Assign oxidation number to the underlined elements in each of the following species:
(a) NaH2PO4
(b) NaHSO4
(c) H4P2O7
(d) K2MnO4
(e) CaO2
(f) NaBH4
(g) H2S2O7
(h) KAl(SO4)2.12 H2O
(a) NaH2PO4
Let's assume oxidation number of P is x.
We know that,
Oxidation number of Na = +1
Oxidation number of H = +1
Oxidation number of O = -2
Then we have
1(+1) + 2(+1) + 1 (x) + 4(-2) = 0
⇒ 1 + 2 + x - 8 = 0
⇒ x - 5 = 0
⇒ x = + 5
Hence, oxidation number of P is +5
(b) NaHSO4
Let's assume oxidation number of S is x.
Oxidation number of Na = +1
Oxidation number of H = +1
Oxidation number of O = -2
Then we have:
1(+1) + 1(+1) + 1 (x) + 4(-2) = 0
⇒ 1 + 1 + x - 8 = 0
⇒ x-6 = 0
⇒ x = +6
Hence, oxidation number of S is +6
(c) H4P2O7
Let's assume oxidation number of P is x.
Oxidation number of H = +1
Oxidation number of O = -2
Then we have:
4(+1) + 2(x) + 7 (-2) = 0
⇒ 4 + 2x - 14 = 0
⇒ 2x - 10 = 0
⇒ 2x = +10
⇒ x = +5
Hence, Oxidation number of P is +5
(d) K2MnO4
Let's assume oxidation number of Mn is x.
Oxidation number of K = +1
Oxidation number of O = -2
Then we have:
2(+1) + 1(x) + 4 (-2) = 0
⇒ 2 + x - 8 = 0
⇒ x - 6 = 0
⇒ x = +6
Hence, Oxidation number of Mn is +6
(e) CaO2
Let's assume oxidation number of O is x.
Oxidation number of Ca = +2
Then we have:
1(+2) + 2(x) = 0
⇒ 2 + 2x = 0
⇒ 2x = -2
⇒ x = -1
Hence, Oxidation number of O is -1
(f) NaBH4
Let's assume oxidation number of B is x.
Oxidation number of Na = +1
Oxidation number of H = -1
Then we have:
1(+1) + 1(x) + 4(-1) = 0
⇒ 1 + x -4 = 0
⇒ x - 3 = 0
⇒ x = +3
Hence, Oxidation number of B is +3.
(g) H2S2O7
Let's assume oxidation number of S is x.
Oxidation number of O = -2
Oxidation number of H = +1
Then we have:
2(+1) + 2(x) + 7(-2) = 0
⇒ 2 + 2x - 14 = 0
⇒ 2x - 12 = 0
⇒ x = +6
Hence, Oxidation number of S is +6.
(h) KAl(SO4)2.12 H2O
Let's assume oxidation number of S is x.
Oxidation number of K = +1
Oxidation number of Al = +3
Oxidation number of O = -2
Oxidation number of H = +1
Then we have:
1(+1) + 1 (+3) + 2(x) + 8(-2) + 24(+1) + 12 (-2) = 0
⇒ 1 + 3 + 2x -16 +24 -24 = 0
⇒ 2x - 12 = 0
⇒ 2x = +12
⇒ x = +6
Hence, Oxidation number of S is +6.
Balance the following redox reactions by ion – electron method :
(a) MnO4 – (aq) + I – (aq) → MnO2 (s) + I2(s) (in basic medium)
(b) MnO4 – (aq) + SO2 (g) → Mn2+ (aq) + HSO4– (aq) (in acidic solution)
(c) H2O2 (aq) + Fe 2+ (aq) → Fe3+ (aq) + H2O (l) (in acidic solution)
(d) Cr2O7 2– + SO2(g) → Cr3+ (aq) + SO42– (aq) (in acidic solution)
What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results ?
(a) KI3
(b) H2S4O6
(c) Fe3O4
(d) CH3CH2OH
(e) CH3COOH
Justify that the following reactions are redox reactions:
(a) CuO(s) + H2(g) → Cu(s) + H2O(g)
(b) Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
(c) 4BCl3(g) + 3LiAlH4(s) → 2B2H6(g) + 3LiCl(s) + 3 AlCl3 (s)
(d) 2K(s) + F2(g) → 2K+F– (s)
(e) 4 NH3(g) + 5 O2(g) → 4NO(g) + 6H2O(g)
Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
(a) P4(s) + OH – (aq) → PH3(g) + HPO2 – (aq)
(b) N2H4(l) + ClO3 – (aq) → NO(g) + Cl–(g)
(c) Cl2O7 (g) + H2O2(aq) → ClO – 2(aq) + O2(g) + H + (aq)
Fluorine reacts with ice and results in the change:
H2O(s) + F2(g) → HF(g) + HOF(g)
Justify that this reaction is a redox reaction.
Write the formulae for the following compounds:
(a) Mercury(II) chloride
(b) Nickel(II) sulphate
(c) Tin(IV) oxide
(d) Thallium(I) sulphate
(e) Iron(III) sulphate
(f) Chromium(III) oxide
While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?
Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions:
(a) 2AgBr (s) + C6H6O2(aq) → 2Ag(s) + 2HBr (aq) + C6H4O2(aq)
(b) HCHO(l) + 2[Ag (NH3)2]+(aq) + 3OH-(aq) → 2Ag(s) + HCOO-(aq) + 4NH3(aq) + 2H2O(l)
(c) HCHO (l) + 2Cu2+(aq) + 5 OH-(aq) → Cu2O(s) + HCOO-(aq) + 3H2O(l)
(d) N2H4(l) + 2H2O2(l) → N2(g) + 4H2O(l)
(e) Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)
The compound AgF2 is an unstable compound. However, if formed, the compound acts as a very strong oxidizing agent. Why?
Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, Cr2O2- 7 and NO– 3. Suggest structure of these compounds. Count for the fallacy.
How do you account for the formation of ethane during chlorination of methane?
What are hybridisation states of each carbon atom in the following compounds ?
(i) CH2=C=O,
(ii) CH3CH=CH2,
(iii) (CH3)2CO,
(iv) CH2=CHCN,
(v) C6H6
What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30°C?
What are the common physical and chemical features of alkali metals?
Calculate the molecular mass of the following:
(i) H2O
(ii) CO2
(iii) CH4
What is the basic theme of organisation in the periodic table?
Explain the formation of a chemical bond.
Choose the correct answer. A thermodynamic state function is a quantity
(i) used to determine heat changes
(ii) whose value is independent of path
(iii) used to determine pressure volume work
(iv) whose value depends on temperature only.
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
a) What is the initial effect of the change on vapour pressure?
b) How do rates of evaporation and condensation change initially?
c) What happens when equilibrium is restored finally and what will be the final vapour pressure?
Justify the position of hydrogen in the periodic table on the basis of its electronic configuration.
A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.
Explain the physical significance of Van der Waals parameters.
Arrange the bonds in order of increasing ionic character in the molecules: LiF, K2O, N2, SO2 and ClF3.
Which one of the following statements is incorrect in relation to ionization enthalpy?
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration.
(c) End of valence electrons is marked by a big jump in ionization enthalpy.
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.
In some of the reactions thallium resembles aluminium, whereas in others it resembles with group I metals. Support this statement by giving some evidences.
How can you explain higher stability of BCl3 as compared to TlCl3?
Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
N2(g) + H2(g) → 2NH3(g)
(i) Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts with 1.00 × 103 g of dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?
How many significant figures should be present in the answer of the following calculations?
(i)
(ii) 5 × 5.364
(iii) 0.0125 + 0.7864 + 0.0215
The equilibrium constant expression for a gas reaction is,
Write the balanced chemical equation corresponding to this expression.
An atom of an element contains 29 electrons and 35 neutrons.
Deduce (i) the number of protons and (ii) the electronic configuration of the element.
Gud to disclose all the answers .anyway thanks