# Inorganic chemistry questions

### Study inorganic chemistry questions online

Problems on inorganic chemistry appear really as problems for most of the students study chemistry in different Schools, Colleges and Universities.

To overcome this problem to achieve a good performance in examinations I will discuss some of these problems topic wise. The content published on this page updated regularly so keep in touch with this page.

#### The polarity of bond in a molecule

Question
How can convert 1 Debye to the coulomb-meter?

Dipole noment in CGS unt,
µ = 4.8 × 10⁻¹⁰ × 10⁻⁸ esu cm
= 4.8 D.

Dipole moment in SI unit
µ = 1.6 ×10⁻¹⁹× 10⁻¹⁰ coulomb × meter
= 1.6 × 10⁻³⁰ coulombs-meter.

∴ 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.
Question
The dipole moment of hydrochloric acid 1.03 D and bond length 1.27 Å. Calculate the charge and percentage of the ionic character of the bond in hydrochloric acid.

μobs = 1.03 D = 1.03 × 10⁻¹⁸ esu cm and length = 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 the bond in hydrochloric acid
= (μobs/μionic) × 100
= {(1.03×10⁻¹⁸)/(4.8×10⁻¹⁰×1.27×10⁻⁸)} × 100
= 16.89 %
Question
What are the unit and dimensions of the dipole moment of a molecule?

Unit of dipole moment = unit of charge × unit of length.

∴ CGS unit of the dipole moment = esu-cm.

Coulomb’s law
F = q₁q₂/r²
or, q₁q₂ = F × r².

(esu)² = dyne × cm²
= gm cm sec⁻² × cm²
∴ esu = gm½ cm3/2 sec⁻¹.

CGS unit of the dipole moment
= gm½ cm3/2 sec⁻¹ × cm
= gm½ cm5/2 sec⁻¹.

SI unit of the dipole moment
= kg½ m3/2 sec⁻¹ × m
= kg½ m5/2 sec⁻¹.

Dimension of dipole moment
[M½ L5/2 T ⁻¹].

#### The polarity of carbon, oxygen, and hydrogen sulfide

Question
Why electronegativity difference of carbon and oxygen large in carbon monoxide molecule but dipole moment of carbon monoxide is very low?

In carbon monoxide polarity of the carbon-oxygen bond very low because of charge on the oxygen atom back-donated to the carbon atom.
This explains by forming a coordinate covalent bond directing towards carbon atom.

Question
Dipole moment and bond angle in hydrogen sulfide 97° and 0.95 D respectively. What is the bond moment of the sulfur - hydrogen link? (Cos970 = -0.122).

Dipole moment = 0.95 D
bond moment = 97°
m₁ = m₂ = mS - H.

μ² = m₁² + m₂² + 2m₁ m₂ cosθ
∴ μ² = 2m²(1 + cosθ).

(0.95)² = 2 m² (1+ cos97°)
or, m² = 0.9025/ (2 × 0.878)
or, m² = 0.5139
∴ m = 0.72.

The polarity of the mS - H link is 0.72 D.

#### The ionization energy of an atom

Question
How to calculate the ionization energy of a hydrogen atom?

The energy corresponding to the transition from n = ∞ to n = 1 gives the ionization energy of the hydrogen atom.

∴ Ionization energy,
EH = (2π²me⁴/h²) [(1/1²) - (1/∞²)]
= 2.179 × 10⁻¹¹ erg.

= 2.179 × 10⁻¹⁸ Joule.

= (2.179 × 10⁻¹⁸)/(1.6 × 10⁻¹⁹) eV
= 13.6 eV.
Question
How to calculate the second ionization energy of helium? Ionization energy of hydrogen = 13.6 eV.

Electronic configuration of helium atom 1S². The second ionization energy of helium atom means the removal of the second electron from 1S subshell against nuclear charge +2.

∴ The ionization energy of the helium atom,
= (Atomic number of He)² × (Ionization energy of H)
= 2² × 13.6
= 54.4 eV.
Question
How to calculate ionization energy of hydrogen when an electron raised to the 2S subshell? The ionization energy of hydrogen = 21.79 × 10⁻¹⁹ Joule.

The ionization energy of the hydrogen atom is the energy of an electron for the transition from 1S subshell to infinity.

The ionization energy of hydrogen = 21.79 × 10⁻¹⁹ J.

∴ Ionization energy for removal of the electron from the 2S subshell to infinity

= (me⁴/8ε₀²h²)[(1/2²) - (0)]
= (21.79 × 10⁻¹⁹/4) Joule
= 5.45 10⁻¹⁹ Joule.

#### Shielding effect and electronic configuration

Question
Electron filling process in the first transition series 3d subshell below a filling 4S subshell. During ionization which electron loses first? Explain with the reference of chromium.

The atomic number of chromium = 24
Electronic configuration
1S² 2S² 2P⁶ 3S² 3p⁶ 3d⁵ 4S¹

Shielding constant (σ) for 4S electron,
σ = (2 × 1.0) + (8 × 1.0) + (8 × 0.85) + (5 × 0.85)
= 21.05.

∴ Effective nuclear charge
= (24 - 21.05)
= 2.95.

Shielding constant (σ) for 3d electron
σ = (2 × 1.0) + (8 × 1.0) + (8 × 1) + (4 × 0.35)
= 19.4.

∴ Effective nuclear charge
= (24 - 19.4)
= 4.60.
Question
Arrange the following with the increasing order of ionization energy lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine, and neon.

Ionization energy left to right along a period decreases due to increasing the size of the atom. But for beryllium and nitrogen due to the presence of full fill and half-filled subshell removing an electron required higher energy than the normal trend.

Thus ionization energy of beryllium greater than boron and ionization energy of nitrogen greater than oxygen.
LiㄑBㄑBeㄑCㄑOㄑN〈FㄑNe.

Question
How to calculate the effective nuclear charge of the hydrogen atom?

The hydrogen atom has a single 1S valence electron. No other electron to shield it from the nuclear charge of a single proton.

Shielding constant = 0.
Effective nuclear charge = 1.0 - 0
= 1.0.

Hydrogen electron sees the full nuclear charge of the nucleus, that is the electron is totally exposed to the proton.

Question
Find out the shielding 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

(1S)² (2S, 2P)⁸ (3S, 3P)⁸ (3d)⁵ (4S)¹

∴ Shielding constant for the 4S electron
(2 × 1.0) + (8 × 1.0) + (8 × 1.0) + (3 × 0.85) + (0 × 0.35)
= 21.05.

Shielding constant for the 3d electron
(2 × 1.0) + (8 × 1.0) + (8 × 1.0) + (4 × 0.35)
= 19.4.
Question
Estimate the shielding constant for the outermost 4S electron of vanadium.

Vanadium has atomic number 23 and the electronic configuration
(1S)² (2S, 2P)⁸ (3S, 3P)⁸ (3d)³ (4S)²

4S subshell contains two electrons out of these two electrons we consider one electron of 4S subshell for calculating shielding constant.

∴ Shielding constant for 4S electron of vanadium
= (2 × 1.0) + (8 × 1.0) + (8 × 0.85) + (3 × 0.85) + (1 × 0.35)
= 19.7.

#### Electron affinity of an atom

Question
Calculate the electron affinity of chlorine from the Born - Haber cycle, given the following data,
lattice energy = - 774 kJ mol⁻¹
the ionization energy of sodium = 495 kJ mol⁻¹,
the heat of sublimation of sodium = 108 kJ mol⁻¹,
energy for bond dissociation of chlorine = 240 kJ mol⁻¹ and
the heat of the formation of sodium chloride = 410 kJ mol⁻¹.

Born - Haber cycle for the formation of sodium chloride from sodium and chlorine. How to measure electron gain enthalpy?
- 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.

Question
Account for the large decrees in electron affinity from lithium to beryllium despite the increase in nuclear charge.

Electronic configuration of lithium and beryllium
1S² 2S¹

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.

A filled shell or sub-shell leads to some extra stability. Beryllium resists an extra electron more than does lithium.

#### How to calculate the oxidation number of an element?

Question
How to calculate the oxidation number of phosphorus in Ba(H₂PO₂)₂?

The oxidation number of beryllium, hydrogen, and oxygen are +2, +1 and -2 respectively in Ba(H₂PO₂)₂. Let the oxidation number of phosphorous x.

∴ (+2) + 2{2(+1) + x + 2(-2)} = 0
or, 2x - 2 = 0
∴ x = +1.
Question
How to calculate the oxidation number of Iron in [Fe(H₂O)₅(NO)⁺]SO₄?

The oxidation number of water, nitrosyl, and sulfate are 0, +1, and -2 respectively. Let the oxidation number of iron in [Fe(H₂O)₅NO⁺]⁺² = x.
x + 0 + (+1) + (-2) = 0
or, x - 1 = 0
∴ x = +1.

The oxidation number of iron in [Fe(H₂O)₅NO⁺]⁺² = +1.

Question
What is the oxidation number of chlorine in bleaching powder?

The molecular formula of bleaching powder is Ca(OCl)Cl. Thus bleaching powder has two chlorine atoms with two different oxidation numbers.
∴ The oxidation number of two chlorine atom are -1 and +1.
.
Question
What is the oxidation number of chromium in Cr₂O₅?

Due to the peroxy linkage oxidation number of chromium in Cr₂O₅ = +6.

#### Oxidation and reduction reaction

Question
Why sulfur dioxide has properties of oxidation and reduction?

Oxidation and reduction properties of sulfur can explain on the basis of different oxidation numbers of the sulfur atom in different compounds. The highest oxidation number of sulfur is +6 and lowest -2.

The oxidation number of sulfur in hydrogen sulfide, sulfur dioxide, and sulfur trioxide are -2, +4, and +6 respectively. For free sulfur oxidation number zero which is the middle of 0 to +6.

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.

The oxidation number of sulfur increase from +4 to +6 and decreases from -2 to zero.

 +4 SO2 + -2 H2S → 2H2O + 0 3S

#### How to balance the chemical equations?

Question
How to balance the chemical equation for the reduction of permanganate to manganous salt by hydrogen peroxide in an acid solution?

The partial balance chemical equation for the reduction of oxidant permanganate in acid solution
MnO₄⁻ + 8H⁺ + 5e ⇆ Mn⁺² + 4H₂O.

The partial balance equation for the oxidation of reductant hydrogen peroxide
H₂O₂ ⇆ 2H⁺ + O₂ + 2e.

Hydrogen peroxide gives oxygen to an acid solution. First equation multiplying by 2 and the second 5 to balancing the electron in this chemical equation.

2MnO₄⁻ + 16H⁺ + 10e ⇆ 2Mn⁺² + 8H₂O
5H₂O₂ ⇆ 10H⁺ + 2O₂ + 10e

2MnO₄⁻ + 5H₂O₂ + 6H⁺ ⇆ 2Mn⁺² + 8H₂O + 5O₂
Question
How to balance the chemical reaction for the reduction of nitrate ion to ammonia by aluminum in an alkaline solution?

In the alkaline solution, we use water and hydroxyl ion according to convenience. The partial chemical equation for the reduction of oxidant.
NO₃⁻ + 6H₂O + 8e ⇆ NH₃ + 9OH⁻

The parietal equation for the oxidation of aluminum to aluminate ion
Al + 4OH⁻ ⇆ AlO₂⁻ + 2H₂O + 3e.

Multiplying by the right factors for balancing electron on the above chemical equation.

NO₃⁻ + 6H₂O + 8e ⇆ NH₃ + 9OH⁻

Al + 4OH⁻ ⇆ AlO₂⁻ + 2H₂O +3e

3NO₃⁻ + 8Al + 5OH⁻ ⇆ 3NH₃ + 8AlO₂⁻
Question
How to balance the chemical reaction for iodine ion and iodate ion in acid solution to liberate iodine by oxidation number?

The chemical reaction for iodine ion and iodate ion in acid solution
I⁻ (-1) + IO₃⁻ (+5) → I₂ (0).

Iodine ion and iodate ion with oxidation number -1 and +5 respectively form iodine molecule with oxidation number 0. Oxidation number increase and decrease on the above chemical equation 1 and 5 respectively.
Putting the right factors the decrease and increase in oxidation number.
5I⁻ + IO₃⁻ → 3I₂

Acidic solution the charges are unequal on two sides of the above expression. Put six hydrogens ion and write three water molecules to balance the equation.

5I⁻ + IO₃⁻ + 6H+ → 3I₂ + 3H₂O
Question
How to balance the chemical reaction between sulfurous acid and dichromate in acidic solution by oxidation number?

The chemical reaction between sulfurous acid and dichromate in acid solution
SO₃⁻² + Cr₂O₇⁻² → SO₄⁻² + 2Cr⁺³.

The oxidation number of two chromium atom decreases by 2 × (+3) = 6 and the oxidation number of sulfur increases by 2.

Equalizing the above equation increase and decrees in the oxidation number.
3SO₃⁻² + Cr₂O₇⁻² → 3SO₄⁻² + 2Cr⁺³

Acidic solution the ionic charges are not equal on the two sides of the chemical reaction, eight hydrogen ions are added to the left-hand side to balancing the chemical equation

3SO₃⁻² + Cr₂O₇⁻² + 8H⁺ → 3SO₄⁻² + 2Cr⁺³

Eight hydrogen ions form four water molecules in this chemical equation and we write four water molecules in the right-hand side for balancing the above equation.

3SO₃⁻² + Cr₂O₇⁻² + 8H⁺ → 3SO₄⁻² + 2Cr⁺³ + 4H₂O

Study inorganic chemistry questions answer for high school and college students, polarity, shielding effect, ionization energy, oxidation number