GENERAL CHEMISTRY
Isa branch of chemistry which deals with behaviour and characteristics of lecrons
This subtopic deals with electron occupying space in an atom. The electron occurs in the orbit!
shell atthe region called orbital. The orbital i the region where there isa maximum probability of
locating electron. At the same time this topic deals with electron of atom which occurs in a
chemical compound. The general chemistry includes the following aspects or subtopics:-
+ Atomie structure
+ Atomic spectrum H ~ spectrum,
+ Wave mechanics.
= Chemical bonding
1.0 ATOMIC STRUCTURE,
Atomic structure deals with structure and component of an atom. The first scientist discovered that
‘matter is made up of small particles called atoms. The term atom means indivisible particles. But
later, different scientists put forward atomic models. These atomic models account for atomic
structure, There are several atomic models which include the following:~
+ Dalton’s atomic theory.
+ Thompson's atomic theory.
+ Bohr’ atomic model
+ Rutherford atomic model,
+ Wave particles duality nature of matter
+ Heisenberg uncertainity prineiple
DALTON’S ATOMIC THEORY
Dalton’s atomic theory includes the following main points
i. Matter is made up of small indivisible particles called atoms.
ji, Atom is neither created nor destroyed
il, Atoms ofthe same elements are similar especially in mass,
iv. Atoms of different elements are different especially in mass,
¥v. Atoms of different elements when combine they do so in small ratio whole numbers
RECENT MODIFICATION OF DALTON’S ATOMIC THEORY
Dalton’s atomic theory was modified because all points were not vali. This resulted into discover
‘of modern atomic theory. The following include point of moder atomic theory:
i. Matter is made up of small indivisible particles called atoms was not valid due to the
existence of three particles in atom. Matter is made up of a small divisible particle
called atoms.

[Particle [Nature of change [symbol [mass Position)

Proton +1 Pe 1.00. Nucleus
[ Neutron [get 09 Nucleus
a CS
i Atom is neither created nor destoyed was not valid due tthe exienc of radioactivity
therfore Atom canbe eet or denroyed by ether mew fsino:uson
ii Atoms of the same clement re simlr expel ih mae was Not vad ue 10 the
Chistence of Soop, Atoms ofthe sume ements have ihr same or ifrent mass.
iv. Atoms of clement are diferent especially in mass was not ald, Atoms of dret
‘lect have eer ceo eres ns
¥. on of diferent lemeats when consine they dos in anal to of whole ber was
ot valid becuse diffent laments combine by ting varie ai of whole number
HOMPSON's EXPERIMENT (DISCOVERY OF ELECTRONS)
‘Thompson's conduct an experimen! investigate ii conduts etiily. The
flowing out was wed drag te experieat
4 eh
+The emission tube have clsrode at cach cod which is connected othe extemal cc. The
cnisston ie is connected Whe vacuum panp in eee Wo mail te Tow prsure i he
‘The circuit is switched on which resus into the allowing observatio:-
+The bul mite light which indicate that he gs condut let.
+ Theres showing of emission tube or emission of ight
«Thetis Muresenc of emision tbe
«The stream of ay runing om eathoe tthe node. Trough investigation of properties of
cathode rays) using magete Held eet and gol cleetoscope etl.
Vind ves ino papeps en rth om

discover of electron, The cathode rays were the electrons.

EXPLANATION OF THOMPSON'S EXPERIMENT IN TERMS OF ATOMIC
STRUCTURE
Ground state: Is lowest energetic state of an atom. Isa state when an electron filled in the
lowest energy level before filling the highest energy level available. The electron filled in atom
in order of increasing energy level. Ths state make atom to be stable.
Excited state: [sa state of an atom when electron filled in the highest energy level before filling
lowest energy level available, Ifthe electron excited jumps tothe extent thatthe nuclear
tractive foree act upon it result pulling back of electrons. When return back to the ground state
release all amount of energy which was absorbed inform of radiation,
RADIATION: This energy causes glowing of emission tube. When the radiation strikes the
‘emission tube causes florescence of emission tube.
Convergent limit: Is a state of atom when an electron is removed completely from ground state
to the infinite, The convergent limit occurs if atom gain high energy which result electrons to
jump tothe highest energy level where the nuclear attractive force cannot act up on it. This,
electron cannot return back tothe ground state it result the atom let to positively charged. The
convergent limit isa factor which eauses some electrons to move from cathode to anode. ‘These
ere stream of rays called cathode rays which later was electrons,
SIGNIFICANCE OF CONVERGENT LIMIT.
‘These include the following:-
+ Itresulted into discover ofthe ionization energy. This ionization energy used in the
inorganic section
+ Itresulted into formation of ion particles. The ionic particle is more reactive when take part
during chemical reaction,
+I resulted into production of rays. These rays are known as atomic spectrum.
‘THOMSON MODEL OF THE ATOM
After the discovery of electrons and protons, the next question was to know how these particles
fare arranged in an atom, The frst simple model of the atom was proposed by J.Thomson in
1898, The Thomson atomic model is popularly known as the Thomson's "plum-pudding”
‘model ofthe atom,
Thomson considered an atom to be a sphere (radius = 10" m or 10* em) of uniform positive
charge into which the negatively charged electrons were embedded. This model 1s lke plum-
‘pudding dotted with raisins

a Aspere ofuntorm
oor charge
This model oF an atom could not explain many experimental facts. So, it was abandoned
RUTHERFORD'S SCATTERING EXPERIMENT
In 1911 Rutherford performed an experiment which is now known as Rutherford!’ scattering
experiment. In this experiment, he bombarded a thin sheet (0.00006 em thickness) of gold with
alpha a particles. The particles were obiained from a radioactive substance. The particles are
doubly ionized helium atoms (He).
"he sattered -partiles produced tiny ashes on striking with the ne sulphide sereen, These
tiny lashes were observed with a movable microscope. Uhe experimental set up used in the
famous a-scatering experiment is shown in figure below. The following observations were made
from the scattering experiment.
Structure of atom

ac
foun
< Tomi ot
—
Pemterconpemed
nara
ps
sit S. Nana
s, SorSiaretins
5) Sone
uteteumst
pes
The e-seatierng from metal fis The -pticles are produced by a radioactive source Since lead
Abootbe a -patls a lad pat with a oles used to obtain a beam ofa patles Thea
pails satered rom the mal sie the Muorescent zine sulphide) sren and produce
tiny ashes. A movable microscope i sed oviw the ashes
1 Mos ofthe apres pase trough the metal i without any change in ther path
Le, they remained undetected
ir "Some ofthe aprile defected trough small angles.
ii, Onlya few of them (1 in 10,000) were actualy dfeted by as mach as 9°, even
large angles, One in 20,00 prices resumed bak sufering a defection of 180"
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Explanation. The results ofthe scatering experiment could not be explained by the Thomson's
atomic. Caleulations showed that a charge spread over a sphere of radius 10° em could deflect

i. Greater than Es E>
ii, Equal to Es ~ E2
ii, Less than Ex — E>
iv. Greater than E)~ E but les than Es
Smaller than Es E>
Solution:
a) 7=2930A
fee
a
soxtetnys
(ohare
f= 102x107 St
Eeht
E=6.3x 10s x 1.02x 107 S*
E=6426x 10")
by Ex=-136x 10°
a= -15.44x 10°)
E=E: Ei
E-E: E:
E=(-1.36x 10°) (-5.44x 10)
E=4.08x 10"
at
=n
oe
sours“
pata
£=6.15x 10'S"
Hence:

Fee

net
s0eitm/e
errors
= 4.88 10"m
Ci) Aline of wave length 20304
ii) Aline of wave length 8092 A”
iii) 12
rE
Line of 2030
Since re
Boetem/e
fa 2o00+10-m
f= 148x 108"
E=ht
E=63x 10%Isx 148x 10%
s Es 0”
D) Solution
i) Greater than Fs — Fa result transition of electron ftom F) and above Es
ii) Equal Es Es transition take place from E> to Es
{ii) Smaller than Es ~ Ez result transition of electrons from E> but cannot reach instead hang
between E> and Es
iv) Greater than Es -E: but smaller than Es E result transition of electron from E> and above
Er but cannot reach Ex
‘THE QUANTUM THEORY
WAVE PARTICLE DUALITY NATURE OF MATTER
States that: "Matter has particle nature as well as wave nature". This means that matter have dual
properties or two properties such as particle nature and wave nature. The wave particle duality
nature of matter was put forward by De Broglie scientist. De Broglic's derived an expression which
applied to find the De Broglie's wave length,

De Broglie’s wavelength is in terms of mass and momentum,
Hence from Einstein equation;
a

(i) Frequency and wavelength

(i) Energy and frequency.
(ii) Energy and wavelength
(b) Orone (0) protects the earth’s inhabitants from the harmful effects whose wavelength is
2950 Caleulate
(i) Frequency.
(Gi) Energy
for this UV light
(6) Ifthe wavelength ofthe first line of the Balmer series in a hydrogen spectrum is 6563,
calculate the wavelength ofthe first line of Lyman series in the same spectrum. The uncertainty
of the momentum of particle is 3.3 x 10""“ams". Find the accuracy in whic its position can be
determined
Question 11
8) (i) A small object of mass 10g is thrown with velocity of 200 mis given h = 6.626 x 105.
Calculate its wavelength,
(Gi) Kinetic energy ofa sub — atomic particles is 5.65 x 107°, Caleulate the frequency ofthe

particle wave (h = 6.626 x 10

Question 14
(a) (i) What do you understand by dual nature of matter?
(Gi) How does de Broglie equation consider the dual nature of matter?
(b) (i) Calculate and compare the energies of two radiations, one with wavelength 800nm and
the other with 400am
(i) What isthe amount of radiant energy associated with atomic 6.662 x 10Is, velocity of
Tight 3x 10's
(6) the wavelength of a beam of light is 28 x 10°'m. Caloulate it's:
(@ Wavelength in em,
(i) Frequency.
(ii) Energy of one ofits photons,
(@) (i Referring o Boke’s atomic model, what's ionization energy?
(Gi) Calculate ionization energy of hydrogen. Reydberg’s constant RH = 1,097 x 10'm"
Downe oes an ps ppes om won natn

(©) (i) Calculate the frequency of the 3" line of visible spectrum.
(RH=1,097 x 10°m, C=3 x 10'm/s)
(Gi) Given thatthe wavelength of the * line of Balmer series is 6563A. Calculate the
wavelength ofthe 2 line in the spectrum,
INORGANIC CHEMISTRY
'sthe chemistry of all the elements and their compounds with the exception of most carbon compounds
‘ut of which only the oxdes,cyanides and carbonates are considered as inorganic compound. inorganic
‘compound can aso be defined a the study of the elements in the periodic table.
PERIODIC TABLE
Periodic table s the table of all the known elements arranged in order of increasing atomic numbers
“The arrangement reflects the electronics configuration ofthe elements
‘THE PERIODIC TABLE CLASSIFICATION OF THE ELEMENTS (MENDELEEV'S AND LOTHAR MEYER
1869)
The fist comprehensive classification of elements was made independently by Mendeleevs in Russia

and Lothar Meyer in German in 1869:They tabulated all the known elements on the basis of relative
atomic mass. The arrangement of elements inthe early periodic table was according to the ordinary

periodic law which tates that "The properties of elements are periodic function of thee relative atomic
When elements were arranged inorder of increasing atomic masses, elements with similar properties
recurred at regular intervalsThe recurrence or repettin of elements with similar properties of eguat
intervasin the periodic table i known as PERIODICITY.
'Mendelee’s and Lothar Meyer placed elements in horizontal rows (periods) which caused elements
with similar properties to appear in the same vertical column (group). Some element were not yet
tlscovered and hence absent that period tabe eg. Noble gases, alum germanium ete
(nthe bass of relative atomic mass three anomalies appeared the early periodic table (Ordinary
petodic table). The postion of Potassium (391), Argon (39), Cobalt (58.98), Nickel (586), Tellurium
(127.6) andiine (1269) hat be reversed to bring them nt correct placing on chemical grounds.
The str oder of relative atomic mass would have Separated the mentioned elements from closely
{elated element. For example Potassum could have been separated frm other aka metals. These
anomalies showed deat that the relative atomic mass was not realy the tue bas of arranging oF
Classifying elements inthe periodic table
‘The Uwe anomalies stated above wee de to ISOTOPE. For example bath Argo and Potassium exhibit
Isotrpy The principal lotopes of these elements are shawn inthe table below.
ELEMENTS | ISOTOPE | PROTONS| NEUTRONS | RELATIVE AT
| Istisotope | 18 18 36
2nd isotope | 18 22 40
Potassium [st isotope 20 39
2nd isotope 22 al
3rd isotope 21 40
R.A.M = (Mass of 1 x Abundance) + (Mass of 2“ x Abundance)
Inthe case of argon, the heavier topes predominatehas higher abundance) ging an
average atomic mass of 39In Potassium the lighter otope predominates giving an average atomic

elements example A IA, IVA, VA, VA and VIA). The nable gas contains full outermost electrons S
and P-sub shells Since ths configuration i very stable, the noble gases are uncrative.
However they form some compounds with strongly electronegative elements like Oxygen and Fluorine
ample of noble gases include Hellum, Neon, Argon, Krypton etc
3. d-BLOCK ELEMENTS,
The d-biocks consist of elements with partially or fulfilled sub shell. They fil their electrons inthe d-
sub shell ofthe penultimate shell. They use electrons from S and sub shells for bonding. xample
Scandium, Manganese ron, Radium, Nickel, Cobalt, Copper, Zinc ete
4. FLOCK ELEMENT
The F-block consists of elements which are called Lanthanides and Actnides. They are also known inner
transition elements. These elements have two partially fille sub-shells namely (n-}d and (0-2) They al
belong tl & because they ae so similar that itis very difficult to separate one from another example
eg. La, U,Np,Lw, Thetc
Nore;
1. The atoms of all elements of the same period have the same numberof shells which are partially
or fully occupied by electrons.
TL The number ofthe main groups is equal to the number of eleetrons in the outermost shell,
IL, The number of period is equal to the principal quantum number (n) of the outermost shell and to
the total number of electrons shell ofthe elements in a given period.
PERIODIC TRENDS IN PHYSICAL PROPERTIES
1. DOWN THE GROUP
1. L The atomic size(Atomic Radius)
The atomic radius ofan uncombined atom cannot be defined strictly because of the uncertain boundary
of electron clouds. The distance between the nuclei of chemically oF covalently combined atoms can be
measured accurately by xray ditfraction method
Therefore, the atomic radius is defined as half the dstance between the nucel of two similar/ identical
atoms joined bya single covalent or metalic bond. There is signficant regular increase in atomic radi
among elements down the group This trends due to increase of numberof electrons and number and
‘numberof shells down the group.
m2

The newly added electrons or shell must be at greater stance from the nucleus than that ofthe
proceeding element (ea noble gas). Aso the added electron i shielded bythe inner electrons
na

cleus has increased eg. Na=1.574", NA” =p.or4
‘rucleus and as the result the size of an anion increase example 1-099” and!” =1.214° the
ELEMENT [OmTCOREER ATOMIC RADIUS um) | TONTC RADIUS am) ] TON

1 33 0.133 0216 '
'NB: Ionic radius i half the distance between the nuclei of two ions in an ionic eystal
Il, Ionization energy
This is the energy required to remove an electron from a gaseous atom or ln. Hence the fist ionization
energy ofan element, Mis the energy required to remove the frst electron from it.
Mog) Mig + e-
The second ionization energy isthe energy required to remove the second electron from a gaseous lon
Mj) > MZ + e7
The successive ionization energes are defined accordingly The higher this energy the tighter the
electron s bound to the atom orion. The ionization energy decreases down the group because OR due
tothe increase in atomic sie ofthe element down the group, the increase in adi downs a group
correlate with the decreasing ionization energies.
“The nuclear charge increases down the group butt is cancelled by shielding effect and screening effect
ofthe electrons of the inner shells. The two effects increases down the grOUp. Since the nuclear charge
Isaflected by the increase of electrons, the atoms become larger down the group and therefore the
outer electrons) is easly removed as itis losely bound. Thus forthe alkali metals Lito Cs, the outer
‘mast electron is most really removed for Cs whichis the largest element ingroup A. Fluorine having
‘the smallest atomic size ingroup VIA has the highest ionization energy.
IV. Electron affinity (EA)
‘The electrons affinity isthe energy change which occur when an electron is added to a gaseous atom or
ton. The non-metalc element realy gain an electron to give a negative ion.
Xgy te" XG) 4H =Ky/mot
Te energy associated with this process i termed asthe electron afity. The process may ether absorb
lor evolve energy example endothermic and exothermic respectively). Some elements evolve large
{quantities of energy Le their electron afities are large and negative. The more negative the electron
affinity, the more the electron is attracted by the nucleus. Thus the fist electron affinity i the energy
change which involve the addition ofthe fist electron to a gaseous atom
us

Mate >My ‘Negative electron affinity (in| wry
a |
a a ci
a a |
a
Og) +e" 0%) 4H = -148kJmol”
0%) +e 70%, AH = *850k/mol-
IV, Electronegativity

—
[PeredSetement JF TP TP TE TFT? TE |

Pommr PEP PP EP |

erenceoviowaven TP PP |
a dl Ml
Be = 1572S?

B= 15%25%2pt

Be has two paired electrons in its 25 orbital. t needs energy frst to split 25 electron pair and secondly to
effect the electron removal. But for B there is only one electron inthe 2P orbital which Is easy to remove
asitismore distant from the nucleus
‘A comparable explanation accounts forthe break between N and O. Their electronic configurations are
asfollows;
N= 15%25"2p*
0 = 15%25"2P
For Nitrogen theres extra stabilty due to half-filedP-sub shell. n Oxygen the 2 sub evel has. a pair of
electrons which shields the “7 and 7 electrons rendering them easier to remove, hence an
abnormally lower ionization energy for Onygen. We may take almost similar account fr the elements in
Period 3 where there's break between Mg and Aland also between P and S.
Mg = 15%25%2p835?
Al = 15%25*2p*35?3pt
P = 15%2572P35737?
$= 15*2S*2p*3s*3pt
FACTORS AFFECTING IONIZATION ENERGY
1). Enfective nuclear energy; The greater the effective nuclear charge the greater the ionization
energy
2). Shielding Effect and Sereening Effect; The greater the shielding and screening effect, the less
the ionization energy
3) Radius ; The greater the distance between the nucleus and the outer electrons of an atom, the
less the ionization energy
4) Sub level; An electron from full or half filled sub level requires additional energy to be
removed
tm, 1ONIc RADH
a2

tonic radi ke atomic vary perioicaly with atomic number. The anions are much lager than the
corresponding atoms while cations ae usually much smaller. The explanation of the trends in atomic
ai applies also tothe very similar trends in tonic radi
Example Na” = 0.095, g?* = 0.065,AI"* = 0.050, P* = 0.212,5** = 0.184,cI” = 0.181
In period 4 the bivalent cation from Tito 2* show a d-blocks contraction similar to but larger than the
corresponding contraction for atomic radius. The same i true ofthe Lanthanide contraction fo the
trivalent cations ta" to Luin period 6
IV. ELECTRONIC AFFINITY
Electron affinity along the period can be dscussed in terms of non-metal and metas. tis generally
‘observed that non-metal have higher values of electrons affinities than metals Non-metals have higher
values because they can realy gain an electron to give negatively charged ion. The process involving
addition of electrons to neutral atom is exothermic because energy must be lost in order that the
formed ions stable.
Generally there i increase of electrons affinity across the group due tothe increase of electronegatvty
which i result of decrease in atomic radi ofthe respective elements. Als increase of effective nuclear
charge increases electron affinity
V. ELECTRONEGATIVITY
Electronegatvty ofthe main group elements increase across each period. Tiss due to the increase in
affective nuclear charge as wel as decrease in atomic size. Therefore the halogens atom in every period
has the highest electronegativity value than the rest members. Alkali are the least electronegative
elements in each period
VL MELTING POINT.
‘The melting point of an element i the measure ofthe amount of energy (heat) which must be supplied
tobreakdown the reguar arrangement of atoms or molecule in crystal
It'sa temperature at which a substance changes from solid into liquid. There are diferent types of force
hich holds atoms and molecules together, example due to such variation in forces the melting point of
the elements in a period do not change uniformity e.g Period 3 elements. Melting point increase sharply
from Na" to Mg, Sodium atom to contribute only one electron tothe metallic crystals, but Mg
contributes two electrons. This accounts for the lower Melting point of Sodium compared to
Magnesium.
Allhas three electrons inthe outer valence shell but contributes only two of them to the “electrons
a2

sea". Thsis why ithas meting point close to that of Magnesium. The thi electron is held fimly tothe
eaten that isnot contribute to the “sea electrons”

Siicon is non-metal with some metalic properties tik luster, electric conductivity and ability to form
alloy with metals. Silicon has the highest Melting point in period 3 due tots “giant covalent structure”
The slcon giant structure is comparable to that of Diamond
Phosphorus and Sulphur have relatively low Melting pont because their molecules are held together by
weak Van Der Waa’ forces. Sulphur has higher Melting point than Phosphorus due to the differences
insizes oftheir molecuesie ®s 2°24 Ss
Chlorine diatomic ands a gas at room temperature. ts melting points very low due toa very low Van
Der Wal forces holding the molecules.
PERIODIC TRENDS IN CHEMICAL PROPERTIES ACROSS PERIOD THREE (Na to Ar)
‘Chemical properties along the period depend on the change ofthe elements from strongly metallic to
non-metal Sodium and Magnesium are strongly metalic while Aluminum is weakly metalic element.
Slicon, Phosphorus, Sulphur and Chlorine ae non-metalic elements. Thus metal properties decrease
acros the period from left side ofthe periodic table tothe right
a) HYDRIDES
“The hydrides of period 3 elements include NaH₂ Mg, Aly, SM PH₂ HS and HCI. Sodium hydride is
strongly ionic while Magnes hydride sage ionic but the bonds re covalent. Aluminium yeti is
covalent
REACTION OF HYDRIDES WITH WATER
Sodium. Magnesium, and Aluminum yields hydrogen gas and metal hydrides, they are basic in nature
because they react with water to form base
NaH) + Hz0q) —> NaOH aq) + Ho)
Malay + HzO) — MgOH)ae) + Hag)
Siicon hydride (sane) evolve hydrogen with water in alkaline medium (catalyses the reaction)
. eer
Sill) +Hs0(y +20Hjaq) —* SiOF- (44) + 442.9)

Phosphine (Pfs non-polar covalent compound and hence doesnot reac wth water. Phosphine isa
non-polar covalent compound due to small sitferenceneletronegativty values between Hydrogen
and Phosphons 1 = 22and P = 21)
“Te hydrides of Sulphur (HS) and Chorin (HC!) ar polar covalent compounds. They hydrolyze in water
to form acid
HS, + HO @ Hy 0faq) + HSiza)
HCl + HzO) H503s9) + Cling)
Therefore the hyaides ofthe strongly metalic elements tend to form alkaline solution with water while
those of the non metas form acide solitons.
b) CHLORIDES
‘The Chiories of period 3 elements include NaI, MgCl₂ AICh Sil, Pls PCh, SiC, The Sodium and
Magnesium chlorides are ionic saltsThe rest ofthe chlorides are covalent ia nature. As metalic
character decreases along the period ionic character of the chiovdes decreases.
REACTION OF CHLORIDES WITH WATER
When the ionic chloride is added to water, there isan immediate traction of polar water molecules for
ions in the chiovides. The sold chloride ether dsolves to form free ions
exampte N@ 2a 94 Cline) «react form new substances.
“The hydrolysis of ehordes varies across the period Sodium chore is nat hyolyzd in water
probability due toa large siz of Na ion leading to low poaritng power fr water molecules. The
Chlorides of the rest elements have covalent characters. As metalic character decrease along the period,
Toni character ofthe chides decreases as wel
The extent of hydrolss of chlorides varies across the period. Magnesium chloride isnot hydrolyzed but
its hydrate crystals undergo hydrolysis when heated to give hydrogen chiride and abasic Magnesium
chlorides.
MgCl₂.6H₂O,) > Mg(OH)Cl;,, + HCl) + SHO,
‘Aluminum chloride is easy hydrolyzed by water to produce an acidic solution. A in is very smal in
size and itis highly charged. Thus its polarizing power is igh, due tots high polarizing power it forms

hydrated ions (A(H:0})". The AP* in strongly plarzes the O-H bond ofthe water molecules in the
[AKH:0}.P* fon tothe extent that the bonds break to release the hydrogen protons. The solvent water

molecules abstract proton rom polarized water moleciles to frm the acc hyroxoniim ion Hi,
The HO" is formed 8 flows
Alc) +6830 ¢9 — (AIH₂ ) eI) + $C)
[AIGO) eT) + H2Oco * [ALHO) SOM + Hahn
(AICO), Oy + H₂Oc0 # (Al(H₂O)«(OH) Tag) + Hikes
[AUCHO) 0H] + #2010 * [ALCHO)a(OHalin + HaOtae)
[AUG.O), (OF) s]eg + #200 * AUHO)s 342049 + Hoe)
The chloride of Silicon Phosphorus and Sulphur hydrolyses completely in water to form acidic solution or
sold
SiClyg + 2Hz0g > SiOrg) + 4HClee)
PClsq) + 3H₂Oq) — H3POxqq) + 3HCliag)
PClggy + 41,0(9 > HyPOgag) + SHClag)
SaClaqy + H₂Oqy — 2HChagy + H2Sg) + H2503;09)
Note: Asmetalie character decreases along the prod onic nature of the horide decrease while the
entent of hyo increases
¢) THE HYOROXIDES
“The hydroxides in period 3 include NaOH₂ Mg(OH},A\#)»"sio(oH),(OH)*” PO(OH },SOOM), and
O,(OHLCIOM). The hydroxides fom sion to chlorine arena true hyéroxide but they are on acids
Sexium and Magnsiom hysroxdes are thet hytong. They have strong tendency of releasing the
(OF roup. Aluminum hydroxide is Amphatere.
‘Te aii character of non-metal hydroxides is det the tendency of leasing or donating Hon
ra

when dissolved in water. Silcon, Phosphorus, Sulphur and Chlorine ae electronegative enough to

withdraw electron by Inductive effec from the OH bond ths, aclitating the release of hydrogen as 2
proton i.
The act ofthe hydronde of, and lincreass with the increase in the electro negativty of
‘Al(OH);
respective elements Solubity decrease rom NaOH to 4!(OFs que to decrease in metalic character
orion character of the hydroxides across the period
4) oxioes
Sostium and Magnesium are strongly metalic and the oxides are on. Aluminum oxides onc bu nt
bases these sodium and magnesium
The oxides of sodium and magnesium are strong bases while aluminum oxides are amphoteric. The
cides ofthe remaining elements are all acid
‘ONIDES Ao, —] 5:0, PO, [50,0
7.0 |50, | cl,0,
REACTION OF OXIOES WITH WATER
Solubility ofthe oxdes of period 3 elements decreases along the period as metallic properties
decreases. The oxides of sodium and magnesiom frm hydroxide with water or steam because of the
protective fm of onde. The oxides of phosphorus sulphur and chlorine reacts with water to form ace
solution. Sica (SiO) does not react with water butt’ aciaie
eq N20 + HaOq) — 2NAOH aq)
MgQq + H₂Oq — Mg(OH),
PiOgs) + 620) —> 4HsPO3(aq)
S05¢q) + H₂Ou) — H2S0s(a)
Cl₂05%q) + H0(q) > 2HCO (aq)

DIAGONAL RELATIONSHIP BETWEEN THE ELEMENTS
The fist element in every group isthe smallest and has the highest electronegativity compared to the
rest group member asa result the fist elements have properties which differ from the rest group
‘member but similar to those tthe next lower elements diagonally. The kindof relationship in which the
clements whic are diagonally located in te perio’ table have similar properties i known as diagonal
relationship.
Diagonal relationship may aso be explained in terms of polriting power ofthe diagonal elements The
polarizing power of an element is the abilty of postive fon to polarize the negative ion When positive
and negative ions approach each other their shape are distorted. The extent by which the fons able to
"undergo distortion i called polarizability. The effec is polarization is as follows;
'NB; Polarization ithe lstortion or deformation of an electron coud ofan anion by a cation
If polarizations quite small the ionic bond results and its high the electronsin the anion are drawn
towards the cation tothe extent that a covalent bond is formed The polarizability and polarving power
cof an ion is affected by;
i, Thesizeof the ion
i, The charge on the ion
Thus polarizing power i igh fora smallion which has high effective nuclear charge. On moving across
period from lef to right, onic radi decreases and effective nuclear charge increases. On descending 2
‘group ionic radi increases while the effective nuclear charge decreases. Hence the diagonal elements
have similar polarising power. Due to these elements have similar properties. The diagonal relationship
can be represented as follows
Element uo 8 CN OF
Electronegatvty 10 15 20 25 30 35 40
Element NoooMg A Si P oS a
Electronegativty 09 12 «15 18 21 25 30

The relationships most significant inthe following pars Land Me, Be and A, Beand ite
a3

LUTHIUM AND MAGNESIUM.
Lthium resembles magnesium and fers frm the other akall metals s follows
7 Both Lind Mg have small atomic and ionic | Have relatively large atomic and onic radii
radii
fi, Formionic nitrides LIN and Mg,N, when | No reaction with nitrogen gas
{ik “Form monoxide (normal oxides) on burning in | Form peroxide example Na;0, or superoxide e
sir example Li,0 and MgO a
We. Bicarbonate are known only ina solution (not _ | Solid Bicarbonates can be made (stable)
stable in a solid)
'. Hydroxides, carbonates and nitrates decompose | No similar decomposition
‘on eating into oxides example A
- aon 5 sublimes
2L10H 5 11,0 + H₂O
«ov! Na;C0,—+ No decomposition
eos 4 ort
‘Ma(OH).— MgO +H₂O (Reason; They are more stable).
yen, Soin tk) Nitrates decompose to form nitrites and oxyger
Nitrates decompose to oxides, nitrogen dioxide and | 2WaNO, + 2NaNO, +0;
oxygen
4 2KNO, + 2KNO, + 0;
ALINO, > 2L1,0 + CO₂ "
| Capen ‘are more stable)
Vi. Phosphates, carbonates, fluorides and, 70" ~ | Corresponding compound much more soluble
hydroxides only slightly soluble in water water
Vii, Halides(except fluorides)are soluble in ‘Corresponding compound are much less soluble
‘organic solvent
‘ili. Compound have covalent character example | Compounds predominantly ionic NaCl
LiCl₂ Mech i
[Eaomeie [Ra

BERYLIUM AND ALUMINIUM

ca ie
cnet
2) eas ect wih cencetaton laces ala a Toes Hy | No waa
04 hyde complense
re ‘
Bey + 206! P2H₂O + BeCOW)Fza) + Macy) quo oi?
Me songs ayag Tamera egg 4
5) Oxides and prologue SS nd | One and Trion ae Wane
bases) ‘eg MgO, Ca(OH), Ca0 ate
Aveta
Alyy) + 2NQOH ug) +3409 # 2NGAU(OH) ag) vosina'®
Be0,+ 2NeOW a) + H2Oq) + N0zBe(OM aug)
Asbuses
eo + 2Hc1—+Becl.+H.0
M0, + 3H1C1— ale, +310
1 cores are ei hyayzed nd exist a dimers (Bec, and | NOTHIN aos example ACT,
‘AlCl the vapour state “The hydrated chloride of magnesium hydoly:
<< a ? Hae) + HCl
ra Mac 8,04) MaLOH ng) + HCl +
Dimer of ei conte
e x a
ui
a a~ _ a
Dias of Amino crite “eachenac
|5) Form fluoro-complenes e.g [BeF,]*~ area

QUESTION
‘outline factors that enable elements to have diagonal similarities
AA. similar eletronegativity
B. similar atomie and ionie sizes.
C have ions with similar polarization power
ANOMALOUS BEHAVIOUR OF THE FIRST ELEMENT IN A GROUP OF THE PERIODIC TABLE
‘The frst element in every group ofthe period show some properties which are not shown to other
elements in the respective group The element said to show anomalous behavior when its properties
itfer with those ofthe rest group member. The anomalous behavior ofthe ist element in a group is
due to the folowing factors,
|A. The first element in a group has the smallest atomic and ionic size when compared withthe
rest group members
B. The first clement in a group has the highest ionization energy
C. ‘The first element in a group has the highest eletronegativity
'D. The frst clement ina group has the higher electronic affinity
ANOMALOUS BEHAVIOUR OF UTHIUM
1. Lithium forms covalent compounds while other alkali metals form
onic compound example 1iClis a covalence compound while NaCl is ionic
1 Lithium reacts with nitrogen gas on heating to form ionic nitride while other alkali
metals do not react
6Ligg + Nagy) > LEN)
IL Lithium reacts slowly with cold water while other alkali metals react vigorously
IV. Lithium forms hydrated chloride while the rest group member form anhydrous
chloride example LiCL2H:0 and NaCl (Li can polarize water due to high polarizing
power)
'V. When burnt in air lithium gives the monoxide while other alkali metas form
peroxide and super oxide

Lig) + Oxy) = LE On
Lithium monoxide
Nagy + O3(g) > Naz0,u
Soaun pee
Keay + Org) > K20e)
VE Lithium does ot farm acti with eye acetylene) while ote metals
farm acl wit tyne
HC me~ yy + Hayy NaC” mC“Nat + yy
N= €#C~ My) +L 5 No reaction
VIL Lithium is only alkali metal whose salts may undergo hydrolysis
Lely) +4049 + LOM + HCl
NaC +H0qy— No reaction
Vill. he hydroxides of ihm decomposes on Retin to monoxide and water wile
svar afeoralkal meats tbe unscoraponed
Lory Usp + #0
NeOtyy i subtme undecomposed
1X. Lithium nitrate decomposes on heating in Lithium monoxide, nivogen dioxide and
oxygen whic henuiceafeterakalimeas decompose Init an
—_
LINO y 4) 5 Ltz0(q) + 4NOxig) + Org)

NaNOyiy > NaNO x) + Ox
KNO0y) > KNO x9 + Orc
X. Lithium carbonate decomposes gently heating into monoxide and carbon
dioxide while the carbonates of othe alkali metals are stable They decompose at
higher temperature
eg, HiC Px) 9 L0%y + CO)
NasCOxn sige NAO + COrc)
XL Lithium hydroxide is less soluble in water and hence « much weak base than
sodium hydroxide a and potassium hydroxide
XIl, Lithium chloride is dliquescent while chloride at he rest group members arent
detiquescent
XII, Lithium sulphate do not form alums while the sulphate ofthe rst group form
alums
ANOMALOUS BEHAVIOUR OF BERVLIUM
1. Beryllium react with concentrated solution of eal to form hydroxo-complexes and
hydrogen gas while other alkaline eath metal do not react
tog, Bey + NaOH ag +240) + NABECOH yap + Hayy
Cai) + NAOH aH.) + No reaction
2. The oxides and hydroxides of beryllium are amphoteric while those of other alkane earth
imetals are base
3. Beryllium chlorides hydrolyze in water while the chlorides of the rest group members do not
BeCls + Hs0 + Be(OH)>+2HCI (ydroysis)
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MgCl₂ + H₂O — No reaction
4. Beryllium chloride dimerize in vapour state while the chloride of the rest group member do
not dimerize.
a. a
a * > *
ot cl
Beryllium chloride Dimer
5. Beryllium form fuoro-complexes while other alkali earth metals do not
6. The chloride of beryllium readily dissolved in organic solvents while the chlorides of other
alkali earth metals do not readily dissolved in organic solvents,
7. Beryllium do not react with water or steam while other group members can react with either
cold or boiling water or steam.
8. Beryllium oxide do not react with water while the oxide of the rest group members react
‘with water to form hydroxide,
CaO + H₂O — Ca(OH) x09)
+, Beryllium do not react with either dilute or concentrated nitie acid “#¥ 0S) white
the rest group member react with both dilute and concentrated "NOs
10, Beryllium carbide hydrotyses in water to form methane while the carbides of
other group members give ethyne
BesCiy + H₂Oq) > Bey +CHyg) _(C* Oxidation state,
CaCip)+2H:O pag > CalOH) + H-C=C-H ((C=C)* Oxidation state)

11, Berylium chloride fumes in moist air while chloride of the rest group members do not.
BeClaeg + Haq) > BeCOH) acy + HCl)
(From air)
The fumes are due to hydrolysis of 8¢C1> in water where MC! is given out
ANOMALOUS BEHAVIOUR OF FLUORINE
Uke other Fist elements in every group fluorine exhibit some properties which cffer withthe rst group
members
“The anomalous behavior of fuorine is due to;
4) Its smaller atomic and ionic size
ii) Absence of d-orbital
il) Is highest clectronegativity value
i) Itshigher electron affinity
“The Anomalous behaviors of flusrine area follows
|. Fluorine is monovalent while other halogens show covalencics of 3 and 5. Chlorine and
Iodine also show a valence of 7. Fluorine is a monovalent because it has no orbital while
other members have d-orbital
2. The elements inthe periodic table show their highest oxidation states when combined with
‘uorine, example "and SF. This is due to highest electroncgativity value of fluorine
3. Fluorine forms hydrogen fluoride molecules which are strongly hydrogen bonded
Te hydrides ofthe rest halogens do not form hydrogen bonds between their molecules
H-F__----—H- Fo H-F

4. The solubility of aries olen difer markedly from the solubilities of other halides of the
same metal, example the allaline earth metal halides are very soluble in water exept the
Tiondes which are nsluble On the ther hand siver ord sth only soluble stve aie
The Murids of ahalne car meal have higher late energies than hydration energies
5, Metals show teirhighest degree of onic character when combined wth fuorne example
AIF, and 5% ar onc but AI and 5 are covalent compound
6, Hydrogen Moore form aii salts contaning the bifluride om (HF), The lide
of est halogens form
pormal salts only, example N@#F; (Actdte salt), NaCLNaBY.KT oy.
7, Fuorne is the ony halogen more cletronegaivity than oxygen and it oen behaves
dierent to ther balogens in reaction wih oxygen cooaining compounds example water F
Gisplaces O fom water unlike oes). HuorineHberaes oxygen fom water whl other
halozens donot
2,0 + 2F, 74H? + 4F- +0,
or L042, AHP + 0,
0 + Cl₂ H+ C1" + HoCt
oe AO#Cls + Clay + HOC
8. Fluorine evolve oxygen from hot concentted alkalis while other halogens form
chloride and chlorate (v)
B+ 40Nag SAP +0,4285,0
Cly-+ 60H age) > Cling) + ClO5ing) + 3H₂O(ae)
(Ctiorate ion)
Clg + Naty NaC NeCtO, +31,0
With cold ite alalisforne given afore and diuorine monoxide while chlorine foam
chloride and hypochlorite,
Fy $20Hiy) 7 2Fzg) + HO +H₂O
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(Dittuorine monoxide)
Clz+ 20H;jxg) > Cling) + C10~ + 420
(Hypochlorite ion)
Clz+ NaOH — NaChigg) + NACIOgq) + HzO
9. Fluorine combines directly with carbon while other halogens have no effect on it
SELECTED COMPOUNDS OF METALS.
(Le COMPOUNDS OF Na, Mg, Ca, Al, Fe, Zn, Cu AND Pb)
METAL OXIDES
Definition; An oxide is binary compound made up of oxygen and other elements, example MgO,
PbO, ALOs, 0x, NO», S02 ete
‘Therefore metal oxides are binary compounds made up of oxygen and metal, example PbO, FeO,
sete
NOTE: The binary oxygen — fluorine compounds are not called oxides of fluorine, but are
called fluorides of oxygen since Muorine is more electronegative than oxygen (example OF
oxygen dilluorine)
GENERAL METHODS IN PREPARATION OF METAL OXIDES.
“There are two methods of preparing metal oxides
) DIRECT METHOD
by INDIRECT METHOD
A. DIRECT METHOD OF PREPARATION OF METAL OXIDES
In this method a metal reacts with a reagent or oxygen ora to give metallic oxide
1) A metal oxide may be prepared by burning a metal in air or oxygen
2M gp + Og) > 2MIO%
$Wa¢9 + Oxy, > 200.)
4K + Ong) 7 2420)

11) Ametal oxide may be prepared by passing steam on red heat metal
3Feqy) + 40g) > Fes Oxy) + 4Haig0
1M} A meal oxide may be prepared by reacting. a metal with an oxidizing agent ike HNOs
Sm) + Cone HNO yg) —* SNOxy + 4NO x) + 2200
B. INDIRECT METHOD OF PREPARATION OF METAL OXIDES,
Tn this method, the metal oxide is obtained by eating carbonates, hydroxides and nitrates ete
Example;
CaCOyey * CaO + C09)
CulOH) acy > CUO + #20
2P8(NO,)ai > 2PO0 5 + 4NO zr) + Op
ALz(SO4)s¢4) > Al2Ox¢a) + 3505¢9)
F050, THO) > FeO¢n + 50x) + THO
MgCOyey > May + COxe)
TYPES OF METALLIC OXIDE
The metal oxides may be classified as flows:
1) BASIC OXIDE
2) ACIDIC OXIDES
3) AMPHOTERIC OXIDES,
4) PEROXIDES

5) SUPEROXIDE
(6) MIXED OXIDES
1. BASIC OMDES
‘These are oxides which react wih acids to form salt and water nly. They also combine with
acidic oxides to form salts. Basic oxides may be tonic or covalent
CuO (a) + HzSOy(aq) > CUSO gag) + Hz)
FeO (4) + HzS0uaq) — FeSOyagy + H2%)
Ma0 y+ HCligg > MoClag) + #200
Ca ¢y + S10) + CASIO 5)
Naj0) + COxgy —* NasCOyy
PbO + 504) PbS gy
Al,Oyy) + 6HClag) —* 2AICly(ag) + 340)
2. ACIDIC OXIDE
‘Acidic oxide are formed by metas in ther higher oxidation sates,
sample CFOs MBO, S80, FeO, ete
These oxides are generally covalent in nature, They dissolve in wate to form oxy-acids and
hence are called acid anhydrides.
‘Sn, +10 —+ HySn0,,(Stannic acid)
Cr0,+ HO —+ Cr0, (Chromic acid)
‘Mn,0, + H₂O —+ 2HMnO, (Permanganic acid)
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3. AMPHOTERIC OXIDE
“These are oxides with both basic and acidie properties, They react with both acids and bases,
Amphoteric oxides include Zn0, AlOs, BeO, PbO, SnO:,CrOs,
Fe,0,,Mn0, ete
{) AS BASES: They react with acids to form salt and water only
PbO») + 2HCligg) > PBClayaq) + H₂O(0)
AL;Oyi4) + 6HClcgq) > AlCl 99) + 34200)
Cr, Oa¢q) + HzSOgaq) — C¥e(SO) (aq) + 342%
ZnO, + HzSOxaq) — 20S Oya + #209
fi) AS ACIDIC OXIDES: ‘These react with bases to form salt and water
‘S10 zi) + 2NGDH aq) — NazSnO gag) + H₂O¢0
Sodium metastannate
PbO jp) + 2NGOH; gq) > NagPBO xing) + HO)
ZnO) + 2NGOH₂gq) + NazZnO sug) + H2 0)
Al;O4iq) + 2NGOH aq) > 2NGAIO 3.05) + HO)
Fe:0y4) + 2NGOH gq) + 2NGFeOsiug) + H₂O()
‘Sodium ferrite
4, PEROXIDES
Peroxides are compounds containing the peroxide fon Na:Oz Peroxides of alkali metals and
alkaline earth metals can be prepared,
8) By heating the metal in the presence of excess oxygen or ait
2Na+ Oscescen) > Na;

'b) By heating the monoxides ofthe metal alone or in the presence of oxygen/air.
2Na,0 > Na,0; + 2Naig)
2Ba0 + 0; ~~» 2Ba0;
©) By the action of oxygen or air on the metal dissolved in liquid ammonia, This method is for
the preparation of K202, RbaOe and C302
44) By the action of H:O: on metallic salt solution in the presence ofan alkali.
BaCl₂ + H₂O; + 2KOH + BaO, + KCI + 2H₂O
PROPERTIES OF PEROXIDES
‘+ Stability of peroxides increases with increasing ofthe electropositive character ofthe metal
‘+ Peroxides are more stable in ry state than when they are in solution form
‘+ Many peroxides are hight hydrated due to hydrogen bonding,
‘example Ns,0; 8,0, C30: 8H:0, 88052440 ete
‘+ They alssolved in water to form alkaline solution and hydrogen peroxide
Na,0, +2H₂O —+ 2NaOH + H₂O,
(ce-cold water)
‘+ When treated with diste mineral acids peroxides give H:0:
NazOz¢9) + 2HClaq) > 2NAC lag) + H₂O2(00)
BaQ4y) + H2S0 aq) — BASO ys) + H₂Oz(0g)
‘+ Peroxide give 0; on heating and hence act as oxidizing agents
Na,0, +2Cr(OH); + 2Na,CrO, + 2NaOH + 2H₂O
NB: PbO» is not peroxide since this oxides does not give water when treated with mineral
acids

5, suPeRonoes
These oxides of metas containing the on the know superoxide ae these of potassium
(KOs, Riu (RHO: and Cacia (3:03
Preparation
M+0,5M0, (N= K,Rbor C3)
Sproides of KR, and Can be rear by burning te metal in exces nye Fir
PROPERTIES OF SUPEROXIDES,
+ They are ylow sos
+ Thestailty of these superoxide inthe order KO: 01
Theyaestongonding opens
{There hyrted by water forming iO and nen
Ko, +0 —+2K0H + #0, + 0,
oily
205 + 28,0 —+ 2017 + 1,0, +0,
Sere ae paramagetc laste otha presen one peed earn nn
6. MixeD ones
These re oxides composed of two simple oxides. The wo simple oxides may be ofthe same
Intl or diferent metal ndifeem onan sates, example Red ead (PhyO.) is combination
OF 20 and: Duetothis can be writen oily
‘Also magnate (F209 isa combination of FeO and FeO
The mined ones with dierent metal ar allows
Magnesium ferrite (MgFexO4) MgOFe";03
‘ZnFexOs nOFex**Os (Zine ferrite)
Prorenries
148d rout dot soa om acai conn

+ Miblatetaes cheney atewne ser cet ofa mono h0) sdead
asta io far tong mensch cto on wamey poe
2PbO.PbO,,, + 4HNOx <q) — 2PD(NO3) (09) + PbOx,) + 2H₂On,
USES; Reeds ed pigment nips
FRRNOUS-FEIDO ONDE Fex0d FeO.
‘FexOs occurs naturally as Magnetite. It may be prepared by heating iron with oxygen or steam
Pabatigiag 5 Peps 6
rnorenies
~The compounds ack incl
-Var sol faromapcti
the compounds inactive chemically
- Reo with acids sdb oxen min of eos and feiss in sluton
Fe30y5) + 6H ing) — Fefzq) + 2Fe fey) + 44200,
Tes ar compounds of mth which cons hyo fons (OH he
cat ncaincly rgd an cample NaOH₂ MgO ZAHM POH
PREPARATION OF METAL HYDROXIDES
Thee re two mato of eparion of metal yids
A) DIRECT METHOD OF PREPARATION OF METAL HYDROXIDES
The ysis which cane prepare by his math ae hse compose of stony
evoostne meals sample OH₂ NaOH and CaOtts

Example
Nag + H30(y + NAOH ae) + 1/> Haig)
Cag + 2Hz0cn —* Ca(OH) x09) + Haig)
2B) INDIRECT METHOD OF PREPARATION OF METAL HYDROXIDES
In this method the metal hydroxide is prepared by
4) The action of water onthe metal oxide;
example
Nay + H₂O(y + NaOH aq) + 1/2 Hag)
a) +230 (9 > Ca(OH) 0g) + Hac
Ba0(.) + Hz0() > Ba(OH)x¢oq)
[Nis the metal hydroxides which are prepared by action of water on metal oxides are soluble in
water
+) Action of calcium hydroxides (milk of lie) on a solution of carbonate, example preparation
Of NaOH and KOH. These metal hydroxides are prepared by precipitation the unwanted ions and
layering a solution ofthe ala, For instance when poessium carbonates and (Ca(OH)
solution are mixed andthe allowed to ete a solution of potasium hydroxides. may
be decanted
Ca(OH) aq) + K2COs(04) 4 CACO sg —e) + 2KOH a4)
©) Precipitation of a metal hydroxide by adding ammonia solution or sodium hydroxide
Solution toa solution of sal ofthe metal
AICL, + NaO Hag) + AI(OH) + 2a ag)
AlClyag) + 3NH OH gq) + AICOH) yy) +3NH₂ Clog)
FeSO qiaq) + NaOH gq) —> Fe(OH) 4) + N43S0 04)
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CuClaigg) + 2NHjOH gq) > Cu(OH) arp) + 2NHCleag)
‘Pb(NOs) zag) + 2NAOH oq) — Pb(OH)ais) + 2NANOsag)
4) Electrolysis of a solution ofthe metal chloride, example preparation of NaOH. Alkali metal
chlorides form conducting solutions and since these metals are highly in the
electrochemical series, their ions remain in solutions during electrolysis and
hydrogen evolved at the cathode, Preferential discharge of chlorides ions enable hydroxyl ions
formed by ionization ofthe water to accumulates inthe solution. As a result dilute solution of |
the metal hydroxide is produced
PROPERTIES OF METAL HYDROXIDES OF THE SELECTED METALS
1, ALKALI METAL HYDROXIDES (MOH)
PHYSICAL PROPERTIES
The hydroxides of group IA metals are white crystalline solids
They melt at moderate temperature without decomposition except (10H)
They are deliquescent solids
They are very soluble in water (Joem alkali. solutions)
CHEMICALS PROPERTIES
~The basic strength ofthe alkali inreases down the group, example calcium isthe strongest base
-When cold and dilute alkali’s reacts with chlorine to form metal chloride and hydrochlorite
2NaOH eq) + Clzyg) > NaCl+ NaClO +H₂O
-When hot and concentrated alkali’s react with chlorine to form metal chloride and chlorate (¥)
NaOH aq) + 3Clazp) + SNACleag) + NACIOs( a9) + 3420)
KOH (oq) + 3Clayg) > SKCl;a9) + KC1Os(09) + 320

Inthe two reactions chlorine undergoes disproportion
Most NaOH and KOH absorbs CO> fom the ar whereby a metal carbonates is formed
NaOH + C09) —* NasCOyy + HO,
= The hydroxides of Group IA metal reacts with acids to form salts and water only, example
undergo neutralization reaction
KOH aq) + HNO sag) — KNOs(ag) + H₂O(0
USES OF HYDRONIDES OF Na AND K
1. Owning tothe highly bask charactr aa metal hyroxdes ar used to absorb aii ese,
fxampie CO;
2. tat metal hyoxdes ae wed in neirzation ection
Fxample
Hlag) + OH(a9) > HQ,
3. Akal metal hyvonides are usedn precitation reaction
Example
Fell + 30H) > FECOM xn
nize) + 30H faq) > ZA(OH) x
4 Caustic oda (a0 used inthe manufactur fk paper and 099
5 Custc potash KOH) s used to manufacture sot soaps
2. ALKALINE EARTH METAL HYDROXIDES (M(OH):,
PHYSICAL PROPERTIES,
= They are white eryline solids
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Solubilities increases considerably down the group from beryllium hydroxides (Be) to barium
hydroxides (Ba). Beryllium hydroxide is insoluble in water.
Solubility of calcium hydroxide dereases with ris in temperature, the others increase,
‘magnesium slightly but strontium and barium hydroxide really. increase in solubility down
the group is due to the fact that atice energy decreases faster than hydration energy (BeQ) is
essentially covalent because ofthe high polarizing effect ofthe small
~ Group IIA hydroxides are much less soluble
Na Hag) + HzS0 gag) > NA:S0yaq) + #209
The hydroxides of Na and K precipitates some metals from their soluble salts (example,
“Aqueous solutions oftheir sls) as hydroxides
Fefiq) + 20H(aq) — Fe(OH) x4)
Cu) + 20H29) + Cu(OH) ca)
Pic) + 20H jeg) > PO(OH) xg)
Pref) + Oia) — Fe(OH) x00)
- Hoth NaOH and KOH iberates ammonia gas when added to ammonium salts
Ot) + Nig) —* NH) + Hy
NaOtegg)+ NH Cli > NaCliag) + NHyey + HO
REACTION WITH AMPHOTERIC METALS
“Zine, Aluminium, Lead and Tin react with hydroxides of sodium and potassium to form
complexes, example aluminate, plumbate, zncate and sturmmate.
Aly + 2OH;zq) + 61,0, —> 2[AI(OH) Vag) + 349)
Alumiate ion
Zn) + 20Heg) + 2H₂Oqy — [Zn(OH) lize + Ha)

Zincate ion
Phy) + 2OH;aq) + 2H₂Ocy — [PH(OH)a]fc—) + Hace)
Plumbate ion
REACTION WITH CARBONDIOXIDE,
When COs is bubbled through aqueous solutions of the NAOH and KOH the carbonates are
formed, With excess ofthe COs the hydrogen carbonates are formed,
NaOH gq) + COsiq) + NQ;COspaq) + H2)
NagCO 0g) + COz¢9) + HzO — NAHCO x09)
Cf group IA clements due tothe decrease in metallic character ofthe elements (example
Group 1A elements are more eleetropositive than their corresponding Group ILA elements).
Also the decrease in solubility may be due to decrease ionie characte ofthe hydroxides from
Group 1A to Group ILA,
NB; A Suspension of slaked lime(caleium hydroxide) in water is called Milk of lime
(CHEMICAL PROPERTIES,
1. ACTION WITH ACIDS AND ALKALIS
‘+ _Berylium hydroxide is amphoteric. reacts with excess sodium hydroxide forming a
solution of sodium berylate
Be(OH) sag) + 2N@OH (aq) > Na;Be(H) 00)
Sodium beryllate
The other hydroxide of group ITA metals do not react with alkalis but react with acids
to form salt and water only
Ca(OH) x¢0q) + HCliaq) > CaClacag) + H2%)
M9(OH)acaq) + 2HNOs(0q) — MG(NOs)iaq) + 22%
2. ACTIONS WITH CARBON DIOXIDE

4+ Moist hydroxides absorb CO rom aie forming carbonates
Ca(OH) x) + COxg) + CACO sag) + H2M%0
= When is bubbled through lime water (Ca(OH):) white precipitate of CaCOs are formed. This
causes the lime water to turn milky. The milky colour disappears when excess CO>in bubbled
through it. The milky colour disappears because calcium carbonates is converted into
caleium hydrogen carbonate which is soluble in water
Ca(OH) acag) + COxg) —* CACO aq) + 200
(Clean solution White precipitate (milky)
CaCOy0¢) + COaig) + HO) + CALHCO₂) 509)
(Milky) (Fx) Clear solution
3. ACTION OF HEAT
The temperature at which the hydroxides begin to decompose increases down the group from
about 300°C for beryllium hydroxide and magnesium hydroxide to about 700°C for barium
hydroxide
Frample
M(OH) acy > MO + HO
Ma(OH) sq) > MOO + HO
4, ACTION WITH AMMONIUM SALTS
All the hydroxides except Be(OH)> react with aqueous ammonium salts o give ammonia gas.
The ammonia gas is easily identified because it turn alkaline to litmus paper.
Example
OH jag) + NHiaqy —? Nagy + H₂O(y
NaOH) + NH₂CI(aq) + NaCligg) + NHyy) + HO
5, ACTION WITH SULPHUDIOXIDE

Sulphur dioxide tur ime water milky due ocaleium sulphite formed. When excess SO₂is
added the milky colour disappears (example, clear solution is formed), The milky colour
disappears due to the formation of calcium bsulphite which i soluble in wate.
calorslagsois) —e casouis) + 1:00)
(Clears) Wate ppt mk esour
uses
1. Ue water is vied totes for carbon dioxide
2. Asuspenson of Magnesium hydrosde in water (nik of magnesium used as an ant ci
3. Ca(OH) isused in mating bulders mortar mistre of sakes ime, sand and water).
4. Amisture of Ca(OH) is used n making bleaching powder.
5. Ca(OH): Isused for neutraizng acim the so
6.Amiture CalOH) and water ite wash isused for coating walls and cel,
7. Ca(OH): used in water softening.
Mg(HCO3)a(aq) + Ca(OH)a(aq) + CaCO3(s) + MgCO; + 2H2,0
8. Ca(OH) suse in sugar refoing tered
ORGANIC CHEMISTRY 1
[ALIPHATIC HYDROCARBONS:

Iniong chain, a stable carbocation will be formed when the carbon fs hounded by many alkyl groups
(since the alky group wil be supply eletrons
(CHy-CH=CH-CH3-CHs+IH——* CH; CH: CH CH CH
|
| | :
Supplies less electron Supp more electron 3-iodo pentane
Note:
Hydrogen s added to the more stable carbon
2. Hydration of alkenes:
Hydration means adstion of water.
= This is addition of water in the presence of mineral acids. The most prefered acid is cone
1:80. The mixture should be heated in order to form alcoho.
CH₂ CH=CH₂+oH₂ “““%» — cH.cHCH₂
5 Lt0H₂ sees
oH
Carbocation willbe formed in CH
Home Work:
‘Anti -markovnikov's rule (organic peroxide HB. In 1933 the American chemist M.S. Kharasch
discovered thatthe addition of Hr to unsymmetrical alkenes in the presence of organic peroxide (RO
=0-R) takes a course opposite to that suggested by Markovnikovs rule
Ee
Peroxide CH CH3 CH3 Br(anti—~ Markovnikov's product)
CH CH = CH + HBr Br
No peroxideCHs ck CH (_— Markovnikovsproduct)

Nore:
Its striety works using HBr with organic peroxide.
‘Mechanism:
1, Peroxide dissociates to give alkoxy free radicals,
a
R-O-O-R > 2R-O
2. Alkcony fee radical combines with HBr to give bromine atom (a free radical).
R-O+HBr —__» R-OH+Br
3. The Bromine atom attacks propane to give a primary fee radical anda secondary fre radical
Br
|
Br+CH₂CH=CH₂ ————*CH₂ CHCH₂ _(less stable 1" free radical)

Br+CH₂CH=CH₂ ———*CH₂CHCH₂ (2 free radical more stable)
4. The more stable 2° free radical attacks the HBr molecule to form anti ~_markynikov product and bromine atom,
CH₂ CHCH₂Br HBr = ———* CH₂ CH₂ CH₂ Br+ Br Bromopropane)
Weeki Test:
1. (0) Naw + 3H) 7 2NHsw)
Given 3°41, =°92K
Since Hryou form imole
he
Wee

CHS CHs
CHsCH=C-CHs+HBr "cH; GH CH CHs
Br
=i
He =>
=22
ma
He =- 46K Jmol
3. (a)(ii) CaC2 + 2H₂O —————* CH;COOH + Ca
Br
reste os ca!
CHs C= CH-CHs + HBr “""S cHs CH CH CHs
CH CH
3. Halogenation
addition of halogens to alkenes
= This reaction is best carried out by simply mixing halogens inthe inert solvents such as carbon
tetrachloride (CCL)
o£
R-CH=CH-R+X, “$3 R-CH-CH-R

oO
v
H’-0-S-0-H
I
°
cole
Eg. CH CH=CH₂ + Cone Hz S04 —* CH CH-CH
OSOsH
2- Propylhydrogensulphate
CH₂ (HCH₂ +H₂O + CH; CH CH; +H₂ SO₂
OSO3H OH
‘7 Oxidation reactions of Alkenes
Alkenes react with oxidizing agent to form diols. Oxidizing agent can be cither KMnOx,
KaCr0x.
= With cold dilute KMnOs or cold alkalineK MnOx you form diols.
OH OH
wnnor |
R-CH=CH-R ——““— R-CH-CH-R
Diol
NOTE: Dio! means two OH

cH=cn HS i
b= CH; ———> CH)=CH)
fart
Of OH Ethylene glycol
411 atmo OH
CH; CH=CH-CH; ————>+ CH; CH- CH CH;
it
OH OH
(0) When hot concentrated accified MnO, or K:C1.0sis used. lkenes are oxidized to carboxylic acid or
ccied to ketones or both
R-CH=CHR @@%*, R_C_OH+R-C-OH
+ Itis broken down to carboxylic acid
i i
2g CH₂CH₂-cHcH₂ cH₂ !°2"'s cH₂ C-oH+CH₂ cH₂C-O8
i
ramon
ees 00/5 CH₂ COOH + CH₂ C- CH₂
CH
Note:
HY double bonds branched, you cant form carbonyl acd

ca,cxt,cxt= cus, 2 cu, cx,c00H +C0,+H₂O
‘asl, ecompores to CO₂ and H₂O
OZONOLYSIS (0;)Oz0N0 ~ ozone
tysis. > Breaking
This isa leaage or breaking eon, aon double and by sing one
«Iosomyis C=C a compet keno pod acy or eons obo depenting oa
che prary str aos
wom:
Corals the best method of cat the poston doe bond inuaown ae
‘he onyeratedcaron In cron| compound cand by conoh isthe oe hat were hed by
dou bonds te oral ees.
ronayshas2 mar steps:
sep
Dowload ree notes ad pat papers tom waw.wazaeiom

Calculate the enthalpy change for the reaction "®2 + CO Fe + CO₂
Solution
Page
1363 ot
Required manpuation KJ mot
4/) (Fe,0, +3CO ——+ 2Fe + 3C0, -28)
1/, 2Fe, 0, + CO₂ ——+3Fe,0,+CO 59
1/y(3Feo+ C0.) —+ e040 -38,
4/,Fe,0,+3/co ——+ Fe+3/,c0, 14
1/, Fe,0,+ 1/,C0; — 1/,Fe,0,+ Veco 983
Feo+4/3co,—+ 1/3 Fe,0,+1/,c0 - 1267
Feo+CO —~+ Fe+ C0, ~ 1684
{The enthalpy change is 16.84K}mol"*
2. CALCULATION OF ENTHALPIES
[BASED ON BOND ENERGIES
BOND ENERGIES
's the energy change which obtained when one mole covalent bonis formed or broken ofan atom
Any rection involves bond breaking and bon formation Reactant bonds are normaly broken while
roduts bonds reformed.
‘Since the bonds energies are known then “““of the reaction can be calculated as the difference
between broken bond energies and formed bond energies

SH. ppe-FB.e
Page Where by;-
| 364
4H Is the heat change of reaction.
B.B.E is the broken bond energies.
F.B.E is the formed bond energies.
Example 1
(a) Define
(i) Bond energy.
Is the energy which is obtained when one mole of covalent bond is formed or broken of an atom.
(ii) Enthalpy of neutralization.
Is the heat given out when one mole of water is formed from the reaction between acid and base at
standard state.
(b) Calculate the heat of formation of ethane given that:
C-H = 413 kJ mol⁻¹
H —H = 436 kJ mol⁻¹
Solution
Required equation
SH'F
C+2H₂ ———>CH₂
i.
c+2(@#-H)——-H ~ C —-H
H

Broken bond energies
= 2(H —H)
Page
| 365 = 2X 436
= 872 kJ mol*
=4(C-H)
= 4X 413
= 1652 kJ mol*
AH°F = BBE — FBE
= 872 — 1652
= -780 kJ mol*
« The heat of fprmation of methane — 780kJ mol⁻¹
Example 2
Calculate the enthalpy of hydrogenation of ethane to ethane.
Given
(C=C = 612 kJ mol⁻¹
(ii)C-—H = 416 kJ mol
(iii) H- HH) = 436 kJ mol
(iv) C—C = 348 kJ mol

Solution
A, H H
Page ; a +H₂—— thn
| 366 H H ” HA
B.B.E F.B.E
= 4(C —H) 6(C — H)
=4x 416 =6x 416
= 1664 = 2496
c-Cc
= (H —H)= 348 =348
= 436
Total FBE = 2496
= (C=C) 348+
miele = 2844
Total BBE = 436
612 + 1664
=2712
AH. pee — Fee
= 2712-2844
<1
= -137-kJ mol

xamplea
Poae
ws
Download fe ote ad pat papers rom wan. wazalmcom

=
2al (2) + 3Cl; (g) —— Al; Cl₂
fe |e
2Al(g) 6Cl(g) +2
El Eaff
LE
i +38
24l** +6CI~
o AHF = Es + Ei + Eat + Ear + Ex
From the data,
2Al(s) +6HCI(aq) —> 2AlCl₂+ 3H₂ — 1033kJ mol*
H₂(g) + Cl₂(g) ——> 2HCl(g) — 184 kJ mol⁻¹
HCl +aq ——> HCl+aq —724 kJ mol⁻¹
AlCl₂ +aq ——> 2AlCl; — 643 kJ mol
Data manipulation
2Al(s) +6HCI°C99) —— 2AlCI3 + 3H₂ —1003kJ mol~*
3H₂(g) + 3Cl₂(g) ——— 6HCI(g) —552 kJ mol⁻¹
—2
2aIcl3. —— AlCl₂ + aq 643 kJ mol
6HCI(g) +aq ——> 6HCI(aq) — 4344 kJ mol⁻¹
Req eqn.
2Al(s) + 3Cl₂(g) ——Al, Cl₂ — 5256kJ mol
CALCULATIONS OF ENTHALPIES BY USING ATOMIZATION DATA

Ding the reacton restart ae stomizton (changed atoms) whe he product energy. Calelte
frombond ener.
Downlosd re nots and past apes rom war wataeicom

=
AH of reaction = Eat-F.B.E
Example1
SH'F
Cl(s)+2H₂ ——> CH₂
tt ra
AH°F = Eatof C+EatofH) — FBE
I
But. FBE= H — | - H- = 4(C-H)
H
Example 2
Calculate the following of formation of CH₂ given that,
Enthalpy of atomization of carbon = 715 kJ mol
= “1
Enthalpy of atomization of hydrogen _ 218 kJ mol
C —C = 348 kJ mol⁻¹
C —H = 416 kJ mol⁻¹
H —H = 436 kJ mol⁻¹

saan
ane al
Clg) 4H
Eat =715 + 4(128)
= -77K}mot⁻¹
we
sito
Eat = 2(715) + 6(218)
soe

2730 — (2496 + 348)
= -144kJ mol
cuales
hydrogen atom. if the heat of formation of methane is. TIKI yen) that,
C—H = 416 kJ mol*
nee
c+2u, “cn,
-77 = 715 + 4X — 4(416)

Scarronce rene rr cxomseemt
Pecan or Ror mR eanon EY SommRY
none
Ca(OH) , + H₂50, —+ CaSO₂ +H₂O
Pw * Pratt solution
1 ="
. 396

=
m _m
vol
«. Mass in(g) = volume in cm®
Quantity of that in this method can be calculate by using heat capacity ( C )and by using specific heat
capacity (c).
Let the quantity of heat be Q.
By using heat capacity
Q=CAe
C= heat capacity.
By using specific heat capacity.
Q=mc Aé
Where by AQig change in temperature.
c= specific heat capacity.
If quantity of heat calculated is the same and 465 also the same, then the relationship between c and C
can be,
Q, =CAd
Q, =mcA@
Qi = Q
cae _ meae
Ao Ae
Cc =mc
Enthalpy of neutralization is calculate with respect to the number of moles of water producer.
Reaction.

=
H+0H — > H₂O
But number of moles of water depends on the moles of limiting reagent
QUESTION
What is limiting reagent?
Limiting reagent is the reactant compound in the neutralization reaction which have small number of
moles.
Example 1
250c”” of NaOH of 0.4M were added to 250cc of the HCI of 0.4M ina calorimeter. The temperature of
°
the two solutions and the calorimeter wast7-05 7 The mass of calorimeter was 50g and its specific
-ig-t °
heat capacity was 400K]g" K. after the reaction the temperature rose to 19.5° c
-1p-1
Assuming the specific heat capacity of all the solution ig #200). Kg" *k . Calculate the standard
enthalpy of neutralization.
Solution
Data
Vb = 250
Mb = 04M
Va =250
Ma =04M
Q, = 17.05
Q@, = 19.55
4Q = 19.55-17.05
= 2.5kg

(The para newaizaton done by postive ndute ett xed by ly group) Hence
x
k }
( oR Cc
yey
pilsdnrsrotsnalpes ainsi ir ear mu linahaineaiiieei
fare &
i ‘t
Lien
5 4
sos
sewn
ws

ws

=
But if the substituent in halogen, ortho product become major product why?
Reason
Halogens like Cl have very small atomic size, Thus they exert very small steric hinderance thus make
incoming electrophile to substitute first at ortho carbons (for every two ortho) carbons there is only one
para carbon).
Deactivators with exceptional of halogens directs incoming eletrophile at meta position i.e. Deactivator
(with exceptional halogens) are meta directors.
This can be explained by considering
i. Position of carbonium ion
ii. Stability of intermediate carbonium ion.
1_POSTTION OF CARBONIUM ION
. To understand this consider mesomerism (-M) in benzoic acid
fe)
ret o® 0° qo " a
u C-0H c-Ol
C-OH Cou G-08 i ) 1
Bi. Pp . ‘s SN
a ~) ‘ — —
LY YY ©
/
From the above shown mesomerism it can be seen that despite the fact that carboxylic group (-COOH)
deactivate the whole benzene ring ortho and para positions are more effected and hence meta carbon
somehow become pereferd position for incoming eletrophile.
Tl. STABILITY OF INTERMEDIATE CARBONIUM ION
. Consider the electrophilic substitution reactions in benzoic acid.
1% CASE

=
0 Reduction of Cu?* to Cu*
n we
NaOH
1) RCH + Cu(OH), “hh cote Gia + H₂O
Brickred ppt
oO
s " a NaOH 2 “ 3 .
ii) RC R'+ Cu(OH), ———no reaction{ There is no formationof brick
red brick ppt)
NOTE.
Benzaldehyde do not react with Benedict/ Fehlings solution
oO
le ll
ie
(0) + Fehlings / Benedict solution —---———~ No reaction
2) BY USING TOLLEN’S REACTION( silver mirror test)
+
With tollen’s reagents aldehyde being reducing agent reduce Ag to AT ie precipitate of AGS) which
appear as silver mirror hence the name mirror test) Aldehyde react with tollen’s (Ammoniacal silver
nitrate) form white ppt of silver which appear like mirror
Ketones being poor reducing Agent give negative mirror test
Le
oO
W
RCH + Ag(NH3),0H——>+RCOONH₂ + NH; + H₂O + Ag)
(Aldehyde) (Silver mirror)
WHILE

=
(e]
I
RC R’Tollens reagent —————> No reaction (Give negative test) ketone
OTHER CHEMICAL REACTION.
IODOFORM TEST.
¢ This is the test for presence of terminal methyl group is directly bonded to carbonyl group by
giving yellow ppt of iodoform CHl,
e For aldehyde only ethanal give iodoform test.
Example
ll
NaOH(aq)
i) CH₂ CCH3+ I, ———>CHI3+ CH3COONa+ H₂O
; NaOH(aq)
ii) CH3CH₂ C CH3+ 21, ———*_ CHI3 + CH3CH₂ COONa+ Nal + H₂O
A
(Yellow ppt)
Generally
i]
NaOH (aq)
CH₂ CR+ Il, ——— RCOONa+ CHI,;+ Nal+ H2,0
A
(Methylketone)
NECTA 2000 PP; QN 14 (a)
QN. Compound A which has an unbranched carbon chain, react methyl magnesium bromide to give
after hydrolysis compound B. chromic acid oxidation of B gives C (CsH:00) or which gives crystal product
with 2, 4- dinitrophenylhydrazine and a positive iodoform test.

os

*
F
where [hs oxidising agent
.
:
:
°
sore
mvncnors
oe

a t
os

=
oO
Example i]
CH; C-—Cl + H₂O ———~ CH;,COOH + HCl
Ethanol chloride
b/ Acidic hydrolysis of Nitrile
Generally
xt
RNC + H₂O ——— RCOOH + other products
Example
CH;CH₂CN + dil.HCl ——> CH3,CH₂COOH
Ill FORM GRIGINARD REAGENT
Carbon dioxide reacts with Grignard reagent followed by acidic hydrolysis to form carboxylic acid.
Generally:-
o o
i] i] Ht HLO
c + RMgX __, RC-~OMgX ——+ RCOOH + Mg(OH)X
i
(Coz)
Example
H* H₂O
CO₂ + CH;CH₂MgCl ——+ CH;CH₂COOH + Mg (OH)Cl
PHYSICAL PROPERTIES OF CARBOXYLIC ACIDS

=
Carboxylic acid has highest boiling point among alcohol phenol carbonyl compound or easy other
hydrocarbons with comparable molecular weight due to stronger hydrogen bonding existing between
molecules of carboxylic acids.
Alcohol phenol carbonyl compound or any other hydrocarbons because they come capable of
making strong hydrogen bonding with water and high polarity of carboxylic group.
Le
o-
"
—- C-0O--H*
Carboxylic acid is capable of undergoing dimerisation when it is in the hydrocarbon solvent it or any
other similar solvent without hydrogen bonding
le y, g-—c 6B \
R—C al —y
——— | n°) “a
Dimer
ACIDIC BEHAVIOUR OF CARBOXYLIC ACID
Carboxylic acids have higher acid strength than alcohol and phenol.
For aliphatic carboxylic acids strength depend on:-
i/ length of carbon chain
ii/ Type of substituent in carboxylic acid
The acidic strength decrease with increase in length of carbon chain due to strong positive effect
exerted by longer alkyl group.
When substituent carboxylic acid is stronger electronegative element like halogens which exert -I acidic
strength is increase due to following reason:-

=
BUT
o
COOH i
et 0 ONa c -o
+ NaOH Oo 1@)
—_—
CHEMICAL TEST OF CARBOXYLIC ACIDS
a/ REACTION WITH ALCOHOL
Carboxylic acid reacts with alcohol in presence of acid
ton 22 ,
RCOOH + R'OH——> RCOOR'+H₂O
The reaction is very important in distinguishing carboxylic acid from phenol is phenol do not react with
alcohol verifying that carboxylic acid is more acidic than phenol
b/ REACTION WITH SODIUM BICSRBONATE
Carboxylic acid reacts with sodium bicarbonate to give effervescence of carbon dioxide gas which turn
lime water milky
- COOH + NaHCO₂ ———~ —COONa + CO₂ + H₂O
Turn lime
water milky
The reaction gives another difference between carboxylic acid and phenol as phenol do not give
effervescence acidic than phenol.
esreaction with Pels(lron(II!) chloride) osric chloride test
carboxylic acid reacts with Fecl; to give Iron (III) carboxylic (alkanoate ) which appear as buff colourled
compound

=
ie. 3 RCOOH + FeCl —— (RCOO)3Fe+3HCl
Buff coloured
The reaction is important in distinguish between alcohol, phenol and carboxylic acid
« Alcohol give no change indicating that there is no reaction between alcohol and Iron (III)
chloride (Fecl;)
* Phenol give purple/violet coloured compound
e Carboxylic give buff coloured compound
.
f/ FORMATION OF ESTER
Acyl chloride reacts with phenol in presence of NaOH., form ester
Le.
R
i]
OH o o-c —®
i]
Og eb -aneee CY, we
Example
o
iH I
o-—c —CH₂
Acyl chloride also reacts with alcohol to form ester although with alcohol there is no need of NaOH(aq)
Le.
RCOCI + R'OH ——> RCOOR' + HCl
E.g.
oO
I
CH; C -—CH + CH3CH₂OH ——> CH; COOCH₂CH; + H₂O

oe