- 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,

Âµ = 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

Âµ = 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 Âµ = Unit of Charge × Unit of Length.

Thus, the unit of Âµ = esu × cm in the CGS system.

Thus, the unit of Âµ = 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 = gm

F = q₁q₂/r²

or, q₁q₂ = F × r²

∴ (esu)² = dyne × cm² = gm cm sec⁻² × cm²

Hence, esu = gm

^{½}cm^{3/2}sec⁻¹
∴ Unit of Âµ in CGS system, gm

= gm

^{½}cm^{3/2}sec⁻¹ × cm= gm

^{½}cm^{5/2}sec⁻¹
∴ Unit of Âµ in SI system, kg

= kg

^{½}m^{3/2}sec⁻¹ × m= kg

^{½}m^{5/2}sec⁻¹
Thus the dimension of Âµ,

[M

[M

^{½}L^{5/2}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, Î¼obs = 1.03 Debye = 1.03 × 10⁻¹⁸ esu cm and length l = 1.27 × 10⁻⁸ cm.

- Charge on the constituent atom(q), = Î¼obs/â„“

= (1.03 × 10⁻¹⁸ )/( 1.27 × 10⁻⁸)

= 0.8 × 10⁻¹⁰ esu - Percentage of the ionic character of HCl, (Î¼obs/Î¼ionic ) × 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 H

_{2}S is 97

^{0}and dipole moment = 0.95 D. Find the bond moment of the S-H link. ( Given, Cos97

^{0}= -0.122)

*Answer*

We have, Î¼ = 0.95 D and Î¸ = 97°

From the equation, Î¼² = m₁² + m₂² + 2m₁ m₂ cosÎ¸

For, H₂S, m₁ = m₂ = mS-H

∴ Î¼² = 2m²(1 + cosÎ¸)

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.

From the equation, Î¼² = m₁² + m₂² + 2m₁ m₂ cosÎ¸

For, H₂S, m₁ = m₂ = mS-H

∴ Î¼² = 2m²(1 + cosÎ¸)

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,

= E

= 2.179 × 10

= 2.179 × 10

= (2.179 × 10

= 13.6 eV

= E

_{H}= 2Ï€^{2}me^{4}/h^{2}[(1/1^{2}) - 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 1S

^{2}. The second ionization potential means the removal of the second electron from the 1S orbital against the nuclear charge of +2.

- Thus ionization potential, = (2Ï€²mZ²e⁴/h²) [(1/n₁²) - (1/n₂²)] = Z² × E

_{H}

- ∴ 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):
E

_{H}= (me^{4}/8Îµ_{0}^{2}h^{2})[(1/n_{I}^{2}) - (1/n_{II}^{2})]**= 21.79 × 10**

^{-19}J
where

**n**refers to the 1S orbital (_{I}**n**) and_{I}= 1**n**refers to infinity._{II}
So the ionization potential for removal of the electron from the 2S orbital to infinity is given by:

**E**= (me

_{H}^{4}/8Îµ

_{0}

^{2}h

^{2})[(1/2

^{2}) - (0)]

= 21.79 10

^{-19}/4**J****= 5.45 10**

^{-19}J###
__Slater's Rules:__

__Slater's Rules:__

__Problem 9:__

In first transition series electron filling up processes begins in the

**3d**level below a filled**4S**level. During ionization process will a^{2}**4S**electron or a**3d**electron be lost first? Explain with reference to**Chromium**.__Answer:__

Electronic Configuration of

**Chromium (Cr)**:24 | 1S^{2} 2S^{2} 2P^{6} 3S^{2} 3p^{6} 3d^{5} 4S^{1} |

Screening constant (Ïƒ) for 4S electron is,

**Ïƒ**= (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 (Ïƒ) for 3d electron is,

**Ïƒ**= (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:

Liã„‘Bã„‘Beã„‘Cã„‘Oã„‘N〈Fã„‘Ne |

- 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,

**Ïƒ**= 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 (Ïƒ)**for the**4S**electron is:**Ïƒ**=(2×1.0)+(8×1.0)+(8×1.0)+(3×0.85)+(0×0.35)

**= 21.05.**

And the

**Screening Constant (Ïƒ)**for the**3d**electron is:**Ïƒ**= (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 (Ïƒ)**

= (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:Born-Haber Cycle applied to NaCl. |

From the above Born - Haber cycle we can written as:

-U_{NaCl} - INa + ECl - SNa - 1/2DCl - Î”Hf=0or, ECl =U_{NaCl} + INa+ SNa+ 1/2DCl + Î”Hf |

∴

**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 : 1S**

^{2}2S^{1}**Be: 4: 1S**

^{2}2S^{2}**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(H**is - (a)+3, (b)+2, (c) +1, (d) -1._{2}PO_{2})_{2}__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(H**._{2}O)_{5}(NO)^{+}]SO_{4}__Answer:__

**H**is neutral thus the oxidation number is

_{2}O**zero**, the oxidation number of

**NO**is

^{+}**+1**and the oxidation number of

**SO**is

_{4}**-2**.

Let the oxidation number of

**Fe**in [Fe(H_{2}O)_{5}NO^{+}]^{+2}is**x**.
∴ x + 0 + (+1) + (-2) = 0

or, x-1 = 0

**or, x = +1**

Thus, the oxidation number of

**Fe**in [Fe(H_{2}O)_{5}NO^{+}]^{+2}is**+1**.__Problem 18:__

Calculate the oxidation number of the element marked with blue in the following compounds, (i)

**K**, (ii)_{2}CrO_{4}**HO****Cl**, (iii)**Ba****O**, (iv)_{2}**Cl****N****O**, (v)**Na****N****H**, (vi)_{2}**Na****N**, (vii)_{3}**CH**, (viii)_{2}Cl_{2}**Ca(O****Cl****)****Cl**, (ix)**Ba(****Mn****O**(x)_{4})_{2}**Ca****H**._{2}__Answer__

Compound | Element | Oxidation Number |

K_{2}CrO_{4} | Cr | +6 |

HOCl | Cl | +1 |

BaO_{2} | Ba | +2 |

ClNO | N | +3 |

NaNH_{2} | N | -3 |

NaN_{3} | N | -1/3 |

CH_{2}Cl_{2} | C | 0 |

Ca(OCl)Cl | Cl | +1 |

Ba(MnO_{4})_{2} | Mn | +7 |

CaH_{2} | Ca | -1 |

__Problem 19:__

What is the Oxidation state of chromium in

**Cr**?_{2}O_{5}__Answer:__

Due to the peroxy linkage oxidation state of

**Cr**in**Cr**is_{2}O_{5}**+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

**H**,_{2}S**SO**, and_{2}**SO**are_{3}**-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**SO**, the oxidation state of sulfur is_{2}**+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**.+4SO_{2} | + | -2H_{2}S | → | 2H_{2}O | + | 03S |

__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:

MnO

_{4}^{-}+ 8H^{+}+ 5e ⇆ Mn^{+2}+ 4H_{2}O
(Reduction of Oxidant)

H

_{2}O_{2}⇆ 2H^{+}+ O_{2}+ 2e
(Oxidation of Reductant)

In acid medium

**H**will give_{2}O_{2}**O**and the partial equation being: First equation is multiplying by_{2}**2**and the second is**5**to have electron balanced. We have,2MnO_{4}^{-} + 16H^{+} + 10e ⇆ 2Mn^{+2} + 8H_{2}O5H _{2}O_{2} ⇆ 10H^{+} + 2O_{2} + 10e |

2MnO_{4}^{-} + 5H_{2}O_{2} + 6H^{+} ⇆ 2Mn^{+2} + 8H_{2}O + 5O_{2} |

__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 H

_{2}O and OH^{-}ion according to convenience. The partial equation with charge balance is:
NO

_{3}^{-}+ 6H_{2}O + 8e ⇆ NH_{3}+ 9OH^{-}
Metallic aluminium will go over to aluminate ion, the partial equation being:

Al + 4OH

^{-}⇆ AlO_{2}^{-}+ 2H_{2}O +3e
Multiplying by right factors for electron balance we have the balanced equation:

NO_{3}^{-} + 6H_{2}O + 8e ⇆ NH_{3} + 9OH^{-} Al + 4OH ^{-} ⇆ AlO_{2}^{-} + 2H_{2}O +3e |

3NO_{3}^{-} + 8Al + 5OH^{-} ⇆ 3NH_{3} + 8AlO_{2}^{-} |

__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

^{-}+ IO_{3}^{-}→ I_{2}
Representation of the above equation with an oxidation number of iodine,

-1I ^{-} | + | +5IO _{3}^{-} | → | 0I _{2} |

Since

**I**(-1) oxidation number of iodide increases by 1 and^{-}**IO**(+5) oxidation number decreases by 5._{3}^{-}
Putting the right factors the decrease and increase in oxidation number balanced:

5I

^{-}+ IO_{3}^{-}→ 3I_{2}
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

^{-}+ IO_{3}^{-}+ 6H^{+}→ 3I_{2}+ 3H_{2}O__Problem 24:__

Use the

**oxidation number method**to balance the reaction between**Sulphurus acid**and**dichromate in acidic medium**.__Answer:__

The reaction represented as,

SO

_{3}^{-2}+ Cr_{2}O_{7}^{-2}→ SO_{4}^{-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,

3SO

_{3}^{-2}+ Cr_{2}O_{7}^{-2}→ 3SO_{4}^{-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.^{+}
3SO

_{3}^{-2}+ Cr_{2}O_{7}^{-2}+ 8H^{+}→ 3SO_{4}^{-2}+ 2Cr^{+3}
Since there are eight hydrogen ion on the left and none on the right, four H

_{2}O added to make atom balance. Thus the final equation is:3SO_{3}^{-2} + Cr_{2}O_{7}^{-2} + 8H^{+}→ 3SO_{4}^{-2} + 2Cr^{+3} + 4H_{2}O |