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.
Dipole moments:
 Problem 1:

How can convert 1 Debye to Coulomb Meter.
 Answer:

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 
 Problem 2:

Find out the Unit and Dimension of Dipole Moments.
 Answer:

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} 
Application of Dipole Moment:
 Problem 3:

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.
 Answer:

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

Given, Î¼obs = 1.03 Debye = 1.03 × 10⁻¹⁸ esu cm and length â„“ = 1.27 × 10 ⁻⁸ cm.

(a). Charge on the constituent atom(q),
= Î¼obs/â„“ = (1.03 × 10⁻¹⁸ )/( 1.27 × 10⁻⁸)
= 0.8 × 10⁻¹⁰ esu

(b) 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 4:

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

See The topic
Application of Dipole Moment 
 Problem 5:

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^{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.
Ionization Potential:
 Problem 6:

Calculate the ionization potential of 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_{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 7:

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 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
 Problem 8:
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?
 Answer:
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
Slater's Rules:
 Problem 9:
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.
 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:
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.
 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 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.
 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},
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 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
 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 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.
 Problem 16:
Oxidation number of P in Ba(H_{2}PO_{2})_{2} is  (a)+3, (b)+2, (c) +1, (d) 1.
 Answer:
(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
 Problem 17:
Calculate the oxidation number of Iron in [Fe(H_{2}O)_{5}(NO)^{+}]SO_{4}.
 Answer:
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.
 Problem 18:
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}.
 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 peroxy linkage oxidation state of Cr in Cr_{2}O_{5} is +6.
 Problem 20:
Why sulphur dioxide has properties of Oxidation and reduction?
 Answer:
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 
 Problem 21:
Express by ion electron method the reduction of permagnate 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_{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} 
 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 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
 Problem 24:
Use 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 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 