TRENDS :-Arrange the following
Q3
in increasing order of given property
[Bond angle of ONO bonds]
Ans
-ANS-
NO2- < NO2< NO2+
Q2
Q4 i O2 F2, O F2, Cl2O Br2O [thermal stability]
ii Cl2O, ClO2, Cl2O6, Cl2O7 [thermal stability]
iii H2O, O F2, Cl2O Br2O [bond angle]
iv H2O,HF,NH3,NF3 [ dipolemoment]
ANS-
v
vi
bond strength
bond strength
vii
bond length
Ans i O2 F2< O F2<Cl2O< Br2O
iiii ,
v
ii Cl2O, <ClO2< Cl2O6<Cl2O7
O F2<H2O <Cl2O <Br2O [bond angle ivi
vi
vii
10
REASONING BASED QUESTIONS FROM P-BLOCK ELEMENTS
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
GROUP-15
Nitrogen does not form pentahalide although it exhibit +5 oxidation state.
Due to absence d-orbitals N can not extend its valency beyond four
PH3 has lower B.pt than NH3
N is more electronegative than P so in NH3 there is intermolecular H-bonding hence it has high b.pt
NH3 acts as Lewis base
Because N has a lone pair electron so NH3 acts as a Lewis base
NO2 dimerises
NO2 has an odd electron so it dimerises to pair up electron and to achieve octet configuration
NH3 is stronger base than PH3
Due to smaller size of nitrogen there is high electron density on nitrogen so electron pair is easily available.
PCl3 fumes in moisture
PCl3 undergoes hydrolysis and gives fumes of HCl.
PCl3 + 3 H2O  H3PO3 + 3HCl
All the five P-Cl bonds are not equal in PCl5
The two axial bonds suffer more repulsion from equatorial bonds and hence are elongated.
H3PO2 has reducing character
Since it has two P-H bonds
H3PO3 is dibasic (diprotic) but H3PO4 is tribasic
In H3PO3 only two H atoms are linked to O which are ionisable the third H is attached to P and not
ionisable because P is less electronegative. In H3PO4 all the three H atoms are with O and ionisable
PCl5 is ionic in solid state
It is due to the following conversion :
2PCl5  [PCl4]+[PCl6]NO is paramagnetic in gaseous state but diamagnetic in liquid and solid state
NO(g) has odd number of electrons so it is paramagnetic but in liquid and solid state it exists as dimmer so
there is no unpaired electron and it will be diamagnetic
NCl3 hydrolysed but NF3 does not
In NCl3 Cl has vacant d-orbitals to accept the lone pair from H2O but in NF3 F has no d-orbitals
NCl3 + 3H2O  NH3 + 3 HOCl
Nitrogen shows little catenation but phosphorous distinctly shows catenation property
Due to smaller size of N there is repulsion between the lone pairs and N-N single bond is weaker than P-P
+5 oxidation state of Bi is less stable than +3
Because inert pair effect is very prominent in Bi , so +5 oxidation state is not stable
Bi in +5 oxidation state is strong oxidizing agent
Because inert pair effect is very prominent in Bi so Bi5+ can be easily converted into Bi3+
NO(nitric oxide) becomes brown when released to air
It oxidizes to NO2
NH3 is a good complexing agent/ NH3 acts as a ligand
It has lone pair of electron on N-atom and can be donated for the coordination bond.
Bi2O3 is not acidic
The size of Bi3+ is very large and so there is very weak +ve electric field around it so it does not interact
with water to release H+
BiH3 is the strongest reducing agent among the group-15 hydrides
Since Bi-H bond is the weakest among pr-15 hydrides so H2 gas is evolved which is reducing
N2 is less reactive at room temperature
Due to having triple bond and hence high bond dissociation energy(946 kJ/mol)
Bond angle in PH4+ higher than in PH3
In PH3 there is lp-bp repulsion so bond angle is less where as in PH4+ there is no lp-bp repulsion
1
22 NH3 has greater bond angle than PH3
N is more electronegative so it attracts the bond pair electron and hence there is greater bp-bp repulsion in
NH3 and hence greater bond angle
23 R3P=O exists but R3N=O does not
N due to absence of d-orbitals can not form pπ-dπ multiple bond
24 N exists as N2 but P exists as P4
Due to smaller size N can form pπ-dπ multiple bonding and exists as discrete N2 molecule but P can not
form pπ-pπ multiple bonding.
25 PCl5 can not act as reducing agent
In PCl5 P has +5 oxidation state. P has five valence electron in its valence shell so it can not increase its
oxidation state beyond +5, so it can not act as reducing agent.
26 Phosphorous is kept under kerosene
It is highly reactive and easily catches fire due to air oxidation
27 H3PO3 is syrupy liquid
Due to intermolecular H-bonding
28 PH3 bubbles but NH3 dissolves in water
NH3 forms H-bonding with water but PH3 can not form so NH3 dissolves but PH3 bubbles out
29 Only a small increase in radius is observed from As to Bi
Due to poor shielding effect of d and f obitals.
30 Nitrogen is gas where as phosphorous is solid at room temp.
Nitrogen is diatomic molecule having weak van der Walls attraction where as phosphorous is tetra atomic
so it has strong van der Walls attraction.
31 N-N bond is weaker than P-P bond
Due to shorter bond length there is greater repulsion between lone pairs in N2
32 Maximum number of covalent bond formed by N is four
Because it has three unpaired electrons and one lone pair.
33 P2O5 can not be used for drying ammonia gas.
P2O5 is acidic it reacts with ammonia in presence of moisture and form (NH4)3PO4
34 NO2 is coloured but its dimmer N2O4 is colourless
Because NO2 has unpaired electron so it can absorb light from VR
35 Acidity of oxyacids of nitrogen increases with increase in oxidation number of N
Because non metallic character increases with oxidation number
36 White phosphorous is more reactive than red phosphorous
White phosphorous consists of discrete P4 molecules which is tetrahedral so reactive but in red
phosphorous the P4 molecules are linked in extended chain structure so it is less reactive.
37 Phophinic acid ( H3PO4) is mono basic / mono protic
Only one H atom is linked with O which is ionisable
38 N2 has higher bond dissociation energy than NO
Because N2 has higher bond order
39 N2 and CO have same bond order but CO is more reactive
CO is polar molecule
40 (CH3)3N is pyramidal but (SiH3)3N is planar
(CH3)3N is pyramidal due to sp3 hybridisation and one lone pair on N but (SiH3)3N is planar due to sp2
hybridisation and its lone pair is donated to vacant d orbital of Si for pπ-dπ overlap
41 The first IE of N is greater than that of O
It is due to half filled and hence stable electronic configuration of N
42 HNO2 disproportionates
In HNO2 the N is in +3 oxidation state which may increase as well as decrease
43 PCl5 can not act as reducing agent
In PCl5 phosphorous is in +5 oxidation state that is the highest oxidation state of P.
2
GROUP-16
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
Group 16 elements have lower I.E with compare to corresponding group 15 elements
Because group 15 elements have stable half filled p-sub shell(ns2np3) so electron can not be removed easily
H2S is less acidic than H2Te
In H2Te there is lower bond energy of H-Te bond due to larger size of Te
H2S acts as reducing agent while SO2 acts as both
In H2S , S has its minimum oxidation state -2 where as in SO2 it is +4 so it can be decreased up to -2 or
increased up to +6 , So H2S is only reducing but SO2 is both.
H2S is acidic while H2O is neutral
H-S bond is weaker due to larger size of S so proton release easier in H2S
SF6 is known but SH6 does not exist
Fluorine is the strongest oxidizing agent so it can oxidizes S to its maximum oxidation state +6 , H can not
Compound of F & O is fluoride of oxygen not oxide of fluorine
F is more electronegative than O
SCl6 is not known but SF6 is known
F is strongest oxidizing agent so it can oxidizes S to its maximum oxidation state +6 . Cl can not. Again Cl
has larger size so steric repulsion is there in SCl6
SF6 is used as gaseous electrical insulator
It is thermally stable and chemically inert
SF6 is not easily hydrolyzed
It is sterically protected by six F atoms hence does not allow H2O molecules to attack the S atom
S exhibits catenation properties but not Se
Due to smaller size of S than Se. S-S bond is much stronger than Se-Se bond
S disappears when boiled with Na2SO3
It forms sodium thiosulphate . Na2SO3 + S  Na2S2O3 ( soluble)
H2O is liquid but H2S is gas
O is electronegative so there is intermolecular H-bonding in water so it is liquid.
Ozone is powerful oxidizing agent
It decomposes to form nascent oxygen
Ka2 is less than Ka1 , for H2SO4 in water
The 2nd proton releases from HSO4- which is difficult. So Ka2 is less than Ka1
O2 is gas but sulphur is solid
Due to smaller size O can form pπ-pπ multiple bond and exists as discrete diatomic molecule.
Group 16 elements are called chalcogens
Chalcogen means ore forming elements. They form several ores
Positive oxidation states of O are generally not found
Due to high electro negativity of O
Thermal stability decreases from H2O to H2Te in group 16`
Due to increase in atomic size from O to Te the bond dissociation energy decreases
O does not show +4 & +6 oxidation states like S
Due to absence of d-orbital in oxygen
The magnitude of electron gain enthalpy of oxygen is less than that of sulphur
Due to very small size of O there is inter electronic repulsion
Among the hydrides of group 16 water shows unusual properties
Due to H-bonding in water the molecules get associated
S exhibits catenation but not O
Because S-S bond is stronger than O-O bond
3
23 Tendency to show -2 oxidation state diminishes from O to Po in group 16
Due to decreases in electronegetivity moving down the group
24 O2 is paramagnetic although it has even number of electrons
Due presence of unpaired electrons in anti bonding molecular orbitals
25 Sulphur in vapour state exhibit paramagnetism
In vapour stare sulphur partly exists as S2 molecule and like O2 it has unpaired electrons in π* orbitals
26 SF6 is less reactive than SF4
In SF6 sulphur atom is sterically hindered due to six F atoms
GROUP-17
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
Halogens have maximum negative electron gain enthalpy(∆egH)
Because they have smallest size in their respective periods
F has less electron gain enthalpy than that of Cl but fluorine is stronger oxidizing agent than chlorine
F has very small size so there is interelectronic repulsion. F is stronger oxidizing agent due to its low bond
dissociation energy and high hydration energy
F exhibits only -1 oxidation state , other halogen shows +1, +3, +5, +7 oxidation states
F is most electronegative element and due to absence of d-orbitals it can not expand its octet so it does not
exhibit positive oxidation state.
Iron reacts with HCl gives Fe(II)chloride and not Fe(III)chloride
Fe + 2 HCl  FeCl2 + H2
H2 liberated prevents the oxidation of FeCl2 to FeCl3
Bond dissociation energy of F2 is less than Cl2
Due to very small size of F there is interelectronic repulsion in F2 so it has low bond dissociation energy
Fluorine does not undergo disproportionation
Disproportination means simultaneous oxidation-reduction. F being the most electronegative element
undergoes only reduction but not oxidation.
NO dimerises but Cl2O does not
NO is odd electron species so it complete its octet by dimerisation
Bleaching by Cl2 is permanent but by SO2 is temporary
Cl2 bleaches by oxidation while SO2 does it by reduction. The reduced product gets oxidized again in air
and the colour returns
HF has lower acid strength than HI
Due to larger size of I , the H-I bond is weaker than H-F bond so HI is stronger
I2 is more soluble in KI than in water
I2 forms complex with KI i.e K+I3HClO is stronger acid than HIO
ClO- is more stable than IO- because Cl is more electronegative, so HClO is stronger
HClO4 is stronger acid than HClO3
ClO4- is more stable than ClO3- due to more resonance
OF2 should be called fluoride of oxygen and not oxide of fluorine
Because F is more electronegative than O
Interhalogens are more reactive than halogens
They are polar
HF is stored in wax coated glass bottle
HF reacts with alkali present in glass.
MF is more ionic than MCl ( M is alkali metal)
Because F- is smaller than Cl- and hence it is less polarisable.
Cl2 + KI  brown, but excess Cl2 turns it colourless
18 HClO4 is stronger than H2SO4
4
19
20
21
22
23
24
25
26
27
28
Because the conjugate base ClO4- is stable due to resonance
ClF3 exists but FCl3 does not
F is smaller in size and can not accommodate three chloride ions due to steric factor.
HF is less volatile than HCl
In HF there is intermolecular H-bonding so the HF molecules get associated
F form only one oxo acid , HOF
Due to absence of d-orbital it can not exhibit higher oxidation states
O form hydrogen bonding , Cl does not
O is more electronegative and small in size than Cl
Halogens are coloured
Due to absorption of radiation from VR
Iodine forms I3- but fluorine does not form F3Due to small size of fluorine
HI can not be prepared by heating KI with conc. H2SO4
The magnitude of electron gain enthalpy of F is less than that of Cl
Due to very small size of F there is inter electronic repulsion.
Fluorine is stronger oxidizing agent than chlorine though it ha lower electron gain enthalpy
Fluorine has higher reduction potential value due to its low bond dissociation energy and high hydration
energy with compare to chlorine.
Acid strength increases in the order HF< HCl < HBr < HI
As size increases from F to Cl the bond dissociation energy decreases from HF to HI
29
GROUP-18
1
He , Ne do not form compound with F
Due to high IE
2 Noble gases have very low b.pt
Because there is only weak dispersion force between their atoms.
3 Hydrolysis of XeF6 is not a redox reaction
Because in the products formed XeOF4 and XeO2F2 the Xe has the same oxidation state (+6) as in XeF6
4 Ne used as warning signal
Because Ne – light has high fog penetration power
5 Noble gases form compounds only with fluorine and oxygen
Because F & O are the most electronegative elements
6 Xe does not form XeF3 or XeF5
Xe has all paired electrons so promotion of one, two or three electrons will give rise to two, four or six
unpaired electrons hence can not form XeF3 and XeF5
7 Out of noble gases only Xe forms compounds
Because Xe has comparatively low IE and vacant orbitals for promotion of electrons
8 Noble gases are mostly inert
Because they have completely filled valence orbitals i.e octet configuration
9 He is used as diving apparatus
Because it is less soluble in blood with compare to nitrogen
10 It is difficult to study the chemistry of Rn
Because Rn is radioactive and hence very unstable
11 Noble gases have comparatively large atomic size
They are mono atomic so their van der Walls radii measured which is longer than covalent/ionic or
metallic radii.
5
REACTION OF P BLOCK ELEMENTS
1.
2.
3.
4.
5.
3HNO2 → HNO3 + H2O + 2NO
41. Al2O3 + 6HCl + 9H2O → 2 [ Al(H2O)6 ]3+ + 6Cl−
NH4CI(aq) + NaNO2(aq) → N2(g) +2H2O(l) + NaCl (aq) 42.
8NH3 + 3Cl2 → 6NH4Cl + N2; NH3 + 3Cl2 → NCl3 + 3HCl
43. PbS(s) + 4O3(g) → PbSO4(s) + 4O2(g)
Ba(N3)2 → Ba + 3N2
44. 2I–(aq) + H O(l) + O (g) → 2OH–(aq) + I (s) + O (g)
2
6.
7.
8.
3
2
2
45. NO ( g ) + O ( g ) → NO ( g ) + O ( g )
3
2
2
46. 4FeS (s ) + 11O ( g ) → 2Fe O3 ( s ) + 8SO ( g )
2
2
2
2
47. SO2(g)+Cl2 (g) →SO2Cl2,SO2 + Ca(OH)2 + H2O +CaSO3 milky
48. 2Fe3+ + SO + 2H O → 2Fe2+ + SO2 − + 4H+
2NH4Cl + Ca(OH)2 → 2NH3 + 2H2O + CaCl2
(NH4)2 SO4 + 2NaOH → 2NH3 + 2H2O + Na2SO4
2
2
4
49. 5SO2+ 2MnO4 + 2H2O → 5SO42− + 4H+ + 2Mn2+
9. ZnSO4( aq )+2NH4OH ( aq ) → Zn(OH)2 ( s ) + ( NH4)2 SO4 (aq )50. SO + H SO → H S O
3
2
4
2 2
7
10. FeCl3+NH4OH ( aq ) → Fe2O3 .xH2O (s) +NH4Cl ( aq )
51.
11. Cu2+(aq) + 4NH3(aq) ‡ [Cu(NH3)4]2+ (aq)
(blue)
(deep blue)
52.
12. AgCl ( s ) + 2NH ( aq ) → Ag ( NH ) Cl ( aq )
3
3 2
53. Cu + 2 H SO (conc.) → CuSO + SO + 2H O
(white ppt)
2
(colourless)
4
4
2
2
13.
54. 3S + 2H SO (conc.) → 3SO + 2H O
2
4
2
2
55. C + 2H SO (conc.) → CO + 2 SO + 2 H O
14. 3NO ( g ) + H O ( l ) → 2HNO ( aq ) + NO ( g )
2
2
3
15. 3Cu + 8 HNO (dilute) → 3Cu(NO ) + 2NO + 4H O
56. CaF + H SO → CaSO + 2HF
2
2
4
4
57. F + 2X– → 2F– + X (X = Cl, Br or I)
2
3
3 2
2
16. Cu + 4HNO (conc.) → Cu(NO ) + 2NO + 2H O
3
3 2
2
2
17. 4Zn + 10HNO (dilute) → 4 Zn (NO ) + 5H O + N O
3
3 2
2
18. Zn + 4HNO (conc.) → Zn (NO ) + 2H O + 2NO
3
3 2
2
2
19. I + 10HNO → 2HIO + 10NO + 4H O
2
3
3
2
2
20. C + 4HNO → CO + 2H O + 4NO
3
2
2
2
21. S + 48HNO → 8H SO + 48NO + 16H O
8
3
2
4
2
2
22. P + 20HNO → 4H PO + 20NO + 4H O
4
3
3
4
2
2
23. BROWN RING TEST
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
2
4
2
2
2
2
2
58. Cl + 2X– → 2Cl– + X (X = Br or I)
2
2
59. Br + 2I– → 2Br– + I
2
2
60. 2F (g) + 2H O (l) → 4H+ (aq) + 4F− (aq) + O (g)
2
2
2
( where X = Cl or Br )
61. 4I− (aq) + 4H+ (aq) + O (g) → 2I (s) + 2H O (l)
2
2
2
62. MnO + 4HCl → MnCl + Cl + 2H O
2
2
2
2
63. 4NaCl + MnO + 4H SO → MnCl +4NaHSO +2H O+Cl
2
2
4
2
4
2
2
64. 2KMnO + 16HCl → 2KCl + 2MnCl + 8H O + 5Cl
4
2
2
2
65. Cl + H O → 2HCl + O, Fe + 2HCl → FeCl + H
2
2
2
2
66. H S + Cl → 2HCl + S,
NO3- + 3Fe2+ + 4H+ → NO + 3Fe3+ + 2H2O
[Fe (H2O)6 ]2+ +NO→ [Fe(H2O)5 (NO)]2++ H2O
67.
P4 + 3NaOH + 3H2O → PH3 + 3NaH2PO2
68.
P4 + 5O2 → P4O10
69.
Ca3P2 + 6H2O → 3Ca(OH)2 + 2PH3
70.
Ca3P2 + 6HCl → 3CaCl2 + 2PH3
71.
PH4I + KOH → KI + H2O + PH3
72.
3CuSO4 + 2PH3 → Cu3 P2 + 3H2SO4
73.
3HgCl2 + 2PH3 → Hg3P2 + 6HCl
74.
P4 + 8SOCl2 → 4PCl3 + 4SO2 + 2S2Cl2
75.
3CH3COOH + PCl3 → 3CH3COCl + H3PO3
76.
P4 + 10SO2Cl2 → 4PCl5 + 10SO2
77.
PCl5+H2O → POCl3+2HCl, POCl3+3H2O →H3PO4+ 3HCl
78.
S8 + 4Cl2 → 4S2Cl2
79.
2Ag + PCl5 → 2AgCl + PCl3
80.
PCl3 + 3H2O →H3PO3 + 3HCl
81.
4H3PO3 → 3H3PO4 + PH3
2
2
Au + 4H+ + NO3− + 4Cl− → AuCl−4 + NO + 2H2O
3Pt + 16H+ + 4NO3 + 18Cl− → 3PtCl6− + 4NO + 8H2O
(cold and dilute)2NaOH + Cl2 → NaCl + NaOCl + H2O
(hot and conc.)6 NaOH + 3Cl2 →5NaCl +NaClO3 + 3H2O
2Ca(OH)2 + 2Cl2 → Ca(OCl)2 + CaCl2 + 2H2O
I2 + 6H2O + 5Cl2 → 2HIO3 + 10HCl
Na2SO3 + Cl2 + H2O → Na2SO4 + 2HCl
2FeSO4 + H2SO4 + Cl2 → Fe2(SO4)3 + 2HCl
SO2 + 2H2O + Cl2 → H2SO4 + 2HCl
XeF4 + O2 F2 → XeF6 + O2
2XeF2 (s) + 2H2O(l) → 2Xe (g) + 4 HF(aq) + O2(g)
6XeF4 + 12 H2O → 4Xe + 2XeO3 + 24 HF + 3 O2
XeF6+H2O→XeOF4 + 2 HF,XeF6+2H2O →XeO2F2 + 4HF
XeF6 + 3 H2O → XeO3 + 6 HF
2NH3 + 3NaClO →N2 + 3H2O + 3NaCl
82. 2K2HgI4 + NH3 + 3KOH→ Brown ppt. of HN HgOHgI
2
83. 2NH3 + NaOCl →NH2.NH2 hydrazine
Al2O3 ( s )+6NaOH(aq)+3H2O ( l ) → 2Na3[Al(OH)6](aq)
REACTION OF P BLOCK ELEMENTS
1.
2.
3.
4.
5.
3HNO2 → HNO3 + H2O + 2NO
41. Al2O3 + 6HCl + 9H2O → 2 [ Al(H2O)6 ]3+ + 6Cl−
NH4CI(aq) + NaNO2(aq) → N2(g) +2H2O(l) + NaCl (aq) 42.
8NH3 + 3Cl2 → 6NH4Cl + N2; NH3 + 3Cl2 → NCl3 + 3HCl
43. PbS(s) + 4O3(g) → PbSO4(s) + 4O2(g)
Ba(N3)2 → Ba + 3N2
44. 2I–(aq) + H O(l) + O (g) → 2OH–(aq) + I (s) + O (g)
2
6.
7.
8.
3
2
2
45. NO ( g ) + O ( g ) → NO ( g ) + O ( g )
3
2
2
46. 4FeS (s ) + 11O ( g ) → 2Fe O3 ( s ) + 8SO ( g )
2
2
2
2
47. SO2(g)+Cl2 (g) →SO2Cl2,SO2 + Ca(OH)2 + H2O +CaSO3 milky
48. 2Fe3+ + SO + 2H O → 2Fe2+ + SO2 − + 4H+
2NH4Cl + Ca(OH)2 → 2NH3 + 2H2O + CaCl2
(NH4)2 SO4 + 2NaOH → 2NH3 + 2H2O + Na2SO4
2
2
4
49. 5SO2+ 2MnO4 + 2H2O → 5SO42− + 4H+ + 2Mn2+
9. ZnSO4( aq )+2NH4OH ( aq ) → Zn(OH)2 ( s ) + ( NH4)2 SO4 (aq )50. SO + H SO → H S O
3
2
4
2 2
7
10. FeCl3+NH4OH ( aq ) → Fe2O3 .xH2O (s) +NH4Cl ( aq )
51.
11. Cu2+(aq) + 4NH3(aq) ‡ [Cu(NH3)4]2+ (aq)
(blue)
(deep blue)
52.
12. AgCl ( s ) + 2NH ( aq ) → Ag ( NH ) Cl ( aq )
3
3 2
53. Cu + 2 H SO (conc.) → CuSO + SO + 2H O
(white ppt)
2
(colourless)
4
4
2
2
13.
54. 3S + 2H SO (conc.) → 3SO + 2H O
2
4
2
2
55. C + 2H SO (conc.) → CO + 2 SO + 2 H O
14. 3NO ( g ) + H O ( l ) → 2HNO ( aq ) + NO ( g )
2
2
3
15. 3Cu + 8 HNO (dilute) → 3Cu(NO ) + 2NO + 4H O
56. CaF + H SO → CaSO + 2HF
2
2
4
4
57. F + 2X– → 2F– + X (X = Cl, Br or I)
2
3
3 2
2
16. Cu + 4HNO (conc.) → Cu(NO ) + 2NO + 2H O
3
3 2
2
2
17. 4Zn + 10HNO (dilute) → 4 Zn (NO ) + 5H O + N O
3
3 2
2
18. Zn + 4HNO (conc.) → Zn (NO ) + 2H O + 2NO
3
3 2
2
2
19. I + 10HNO → 2HIO + 10NO + 4H O
2
3
3
2
2
20. C + 4HNO → CO + 2H O + 4NO
3
2
2
2
21. S + 48HNO → 8H SO + 48NO + 16H O
8
3
2
4
2
2
22. P + 20HNO → 4H PO + 20NO + 4H O
4
3
3
4
2
2
23. BROWN RING TEST
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
2
4
2
2
2
2
2
58. Cl + 2X– → 2Cl– + X (X = Br or I)
2
2
59. Br + 2I– → 2Br– + I
2
2
60. 2F (g) + 2H O (l) → 4H+ (aq) + 4F− (aq) + O (g)
2
2
2
( where X = Cl or Br )
61. 4I− (aq) + 4H+ (aq) + O (g) → 2I (s) + 2H O (l)
2
2
2
62. MnO + 4HCl → MnCl + Cl + 2H O
2
2
2
2
63. 4NaCl + MnO + 4H SO → MnCl +4NaHSO +2H O+Cl
2
2
4
2
4
2
2
64. 2KMnO + 16HCl → 2KCl + 2MnCl + 8H O + 5Cl
4
2
2
2
65. Cl + H O → 2HCl + O, Fe + 2HCl → FeCl + H
2
2
2
2
66. H S + Cl → 2HCl + S,
NO3- + 3Fe2+ + 4H+ → NO + 3Fe3+ + 2H2O
[Fe (H2O)6 ]2+ +NO→ [Fe(H2O)5 (NO)]2++ H2O
67.
P4 + 3NaOH + 3H2O → PH3 + 3NaH2PO2
68.
P4 + 5O2 → P4O10
69.
Ca3P2 + 6H2O → 3Ca(OH)2 + 2PH3
70.
Ca3P2 + 6HCl → 3CaCl2 + 2PH3
71.
PH4I + KOH → KI + H2O + PH3
72.
3CuSO4 + 2PH3 → Cu3 P2 + 3H2SO4
73.
3HgCl2 + 2PH3 → Hg3P2 + 6HCl
74.
P4 + 8SOCl2 → 4PCl3 + 4SO2 + 2S2Cl2
75.
3CH3COOH + PCl3 → 3CH3COCl + H3PO3
76.
P4 + 10SO2Cl2 → 4PCl5 + 10SO2
77.
PCl5+H2O → POCl3+2HCl, POCl3+3H2O →H3PO4+ 3HCl
78.
S8 + 4Cl2 → 4S2Cl2
79.
2Ag + PCl5 → 2AgCl + PCl3
80.
PCl3 + 3H2O →H3PO3 + 3HCl
81.
4H3PO3 → 3H3PO4 + PH3
2
2
Au + 4H+ + NO3− + 4Cl− → AuCl−4 + NO + 2H2O
3Pt + 16H+ + 4NO3 + 18Cl− → 3PtCl6− + 4NO + 8H2O
(cold and dilute)2NaOH + Cl2 → NaCl + NaOCl + H2O
(hot and conc.)6 NaOH + 3Cl2 →5NaCl +NaClO3 + 3H2O
2Ca(OH)2 + 2Cl2 → Ca(OCl)2 + CaCl2 + 2H2O
I2 + 6H2O + 5Cl2 → 2HIO3 + 10HCl
Na2SO3 + Cl2 + H2O → Na2SO4 + 2HCl
2FeSO4 + H2SO4 + Cl2 → Fe2(SO4)3 + 2HCl
SO2 + 2H2O + Cl2 → H2SO4 + 2HCl
XeF4 + O2 F2 → XeF6 + O2
2XeF2 (s) + 2H2O(l) → 2Xe (g) + 4 HF(aq) + O2(g)
6XeF4 + 12 H2O → 4Xe + 2XeO3 + 24 HF + 3 O2
XeF6+H2O→XeOF4 + 2 HF,XeF6+2H2O →XeO2F2 + 4HF
XeF6 + 3 H2O → XeO3 + 6 HF
2NH3 + 3NaClO →N2 + 3H2O + 3NaCl
82. 2K2HgI4 + NH3 + 3KOH→ Brown ppt. of HN HgOHgI
2
83. 2NH3 + NaOCl →NH2.NH2 hydrazine
Al2O3 ( s )+6NaOH(aq)+3H2O ( l ) → 2Na3[Al(OH)6](aq)
7
REACTION OF P BLOCK ELEMENTS
1.
2.
3.
4.
5.
3HNO2 → HNO3 + H2O + 2NO
41. Al2O3 + 6HCl + 9H2O → 2 [ Al(H2O)6 ]3+ + 6Cl−
NH4CI(aq) + NaNO2(aq) → N2(g) +2H2O(l) + NaCl (aq) 42.
8NH3 + 3Cl2 → 6NH4Cl + N2; NH3 + 3Cl2 → NCl3 + 3HCl
43. PbS(s) + 4O3(g) → PbSO4(s) + 4O2(g)
Ba(N3)2 → Ba + 3N2
44. 2I–(aq) + H O(l) + O (g) → 2OH–(aq) + I (s) + O (g)
2
6.
7.
8.
3
2
2
45. NO ( g ) + O ( g ) → NO ( g ) + O ( g )
3
2
2
46. 4FeS (s ) + 11O ( g ) → 2Fe O3 ( s ) + 8SO ( g )
2
2
2
2
47. SO2(g)+Cl2 (g) →SO2Cl2,SO2 + Ca(OH)2 + H2O +CaSO3 milky
48. 2Fe3+ + SO + 2H O → 2Fe2+ + SO2 − + 4H+
2NH4Cl + Ca(OH)2 → 2NH3 + 2H2O + CaCl2
(NH4)2 SO4 + 2NaOH → 2NH3 + 2H2O + Na2SO4
2
2
4
49. 5SO2+ 2MnO4 + 2H2O → 5SO42− + 4H+ + 2Mn2+
9. ZnSO4( aq )+2NH4OH ( aq ) → Zn(OH)2 ( s ) + ( NH4)2 SO4 (aq )50. SO + H SO → H S O
3
2
4
2 2
7
10. FeCl3+NH4OH ( aq ) → Fe2O3 .xH2O (s) +NH4Cl ( aq )
51.
11. Cu2+(aq) + 4NH3(aq) ‡ [Cu(NH3)4]2+ (aq)
(blue)
(deep blue)
52.
12. AgCl ( s ) + 2NH ( aq ) → Ag ( NH ) Cl ( aq )
3
3 2
53. Cu + 2 H SO (conc.) → CuSO + SO + 2H O
(white ppt)
2
(colourless)
4
4
2
2
13.
54. 3S + 2H SO (conc.) → 3SO + 2H O
2
4
2
2
55. C + 2H SO (conc.) → CO + 2 SO + 2 H O
14. 3NO ( g ) + H O ( l ) → 2HNO ( aq ) + NO ( g )
2
2
3
15. 3Cu + 8 HNO (dilute) → 3Cu(NO ) + 2NO + 4H O
56. CaF + H SO → CaSO + 2HF
2
2
4
4
57. F + 2X– → 2F– + X (X = Cl, Br or I)
2
3
3 2
2
16. Cu + 4HNO (conc.) → Cu(NO ) + 2NO + 2H O
3
3 2
2
2
17. 4Zn + 10HNO (dilute) → 4 Zn (NO ) + 5H O + N O
3
3 2
2
18. Zn + 4HNO (conc.) → Zn (NO ) + 2H O + 2NO
3
3 2
2
2
19. I + 10HNO → 2HIO + 10NO + 4H O
2
3
3
2
2
20. C + 4HNO → CO + 2H O + 4NO
3
2
2
2
21. S + 48HNO → 8H SO + 48NO + 16H O
8
3
2
4
2
2
22. P + 20HNO → 4H PO + 20NO + 4H O
4
3
3
4
2
2
23. BROWN RING TEST
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
2
4
2
2
2
2
2
58. Cl + 2X– → 2Cl– + X (X = Br or I)
2
2
59. Br + 2I– → 2Br– + I
2
2
60. 2F (g) + 2H O (l) → 4H+ (aq) + 4F− (aq) + O (g)
2
2
2
( where X = Cl or Br )
61. 4I− (aq) + 4H+ (aq) + O (g) → 2I (s) + 2H O (l)
2
2
2
62. MnO + 4HCl → MnCl + Cl + 2H O
2
2
2
2
63. 4NaCl + MnO + 4H SO → MnCl +4NaHSO +2H O+Cl
2
2
4
2
4
2
2
64. 2KMnO + 16HCl → 2KCl + 2MnCl + 8H O + 5Cl
4
2
2
2
65. Cl + H O → 2HCl + O, Fe + 2HCl → FeCl + H
2
2
2
2
66. H S + Cl → 2HCl + S,
NO3- + 3Fe2+ + 4H+ → NO + 3Fe3+ + 2H2O
[Fe (H2O)6 ]2+ +NO→ [Fe(H2O)5 (NO)]2++ H2O
67.
P4 + 3NaOH + 3H2O → PH3 + 3NaH2PO2
68.
P4 + 5O2 → P4O10
69.
Ca3P2 + 6H2O → 3Ca(OH)2 + 2PH3
70.
Ca3P2 + 6HCl → 3CaCl2 + 2PH3
71.
PH4I + KOH → KI + H2O + PH3
72.
3CuSO4 + 2PH3 → Cu3 P2 + 3H2SO4
73.
3HgCl2 + 2PH3 → Hg3P2 + 6HCl
74.
P4 + 8SOCl2 → 4PCl3 + 4SO2 + 2S2Cl2
75.
3CH3COOH + PCl3 → 3CH3COCl + H3PO3
76.
P4 + 10SO2Cl2 → 4PCl5 + 10SO2
77.
PCl5+H2O → POCl3+2HCl, POCl3+3H2O →H3PO4+ 3HCl
78.
S8 + 4Cl2 → 4S2Cl2
79.
2Ag + PCl5 → 2AgCl + PCl3
80.
PCl3 + 3H2O →H3PO3 + 3HCl
81.
4H3PO3 → 3H3PO4 + PH3
2
2
Au + 4H+ + NO3− + 4Cl− → AuCl−4 + NO + 2H2O
3Pt + 16H+ + 4NO3 + 18Cl− → 3PtCl6− + 4NO + 8H2O
(cold and dilute)2NaOH + Cl2 → NaCl + NaOCl + H2O
(hot and conc.)6 NaOH + 3Cl2 →5NaCl +NaClO3 + 3H2O
2Ca(OH)2 + 2Cl2 → Ca(OCl)2 + CaCl2 + 2H2O
I2 + 6H2O + 5Cl2 → 2HIO3 + 10HCl
Na2SO3 + Cl2 + H2O → Na2SO4 + 2HCl
2FeSO4 + H2SO4 + Cl2 → Fe2(SO4)3 + 2HCl
SO2 + 2H2O + Cl2 → H2SO4 + 2HCl
XeF4 + O2 F2 → XeF6 + O2
2XeF2 (s) + 2H2O(l) → 2Xe (g) + 4 HF(aq) + O2(g)
6XeF4 + 12 H2O → 4Xe + 2XeO3 + 24 HF + 3 O2
XeF6+H2O→XeOF4 + 2 HF,XeF6+2H2O →XeO2F2 + 4HF
XeF6 + 3 H2O → XeO3 + 6 HF
2NH3 + 3NaClO →N2 + 3H2O + 3NaCl
82. 2K2HgI4 + NH3 + 3KOH→ Brown ppt. of HN HgOHgI
2
83. 2NH3 + NaOCl →NH2.NH2 hydrazine
Al2O3 ( s )+6NaOH(aq)+3H2O ( l ) → 2Na3[Al(OH)6](aq)
8