01 MARK
1) Why does PCl3 fume in moisture?
Ans: PCl3 hydrolyses in the presence of moisture giving fumes of HCl.
PCl3 + 3H2O - H3PO3 + 3HCl
2) PH3 forms bubbles when passed slowly in water but NH3 dissolves. Explain why?
Ans:- Due to high electronegativity & small size of N, NH3 forms H- bonds with water and hence it is water soluble. On the other hand, due to its lower electronegativity and its bigger size.
PH3 does not form H-bonds with H2O. As a result, it does not dissolve in H2O and hence escapes as bubbles.
3) Solid Phosphorus pentachoride behaves as an ionic compound. Explain why?
Ans In the solid state PCl5 exists as an ionic solid [PCl4]+ [PCl6]- in which the cation [PCl4]+ tetrahedral and the anion [PCl6]- is octahedral.
4) Sulphur in vapour state exhibits paramagnetic behavior. Give reason.
Ans In vapour state, sulphur partly exists as S2 molecule which has two unpaired electrons in the antibonding π-orbitals like O2 and hence, exhibits paramagnetism.
5) In solution of H2SO4 in water, the second dissociation constant, Ka 2 is less than the first dissociation constant, Ka 1. Explain.
Ans Ka2 is less than Ka1 because the negatively charged HSO4- ion has much less tendency to donate a proton to H2O as compared to neutral H2SO4.
6) Draw the structure of SO2 molecule. Comment on the nature of two S-O bonds formed in it. Are the two S-O bonds in this molecule equal?
Ans Structure of SO2 molecule is given below:
In SO
2,
sulphur is sp2 hybridised. The molecule of SO2 is angular. It is resonance hybrid of the above two
canonical forms. In SO2, both the S-O bonds are covalent and have equal strength due to resonance.
7) Write the balanced chemical equation for the following reaction : Excess of SO2 reacts with sodium hydroxide solution.
Ans Chemical equation for sodium hydrogen sulphite.
2NaOH + SO2 - Na2SO3 + H2O Na2SO3 + H2O + SO2 - 2NaHSO3
Sodium hydrogen sulphite
8) Name two poisonous gases which can be prepared from chlorine gas.
Ans Phosgene (COCl2), tear gas (CCl3NO2), mustard gas (ClCH2CH2SCH2CH2Cl) all are obtained from chlorine gas..
9) Write two uses of ClO2
Ans i) ClO2 is a powerful oxidizing agent & chlorinating agent ii) It is an excellent bleaching agent.
10) What inspired N Bartlett for carrying out reaction between Xe and PtF6?
Ans Bartlett found that the first ionization enthalpy of molecular oxygen is almost similar with that of xenon. Thus, after preparing red coloured compound O2+ [PtF6]- , he got inspired for
Ans a) On moving down the group, size of the elements increases. As the size increases, tendency to attract a proton decreases and thus, the basic character decreases. Hence the
increasing order of basic strength is BiH3 < Sbh3 < AsH3 < Ph3 < NH3
b) The stability of hydrides decreases from NH3 to BiH3 which can be observed from their bond dissociation enthalpy. Thus the reducing character of hydrides increases in the order – NH3 < PH3 < AsH3 < SbH3 < BiH3
2) Explain the following :
i) NO2 readily forms a dimer.
ii) BiCl3 is more stable than BiCl5.
Ans i) NO2 is an odd molecule due to the presence of odd number of valence electrons.
Hence, it readily dimerises to give more stable N2O4 molecule with even number of electrons.
ii) Due to inert pair effect, +3 oxidation state of Bi is more stable than its +5 oxidation state. Thus, BiCl3 is more stable than BiCl5.
3) Draw the structures of white phosphorus and red phosphorus. Which one of these two types of phosphorus is more reactive and why?
Ans Structure of white phosphorus is given below :
Structure of red phosphorus is given below:
White phosphorus is more reactive because of its discrete tetrahedral structure and angular strain.
4) Write the conditions to maximize the yield of H2SO4 by contact process.
Ans The key step in the production of H2SO4 is the oxidation of SO2 to SO3. SO2(g) + O2(g) 2SO3 (9) ∆ ƒ
Ho = - 196.6 KJmol-1
The reaction is exothermic & reversible and the forward reaction proceeds with decrease in volume. Therefore in accordance with Le chatliers; principle, to maximize the yield of SO3 &
hence of H2SO4, a low temperature (720 K), a high pressure (2 bar) & V2O5 is used as a catalyst.
5) Assign reasons for the following.
a) H2S is more acidic than H2O.
b) Sulphur has a greater tendency for catenation than oxygen.
Ans a) Due to decrease in (E-H) bond dissociation enthalpy down the group, acidic character increases. Thus, H2S is more acidic than H2O.
b) Bond energy of S-S bond (213 kJ mol-1) is greater than O-O bond (138 kJ mol-1).
Due to small size of oxygen atom, there is greater lp-bp repulsion in O-O, resulting in weakening of O-O bond more than in S-S bond. Therefore, the tendency of catenation in oxygen is lower than sulphur.
6) Draw the structure of each of the following.
a) H2S2O8 b) H2SO4
Ans a) Structure of H2S2O8 is given below:
b) The structure of H2SO4
7) Complete the following chemical equations:-a) NaOH (hot & Conc.) + Cl2 →
b) Cl2 + F2 (excess) →
Ans a) 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O (hot & conc)
b) Cl2 + 3F2 (excess) → 2ClF3
8) a) Which neutral molecule would be isoelectronic with ClO-?
b) How are interhalogen compounds formulated and how are they prepared?
Ans a) ClO- has 26 electrons. A neutral molecule with 26 electrons is OF2. b) Interhalogen compounds are formulated as
XX| XX3| XX5|
, etc.
The interhalogen compounds can be prepared by direct combination or by the action of halogen on lower interhalogen compounds. The product formed depends upon some specific conditions, eg:
Cl2 + F2
K 437
2ClF (Equal volume)
Cl2 + 3F2
K573
2CIF3
(Excess)
9) a) How are the following compounds prepared from XeF6? i) XeOF4
ii) XeO3
Ans i) XeF6 + H2O → XeOF4 + 2HF ii) XeF6 + 3H2O → XeO3 + 6HF
9) b) XeF2 is linear molecule without a bent. Explain.
Ans XeF2 is linear molecule. According to VSEPR theory, the three lone pairs will occupy the equalorial positions and two bond pairs will occupy axial positions to minimize lp-lp and lp-bp repulsions.
10) Explain Why –
a) Noble gases have comparatively large atomic sizes.
b) Noble gases form compounds with fluorine & oxygen only.
Ans a) It is due to the reason that noble gases have only vander waals radii while others have covalent radii & by definition, Vander waals radii are larger than covalent radii.
b) Fluorine and oxygen are the most electronegative elements and hence, are very reactive. Therefore, they form compounds with noble gases particularly with xenon.
03 MARKS:
1) What happens when –
a) White phosphorus is heated with conc. NaOH solution in an inert gas atmosphere?
b) Orthophosphorous acid is heated?
c) PCl5 is heated.
Ans a) When white phosphorus is heated with Conc. NaOH solution in an inert gas atmosphere, phosphine gas is produced.
P4 + 3NaOH + 3H2O → PH3 + 3NaH2PO2
b) On heating, H3PO3 disproportionate into H3PO4 and phosphine.
4H3PO3 → 3H3PO4 + PH3
c) When PCl5 is heated, it decomposes into PCl3 and Cl2. PCl5 → PCl3 + Cl2
2) Complete the following chemical reaction – i) I2 + conc.HNO3 →
ii) HgCl2 + PH3 → iii) AgCl(s) + NH3 (aq) →
Ans i) I2 + 10HNO3 → 2HIO3 + 10NO2 +4H2O ii) 3HgCl2 + 2PH3 → Hg3P2 + 6HCl
iii) AgCl(s) + 2NH3 (aq) → [Ag(NH3)2] cl (aq)
3) On heating compound (A) gives a gas (B) which is a constituent of air. This gas when treated with 3 mol of hydrogen (H2) in the presence of a catalyst gives another gas (C) which is basic in nature. Gas C on further oxidation in moist condition gives a compound (D) which is a part of acid rain. Identify compounds (A) to (D) and also give necessary equations of all the steps involved.
Ans Compounds (A) to (D) are as follows : (A) = NH2NO2 (B) = N2
(C) = NH3 (D) = HNO3
The reactions are given as under : i) NH4NO2
4) On reaction with Cl2, phosphorus forms two types of halides A and B. Halide A is yellowish-white powder but halide B is colourless oily liquid. Identify A and B and write the formulas of their hydrolysis products.
Ans A is PCl5 (It is yellowish-white powder)
P4 + 10Cl2 → 4PCl5
5) An amorphous solid “A” burns in air to form a gas “B” which turns lime water milky.
The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified aqueous KMnO4 solution and reduces Fe3+ to Fe2+. Identify the solid “A” and the gas
“B” and write the reactions involved.
SO2 is produced as a by product during roasting of sulphide ore. The sulphides are converted into oxides with the evolution of SO2.
6) Complete the following chemical equation i) C + H2SO4 (conc.) →
ii) SO3 + H2SO4 (conc.) → iii) O3(9) + I-(aq) + H2O (e) →
Ans i) C + 2H2SO4 (conc.) → CO2 + 2SO2 + 2H2O ii) SO3 + H2SO4 (conc.) → H2S2O7
Oleum
iii) O3 (g)+ 2I- (aq) + H2O(l) → 2OH-(aq) + I2(s) + O2(g) 7) Account for the following:
a) Iron on reaction with HCl forms FeCl2 & not FeCl3.
b) O2 & F2 both stabilize higher oxidation states of metals but O2 exceeds F2 in doing so.
c) ICl is more reactive than I2?
Ans a) HCl reacts with Fe and produces H2. Fe + 2HCl → FeCl2 + H2
Liberation of hydrogen prevents the formation of ferric chloride
Ans b) O2 and F2 both stabilize higher oxidation states of metal but O2 exceeds F2 in doing so due to ability of oxygen to form multiple bonds with metals.
Ans c) Interhalogen compounds are more reactive than halogens (except fluorine) because
XX|
bond (I-Cl bond in question) in interhalogens is weaker than X-X bond (I-I bond) in halogens, except F-F bond. In other words, I-Cl bond is weaker than I-I bond. That’s why, ICl is more reactive than I2.
8) a) Draw the structure of i) XeOF4 Molecule.
ii) HClO4 Molecule iii) BrF3
Ans i)
(ii)
(iii)
9) How are xenon fluorides XeF2, XeF4 and XeF6 obtained ?
Ans These xenon fluorides are obtained by direct reaction between Xe and F2, under different conditions as shown below:
Xe (g) + F2 (g) 673 k , 1 ¯¿
¿ → XeF2 (s) (excess)
Xe (g) + 2F2 (g) 873 k , 7 ¯¿
¿ → XeF4 (s) (1:5 ratio)
Xe (g) + 3F2 (g) 573 k , 60−70 ¯¿
¿ → XeF6 (s) (1: 20 ratio)
10) Account for the following –
a) Helium is used in diving equipments.
b) Structure of xenon fluorides cannot be explained by valence bond approach.
c) Bleaching of flowers by chlorine is permanent while that by sulphur dioxide is temporary.
Ans a) Helium is used as a diluent for oxygen in modern diving apparatus because of its very low solubility in blood.
b) According to the valence bond approach, covalent bonds are formed by the overlapping of half-filled atomic orbitals. But xenon has fully-filled electronic configuration.
Hence, the structure of xenon fluorides cannot be explained by VBT.
c) Cl2 bleaches coloured material by oxidation. Cl2 + H2O → 2HCl + [O]
hence bleaching is permanent.
On the other hand, SO2 bleaches coloured material by reduction & hence bleaching is temporary SO2 + 2H2O → 2H2SO4 + 2[H]
05 MARKS
1) On heating lead (II) nitrate gives a brown gas “A”. The gas “A” on cooling changes to colourless solid “B”. Solid “B” on heating with NO changes to a blue solid ‘C’. Identify ‘A’, ‘B’
and ‘C’ and also write reactions involved and draw the structures of ‘B’ and ‘C’.
Ans The gas A is NO2
The reactions are explained as under : 2Pb(NO3)2
B C
(Brown colour solid)
Structure of N2O4
Structure of N2O5
2) i) How would you account for the following ?
a) NF3 is an exothermic compound whereas NCl3 is not. Explain.
b) All the P-Cl bonds in PCl5 molecule are not equivalent. Explain why?
c) Why nitrogen gas is very unreactive ?
Ans a) In case of nitrogen, only NF3 is known to be stable. N-F bond strength is greater than F-F bond strength, therefore formation of NF3 is spontaneous. In case of NCl3, N-Cl bond strength is lesser than CL-Cl bond strength. Thus, energy has to be supplied during the formation of NCl3.
Ans b) PCl5 has a trigonal bipyramidal structure and the three equatorial P-Cl bonds are equivalent while the two axial bonds are different and longer than equatorial bonds. This is because the axial bond pairs suffer more repulsion as compared to equatorial bond pairs.
Ans c) Nitrogen is chemically less reactive. This is due to the presence of a more stable triple bond in N2 molecule whereas phosphorus forms only P-P single bond. Therefore,
phosphorus is more reactive than nitrogen.
ii) Draw the structures of the following a) (HPO3)3
b) H4P2O7
Ans a) Structure of (HPO3)3 is given below:
b) Structure of H4P2O7 is given below:
3) a) P4O6 reacts with water according to equation P4O6 + 6H2O → 4H3PO3, Calculate the volume of 0.1 M NaOH solution required to neutralize the acid formed by dissolving 1.1 g of P4O6 in H2O.
Ans P4O6 + 6H2O → 4H3PO3
H3PO3 + 2NaOH → Na2HPO3 + 2H2O] x 4
Molarity of NaOH solution is 0.1 M 1.1 mol NaOH is present in = 1 litre 0.04 mol NaOH is present in = 1
0.1 x 0.04 litre = 400 ml.
b) Give an example to show the effect of concentration of nitric acid on the formation of oxidation product.
Ans Dilute and concentrated nitric acid gives different oxidation products on reaction with copper metal.
3Cu + 8HNO3 (dil.) → 3Cu(NO3)2 + 2NO + 4H2O Cu + 4HNO3 (conc.) → Cu(NO3)2 + 2NO2 + 2H2O
NO and NO2 are the oxidation products obtained wilh dil. HNO3 and conc. HNO3 respectively.
4) Give reasons for the following.
i) Decomposition of O3 molecule is a spontaneous process.
ii) SF6 is inert towards hydrolysis.
iii) (CH3)3 P=O exists but (CH3)3 N=O does not.
iv) Oxygen has less electron gain enthalpy with negative sign than sulphur.
v) SO2 is an air pollutant.
Ans i) Decomposition of O3 is an exothermic process (ΔH = -ve) and occurs with increase in entropy (ΔS = +ve). These two effects reinforce each other which results in large negative Gibbs energy change. It favours its decomposition into oxygen.
ii) In SF6, S is surrounded by 6 F- octahedrally. Therefore, attack of water molecule on S is sterically hindered.
iii) Due to absence of d-orbitals,N cannot form pπ-dπ multiple bonds.As a result,N cannot expand it’s covalency more than four,but in R3N=O,N has it’s covalency 5,therefore,the compound R3N=O doesnot exist.On the other hand,due to the presence of d-orbitals in P,it forms pπ-dπ multiple bonds,therefore can expand it’s covalency more than four.As a result P can form (CH3)3 P=O.In this compound the covalency of P is 5.
iv) The electron gain enthalpy of oxygen is less negative than sulphur due to its compact size. As a result of which, the electron repulsions in the relatively compact 2p subshell are comparatively large hence the incoming electrons are not accepted with the same ease as in case of sulphur.
v) SO2 is water soluble, therefore it dissolves in rainwater causing acid rain.
Moreover, when released in air, it mixes with it and leads to several diseases like eye irritation, redness, asthma, bronchitis, etc. Thus, it is considered as an air pollutant.
5) a) Name the two most important allotropes of sulphur. Which one of the two is stable at room temperature? What happens when the stable form is heated above 370 K?
Ans Two most important allotropes of sulphur are i) Rhombic sulphur (α-sulphur)
ii) Monoclinic sulphur
The stable form at room temperature is rhombic sulphur which transforms to monoclinic sulphur when heated above 369K
Rhombic sulphur Monoclinic Sulphur
b) Draw the structures of –
i) H4S2O7ii) O3 iii) S8
Ans i)
ii)
370K
iii)
Crown Shape
6) a) How would you account for the following –
i) The oxidizing Power of oxoacids of chlorine follows the order – HClO4 < HClO3 < HClO2 < HClO
ii) The halogens are coloured.
iv) F2 is a stronger oxidizing agent than Cl2
b) Complete the following chemical equations.
i) I2 + NaClO3 →
ii) I2 + H2O + Cl2 →
Ans a) As the oxidation number of halogen atom in oxoacid increases, its oxidizing power decreases. Therefore, HClO is least stable and gives [O] most easily, so its oxidizing power is greater than HClO4.
b) All halogens are coloured due to the absorption of raditions 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.
c) Since Eo for F2/F- eleclrode is higher than that of Cl2/Cl- electrode F2 is more easily reduced than Cl2 & is a stronger oxidizing agent than Cl2.
Ans b) i) I2 + 2NaClO3 → 2NaI+2ClO2 + O2
ii) I2 + 6H2O + 5Cl2 → 2HIO3 + 10HCl Iodic acid.
2) a) How can you prepare Cl2 from HCl & HCl from Cl2. Write reactions only.
b) Write the reactions of F2 & Cl2 with water.
Ans a) HCl can be oxidized to Cl2 by a number of oxidizing agents such as MnO2, KMnO4, K2Cr2O7, etc.
Reaction with MnO2 is given below:
MnO2 + 4HCl → MnCl2 + Cl2+ 2H2O Cl2 can be reduced to HCl by reaction of H2 in presence of diffused sunlight.
H2 + Cl2
Diffused sunlight
220 > 2HCl
b) F2 being a stronger oxidizing agent oxidises H2O to O2 or O3. The reactions are given as under:
2F2 (g) + 2H2O (l) → 4H+ (aq) + 4F- (aq) + O2 (g) 2F2 (g) + 3H2O (l) → 6H+ (aq) + 6F- (aq) + O3 (g)
Cl2, on the other hand, reacts with H2O to form hydrochloric acid and hypochlorous acid as per the following equation :
Cl2(g) + H2O (l) → HCl (aq) + HOCl (aq) Hydrochloric acid Hypochlorous acid
8) a) Give reason for the following.
i) F2 is more reactive than ClF3 but ClF3 is more reactive than Cl2. ii) H3PO2 is a stronger reducing agent than H3PO3.
iii) PCl5 is more covalent than PCl3.
b) Complete the following chemical equation.
i) XeF4 + O2F2 →
ii) XeF4 + SbF5 →
Ans a) i) Fluorine due to its small size, high electronegativity and low bond energy is more reactive than ClF3 but bond energy of C-Cl bond is higher than Cl-F bond, therefore ClF3 is more reactive than Cl2.
ii) The structure of H3PO2 ,H3PO3 & H3PO4 are as follows:
The acids which contain P-H bond, have strong reducing properties. Hypophosphorous acid (H3PO2) contains two P-H bonds, whereas orthophosphorous acid (H3PO3) has one P-H bond. Hence, H3PO2 is a stronger reducing agent than H3PO3.
iii) Since, pentavalent metal ion has higher polarizing power than trivalent metal ion.
Thus, PCl5 is more covalent than PCl3.
b) i) XeF4 + O2F2
XeF6 + O2
ii) XeF4 + SbF5 → [XeF3]+ [SbF6]
-CHAPTER – 9
COORDINATION CHEMISTRY