Inorganic chemistry questions answers

    Problems on inorganic chemistry appear really as problems for 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. Priyam study centre is updated regularly so keep in touch with this page.
    How can convert 1 Debye to the coulomb meter?
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
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]
    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.
    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
    = 16.89%
    The difference between the electronegativity of carbon and oxygen is large but the dipole moments of carbon monoxide are very low - Why?
    However, in CO, there is a large difference in 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.
    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)
We have, μ = 0.95 D and θ = 97°
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.
    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
    Calculate the second ionization potential of helium (given ionization potential of hydrogen = 13.6 eV).
    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

Thus second ionization potential of helium,
= 22 × 13.6
= 54.4 eV
    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?
    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
EH = (me⁴/8ε₀²h²)[(1/n₁²) - (1/n₂²)]
= 21.79 × 10⁻¹⁹ J
where n₁ refers to the 1S orbital and n₂ refers to infinity.
    So the ionization potential for removal of the electron from the 2S orbital to infinity is given by
EH = (me⁴/8ε₀²h²)[(1/2²) - (0)]
= (21.79 × 10⁻¹⁹/4) Joule
= 5.45 10⁻¹⁹ Joule

Slater's rules

    In the first transition series electron filling up processes begins in the 3d level below a filled 4S2 level. During the ionization process will a 4S electron or a 3d electron be lost first? Explain with reference to chromium.
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
    Arrange the following with the increasing order of Ionization Potentials Li, Be, B, C, N, O, F, Ne.
    Calculate the effective nuclear charge of the hydrogen atom.
    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.
    Find out the screening constant of the 4S and 3d electron of chromium atom.
    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
    Estimate the screening constant for the outermost 4S electron of Vanadium.
    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
    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,
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.
    Born - Haber cycle for the formation of NaCl (S) is:
Questions and answers inorganic chemistry
Born - Haber cycle
    From the above Born - Haber cycle we can write 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
    Account for the large decrees in electron affinity from lithium to beryllium despite the increase in nuclear charge.
    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.
    (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
    Calculate the oxidation number of Iron in [Fe(H₂O)₅(NO)⁺]SO₄.
    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.
    Calculate the oxidation number of the Cl in Ca(OCl)Cl.
    The oxidation number of two cl atom is -1 and +1.
    What is the oxidation state of chromium in Cr₂O₅?
    Due to the peroxy linkage oxidation state of Cr in Cr₂O₅ is +6.
    Why sulfur dioxide has properties of oxidation and reduction?
    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 the middle of 0 and +6. 
    Thus, the oxidation state can increases from +4 to +6 and decreases from +4 to 0.
+ -2

+ 0
    Express by ion-electron method the reduction of permanganate to manganous stat by hydrogen peroxide in acid medium.
    This Reaction occurs in acid medium and the partial equation is:
MnO₄⁻ + 8H⁺ + 5e ⇆ Mn⁺² + 4H₂O
(Reduction of Oxidant)

H₂O₂ ⇆ 2H⁺ + O₂ + 2e
(Oxidation of Reductant)
    In an acid medium, H₂O₂ will give O₂ and the partial equation being the first equation is multiplying by 2 and the second is 5 to have electron balanced.
We have,2MnO₄⁻ + 16H⁺ + 10e ⇆ 2Mn⁺² + 8H₂O
5H₂O₂ ⇆ 10H⁺ + 2O₂ + 10e

2MnO₄⁻ + 5H₂O₂ + 6H⁺ ⇆ 2Mn⁺² + 8H₂O + 5O₂ 
    Express by ion-electron method the reduction of nitrate ion to ammonia by aluminum in aqueous NaOH.
    In the alkaline medium, we use H₂O and OH⁻ ion according to convenience. The partial equation with charge balance is:
NO₃⁻ + 6H₂O + 8e ⇆ NH₃ + 9OH⁻
    Metallic aluminum 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⁻
    Use the oxidation number method to balance the reaction of iodide ion and iodate ion in an acid medium to liberate iodine.
The reaction represented as
I⁻ + IO₃⁻ → I2

Representation of the above equation with the oxidation of iodine,
I⁻ + IO₃⁻ → 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
    Use the oxidation number method to balance the reaction between Sulfurous acid and dichromate in acidic medium.
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 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

Inorganic chemistry questions answer for the for high school and college courses with explanation

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