Problems on Inorganic Chemistry appear really as problems to the most of students studying in different Schools, Colleges and Different universities. To Overcome this Problems 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.

How can convert 1 Debye to Coulomb Meter.

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 × metre
= 1.6 × 10^{30} C × m
Thus, 4.8 Debye = 1.6 ×10^{30} C×m
or, 1 Debye = (1.6 × 10^{30})/4.8 C × m
∴ 1 Debye = 3.336 × 10^{30} C×m 

Find out the Unit and Dimension of Dipole Moments.

Unit of Âµ = Unit of Charge × Unit of Length.
Thus, the unit of Âµ = esu × cm in CGS system.
But from the Coulomb’s Low,
F = q_{1}q_{2}/D r^{2}
or, q_{1}q_{2} = F × Dr^{2}
∴ (esu)^{2} = dyne × cm^{2}
= gm cm sec^{2} × cm^{2}
Hence, esu = gm^{1/2} sec^{1} cm^{3/2}
∴ Unit of Âµ in CGS system,
gm^{1/2} × sec^{1} × cm^{3/2} × cm
gm^{1/2} × cm^{5/2} × sec^{1} 
In SI System, gm^{1/2} × cm^{5/2} × sec^{1} 
Thus the Dimension of Âµ gm^{1/2} × cm^{5/2} × sec^{1} 

The displacement of HCl is 1.03 Debye and its bond length is 1.27 A. calculate the (a) charge on the constituent atom and (b) the % of the ionic Character of HCl.

Given, Î¼_{obs} = 1.03 Debye = 1.03 × 10^{18} esu cm and length l = 1.27 × 10^{8} cm.

(a) Hence the charge on the constituent atom,
= (1.03 × 10^{18} )/( 1.27 × 10^{8})
= 0.8 × 10^{10} esu

(b) The % of the ionic Character of the molecule,
= (Î¼_{obs}/Î¼_{ionic} ) ×100
=(1.03 × 10^{18})/(4.8 ×10^{10} × 1.27 × 10^{8}) × 100
= 16.89%

The Difference between electronegativity of Carbon and Oxygen Is large but the dipole moments of carbon monoxide is very low  Why?

See The topic Application of Dipole Moments.

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)

We have, Î¼ = 0.95 D and Î¸ = 97^{0}
From the equation,
Î¼^{2} = m_{1}^{2} + m_{2}^{2} + 2m_{1} m_{2} cosÎ¸
For, H_{2}S, m_{1} = m_{1} = m_{SH}
∴ Î¼^{2} = 2m^{2}(1 + cosÎ¸)
Putting the value we have,
(0.95)^{2} = 2 m^{2} (1+ cos97^{0} )
Here, m = m_{S  H}
or, 0.9025 = 2 m^{2} (1  0.122)
or, m^{2} = 0.9025/ (2 × 0.878)
or, m^{2} = 0.5139
or, m = 0.72
Thus the bond moment of the m_{S  H} link is 0.72 D.

Calculate the ionization potential of hydrogen atom in eV.

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_{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

Calculate the second ionization potential of helium (given ionization potential of hydrogen = 13.6 eV).

The ground state electronic configuration of helium is 1S^{2}. The second ionization potential means removal of the second electron from the 1S orbital against the nuclear charge of +2.

So we have:
Ionization Potential,
IP = (2Ï€^{2}mZ^{2}e^{4}/h^{2}) [(1/n_{1}^{2})  (1/n_{2}^{2})] = Z^{2} × E_{H} 

∴ Second Ionization Potential of Helium,
= 22 × 13.6
= 54.4 eV
The normal ionisation potential of hydrogen atom is 21.79 × 1019 J. What will be the value of ionisation potential when an electron is raised to the 2S level?
The normal ionisation potential of 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_{I} refers to the 1S orbital (n_{I} = 1) and n_{II} refers to infinity.
So the ionisation potential for removal of the electron from the 2S orbital to infinity is given by:
E_{H} = (me^{4}/8Îµ_{0}^{2}h^{2})[(1/2^{2})  (0)]
= 21.79 10^{19}/4 J
= 5.45 10^{19} J
In first transition series electron filling up processes begins in the 3d level below a filled 4S^{2} level. During ionization process will a 4S electron or a 3d electron be lost first ? Explain with reference to Chromium.
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.
Arrange the following with the increasing order of Ionization Potentials
Li, Be, B, C, N, O, F, Ne.
Liã„‘Bã„‘Beã„‘Cã„‘Oã„‘N〈Fã„‘Ne 
Calculate the effective nuclear charge of the hydrogen atom.
Hydrogen atom has a single 1S valence electron. There is no other electron to screen it from the nuclear charge of 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.
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 (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.
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 consider 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.
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 formation of NaCl (S) is:
BornHaber Cycle applied to NaCl. 
From the above Born  Haber cycle we can written as:
U_{NaCl}  INa + ECl  SNa  1/2DCl  Î”Hf=0 or, ECl =U_{NaCl} + INa+ SNa+ 1/2DCl + Î”Hf 
∴ 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 : 1S^{2} 2S^{1}
Be : 4 : 1S^{2} 2S^{2}
Lithium has an incompletely filled 2S subshell while beryllium has the sub shell filled. Thus lithium can receive an electron in its 2S subshell but for beryllium a still higher energy 2P level has to be made of. A filled shell or subshell leads to some extra stability. Hence beryllium resists an extra electron more than does lithium.
Oxidation number of P in Ba(H_{2}PO_{2})_{2} is  (a)+3, (b)+2, (c) +1, (d) 1.
(c) +1
Oxidation number of Ba is +2, oxidation number of hydrogen is +1 and oxidation number of oxygen is 2.
Let the oxidation number of P is x.
∴ (+2) + 2{2(+1) + x + 2(2)} = 0
or, 2x2=0
or, x=+1
Calculate the oxidation number of Iron in [Fe(H_{2}O)_{5}(NO)^{+}]SO_{4}.
H_{2}O is neutral thus the oxidation number is zero, oxidation number of NO^{+} is +1 and the oxidation number of SO_{4} is 2.
Let the oxidation number of Fe in [Fe(H_{2}O)_{5}NO^{+}]^{+2} is x.
∴ x + 0 + (+1) + (2) = 0
or, x1 = 0
or, x = +1
Thus, the oxidation number of Fe in [Fe(H_{2}O)_{5}NO^{+}]^{+2} is +1.
Calculate the oxidation number of the element marked with blue in the following compounds, (i) K_{2}CrO_{4}, (ii) HOCl, (iii) BaO_{2}, (iv) ClNO, (v) NaNH_{2}, (vi) NaN_{3}, (vii) CH_{2}Cl_{2}, (viii) Ca(OCl)Cl, (ix) Ba(MnO_{4})_{2} (x) CaH_{2}.
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 
What is the Oxidation state of chromium in Cr_{2}O_{5}?
Due to peroxy linkage oxidation state of Cr in Cr_{2}O_{5} is +6.
Why sulphur dioxide has properties of Oxidation and reduction?
This can be explain by the oxidation state of sulphur. The oxidation numbers of sulphur in the compounds H_{2}S, SO_{2}, and SO_{3} are 2, +4 and +6 respectively.
Thus the highest oxidation state of sulphur is +6 and lowest is 2. The oxidation state of free sulphur element is 0. In SO_{2}, the oxidation state of sulphur 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 SO_{2} 
+  2 H_{2}S 
→  2H_{2}O 
+  0 3S 
Express by ion electron method the reduction of permagnate to manganous stat by hydrogen peroxide in acid medium.
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_{2}O_{2} will give O_{2} and the partial equation being:
First equation is multiplying by 2 and the second is 5 to have electron balanced.
We have,
2MnO_{4}^{} + 16H^{+} + 10e ⇆ 2Mn^{+2} + 8H_{2}O 5H_{2}O_{2} ⇆ 10H^{+} + 2O_{2} + 10e 
2MnO_{4}^{} + 5H_{2}O_{2} + 6H^{+} ⇆ 2Mn^{+2} + 8H_{2}O + 5O_{2} 
Express by ion  electron method the reduction of nitrate ion to ammonia by aluminium in aqueous NaOH.
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}^{} 
Use Oxidation Number method to balance the reaction of iodide ion
and iodate ion in acid medium to liberate iodine.
The reaction represented as,
I^{} + IO_{3}^{} → I_{2}
Representation of the above equation with oxidation number of iodine,
1 I^{} 
+  +5 IO_{3}^{} 
→  0 I_{2} 
Since I^{} (1) oxidation number of iodide increases by 1 and IO_{3}^{}(+5) oxidation number decreases by 5.
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
Use oxidation number method to balance the reaction between
Sulphurus acid and dichromate in acidic medium.
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 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 