Questions and Answers Inorganic Chemistry

    Problems on inorganic chemistry appear really as problems to most of the students studying in different Schools, Colleges and different universities.
    To overcome this problem to achieve a good performance in examinations I will discuss some of these problems topic wise. This page is updated regularly so keep on touch with this page.
  • Problem
    How can convert 1 Debye to coulomb meter?
  • Solution
The dipole moment in CGS system is,
µ = 4.8 × 10⁻¹⁰ × 10⁻⁸ esu cm
= 4.8 D

In SI system, 1.6 ×10⁻¹⁹ × 10⁻¹⁰ coulomb × meter
= 1.6 × 10⁻³⁰ coulombs × meter

Thus, 4.8 Debye = 1.6 ×10⁻³⁰ coulomb × meter
or, 1 Debye = (1.6 × 10⁻³⁰)/4.8 coulomb × meter
∴ 1 Debye = 3.336 × 10⁻³⁰ coulomb × meter
  • Problem
    Find out the unit and dimension of Dipole Moments.
  • Solution
Unit of µ = Unit of Charge × Unit of Length.
Thus, the unit of µ = esu × cm in the CGS system.

But from the Coulomb’s Low,
F = q₁q₂/r²
or, q₁q₂ = F × r²
∴ (esu)² = dyne × cm² = gm cm sec⁻² × cm²
Hence, esu = gm½ cm3/2 sec⁻¹

∴ Unit of µ in CGS system, gm½ cm3/2 sec⁻¹ × cm
= gm½ cm5/2 sec⁻¹

∴ Unit of µ in SI system, kg½ m3/2 sec⁻¹ × m
= kg½ m5/2 sec⁻¹

Thus the dimension of µ,
[M½ L5/2 T ⁻¹]
  • Problem
    The displacement of HCl is 1.03 Debye and its bond length is 1.27 A. Calculate the (1) charge on the constituent atom and (2) the % of the ionic character of HCl.
  • Solution
    Given, μobs = 1.03 Debye = 1.03 × 10⁻¹⁸ esu cm and length l = 1.27 × 10⁻⁸ cm.
  1. Charge on the constituent atom(q), = μobs/ℓ
    = (1.03 × 10⁻¹⁸ )/( 1.27 × 10⁻⁸)
    = 0.8 × 10⁻¹⁰ esu
  2. Percentage of the ionic character of HCl, (μobs/μionic ) × 100
    =(1.03×10⁻¹⁸)/(4.8×10⁻¹⁰×1.27×10⁻⁸)×100
    = 16.89%
  • Problem
    The difference between the electronegativity of carbon and oxygen is large but the dipole moments of carbon monoxide is very low - Why?
  • Answer
    However, in CO, there is a large difference of electronegativity between C and O but the molecule is the very low value of dipole moment. This suggested that the charge density in O-atom is somehow back-donated to C-atom.
    This explains by forming a coordinate covalent bond directing towards C-atom.
  • Problem
    The bond angle in H2S is 970 and dipole moment = 0.95 D. Find the bond moment of the S-H link. ( Given, Cos970 = -0.122)
  • Answer
We have, μ = 0.95 D and θ = 97°
From the equation, μ² = m₁² + m₂² + 2m₁ m₂ cosθ
For, H₂S, m₁ = m₂ = mS-H
∴ μ² = 2m²(1 + cosθ)
Putting the value we have, (0.95)² = 2 m² (1+ cos97° )
Here, m = mS - H
or, 0.9025 = 2 m² (1 - 0.122)
or, m² = 0.9025/ (2 × 0.878)
or, m² = 0.5139
or, m = 0.72
Thus the bond moment of the mS - H link is 0.72 D.
  • Problem
    Calculate the ionization potential of a hydrogen atom in eV.
  • Answer
    We know that the energy corresponding to the transition from n = ∞ to n = 1 gives the ionization potential of the hydrogen atom.
Thus the Ionization Potential of the Hydrogen atom,
= EH = 2π2me4/h2 [(1/12) - 0]
= 2.179 × 10-11 erg
= 2.179 × 10-18 Joule
= (2.179 × 10-18)/(1.6 × 10-19) eV
= 13.6 eV
  • Problem
    Calculate the second ionization potential of helium (given ionization potential of hydrogen = 13.6 eV).
  • Answer
    The ground state electronic configuration of helium is 1S2. The second ionization potential means the removal of the second electron from the 1S orbital against the nuclear charge of +2.
    Thus ionization potential, = (2π²mZ²e⁴/h²) [(1/n₁²) - (1/n₂²)] = Z² × EH
    ∴ Second Ionization Potential of Helium, = 22 × 13.6 = 54.4 eV
  • Problem 8:
The normal ionization potential of a hydrogen atom is 21.79 × 10-19 J. What will be the value of ionization potential when an electron is raised to the 2S level?
  • Answer:
The normal ionization potential of a hydrogen atom is the energy required to shift the electron from the 1S orbital to infinity and is given by (in SI units):
EH = (me4/02h2)[(1/nI2) - (1/nII2)] 
= 21.79 × 10-19 J
where nI refers to the 1S orbital (nI = 1) and nII refers to infinity.
So the ionization potential for removal of the electron from the 2S orbital to infinity is given by:
EH = (me4/02h2)[(1/22) - (0)] 
= 21.79 10-19/4 J
= 5.45 10-19 J

Slater's Rules:

  • Problem 9:
In first transition series electron filling up processes begins in the 3d level below a filled 4S2 level. During ionization process will a 4S electron or a 3d electron be lost first? Explain with reference to Chromium.
  • Answer:
Electronic Configuration of Chromium (Cr):
24 1S2 2S2 2P6 3S2 3p6 3d5 4S1
Screening constant (σ) for 4S electron is,
σ = (2×1.0)+(8×1.0)+(8×0.85)+(5×0.85)
= 21.05.
∴ Effective nuclear charge,
Z⋆; = (24 - 21.05)
= 2.95.
Screening constant (σ) for 3d electron is,
σ = (2×1.0)+(8×1.0)+(8×1)+(4×0.35)
= 19.4.
∴ Effective nuclear charge,
Z = (24 - 19.4)
= 4.60.
  • Problem 10:
Arrange the following with the increasing order of Ionization Potentials Li, Be, B, C, N, O, F, Ne.
  • Answer:
LiBBeCONFNe
  • Problem 11:
Calculate the effective nuclear charge of the hydrogen atom.
  • Answer:
The hydrogen atom has a single 1S valence electron. There is no other electron to screen it from the nuclear charge of a single proton. Thus, σ = 0 and Z = 1.0 -0 = 1.0.
Thus hydrogen electron sees the full nuclear charge of the nucleus, that is the electron is totally exposed to the proton.
  • Problem 12:
Find out the Screening Constant of the 4S and 3d electron of Chromium Atom.
  • Answer:
Chromium has atomic number 24 and the electronic configuration according to the Slater's Rule is: 
(1S)2 (2S, 2P)8 (3S, 3P)8 (3d)5 (4S)1
Screening Constant (σ) for the 4S electron is:
σ =(2×1.0)+(8×1.0)+(8×1.0)+(3×0.85)+(0×0.35)
= 21.05.
And the Screening Constant (σ) for the 3d electron is:
σ = (2 ×1.0)+(8×1.0)+(8×1.0)+(4×0.35)
= 19.4.
  • Problem 13:
Estimate the screening constant for the outermost 4S electron of Vanadium.
  • Answer:
Vanadium has atomic number 23 and the electronic configuration according to the Slater's Rule is : 
(1S)2 (2S, 2P)8 (3S, 3P)8 (3d)3 (4S)2
We have considered only one electron of the two 4S electrons.
∴ Screening Constant (σ)
= (2×1.0)+(8×1.0)+(8×0.85)+(3×0.85)+(1×0.35)
= 19.7.

  • Problem 14:
Calculate the electron affinity of chlorine from the Born - Haber cycle, given the following date : 
lattice Energy = - 774 kJ mol-1,
Ionization Potential of Na = 495 kJ mol-1,
The heat of Sublimation of Na = 108 kJ mol-1,
Energy for Bond dissociation of Chlorine (Cl₂) = 240 kJ mol-1
and Heat of Formation of NaCl = 410 kJ mol-1.
  • Answer:
Born - Haber Cycle for the formation of NaCl (S) is:
Questions and Answers Inorganic Chemistry
Born-Haber Cycle applied to NaCl.
From the above Born - Haber cycle we can written as:
-UNaCl - INa + ECl - SNa - 1/2DCl - ΔHf=0
or, ECl =UNaCl +  INa+ SNa+ 1/2DCl + ΔHf
ECl = -774 + 495 + 108 + 120 + 410
= 359 kJ mol⁻¹
  • Problem 15:
Account for the large decrees in electron affinity from lithium to beryllium despite the increase in nuclear charge.
  • Answer:
Atomic number and the electronic distribution of lithium and beryllium are:
Li : 3 : 1S2 2S1
Be: 4: 1S2 2S2
Lithium has an incompletely filled 2S sub-shell while beryllium has the subshell filled. Thus lithium can receive an electron in its 2S sub-shell but for beryllium, a still higher energy 2P level has to be made of. A filled shell or sub-shell leads to some extra stability. Hence beryllium resists an extra electron more than does lithium.

  • Problem 16:
Oxidation number of P in Ba(H2PO2)2 is - (a)+3, (b)+2, (c) +1, (d) -1.
  • Answer:
(c) +1
The oxidation number of Ba is +2, the oxidation number of hydrogen is +1 and the oxidation number of oxygen is -2.
Let the oxidation number of P is x
∴ (+2) + 2{2(+1) + x + 2(-2)} = 0 
or, 2x-2=0 
or, x=+1
  • Problem 17:
Calculate the oxidation number of Iron in [Fe(H2O)5(NO)+]SO4.
  • Answer:
H2O is neutral thus the oxidation number is zero, the oxidation number of NO+ is +1 and the oxidation number of SO4 is -2
Let the oxidation number of Fe in [Fe(H2O)5NO+]+2 is x.
∴ x + 0 + (+1) + (-2) = 0
or, x-1 = 0
or, x = +1
Thus, the oxidation number of Fe in [Fe(H2O)5NO+]+2 is +1.
  • Problem 18:
Calculate the oxidation number of the element marked with blue in the following compounds, (i) K2CrO4, (ii) HOCl, (iii) BaO2, (iv) ClNO, (v) NaNH2, (vi) NaN3, (vii) CH2Cl2, (viii) Ca(OCl)Cl, (ix) Ba(MnO4)2 (x) CaH2.
  • Answer
Compound Element Oxidation Number
K2CrO4 Cr +6
HOCl Cl +1
BaO2 Ba +2
ClNO N +3
NaNH2 N -3
NaN3 N -1/3
CH2Cl2 C 0
Ca(OCl)Cl Cl +1
Ba(MnO4)2 Mn +7
CaH2 Ca -1
  • Problem 19:
What is the Oxidation state of chromium in Cr2O5?
  • Answer:
Due to the peroxy linkage oxidation state of Cr in Cr2O5 is +6.
  • Problem 20:
Why sulfur dioxide has properties of Oxidation and reduction?
  • Answer:
This can be explained by the oxidation state of sulfur. The oxidation numbers of sulfur in the compounds H2S, SO2, and SO3 are -2, +4 and +6 respectively.
Thus the highest oxidation state of sulfur is +6 and the lowest is -2. The oxidation state of the free sulfur element is 0. In SO2, the oxidation state of sulfur is +4, this oxidation state is middle of 0 and +6. 
Thus, the oxidation state can increases from +4 to +6 and decreases from +4 to 0.
+4
SO2
+ -2
H2S

2H2O
+ 0
3S

  • Problem 21:
Express by ion electron method the reduction of permanganate to manganous stat by hydrogen peroxide in acid medium.
  • Answer:
This Reaction occurs in acid medium and the partial equation is:
MnO4- + 8H+ + 5e Mn+2 + 4H2O
(Reduction of Oxidant)
H2O22H+ + O2 + 2e
(Oxidation of Reductant)
In acid medium H2O2 will give O2 and the partial equation being: First equation is multiplying by 2 and the second is 5 to have electron balanced. We have,
2MnO4- + 16H+ + 10e 2Mn+2 + 8H2O
5H2O210H+ + 2O2 + 10e

2MnO4- + 5H2O2 + 6H+ 2Mn+2 + 8H2O + 5O2
  • Problem 22:
Express by ion - electron method the reduction of nitrate ion to ammonia by aluminium in aqueous NaOH.
  • Answer:
In alkaline medium we use H2O and OH- ion according to convenience. The partial equation with charge balance is:
NO3- + 6H2O + 8e NH3 + 9OH- 
Metallic aluminium will go over to aluminate ion, the partial equation being: 
Al + 4OH- AlO2- + 2H2O +3e 
Multiplying by right factors for electron balance we have the balanced equation:
NO3- + 6H2O + 8e NH3 + 9OH- 
Al + 4OH- AlO2- + 2H2O +3e 

3NO3- + 8Al + 5OH-  3NH3 + 8AlO2-
  • Problem 23:
Use Oxidation Number method to balance the reaction of iodide ion and iodate ion in acid medium to liberate iodine.
  • Answer:
The reaction represented as,
I- + IO3- I2
Representation of the above equation with an oxidation number of iodine,
-1
I-
+ +5
IO3-
0
I2
Since I- (-1) oxidation number of iodide increases by 1 and IO3-(+5) oxidation number decreases by 5. 
Putting the right factors the decrease and increase in oxidation number balanced: 
5I- + IO3- 3I2
Since the medium is acidic and the charges are unequal on two sides of the above expression we put Six H⁺ and write three H₂O to balanced the equation:
5I- + IO3- + 6H+ 3I2 + 3H2O
  • Problem 24:
Use the oxidation number method to balance the reaction between Sulphurus acid and dichromate in acidic medium.
  • Answer:
The reaction represented as,
SO3-2 + Cr2O7-2 SO4-2 + 2Cr+3
Oxidation of two Cr decreases by 2 × (+3) = 6 and the oxidation number of S increases by 2. Equalizing the above equation increase and decrees in oxidation number,
3SO3-2 + Cr2O7-2 3SO4-2 + 2Cr+3
Because of the medium is acidic and the ionic charges are not equal on the two sides, eight H+ ions are added to the Left-hand side.
3SO3-2 + Cr2O7-2 + 8H+ 3SO4-2 + 2Cr+3
Since there are eight hydrogen ion on the left and none on the right, four H2O added to make atom balance. Thus the final equation is:
3SO3-2 + Cr2O7-2 + 8H+ 3SO4-2 + 2Cr+3 + 4H2O

Dipole Moments, Ionisation Potential or Ionization Energy, Slater's Rules for calculation of Effective nuclear Charge, Electron Affinity, Oxidation Number and Balancing Oxidation Reduction Reactions Questions and Answers.

Inorganic Chemistry

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