1.1-CHEMICAL KINETIC
Chemical kinetics is the study of speed or rate of chemical reaction under different conditions and
mechanisms of the reaction.
The rate of reaction
Rate of reaction is the change in concentration of reactants and products per unit time.
Consider the following hypothetical reaction:
A+3B —2C+2D
Rate of reaction can be expressed in terms of disappearance of A or B and rate of formation of C or D.
But these rate are not the same i.e. B disappears 3 times as fast as A. Therefore rate of disappearance of
B is equal to 3 times rate of disappearance of A.
Rate of disappearance of B= 3 times rate of disappearance of A
Similarity;
Rate of formation of C = 2 times rate of disappearance of A
Rate of formation of D = 2 times rate of disappearance of A
2/

Rate of formation of D= / 3 times rate of disappearance of B
The rates of disappearance are negative (since concentration decreases with time) and rate of formation
are positive (concentration increase with time). But all result in single positive expression for the rate of
reaction.
To remove the effect of stoichiometry, the rate for each species is divided by the co-efficient of the
molecules in the balanced equation

Rate of reaction = —Rate of disappearance of B

Rate of reaction = — 1, Rate of disappearance of B

Rate of reaction = 1/, Rate of formation of C

i

Rate of reaction = 1/, Rate of formation of D

Example: 01
In the above hypothetical reaction, the initial concentration of Ais 1.0M and iminute later was found to
be 0,9982M.
1) What is rate of reaction in moles per litre per second?
“Lepp-i
ii) What is rate of formation of C in ™O!L *sec™*y
Solution;
i)   Rate of reaction = —Rate of disappearance of A
.        -[AA
Rate of reaction = ee
__ -(0.9982-1.0)M
~        60sec
= 3 x 107molL7'sec™ or 3 x 1075Msec
ii)   Rate of reaction = 1, Rate of formation of C
3x 107° x 2 = Rate of formation of C
= 6X 1075molL"'sec™!
Example: 2
In the reaction; cA +B +3D
-14           -1
Reactant A is found to be disappearing at a rate of  ———
a) What is the rate of reaction?
b) What is the rate of formation of D?
Solution;

a)         Rate of reaction = —Rate of disappearance of A
«          —[AA]
Rate of reaction = “a
= 2.6 x 107*molsec7?
b)        Rate of formation of D = 2/; * Rate of disappearance of A
= 2/, X 2,6 X 107 molsec™
= 1.73334 107 molsec™?

RATE LAW
The law states that “The rate of reaction is direct proportional to the concentration of the reactants
each raised to the power of equal to the order of the reaction”

Rate « [Reactants]

Rate = K[Reactants]
Where K = Rate constant (Velocity constant)
By definition;
Rate constant is a constant of proportionality in the rate equation which is the measure of the speed
with which a reaction is taking place at the given temperature
The units of K depends on the order of reaction
Order of reaction this is the number which shows the manner with which the rate of reaction depends
on the concentration of reactants.
Order of reaction can only be obtained experimentally and cannot be deduced from overall balanced
equation. OR order of reaction is the sum of power of concentration of the reactants in
the rate law or rate equation.
Consider the example of the reaction; A+ 2B — Products

=         a      b
The rate law or rate equation is given by; Rate     KA]  [B]
The reaction order with respect to [A] is a and [B] is b and the overall order of reaction is (* + b and
that can be a fraction
1. Zero order reaction
Is the reaction in which the rate of the reaction is independent on the concentration of the
reactants.
Consider the reaction; A — Products
Rate = K[A]°
Rate = K
The rate is constant and independent of the concentration of A
Units of rate constant, K
K = Rate
Rate = [aConcentration]
But                          at
AContentration
K = Uitententration]
at
ot
K  = mo
sec
K = molL'sec™*
Assignment: Find any two examples of zero order reactions
Graph of rate against concentration A

Zeroorder
Rate
concentration
2. First order reaction

Is reaction in which the rate of reaction ts directly proportional to the first power of the
concentration of the single reaction reactant.
Consider the reaction; =

Rate = K[A]*

Rate = K[A]
The units of rate constant, K

Rate = K[A]
K= Rate
[4]
K= molL~*sec™*
molL-1
K =sec™1_or     (time)
Graph of rate against concentration A

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2MNOj(aq) + 101(aq) + 16H(aq) > 2Mnigqy + Slo(aq) + BH20 (2g)                (1)
The iodine produced is titrated with Na25203 to pale yellow, then add starch and continue titrating
until the blue-black colour is discharged.
Starch is not added at the beginning because the concentration of iodine is large. lodine react with
starch to form the blue black complex hence more volume of Naz 20s is required to discharge the blue
black complex.
Reaction with Na2S203

5.05 aa) + iy

I + i + Was)

5,0; + PERE PE scnicmemnninllh
When equation (1) and (2) are combined

12MnO, + 10) -+ 16H*—> 2Mn** + SI? + 8H,0

5P5,02° + 1+5,02> + 217

2Mn0; + YO + 16H ++ 2Mn?*+ + Sf, + 8H,0

105203” + 5f >55,03° +  tef-

2MnO; + 105,02°> + 16H* 4 2Mn** + 55,02° + 8H,0

2:10
1:5
MnO; : 5,02-

b) If the oxidant is acidified K2cr207
Cr,07 — Cr 3, (red)
9e~ + Cr,07> + 14H*+ > 2Cr3*+ + 7H,0

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2 \Cr,02> + Be ~ + 14H* > 2Cr3* + 7H,0

8 I~ +124 2e7

2Cr,07- + 166 - + 28H* —4Cr3* + 14H,0

16]- + 81,+ ge

2Cr702- + 28H* + 16I- > 4Cr3* + 14H,0 + 81,
2 \Cr,02> + 6e— + 14H +> 2Cr3* + 7H,0
6|21-+ 1, + 2e7

2Cr,07- + ye + 28H *— 4Cr3* + 14H,0

12]- + 61, + ye

2€r,02- + 121” + 28H* > 4cr** + 61, + 14H,0

2Cr,02_ + 28H* + 12) ~+4Cr3* + 14H,0 + 61;

25,02- + 1,+5,02° + 2I-

2Cr,02- + 28H* + : -5 4Cr 3+ + 14,0 + gt,
125,02- + of. 65,02- + 12f-

2Cr,07- + 125,07°+ 28H ++ 4Cr3* + 14H,0
2:12
Mole ratio will be
Cr,02- : 5,027
1:6
c} If the oxidant is acidified KIO,
12H* + 210; +1,+ 6H,0

2 |12H* + 2105 + 10e~+1.+ 6H,0

10|21~ = Ingagy + 2e7
24H* + 4105 + Sa + 12H,0
2017 + 10l2¢aq) +  Je
410 + 201 ~ + 24H*-+21, + 12H,0

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From equation (2)5,02- + 15,027 + 2I-
1)4/0;5 + 20]- + 24H*+—2I,+ 12H,0
12|25,02- + 1+S,02- + 21
4105 + 20f- + 24H* +129 + 12H,0
245203” + 12f. 125,03" + 24f-
410; + 245,02- + 24H*-+125,02- + 12H,0 + 4I7-
4:24
1:6
The mole ratio is
105: 5,037
1:6
ELECTRODE POTENTIAL AND ELECTROCHEMICAL SERIES
ELECTRODE POTENTIAL
Metals have a small tendency to dissolve in solution of their ions producing cations leaving their valency
electrons on the metal rod. The metal acquires a negative potential which prevents further release of
cations and equilibrium is established
—      +       -
M = MM +ne   ne = number of electron (s)
As a result the region of solution very close to the rod suffers an increase in charge while the rod carries
a layer of negative charge (electrons). Then an electric double layer is set up and this layer is known as
“Helmholtz double layer”
.                       Solution of its ions (ZnSO)
v
Helmholtz double layer
Whenever there is a separation of negative and positive charges we should be able to measure the
voltage i.e. voltage between the electrode and surrounding solution and this is called Electrode

=
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potential.

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3.The resistivity of 59M KCI solution is 3619 and conductivity cell containing such a solution was found
to
have a resistance of 5500
a.Calculate the cell constant
b.The same cell filled with 0.1M ZnSO, solution had resistance of 720. What is the conductivity of
this solution?
KOHLRAUSCH'S LAW OF INDEPENDENT IONIC MOBILITY
Kohlrausch notes that the difference between molar conductivity at infinity dilution 4°° values for the
two salts which were strong electrolytes and of the same cation and anion was always constant.
Using the he values in D.. em? mol?
|           —________-» 5amie cation
Same cation
KCL(130)       KNO3 (126.3) + 3.7
Naci(108.9)       NaNo; (105.2) + 3.7
4                          L
211                     21.1
This observation shows that each type of ion (caution or anion) contribute a definite amount of molar
conductivity of an electrolyte of infinity dilution independently from other ions present in the solution
i.e. a fraction of current that an ion carry is always constant and it doesn’t depend on the compound in
which it is contained.
Hence Kohlrausch's law
States that;"The molar conductivity of an electrolyte at infinity dilution is equal to the sum of the molar
conductivities of the caution and anion".
OR State that the molar conductivity at infinity dilution of the solution equal to the sum of molar
canductivity at infinity dilution of its components ions.
+        -
Le. co = reo 4 oo
»                                *                -
Eg =  (Nacl)_»,, (Na*), >. ic  )
Aa(Ala(S2#)3) = 2>= (aL) +3>= (594° )

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Application of Kohlrausch’s law

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In direct determination of molar conductivity at infinity dilution for weak electrolytes. Thus by using
strong electrolytes we can easily. Calculate the molar conductivities at infinity. Dilution for weak
electrolytes.
E.g.: The on   (C43co0n) can be determined from *™ of potassium ethanoate (c#2cooK)
hydrochloric acid (HCI) and potassium chloride (KCI)

=
CH;COOH + KCl ~§ CH;COOK + HCI
Au(CHsCOOH) = *20(CHsCOOK) + *20(HCI) - > (cl)

= ?$20(CH3COO") + *29(K*) + 22H") + *20(CI) - *20(K") - *22(Cl)
-s                   a
eel CHa COOH) = ““(CH3COOH) + “*(H")
REVIEW QUESTIONS
1. Calculate c0(NHaQH given that co  of three strong electrolyte NaCl, NaOH and NH.Cl in B...cm* mol
are 126.4, 248.4 , 149.8 respectively.
2.The molar conductivity of 0.093 CH;COOH solution at 298k is 536 x 10“ Sm?mal? ‘The molar
conductivity at infinity dilution of H* and CH3;COO are 3.5 x 10? and 0.41 x 10°Sm’ mol ‘what is are the
dissociation constant of CH;COOH
3..40.05M HF solution has a conductivity of 91.81mol* m‘ at 298k.At the same temperature  co
(NaF),    a %(NaoH) and co (H20) are 493360 and 162Sm’mol* respectively. Calculate dissociation
constant.
4. The }..(Nal), A..(CH;COONa) and A..(CH;COO,Mg) are 12.69,9.10 and 18.785m’ mol” respectively at
25"C. What is the molar conductivity of Mglz at infinity dilution.
ORGANIC CHEMISTRY 1.2- AMINES
i) STRUCTURES OF AMINES.

Amines are derivatives of ammonia in which one or more hydrogen have been replaced by alkyl group
or aryl group.
General formula:

R-NH2

1 amine

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If two hydrogens have been replaced we have;
Fi
H—  si  —R
2? amine
If 3 hydrogens have been replaced we have;
p
R— N —R
3° amine
NOTE:
+
R
R-N-R
R         x
Quatenary salt of amine
ii) (NOMENCLATURE) IUPAC SYSTEM

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CH; NH3           TUPA name: methanamine (methylamine)
CH3 CH: NH? — Ethanamine
CH;3— NH          N- methylamine — 1° Amine .
H
|
CH; N-CH? CH3 ON - methylethylamine — 2° - Amine .
CH;N-CH> CH;
NN - Dimenthyl ethylamine- 3°-Amine
CH;
CH; CH? CH? CH - CH3,
|                   Pent-an — 2 - amine
NH?
CH; CH - CH: CH — CH3
4- Methyl —pentan —2 - amine
CH; NH)

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CH3 CH2 CH? CH — CH3
N-CH3       N_N-—dimethyl pentan — 2 — amine
OR
CH;        2 -(N, N - dimethyl) pentanamine
CH3 N- CH? CH) - CH3
CH,                                        N - Ethyl — N —methylpropananmine
CH;
Hz N - CH2 CH; CH NH3    Propane - 3 - diamine
NH
a
TUPAC name: Benzenamine
<O> NH  <O»            M, N— diphenylamine
NHCH?
N- Methylbenzenamine
o
MHz
CL          2- Methyibenzenamine
CHs
ere eB    N- benzyl — N- methylbenzanamine
oH
NH:
oF                        2. aminopbencl
COOH
OL                      2-tmlnabenole atid
NH:

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CH, CH, CH - CH, NH,
1         1-aminabutan-2-ol
H2N — CH? CH; CH? COOH
.             4-aminobutanoic acid
iii ) PREPARATION OF AMINES
LALKYLATION OF AMMONIA WITH HALOALKANES
R-X+2NH3; het. R-NH> + H-X
(Alc)
(excess)
1 amine can be obtained by reaction of alkyl halide with excess ammonia.
Why excess ammonia?
So as to prevent the formation of 2° amine or 3" amine. Therefore, this is not a suitable for preparation
of amine because the substitution of hydrogen does not stop at the first stage.
R-NH2+R-X—-R-NH-R+ HX.
R-NH-R+R-X—-R-N-R+HX.
|
R
rR          .
f —WN —R |
R-N-R+R-X—* | 4.   a
|
R
2, REDUCTION OF NITROALKANES (R —- NO2)
1" amines can be obtained by reduction of nitroalkane. The reducing agent is LIAIHs or $n/HCl or
Fe/HCl

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LiAIH
R-NO)+3H) ———» R-NH)+H)0
LiAIH
CH; CH) NO} +3H2 ——~*» CH3 CH; NH) + H20
3, REACTION OF AMIDES WITH LiALHg
1 amides can be reduced to 1° amines
R BL ni, Hélhy se CH, NH, +H,0
E.g:
o
CH, CH, ll  —= NH, HAI, § CH, CH, CH, NH, + H,O
ont  —NH, Lally CH, CH, NH, +H,O
4. AMMONIATION OF CARBONYL COMPOUND FOLLOWED BY REDUCTION
WITH HYDROGEN IN THE PRESENCE OF NICKEL
Oo                        H
ul                         |
R-C-H+NH;—+ R-C=N-H+H20
R-C=N-H+H;—‘+ R-CH;NH;
Ez
CH; CHO+NH; —»* CH;CH=NH+H20
CH;CH =NH+H;_*i_, CH;CH;NH;
oO
iI
CH;C -CH; +NH; —» CH;C {(CH;}=N-H   HM.  CH;-C {CH3}-NH?2

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5, REDUCTION OF NITRILES
R-CN £44, R -CH4NH?
CH; CN£“4, CH3;CH; NH?
CH; CH; CN Na:CH3CH;OH2CH; CH CH) NH)
6, REACTION OF ALCOHOLS WITH AMMONIA
Catalytic ammoniation of alcohol gives a mixture of amines.
Af ORC
CH; OH + NH   Highpressure     CH; NH; +H,0
CH; NH2 + CH; OH —CH; NH —- CH; +H20
iL HOFFMAN’S DEGRADATION OF AMIDES
"|  — NH, + Br, +NaQH——+ R-NH, +NaCO, +Na Br
It is important reaction in the conversion of amide to amine with one carbon less.

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Eg: Convert arr  or™
(i) By Hoffman's degradation reaction
AH, + Br, + NOH SNH, +NaCO, Nar,
O                               0
Gi) CH C(O)-NH- CH: to   —
@ii) CH:CHO te CH; CH2 NH:
Gv) CHsBr to CH; CH: NH:
Solution:
@ ont     NH - CHy Liat « CHsCH NH
a                                  |
CH:
n gmong
i) CH, oH cH, cH,
CH; CH; + Cl, Uv-light CH; CH; Cl
CH; CH) Cl + NH2—CH; CH2 NH?
ary
-       —_                 -
(iv) CH; Br+ Na ocher CH; CH; + NaBr
CH; CH3;+Cl) Uv-light CH; CH? Ci+ HCl
CH; CH; Cl+ 2NH; —CH; CH? NH2 + NH, Cl
4.PROPERTIES OF AMINES
a)PHYSICAL PROPERTIES OF AMINES

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Boiling point and melting point
Amines have high boiling and melting points due to their ability to form hydrogen bonding as compared
to hydrocarbons.
H
|
R—N            RK               Rg
;              |
Hydrogen          . i           N— Ha... —H
band                  H               |                 |
'               H                H
R—H —H                Hydrogen bond
,
(i) Its boiling point is lower than alcohol (R - OH)
Reason: In oxygen, there are 2 lone pairs which increase its chance of hydragen bonding.
It also has lower boiling point than carboxylic acid.
1° amines have high boiling point than 2° amines. For 3° amines, there is no possibility of
H-bonding since Nitrogen is not attached to hydrogen.
(ii) Solubility:
Lower aliphatic amines are soluble in water.
Reason:- They can form hydrogen bond with water.
As the molecular mass increases, solubility decreases.
Reason: -The long chain of hydrocarbon (R) is insoluble since it is water phobic. As the alkyl group
becomes bigger, solubility decreases.
Amines containing 6 or more carbon do not dissolve in water.
E.g: aniline is insoluble due to this reason.
b) CHEMICAL PROPERTIES OF AMINES

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1. Basie character

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Amines are the most important organic bases. Aliphatic amines are slightly stronger Lewis bases than
ammonia. Lone pair an the nitrogen is readily available to accept proton in amines than it is an ammonia
R R— Positive inductive effect
| |
R-NH2_ R-N-H_ R-N-R
2° amine the strongest base among the three
The stability of hydrated ion decrease from 1° amine 3° amine
R R
|
o nN R—N—R R—NH: R—-N-H
HOW HAH Strong base
Weak base
Reason:- 3° amines forms a very unstable ammonia salt. Although it can easily donate a lone pair of
electrons as compared to 1° and 2° amines, it forms very unstable hydrated ion.
Note:
The inductive effect on alkyl group tends to make 3° amines more basic while hydration effect tends to
°
make 1 amine more basic. As a result of the combined effect of these two the observed order of the
base strength will be;
2° amine > 1’ amine > amine
OO
R-NH?+ HCl R - NH3 Cl
OO
CH3 NH? + HCI CH3 NH3 Cl
Methylammonium chloride
Q. Between the following amines, which is stronger base

n
P= () RT
P = [X] RT
Pie tet, Poe = (92) RTnnnete
Substitute the partial pressure in equation (2)
Kp =_([H20] RT)?
({02] RT) ({H2] RT)?
to) _(RT)?
K , =la]la] , (RT) (RT)?
K )=K _ (RT) (240420)
Kp=Ke (RT)*
2. a) Derive the relationship between K , and K, for ammonic synthesis.
<_
No (g) + 3H2(g) 7” 2NH3 ig)
(N H3]?

n
P= Cer
P =[X] RT
Pri, - (NHy RT, Py,. (72) RT, Py, = (Nz RT
Substitute the partial pressure in equation (2)
({NH3]RT)?
Ko= ([H2]RT)3 ((N2]RT)
(NH)? [rT]?
kp =#2]7 [Na] (RT) 3 [RT]
2-(3+1)
Ky = Kex (RT)O-G*Y)
K)=K (RT)?
b) If K-= 0.105 mol?dm* at 472°C. Calculate K ,
[R=8.31 dm? KPa mol? K?]
Solution:
K ,=K_x (RT)?
= 0.105 x (8.31x 745)?
= 0.105 x (6190.95)?
= 2.7395x 10°(KPa)?
DETERMINATION OF EQUILIBRIUM CONSTANT

Consider the reaction:-
CH3CH20H + CH3COOH ~ CH 3COOCH> CH3+H20
At start a b oO oO
Equilibrium (a—x) (b-x) e) 5)
a-x b-x
:. [CH3CH2 OH] = > ; [CH3COOH] = —..............ete
v
2. Ke = [CH3 COOCH2CH3]H20
[CH3CH20H] [CH3COOH]
=" —
ie)
x?
Ke- Gem

Example:
1. An equilibrium system for the reaction between H2 and |2, to form HI at 670K in 5I flask
contains 0.4 moles of H2, 0.4 moles of |, and 2.4 moles of HI. Calculate the equilibrium
constant K ¢.

Ho ig) tl2ig) 7 2Hl ig)

(Hr)?
k= Hell]

= -1
[HI]= 5 =0.ag Moll
0.4
a -1
[H:)= 5 =0.03 MlL
[hb] = 5 =0.08 molL
(0.48)?
“= (0.08)(0.08)
c=
K.=36
2. A mixture of 1.0 x10? moldm?H 2 and 2.0 x 10? moldm? |; are placed into a container at
450°C. After equilibrium was reached the HI concentration was found to be 1.87 x
-3
10 moldm *Calculate the equilibrium constant.
Hp ig) laig) 7 2HI ig)
At t=0 a b it)
Equilibrium: a-x —_b-x 2x
* 2x = 1.87 x10?
X= 9.35 x 10¢moldm *
a=1.0 x 10°
a-x = 1.0 x 10?- 9.35x 10%
= 6.5 x 10°moldm*

i) Temperature

ii) Concentration

iii) Pressure

e The first three affects both rates and position of chemical equilibrium (i, ii and iii)
e = The other three affects the rate of chemical equilibrium
1) Temperature
a) Increasing the temperature, increases the rate of reaction because usually at high
temperature the collision factor increases also the number
of molecules having necessary activation energy is large.
b) Effect on the position of equilibrium is explained by using Le-Chateliers principle which
states that “when a system at equilibrium is
subjected to a change, processes occur which tend to counteract the change” (If a system in
equilibrium is disturbed (change in temperature and pressure) the system adjusts itself so as
to oppose the disturbance).
Consider the reaction
2SO2 (g) + O2 (g) = 2503 (e) + Heat (negative)
Reactants S Products
If temperature is increased in the system, the equilibrium moves in a direction where there

is a absorption of heat and if the temperature is decreased in the system, the equilibrium
moves in a direction where there is release of heat.

Effect of temperature, on the position of equilibrium can be explored by Vant Hoff's law of
mobile chemical equilibrium which states that

“For any system in equilibrium high temperature favours endothermic reactions and low
temperature favours exothermic reactions.
The way in which equilibrium constant changes with temperature is found both theoretically
and experimentally governed by the following
relationship;
f d(In K) f Am

dT -~ RT
a°t Hm = change in molar heat
K = Equilibrium constant
On intergrating the equation above;

= —4Am
InK= ae re
Where c = constant

If K, and K2 are equilibrium constants corresponding to T; and T2 the constant term can be
eliminated from the equation above so as to give Vant Hoff's equation i.e.

xAHm 4 ¢

—AH,

=4Hm +-C
Subtracting equation (1) from (2) i.e.

—aHm —aHm
"+0 -("+e
InK,—InK,= ¢ RT, ) ¢ RT, )
Kz) _ Aim (- _ +)
But 1 In = 2.303log
1 1
K, som ([— — —
log{ = asm ( )
a(n) = 2308R Ty TA enreeed
1 1 BON
Buti %= Take
Kz\ aHm (72—7:
tog() 22%. (F")
= o8(Z) =2.303R TTF ad
Where 1=2=3
Example1: For the reaction;
NzO, 7 2NO, "tH = 61.5K) mol*
Kp = 0.113 at 298K
i) What is the value of Kpat 0°C?
ii) At what temperature will Kp =1?
Answers:
tog (2) aHm (=)
i) From ky/ -2303r \ TT2

Kp2 = 0.113

(a) Ke[Fe(CN)«]

Potassium hexacyanoFerric (ii)
(b) [Co(NH3)4(NO2]SO.

Tetra aminebromonitrocabalt (iii) sulphate
(c) (Cr(NHs)6(Cr(C2)4)3]

Hexaminechlomium (iii) trioxalatochromium (iii)
(d) [AI(H20) (OH)s]?

Aquopentahydroxoaluminium (iii)
(e) [Pt(en)2Ch]?

Dichlorodimetlydiamine platinum
(f) K[CoCI(H,0)NH5)5H,0

(1) + Cr + (-4) + (0) +5(0)=0

Cr=+3

(e) (Pt(en)2Cl2)

Dichrodimethlyamine platinum
(f) K [CrCL4(H20) NH3] 5H2)

(1) + Cr + (-4) + (0) + (0) (5(0) =)Cr=+3.

Pentahydrate potassium amine aquatetrachlorochromate (iii)

Question
(a) Complex compound [CO(NHs) s(Br)SO.
(i) Name the complex compound above
(ii) What the coordinate number of central atom
(iii) If all ligand placed by chloride ligand what is the charge of complex compound.
(iv) Write isomers of compound
(b) Using hybridization principles prove the following.

Fe(CN); -4 d? sp3 hybridized

CO F6 -3 d’sp® hybridized

Ni(CN)S* dsp? hybridized
Solution
(a) [Co(NHs)sBr]SOq

(i) Diaminebromocobatt (iii) Sulphate

(ii) Co-ordinate number =3
(v) [CO(NH3)Br]"?[COCI3] the complex compound so neutral
(vi) Isomers of complex compound

[CO(NHs)2Br]"?
[CONH3Br2)"?

Isomers of ligand obtain through changing the number of each ligand
(b) [Fe(CN).]}* d? sp3 hybridization
3d
fe] (uu [sds [s ds]
4s 3d
ree CGD DTT
CN CN
4s 3d 4p
CN CN CN
4s 3d 4p
Fe(CN) s~*= [Ar] C4
CN CN CN CN

unpaired electrons in 3d orbitals.
3d

This is because ligands of high spin when approaches the d — orbitals and cause small separation of the d
— orbitals. The energy of separation (AA E) between double degenerate and treble is small, electrons will
spread in accordance to Hund’s Rule in which each electron occupy its orbital singly before pairing.
In that case the central transition metallic ion remains with unpaired electrons in the electronic
structure. Hence the complex combined will be paramagnetic. Consider Fe” ion when complexes with
ligands of high spin like water [H2O]

3d 4s 4p

#H,0
ree TT = Mehe a8
3d
[u[ +] ]
This complex [Fe(H20.)4]”* is a 4sp* — hybrid complex and is paramagnetic. Since Fe’* ion in the complex
contains unpaired electrons in the d— electrons. If on the other hand Fe”* complexes with ligands of low
spin the paramagnetic property will be destroyed. The complex will be diamagnetic. Why?
This is because ligands of low spin split the d — orbitals and cause large separation.The energy of
separation [ AE] between the two degenerates becomes large.If the energy of separation is large

electrons fill in the orbital in accordance to AUFBAU’S PRINCIPLE in which electron pair themselves in

Treble degenerate orbitals, which have low energy according to AUFBAU’S PRINCIPLE. Consider Fe?
complexes with ammonia which is ligand of low spin.This complex is 3d’4sp* hybrid complex and is
diamagnetic since there are no unpaired electrons in electronic structure of Fe?’.All electrons Fe** are
paired in Treble degenerate orbitals.
QUESTIONS
NECTA 2001 P2 Question 6c
Use the configuration of 3d — orbital electron on cobalt (iii) ion to explain why [COf.]* is paramagnetic
while [COCNg] is not paramagnetic? [10%]
Solution
Consider electronic structure of CO and CO* ions given below
3d
vo = wee Cae |
cos = [Ar] 4Se
Pitutat i]s |
Then
Complex [CoF«] > is paramagnetic because fluoride ion is a ligand of high spin hence cause small
magnetic field 3d — orbit of cobalt (iii) ion hence the energy separation is small and filling of electrons in
according to Hund’s Rule, thus cobalt (iii) ion contains unpaired electrons in the 3d — orbitals. Hence a
paramagnetic substance.
Consider
This is 4sp? hybrid complex. And it is paramagnetic due to the in 3d — orbital
But
The complex [COCN«]* is not paramagnetic due to absences of unpaired electrons in 3d — orbital in
cobalt (iii) ion in the complex caused by complexing with ligands of low spin the cyano. This ligands exert
strong magnetic field to the unpaired 3d — orbital hence large separation which cause excitation of be

difficult hence filling of electrons is according to AUFBAU’S PRINCIPLE.
3d 4s 4p
ew ODED D[ = =. =
[ufo fo]
This complex is 3d*4sp* hybrid complex and is diamagnetic since there are no un paired
electrons in electronic structure of Fe”. All electrons Fe” are paired in treble degenerate
orbitals.
QUESTIONS
NECTA 2001 P2 Question.6c
Use the configuration of 3d — orbital electron an cobalt (iii) ion to explain why [COfs]* is
paramagnetic while [COCNg] is not paramagnetic? [10%]
Solution
Consider electronic structure of CO and CO* ions given below.
3d
soos wise Cee Te
cos+ = [Ar] 4S°¢
Lt futads | a |
Then
Complex [CoF«] * is paramagnetic because fluoride ion is a ligand of high spin hence cause small
magnetic field 3d — orbit of cobalt (iii) ion hence splits separates the hence the energy separation is
small and filling of electrons in according to Hund’s Rule, thus cobalt (iii) ion contains unpaired electrons
in the 3d — orbitals. Hence a paramagnetic substance.
Consider.

3d 4s 4p 4a
F F F F F FF
com =ei[t]u] 1] 1]
Le] + [4]
This is 4sp* hybrid complex. And it is paramagnetic due to the in 3d — orbital
But
The complex [COCN«]* is not paramagnetic due to absences of compared electrons in 3d — orbital in
cobalt (iii) ion in the complex caused by complexing with ligands of low spin the cyano. This ligands exert
strong magnetic field to the unpaired 3d — orbital hence large separation which cause excitation of be
difficult hence filling of electrons is according to AUFBAU’S
PRINCIPLE.
3d 4s 4p
CN CN CN CN CN CN
[stsqs]
COs = [Ar]
[u[u] |
Paramagnetic property of transition metals can be destroyed in two ways
a. Temperature/ oxidation process
b. Complex compound formation
a. TEMPERATURE/ OXIDATION
Raise in temperature destroy magnetic property of the elements. Raise in temperature causes
excitations of electrons and ionization of the atoms by losing electron.
lonization to an extend of losing all unpaired electrons destroys magnetic property of the substance.

3s-
Na [Ar] LJ paramagnetic
3s
Nat = [Ar] | diamagnetic
b. COMPLEX COMPOUND FORMATION
Formation of complex compounds by coordination with ligand may also destroy magnetic property of
transition metals. However this will depends on oxidation state of the metals and nature of the ligands
involved in company compound formation whether are ligand of high or low spin.
When the complex compound formed by ligand of high spin such as C:0,?, OH-, F, Br, 1-, Cl- energy
separation between treble and degenerate. Electron filled according to the Hund’s rule this result
electron to be unpaired in the double and treble degenerate the complex remain paramagnetic example
COF.? is paramagnetic substance. The F- filled in 4s, 4p and 4d vacant orbital the 3d remain with
unpaired electron which result paramagnetic
Question
a. Predict the coordination number of Ni’* and state whether the complex will be paramagnetic or
diamagnetic it Ni* complex with
i) Bromine ions, Br ~ [Ligands of high]
ii) Ammonia molecules NHs [ligands of low spin]
b. Fe complexing with NH, of low spin
C) VARIABILITY IN OXIDATION STATES
With exception to zinc, transition metals form more than one stable oxidation state. They have variable
oxidation state. Variability in oxidation state in transition metals is explained by the ability to ionize by
losing electron from both sub energy levels 4; and 3d.The energy present between 4s and 3d is very
small. The gap existing between 4s and 3d is small. The different between the two degenerated is so
small that just normal radiant energy from the sun is enough to excite electrons and so small that
electrons from sub energy level 4s and 3d can be removed by almost the same amount of ionization
energy. So that during ionization, transition
metals give off 4s - electrons first followed by 3d Transition metals depends on the number of electrons
present in 4s — orbital. Chromium has lowest
oxidation state of *2, looses two electrons in 4s — orbital and copper the lowest oxidation state is “1. The
maximum or the highest oxidation state is (achieved) attained after losing all the 3d electrons.

3d 4s
Poeddsd))
Mg =[Ar]
r= [Ar] 3d 4s
CoE) GG)
3d 4s

ee CITTTI ©
The extent of losing all the 3d — electrons is possible only for the first five elements from Sc to Mn.
Elements beyond manganese can ionize to an extent of losing all the electrons because of large nuclear
charge in these elements which cause strong effective charge pull, over the 3d electrons. The only
oxidation state for elements found beyond manganese is ‘3 and for copper stable oxidation state is ‘2
which is attained after loss of the single 4s - electron and another from sub energy level 3d. The nuclear
charge is high in zinc and hence the nuclear pull in 3d electron for zinc is maximum. Hence it is difficult
to remove electrons from sub energy 3d in zinc. Zn has only one stable oxidation state of *2 which is
formed after losing the two electrons. Only one stable oxidation state for zinc is one of the factors which
excludes zinc from transition metals.
NOTE:
For these elements with atomic number 21 to 25 which can ionize to an extent of losing all the d —
electrons, the tendency is that increase in oxidation states accompany with increase in acidic character.
Lower oxidation states have basic character and higher oxidation states are acidic in nature. This is
caused by increase in ionization energy needed to form the ions.The increase in acidic character with
increasing oxidation states can be justified by considering the various oxides of chromium.
The various oxides of chromium include.

Cr,0,CrO Cr, Os Cr, 05 and Cr O3
Base oxides amphoteric oxides Acidic oxides
The first two oxides, chromium (iii) oxide [Cr20] and chromium (ii) oxide [CrO] are basic solutions which
have no chemical reaction with other basic solution even alkaline solutions.
cro + H,0 — Cr [0H],
then  Cr[OH]; + NaOH — No reaction
Cr [0H] 2react with acidic solution

Cr(OH}, + 2HCL —+ (CrCl, + 2H,0

Chromium (ii) chloride
Chromium (iv) oxide [Cr20s] is amphoteric. It dissolve in water and form hydroxides which dissolve in
both alkaline solutions and acidic solutions.
Cr,0; +  H,O — 2Cr [OH],
As base
Chromium (iv) hydroxide react acid solution
Cr[OH]; + 3HCL — CrCl, + 34,0

As acid
Chromium (iv) hydroxide dissolves in alkaline solution and form complex salt

Cr [OH]; + 2HCL — Nasicr (on),

NaCrO, +  3H,0
Sodium chromide (iil)
The last two oxides, chromium (v) oxide Cr0s and chromium (vi) oxide (CrOs) are ACIDIC
OXIDES. They dissolve in water and form strong acidic solutions
CO; . HO? HaCrOx
Chromic acid

Chromic acid is so strong acid that has no chemical reaction with other acidic solution

H.CrOs .  H50. — NO reaction
Chromic acid reacts with basis solutions and form normal salts

H,Cr0, + 2KOH — K,CrO, + 2H,0
Reaction of the various oxides of chromium real justify with increase in Acidic character.
NOTE

e.g 100g (meq /100g)
mass of cation
No of equivalents=-22#22!##£#t---..---------(i)
Equivalent ‘means the mass of cation that will replace (exchange) 1g (mole)
of H*.
Eg
23g of Na’ (i.e Imole of Na*) exchange 1g of 1* and hence Na* has equivalent wt
of 23g.
(#4)
4og of Ca” (I mole of Ca") exchange 2g (2moles of H” which means 20g \ ? /of
2a)
Ca** exchange 1g ( 2 ot H
Hence Ca?* have equivalent wt of 20.
Atomic wt
“equivalent @ = dmount of ioniccharge (iii)
mass of acation
but” azomic st = no of moles of the cation
Number of mole
“number of equivalents = °°" x Amount of ionic charge
Where no of mill moles = No of moles x 1000
Mass of cation
No of mill moles = 4 += x 1000
Mass of cation in milligrams
= ‘Atomic ot

Factors affecting value of cation exchange capacity.

Cation exchange capacity is affected by the following factors:
- Amount of clay (soil texture)
- Type of clay
- Soil organic matter
- P# of the soil.
i) Amount of clay in soil
A high silicate- clay soil hold more exchangeable cations than a low
silicate clay soil and hence have greater CEC value.
Silicate clay (Si— (OH) easily lose H” off Hydroxy group in basic
medium resulting into net negative charge i.e the greater the —ve charge
the greater the CEC
(ii) Clay types
Different clay types have different CEC value due to difference in surface
area.
(iii)Soil Organic Matter
Organic matter in the Soil, also are negative charged. Therefore
- High — Organic Soil have greater CEC value than a low organic Soil.
- On another hand if an organic matter continue to decay, the CEC tends to
decrease with the decomposition.
- They can retain more cations.
Therefore organic matter particles have greater CEC than clay particles.

(iv) P" of the soil.
P" affects CEC of colloids that are charged based on hydroxyl group instead
isomorphous substitution.
Under acidic (low P# ) condition.
H’ are in excess resulting to less ionization of the colloids, less negative charge
and hence low CEC value.
Under basic (High P" ) condition.

OH are in excess resulting to more ionization of the colloids , more negative
charge on the colloids surface and hence high CEC value.
PERCENTAGE BASE SATURATION
Acid cations and Base cations
Acid cations are exchangeable cations (mainly (H* and Al**) which tend to acidify
the soil

- They are also known as exchangeable acids

- In very acidic soil, H* and Al* dominates other adsorbed cations.

- The acidity of AP** is explained by its cationic hydrolysis

according to the following equation:
[Al(#,0),]°° +#,0—+ [Al(H,0)OH}* + ,
Al>* in the soil water solution Determine the acidity of the soil solution
: Base cations (or exchangeable bases) are exchangeable cations which are
capable of neutralizing soil acidity . Common exchangeable bases are Ca", Mg?*,

E.g. Ladmium and lead, below P' 4 get dissolved and may enter plant

and make crops unfit for human consumption
Conclusion:
Determination of P# of agricultural soil is very important for the selection
of suitable crops. Below the P" 4.8 a soil may need liming in order to
avoid aluminum toxicity.
CAUSES OF SOIL ACIDITY
« Soil acidity is caused when heavy rains leach bases like Ca?*, Mg?» K*
and Na* from the soil to the ground water table leave surplus H™ in the
soil.
. Also industrial regions may bring sulphuric (vi) acid and nitric (v) acid
to the soil which increases its acidity.
. Acidic mineral fertilizers like ammonium sulphate (vi) and ammonium
chloride make the soil more acidic due to hydrolysis.
- Also the nitrification of ammonium ions by bacteria produces H*(ag)

NHg(aq)+ 202,» 2H"jaq) +NO;,,.. + H20
sea) £02 (2a) * NO %(aq) * 2% Q
NOTE: Organic acids are produced during the decomposition
of organic matter. Due to this, most soils in humid tropics are Acidic.
Liming & liming materials
Discuss:
- Meaning and significance of liming as treatment to soil P".
- The efficiency of liming materials i.e. neutralizing values of
carbonates, oxides, hydroxides and silicates.
- Beneficial effects of liming

- Detrimental effects of over-liming

Liming:

It refers to the process of adding basic compounds of calcium and
magnesium to acid soil (P* < 5) in order to raise the P" of the soil to
the required level.

The compound of calcium and magnesium (oxides, hydroxides,
carbonates and silicates) are called agricultural limes. Such compounds
are contained in limestone, dolomite, building (slaked) lime, oyster
shells and blast furnace slag which can be used for liming soil if finely
grind.
Significance of Liming:
. To raise the soil P"

Q. 1. Explain similarities and differences between industrial fertilizers and
manure

Q. 2. Tomention the advantages of straight and mixed fertilizers of N, P
and K

Q. 3. Mention examples of straight and mixed fertilizers and classify them

Q. 4. Compare the relative advantages and disadvantages of manures as
compare industrial fertilizers.

How to maintain and improve soil fertility:

A fertile soil provides all essential plant nutrients in amounts which are
suitable for the growth of most plants.
How can a soil be made more fertile
1. Good cropping system

2. Adding manure to the soil
3. Adding industrial fertilizers to the soil
1.GOOD CROPPING SYSTEM:
Crop rotation should be practiced. Every season, another crop should be
planted on a given field. Legumes should be rotated with cereals, shallow
rooted plants with deep rooted ones .
2.ADDING MANURE:
Manures are organic materials that can be added to the soil to increase soil
fertility.
E.g. Kraal manure — from cattle kraal
Farmyard manure — wastes from animals
Compost manure-mixture of soil and decomposing organic matter
Bio gas manure ~ from effluent of bio gas plants
3. ADDING INDUSTRIAL FERTILIZERS:
- Fertilizers are most inorganic compounds which contain one or more plant
nutrients in a concentrated form.
E.g. Nitrogen fertilizers, phosphorus fertilizers, potassium fertilizers.
Similarities between manure and industrial fertilizers.
- They both increase humus to the soil and promote good soil structure
- Both manure and industrial fertilizers contain minerals Nitrogen,
phosphorus and potassium which are essential for plant growth.
Differences between manure and industrial fertilizers

1. It is organic material 1. It is inorganic compound

The Ksp value of the salt can be determined from its solubility in moles per litre (mol/L)
When concentrations are given in any other units such as g/L, they must be converted to mol/L
Example 1
The solubility of Agl is 1.22 x 10® mol/L. Calculate the Ksp for Agl
Solution
Agly = ABiaa + Mag
Each 1 mole of Ag! that dissolves gives 1 mole of Ag’ and 1 mole of I’in solution, concentration of each
ion solution is 1.22 x 10® mol/L.
Hence

Ag) = Age + Cl

1.22x10* —1.22x10*
Ksp = [Ag’] [CI]
=(1.22x 10°)

Ksp = 1.4884 x 10 moPL?
Example 2
PbCl2 dissolves to a slightly extent in water according to the equation
PbCli) = P**(aa)+ 2Cliea)
Calculate the Ksp for PbCI, if (Pb**) has been found to be 1.62 x 10 ? mol I”.
Solution
PbCl, == — Pb** + 2c
162x107 162x107 = (2x 1.62 x 107)

Ksp = [Pb**] [CI]?
= 1.62 x 10 ? moll x 1.0497 x 10% mol’L?
Ksp = 1.7005 x 10 * mol’L*
Example 3

The solubility of Pb(CrO,)is 4.3 x 10 * gt. Calculate the Ksp of Pb(CrO.)

(Pb =207, Cr = 52, O= 16)
Solution
PbCrO. = Pb* + CrO.*
43x10%gI" 4.3x10%gt? 4.3x10%gI*
To calculate the molar mass of (PbCr04)
Pb(207) +Cr(52) +0(16)4=323gmol*
323g > 1mole
43x10 > x
x=1,33x107mol
PbCrOs = Pb* + CrO.?
1.33x 10 71.33x10 7133x107
Ksp = [Pb**] [CrO,7]

= (1.33 x 1077

Ksp = 1.76 x 10“ M?

Example 4
100 ml sample is removed from water solution saturated with MgF, at 18°C. The water is completely
‘evaporated from the sample and 7.6mg of MgF, is obtained. What is the Ksp value for MgF, at 18°C
Solution
V=100 ml
m=0.076g.

62g > 1mole

0.076, > x

x=1.22x 10? mole!
MeF: => Mg* + 2

1.22x10% 1.22x10% (2x1.22x 107)
Ksp = [Mg™] [F]*
= (1.22 x 10 *) (2.44 x 10%)

Ksp = 7.33 x 10° moll
DETERMINATION OF MOLAR SOLUBILITY FROM Ksp VALUE
If the Ksp value is known, the molar solubility can be obtained since Ksp shows the maximum
concentration of ions which exists together in a solution.
Example 1
Calculate the molar solubility of Ag.CrO. in water at 25°C if its Ksp is 2.4 x 10°22.

AgsCrOus T= 2Agr isa) + CIO? jag)
Let the solubility be S
AgiCrOa) 2 Ag’ ion) + CrO2? jn)
s 2s s

From Ksp = [Ag‘]?[CrO.*']
24x10” =(2S)?

24X10 =48° <=> s= spano®
S = 8.434 x 10 * mol L*
Example 2
Calculate the solubility of CaF2in water at 25°C if its solubility product is 1.7 x 10 *° >
Solution
CaF, Ca* + 2F-

s s 2s

pH = 14-pOH
=14-3.801
pH = 10.199

Solution 2
We find concentration of OH- in NaOH

nem

4xi07*
a)

n= 1x10" moles

1x10~4 moles — 10cm3

x — 1000cm?

x = 0.01M

Zn(OH)2 = [Zn 2*] [OH™]}?

Qsp = [2n?*] [OH]?

= (1.6 x 10 “4)(0.01)?

Qsp = 1.6x 10 Em?
Qsp > Ksp Hence precipitation will occurs
PRECIPITATION REACTION IN QUALITATIVE ANALYSIS (ION SEPARATION)
Qualitative analysis refers to a set of laboratory procedures that can be used to separate and test for
presence of ions in solutions. This can be done by precipitations with different reagents or by selective
precipitation.
Consider an aqueous solution which contains the following metals ions Ag’, Pb”",Cd”* and Ni?" which
have to be separated. All the ions form very insoluble sulphides (Ag:S, PbS, CdS, NiS). Therefore sulphide
is a precipitating reagent of all the above metal ions. However, only two of them form insoluble
chlorides (AgCI and PDC!) i.e aqueous HCI can be used to precipitate them while the other two ions
remain in solution.

Tons in solutions
Add HO (4,
a » »
Precipitates AgCl, PbCl2 ed and Ni ions remain
solution
The separation of PbCI, from AgCl is not difficult since PbCl, dissolves in hot water while AgCl remains
insoluble
The separation of Cd ** and Ni?* can be done by selective precipitation with sulphide ions by considering
the Ksp values of the two compounds.
Example Ksp (CdS) = 3.6 x 10? and Ksp(NiS) = 3.0 x 10 7+
The Compound that precipitate first is the one whose Ksp is exceeded first (one with smaller Ksp).
Suppose the solution contains 0.02M in both Cd** and Ni’, the sulphide ions concentration necessary to
satisfy the solubility product expression for each metal sulphide is given by
-). Ke
For Cds needed [S?"] = rod
= 36x10"?
[s**] = 1.8.x 10°27
Concentration of S* can exist in which the solution without precipitation (above which precipitation
occurs)
x
For Nis, needed [S?"] = pei
= 3.0x10~**
[s**] =1.5x10°19M
The much smaller S* concentration is needed to precipitate (CdS than to begin forming NiS thus CdS
precipitate first before NiS.

Just before NiS begins to precipitate, how many Cd”* remains in the solution?
Concentration of S* needs to be slightly in excess of 1.5 x 10 *° M for NiS to begin precipitation. The
[Cd**] that can exist in solution when the concentration of S* ions is 1.5 x 10 “is given by
ep
icd2*]) =
te =
= 36x10"
1sx107**
[cd?*] =2.4x10 710m
To find % of Cd®* which has precipitated.
0.02-2.4x107*°
0.02
= 99.99%
This means that we can separate Cd”* and Ni’* ions in aqueous solution by careful controlling
concentration of S* ions.
Question 1
The Ksp of Agx are (AgCl) = 1.7.x 10%
(AgBr) =5.0x 10-8
(Agi) =8.5x 10 7”
A solution contains 0.01M of each of Cl, Br, and I-. AgNOs is gradually added to the solution. Assume
the addition of AgNO3 does not change the volume.
(a) Calculate the concentration of Ag” required starting precipitation of all three ions.
(b)Which will precipitate first
(c) What will be the concentrations of this ion when the second ion start precipitating
(d) What will be the concentration of both ions when the third ion starts precipitation
Solution

(2) For AgCl needed [Ag*] = _
- 1% 107%?
[Ag+] =1.7x10°8M
For AgBr, needed [Ag+] = 5]
. 5x107**
oot
[Ag+] = 5x 10°24m
For Agi needed [Ag+] = ia
. 8.5x107*"
[Ag+] = 8.5x 105m
(b) Agi will precipitate first because the Ag concentration is very small
(c) When second ion starts to precipitate ie AgBr start to precipitate, concentration of Ag will
be
= = _
lat] = Soom
-25* 10727
sxi07#
[Ag+] = 1.7x 108M
(b)For [I] when AgBr starts to precipitate
- Ksp
w= 1.7x107*
8.5x107*"
* [7xi0-*
(r] =5x10°%™
For [Br-] when AgBr starts to precipitate

[er-) = Ks? _
1.7x10-*
= et o-
17x 107"
[Br-] = 2.94 x 105M
Question 2
Asolution contains 0.01M of Ag’ and 0.02M of Ba”. A 0.01M solution of NazCrOs is added gradually to
it with a constant stirring.
(a) At what concentration of NaxCrO4will precipitation of Ag’ ions and Ba* starts?
(b) What will precipitate first?
(c) What will be the concentration of the first precipitated species when the precipitation of the second
species starts?
(Ksp (Ag2CrO«) 2 x 10? MP, Ksp(BaCrO.) 8.0 x 10 * M2)
Question 3
To precipitate calcium and magnesium ions, ammonium oxalate (NH‘)2 C20s is added to a solution ie
0.02M in both metal ions. If the concentration of the oxalate ions is adjusted properly, the metal oxalate
can be precipitated separately.
(a) What concentration of oxalate ions (C,0 2) will precipitate the maximum amount of Ca”* ions
without precipitating Mg” ions.
(b) What concentration of Ca’* ions remain when Mg” ions just begin precipitation.
{c) The Ksp of two slightly soluble salts, AB; and PQ, are each equal to 4.0 x 10 “8. Which salt is more
soluble?
(d) What is the minimum volume of water required to dissolve 3g of CaSO, at 298K
(Ksp (CaSOs) = 9.1 x 10 * M’)
ANSWERS
Question 2 solution
Given [Ag’] = 0.01 M [Ba*] = 0.02M
For Ag:CrOsto begin precipitating

ii) Damping of solid and liquid waste on the land and into large water masses , engravers,
wells and oceans.
iii) Opening up gaseous effluent from industries into the air.
AGRICULTURE CHEMICALS
- Use of the pesticides and fertilizer.
- This causes a loss in biodiversity and soil destruction .
~ Pesticides causes water pollution , facilitate growth of sea weeds causing oxygen deficiency
for marine organism.
- Pollination decline, pesticides kills organisms which are agent of pollination.
EFFECT IN ORGANIC FERTILIZERS:
- Soil Acidification
- Energy consumption.
- Climate Change.
Suggestions
- Use of organic biodegradable.
~ Manure.
- Avoid monoculture practices.
ii) DAMPING OF SOLID AND LIQUID INDUSTRIAL WASTE OF THE LAND AND INTO
LARGE WATER
- Damping of solid and liquid industrial wastes on land and into large water cause deterioration
of water quality and land in general.
Petroleum product, heavy metals from the industries may leads to environmental destruction
* Destruction of environment from petroleum product is mainly due to accidental spills from the
ships , tanker trucks , pipe lines and leaky of underground storage tanks.
«Heavy metals such as mercury often deposited with sediment in the bottoms of stream . They
may become incorporated into plants.

iti) OPENING UP GASEOUS EFFICIENT FROM INDUSTRIES INTO THE AIR
- Working on the thermal plants and different plants that are used to manufacture different
types of fertilizers or pesticides, also production of building materials can encourage of the
production toxic materials which goes into air in form of smoke.
- Poisonous gases eg. Sulphur dioxide may be emitted also inform of smoke. These cause
destruction of air, causing harmfully diseases to human beings.
ENVIRONMENTAL MANAGEMENT INTERVENTIONS
Environment management and interventions should be done through the following: -
i) Management waste.
ii) Chemical treatment of urban sewage.
iii) Chemical treatment of urban supplies of edible / portable water.
POLLUTION
Environmental pollution is the contamination of air, water, and land form man made wastes.
Pollution leads to depletion of ozone layer global warming and climate change.
1. WATER POLLUTION (AQUATIC POLLUTION)
Water pollution is the contamination of water by foreign matter that deteriorates the quality of
water, It occurs in lakes, oceans and rivers.
- It involves the release of toxic substances, pathogens, substances that require much oxygen to
decompose ,easy — soluble substances and radioactive substances.
i) Sources of water pollution
‘The major forms and the sources of water pollution are: -
a) Oil (Petroleum Product) discharged
- These including manufacture of plastics, lubricants solvents and synthetic fabrics fractional
distillation of clued oil to produce vehicle fuel , paraffin wax , refinery gases for domestic
cooking and bitumen for road surfacing and products is mainly due to accidental spills from the

ships , tanker truckers , pipeline and leakage , from underground storage tanks.

b) Synthetic chemical pesticides
- Pesticides such as herbicides, Fungicides used in agriculture and public health programmers
to control pests are important source of water pollution.
~ They get into water sources through run — off and atmospheric transport and deposition.
- Pesticides accumulates in plants and animals, when they die, they spread to water sources, thus
increasing water toxicity.
) Heavy metals
Heavy metals such as mercury are dangerous pollutants. They are often deposited with sediment
in the bottom of the streams. When deposited on surface they become incorporated in plants food
‘crops and animals. If they dissolve and water is withdrawn for agriculture or human being use,
poisoning can result,
i) Management of water
‘Water management refers to practices of planning developing, distribution and optimum,
utilizing of water resources under defined water polices and regulations.
‘These are: -
a) Management of water treatment of drinking water, industrial water sewage or waste water
b) Management of water resources.
c) Management of irrigation.
Treatment of water maybe divided into two;-
i) Purification of domestic use.
ii) Treatment for specialized industrial application,
2, AIR POLLUTION
Air pollution includes all contaminants found in the atmosphere. These are dangerous
substances can be either in the form of gases or particles.
Sources of air pollution are natural and human ~ based.

CAUSES OF AIR POLLUTION

‘The biggest causes are the operation of fossil fuel , burning power plants, and automobiles that
combust fuel.
‘TYPES OF AIR POLLUTION
i) Smog ~ This is the first type of air pollution.

When smoke present in the atmosphere combine fog present in the air. A mixture formed is
Smog or photochemical smog.

ii) Green House effect — It is formed due to the contamination of several important gases
with the air. These gases are called green house gases.
eg. Methane, sulphur, nitrogen, Carbon monoxide, hydrogen and ozone. These are very harmful
for the human skin and causes cancer.

Effects of air pollution
Air pollution allects human health, vegetation and livestock.

i) Effects on human health
Severe air pollution cause many fatal diseases and disorders some of the effects caused by
inhaling polluted air are: -

‘a) Sulphur dioxide enters soft tissues causing drying of the mouth, scratchy throat and smarting
eyes.

b) Hydrocarbons and many other air pollutants cause skin cancers.

¢) Oxides carbon , sulphur , nitrogen diffuse into the blood and combine with haemoglobin
causing reduction in it is oxygen carrying capacity.
ii) Effects on vegetation

Air pollution has serious harmful effects on vegetation.

Effects of air pollution on vegetation are :-

a) Sulphur dioxide causes chlorosis i.e. loss of effects on tress , plants and vegetation.

b) Oxides of nitrogen and fluorides reduces crop yields.

c) Photochemical Smog bleaches and blazes foliage of economically important of plants and
‘crops.
iii) Effects on Livestock
General effects of air pollution on livestock are the same as in the case of human being.
Various fluorine compounds which fall on foliage plants are eaten by livestock causing abnormal
calcification of bones and teeth, called fluoride toxicity. Fluorosis can causes loss of weight and
frequent diarrhea in animals.
3. LAND POLLUTION
This is contamination of land surface through damping , urban wastes , industrial wastes ,
‘mineral exploitation and misusing the soil by harmful agriculture process.
Causes Of land pollution
+ Increase in urbanization is major cause of land pollution.
+ Construction uses up forest. This leads to the exploitation and destruction of forests.
* Disposal of non— biodegradable wastes included containers, bottles and cans made of plastics,
used cars and electronics goods used to the pollution of land.
EFFECTS OF LANDS POLLUTION
* Makes places look dirty due of tonnes and tonnes domestic wastes dumped without proper
disposal of them.
+ Land pollution affects respiratory system of human being.
* Land pollution has serious effects on wildlife. Flora which provides food and shelter to wild life
destroyed.
Prevention of Land pollution
+ People should be educated and made aware about the harmful effects of littering.
+ Items used for the domestic purpose should be reused or recycled
+ Inorganic matter such as paper , glass, plastics and metals should be reclaimed and then
recycled.
Questions:

1. Industries are among the leading sources of air pollutant.

(a) Name four substances from industries which contribute to air pollution.
(b) Explain two other sources of air pollutants.
(©) Give 3 effects of air pollution
(4) Explain how industrial worker can be protected against harmful chemical
fumes,
2. (@) How is ozone formed?
(b) What are the causes of depletion of ozone layer?
(c) Explain harmful effects of depletion of ozone layer.
3. What is green house effect and what are its effets
4, Write short notes on:
(i) Acid rain
(ii) Smog.
(ii)Environmental effect caused by mining.
5. What do you understand by the term Eutrophication and its causes? ~ How
does it threaten the development of marine life?
ANSWER:
1. (a) Substances;-
= Carbon dioxide.
. Carbon monoxide.
Methane
- Sulphur dioxide.
(b) Other sources:-
Burning fuels from car.

(0) Depletion of ozone layer:~

Spread of air borne diseases E.g. Tuberculosis.
Acid rain
Global warming.
(d) Industrial worker can be protected against harmful chemical fumes by:
sing protective masks to protect them from harmful fumes.
. Good ventilation systems in industries i.e. air circulation.
Close chambers for chemical processes.
2.(a) Ozone layer formation;-
uvdight
°x¢) —— 20@)
26g) +O@y—> 3@)
{(b) Causes of depletion of ozone layer
(a) Natural destruction.
(®) Artificial destruction.
(a) Artificial destruction
. Through gases e.g: CFC’S
uy
cl,CF, — Cl +CICF,
Cl + 0,—*CIO+ 0,
clo+O—» Cl +0,
This is a chain reaction which later leads to the destruction of ozone layer

solution containing 0.01mole of Co c's-5NH5s jeads to immediate precipitation of only 0.02 moles of
silver chloride.
GENERAL QUESTIONS
1. Explain very briefly using equations where possible the extraction of copper from its commercial hal
under the heading of reduction and
purification.
2. (a) Write down:
(i) Four reasons which ustfes the placement of hydrogen in group lof the periodic table
(i) Four reasons which puts hydrogen in group (vl) of periodic table
(0) (i) In which group do you think hydrogen should strictly belong
(i) Give reasons for your answer for(i) above
(c) Describe the ation of water on hydrides of period It
(6) Compare the thermal stability of carbonates of group ! and Ilby using a specific example show
their differences
3. Comment with help of chemical equations where necessary inthe folowing:
(i) tron (i) chloride cannot be prepare by heating iron filing in a steam of chlorine gas
(i) Hydrochloric acid cannot be used as acidic medium during redox titration of KMInO, against Fe
50.
(i) Solid Al (OH)? is soluble in aqueous solution of NaOH.
ANSWER:
1. Extraction of copper
as

steps:
Roasting:
cu Fe? ‘is heated in plenty supply of air
CuF eS) +02 4 CuzS,. + FeS + So; T
Also
Cu,S+0;—» Cu,0 +S0,7
FeS+0, —* FeO+ SO, 7
Now silica is added to Fe to remove FeO
FeO + Sio, —+ Fe Sio,
Slag (removed)
Self— Reduction reaction:
In self-reduction reaction, “25 reacts with C20 to form molter copper which later solidifies
CuS + 2Cu,0 > 6Cucs + S02
Purification:
Purification is done electrolytically.
At cathode
Cu + 26 —+ Cu
(ure copper)
At anode
* os
Cu gy Cu™ 428
(€) Solution.

NaH+H,0 —* NaOH + H,
Basic
MgH, + H,O0 —» Mg(OH), +H,
Basic
‘The hydrides of Na and Mg form basic solution.
‘The hydride of sulphur — chlorine form acidic solutions due to presence of hydroxonium ion.
H,S+H,0 —* H,0°+S>
Acidic
HCl+#,0 —> H,O°+cr
Acidic
‘The hydrides of phosphorus ie. phosphine is amphoteric
(d) Group I:
Carbonates of group | are thermally stable due to the large size of the atoms hence it will be
capable of accommodating the carbonate ion.
Na,CO, — No reaction
AK,CO, -» No reaction
Group II elements are easily decomposed
Mgco, —» MgO +Co,
Caco, +=—* Ca0+C0;
3. (i) Chlorine isa strong oxidizing agent hence it will oxidize iron straight to Iron Il chloride
2Fe + 3Cl, —* 2FeCl;
a7

(i cr may react with Fe™ (titration product) which will form FeC!3 which is red in colour.
This will make change in colour of MnO, be difficult to account for (or difficult to notice)
2Fe + 3Cl, —> 2FeCl,
L
4, (a) Define the following
(i) Atomic radius
(ii) lonization energy
(b) Contrast the action of heat onthe following pairs of compounds
(Nan and ca (n°)2
(i) N93 C93 and pbcs
(©) Describe how you can distinguish chemically the following pairs of compounds
(i) Mg(0H)? and mig cS
ly
(inact and alc
(iy cu and cu (n23)2
2.5. (a) Explain the following by chemical equation. where necessary:
Solid carbonate of iron (iii) and Al have never been isolated
(i) Lime water is used to test the presence of Co2 gas
a8

(i) When sodium hydrogen carbonate is added to copper (i) sulphate effervescence is observed.
(i) Lead hydroxide ppt dissolves in excess sodium hydroxide solution,
(b) By using equation. Where necessary describe:
{Two methods of preparing copper (i) chloride in the lab
(i) Describe how 2n0 reacts with both acid & bases
(6 ist 2 important uses of Ca0 in daily life
6. (2) Define
(i) Heat of reaction
(i) Bond energy
(6) The enthalpy of formation of CC!+is —135.5 KImol and enthalpyies of atomization of graphite and
chloride are 715 and 121.2K) respectively. Calculate the C-CI bond energy
+) wen I H ; ,
(2 when 3g of c Hs and C*+are completely burnt in excess oxygen 32, 143 and 56 Ki of heat is
liberated. Calculate the standard enthalpy of formation of C+
(i) The heat of formation of Fe; 0; is ~824,0KIml?, What will DH of the reaction
2Fe,0; —> 4Fe,, +30.
us @ 2@
ANSWERS:
4. (a) (i) Atomic radius i half the internuclear distance between two similar atoms covalently bonded
bya single bond
(i Ionization energy is the energy required to remove the most loosely held election from the
shell ofan atom,

(b) (i) Naw5 is more stable than Ca (N°)? because sodium is highly electropositive hence bond
energy between sodium and nitration is high hence it comes very stable.
(i) N22¢3 is more stable than PCS since sodium i highly electropositive hence there is high
bond energy between sodium and
carbonate thus is becomes stable.
(6) ( Mgc> will release C72 upon heating which will turn lime water milky.
MgCO,—> MgO +CO, tum lime water milk
(i) aic!> isa Lewis acd, hence it can react witha base to give a salt and HCl. But NaCl cannot
react with a base
AICI; +NaOQH —* Na Al, +HC1
(ii) Cu(N22)2 will form a brown ring due to formation of N°? but Cus does not.
5. (a) (i) Fe C7 and AIC have never been isolated due to the small size of Fe and Al hence they have
high polarization power. Hence they will decompose as soon as they are formed
(i) Lime wateris used to test the presence of C2 since calcium carbonate is insoluble hence the it
becomes milky
(ii) The reaction between sodium bicarbonate and copper (i) suiphate, effervescence is observed
due to formation of carbon dioxide
2NaHCO,+ CuSO, —> CuCO, + Na,SO, + H,0+CO;

(iv) Lead hydroxide ppt dissolves in sodium hydroxide due to the formation of a soluble complex
compound
Pb(OH),+2NaOH —* Na;Pb0, + 2,0
(b) (i) 2" method is by reacting copper carbonate with HCl
CuCOs%q) + IHCLyy — —PCuClayy) CO) + #20)
2" method is by reacting Cu (OH)? with HCI
Cv OH)ageg) + ZHClygy §— —F—CClayag) + 240s
(i) Ca(NO,), can be converted to CaCl, by reacting with HCL
Ca (NOs) (09) THC,» —F CaClaa + 2HNO si)
(iii) ZnO is amphoteric hence it reacts with both acids and bases
+ — 105 + H,
BBQ) * ZNAOH a) Naz Zn02 + #20;
+ — +
F900" 7BCKaa) Zn Cly * Hy
(c) Uses of C20
= Manufacture of cement.
= Used as a drying agent for ammonium
= Lining of furnaces
at

=> Formation of slag in Blast furnace
a2

=> Making Ca©2 which interns is used in making methane
6. (a) (i) Heat of reaction is the enthalpy change when number of moles of reactants as represented by
‘balanced chemical equation change into products.
Bond energy is the amount of energy required to break 1mole of a bond

4H. OH, By, g4Ham

4 (C~Cl) =(-135.5) +715 +4(121.1)

C-Cl= 333.725 KJ mol*

(c) (i) Solution:

a= 2
n= 0.08 3mols
0.083 moles —» 32KJ
Imole —» x
X= 384 KJ mot of ¢
-2
ait, = 3
=0.5 moles
0.Smoles —» 143 KJ
mole, x
X= 286 KI mot! of H
-2
Bee = a
= 0.0625 motes
0.0625 moles —» 56 KI
Imole —» x
X= 896 KJ mot

Cit 2a) > CHugy ~ Required
Cyt Osc COxcg) AH = 384 KI mor!
Hagy* 20xq) > HzO SH=-286KI mot!

- 1
CHyg)+ Orc) > CO) +H:0jp AH = -896 KI mot

Applying Hess's law
Cerra > — COxc) 4H = - 384 KJ moot!
2H) +Oxcq) > 2H20 gy OH=-572KJ mot!
COzig+ HO) > Cys) + Ox) SH=+896 KI mot!
(ii) Solution:

3 mi
2Feq) + 302%) — Fez Os¢aq) SH= -824KI mol
2Fe20saq) AFC) + 30209)

AF yy +30: 2 Fe20xeg) 4H = +824
2Feyy+ 20s) —Fes0y0q) AH 4 = 412KT
2Fe:0x¢0q) AF) +30x%q) SH = 412K mol!
7. Define the following terms
(2) (i) Hydrogen spectrum

Series of arrangement of various wavelength of radiation formed when hydrogen atoms are exposed

under high voltage electric discharge in
a discharge tube.
(i) Convergence Limit
Is the point at which the distance from the nucleus isso large such that electron moving beyond this
point cannot go back to its ground state:
(ii) Degenerate orbitals:
- These are orbitals with equivalent energies
«They share the same azimuthal quantum number
(b) (i) Solution:
__ 276x06F
nn
aA
pes
= -2.176x10%(° 4)
Ae=1.934x10")
Fn, -®
(i) For shortest wavelength, B=, =n =
pF Fs
22176110" =D

4e22.176x10"
a8

hu
2.176 x 10%#==
he
A= 2.176x10-*8
6.63 x 10° 3 x 10
SS
2.176 x 10°
1 29.13x10%m
(c) Solution
n=21=1miz-1ms=- 2
wl 1 |
o> 0 V-shape
Sp hybridization
\VSEPR - Valency shell electron pair repulsion. We use Tines to indicate the bonds
8. (a) Dative bond
Ita type covalent bond in which one atom contributes both electrons

lr \ H .
was +H ——— H-N-H
i |
H
Van Der Waals force is the dipole-dipole interaction between two non polar atoms.
London forces are temporary dipole dipole interactions.
Experiment:
A: Isa solution made by dissolving 4.28g/ 500 cm? of Mi7? of distilled water
8: 1510" ki solution
: Isditute #257
D: Is 0.12M solution Na2 S: 05. SH: 0
S: 1s starch indicator solution
THEORY:
In acidic media ladate ions (1 25) reacts with excess lodine ions (/)to liberate lodine according
to the following reaction
- ——
105 oq +68 ges! > 320432 (a)
The iodine | liberate with a standard solution of thiosulphate ions according to the following
reaction equation.

A rd
u 1 te 1
Ge SHA
Homolytic bond cleavage.
Is the type of bond cleavage in which the covalent bond breaks symmetrically so that are electron
moves to each end.
Heterolytic bond cleavage.
Here the covalent bond breaks unsymmetrical and all bonding electrons are taken by the more
electronegative atom
NO@+NO@ = Org Fast
N202,,.92 » 2Nn9 stow
a= Ko (N202y (02) this sow step will determine rate law
¥r -¥5 iwoy two}
+ -twol
Ry Ky (N22)
At equilibrium Rr= Re
Ky toy? = Ko (N22)
x,
Es oy? = (4202)

Ey
reKo Ks nop 1%}
R= KINO} (72)
Bn@ ——> 2Br
Br + 42 ——> HBr+H@-slow
H@+Brg@ ——> HBr fast
‘The slow step will determine the rate law
R= Kote ey
Consider the equilibrium reaction
Bn —*> 2Br
Bratayy) —> 28r
Re_Ky pty
Ry Ky gq)?
At equation:
Ry Ry
yg) -%> ayy
Lf
(09 =[8r"

—
[Br] = Ky, {er}?
R= Kako? gei Ma
ex od (ay
xy
Kee K,
acoona > cco + nt
CH3C00" + HO —* CH3COOH + OH
start oa 0 0
oie ox x
{em coonston-}
Ky= [CH COO=T
Ol-x
a= 1 x109
[1 cOONal = ¥
Let added volume be x
4s7

.  Wazaelimu.com
{H coona] = >

=
oas
ne0.1x

=
[H.COONa] = 005+ x
ok

[HCOOH] = 054
14. Explain the meaning and significance of colloids
415, Discuss the properties of soil colloids

~ Surface area

~ Electric charge

~ Ton exchange (diagram)
16, Explain the mechanism of ion exchange in soil
17. Calculation of percentage base saturation of a soll sample. With worked examples (how to calculate)
18. Aluminum has high polarizing power hence the compound formed between Al and carbon will be
unstable therefore it will decompose immediately after formation.

[P +(NHs) el + Ag NOs ——+ AgCl + [P+ (NHs) el2] NOs
‘The chloride outside is not part of the complex meaning that itis ionization, Therefore Ag will reac. In [
INH): cL] the chloride is part ofthe complex hence there is no ionizable chloride ion and there no
reaction
[Pt (NH3)2 CL4] + AgNO3 ——> NO reaction

Gi) NH, OH +Cu* ——> Cu (OH);+NH™
4NH5 a + Cu (OH) —* [Cu (NES)] OFaap
(iv) AgCl + 2NH; ——+ [Ag (NEs):]
Soluble complex
sicher
19, (a){i) CHs—C-CHs + CH Mg CL CHa—C- CHa
(ii) CHs-C-C-CHs #NaOH +l, ——® CH + NaQOC-CONa + Nal
Kon
(ill) CHa CH C—NH> 2 CHa CH NH> + KBr +CO +420
‘NaOH
(iv) CH CH NHa+CHels CHa CH: N=C+HCL
20, Distinguish between
oO
0) ll
OH C-H
And
‘When Br:/#:0 is added to phenol, white ppt is formed
OH Br OH Br
\
(+ Br —> on
‘When 8r,/H:0 Is added, no reaction
°
CH
Cy + Brz2 HO ——-+ No reaction

(ii) CHs C- CHs and CH; CH2 C -H
By Iodoform test or by silver mirror test
(CH; CH CH; +h + NaOH ———® CH]; + CH; COONa + H20
(CH: CH2 CH; +h + NaOH ———* No reaction
(ii) CHs CH: OH and CHs CH2 CH2 OH
(CH; CH; OH + NaOH + h ——® CHI; + NaCooNa + H20
Yellow
(iv) CHs CH; OH +: + NaOH ——® Noreaction
22. With the aid of chemical equations explain the following
(i) Mercury(ii) iodide solution but not in potassium iodide solution but in water
(ii) The pink solution of cobalt (ii) chloride turns blue when conc. HCL is added
gelatinous ppf of
(ii) Copper (i) hydroxide turns deep blue in excess ammonia
(iv) ZnO and Pbo dissolves in hot conc sodium hydroxide solution
23. using relevant balanced chemical equation describe the process of extracting copper from copper
paanhhn onan tr
(i) Concentration
(i) Roasting
(ii) Removal of ion impurities
(iv) Self- reduction reaction
ast

a ‘What is meant by the term d- block element
b. Write E.C of Cu, Fe**, Mn**
(i) Explain in terms of E.C why Fe** ions are readily oxidized to Fe** ions but manganese
(ii) ions are not readily oxidized to Mn**
25, 2.5 x 10° moles of a compound with a formula was dissolved in 0.1M a silver nitrate solution. SO cm?
‘were required for complete precipitation of the chloride ions present.
(i) Deduce the ionic formula of the compound
(ii) Draw the structure of the complex ion present and name it
26. Write the formula of the following complexes
(i Tetra ammine copper(ii) Sulphate mono hydrate
(iPotassium
27. Write down a balanced chemical equation for the following
(i) Adding of excess ammonia solution to aluminium ion
(ii) Ion(iii) oxide is heated with aluminium power

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