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ELECTROVALENCY
The number of electrons lost or gained by an atom of any element is termed as electrovalency. The element which gives up the electron(s) to form positive charge or ions are said to have positive valency. While the element(s) which accepts electrons to form negative ions are said to exhibit negatively valency.

Note:- Variable electrovalency of iron exist as Fe2+ and Fe3+ in ferrous sulphate and ferric sulphate .when a compound is formed by the transfer of electrons, the element that loses electron(s) is said to be oxidized and the element that gains electrons is said to be reduced.
Oxidation is a process which involves loss of electrons where as reduction is the process which involves gaining of electrons.
PROPERTIES OF IONIC/ELECTROVALENT BOND
An ionic/Electrovalent bond has the following properties:-
(i) An ionic bond is formed due to the coulombic attraction between the positively and negatively charged ions.
(ii) An ionic bond is non-directional, the strength of interaction between two ions depend upon distance but not the directional.
(iii) An ionic bond gets broken when the substance is dissolved in polar solvent such as water or when the substance is melted.

Typical examples of ionic bond:-
(a) Na + Cl → NaCl

EcoleBooks | CHEMISTRY As LEVEL(FORM FIVE) NOTES - GENERAL CHEMISTRY (2)

EcoleBooks | CHEMISTRY As LEVEL(FORM FIVE) NOTES - GENERAL CHEMISTRY (2)
EcoleBooks | CHEMISTRY As LEVEL(FORM FIVE) NOTES - GENERAL CHEMISTRY (2)
FACTORS INFLUENCING THE FORMATION OF IONIC BOND

The main steps involved in the formation of an electrovalent/ ionic bond are:-
(i) Removal of electrons from one atom, in this stage energy equal to the ionization energy is absorbed.
(ii) Gaining of electrons by the other atoms .In this step energy equal to the electrons affinity is released.

(iii) Combination of cation and anions. These ions are held together by coulombic force of attraction. In this step, energy equal to the lattice energy is released.

(II) COVALENT BOND
A covalent bond is formed between two atoms (similar and dissimilar) by a mutual sharing of electrons .the shared pairs of electrons are counted towards the stability of both the participating atoms.

DEFINITION: – A covalent bond is defined as the force of attraction arising due to mutual sharing of electrons between the two atoms.
When the two atoms combine by mutual sharing of electrons, then each of the atoms acquire stable configuration of the nearest noble gas.

COVALENCY
Is the number of electrons which an atom contributes towards mutual sharing during the formation of a chemical bond.
Example of the Covalency: – Covalency of hydrogen (H2)

EcoleBooks | CHEMISTRY As LEVEL(FORM FIVE) NOTES - GENERAL CHEMISTRY (2)
CHARACTERISTICS OF COVALENT BOND
(i) Mode of formation, covalent bond are formed due to mutual sharing of one or more pairs of electrons.
(ii) Directional character , covalent bonds are directional in nature this is because the shared electrons remains localized in a definite space between the nuclei of the two atoms.This gives a directional character to the covalent bond.
SINGLE COVALENT BOND
A covalent bond formed by mutual sharing of one pair of electrons. A single covalent bond is represented by a small line (-) between the two atoms.
E.g
EcoleBooks | CHEMISTRY As LEVEL(FORM FIVE) NOTES - GENERAL CHEMISTRY (2)
EcoleBooks | CHEMISTRY As LEVEL(FORM FIVE) NOTES - GENERAL CHEMISTRY (2)
MULTIPLE COVALENT BONDS
The covalent bonds developed due to mutual sharing of more than one pairs of electrons are termed as multiple covalent bond.
The multiple covalent bonds are:-
(i) Double covalent bond
(ii) Triple covalent bond
Double covalent bond is the bond formed between two atoms due to the sharing of two electrons pairs. Its simply called Double bond

E.g. O2 → O = O
CO → C = O etc.

Triple covalent bond is the bond formed due to the sharing of three electron pairs
E.g N2 → N ≡ N,
Acetylene H – C ≡ C – H.

(I) FORMATION OF MOLECULES HAVING DOUBLE BOND.
(i) Formation of oxygen (O2) molecule:-
Each oxygen atom has six electrons in its outer most shell. Thus it requires two more electrons to achieve the nearest noble gas configuration.
E.g
EcoleBooks | CHEMISTRY As LEVEL(FORM FIVE) NOTES - GENERAL CHEMISTRY (2)
(ii) Formation of carbon dioxide(CO2) gas
The electronic configuration of carbon and oxygen are :-

C 1s2 2s22p2 2,4
O 1s2 2s22p4 2,6

Thus each carbon atom requires four ,and each oxygen atom requires two more electrons to acquire noble gas configuration.This is achieved as follows
e.g
EcoleBooks | CHEMISTRY As LEVEL(FORM FIVE) NOTES - GENERAL CHEMISTRY (2)
(iii) Formation of molecules having triple bond :-
(a) Formation of nitrogen (N2) molecule.
EcoleBooks | CHEMISTRY As LEVEL(FORM FIVE) NOTES - GENERAL CHEMISTRY (2)

COMPARISON BETWEEN SINGLE,DOUBLE AND TRIPLE COVALENT BONDS
Triple bond length < Double bond length < Single bond length.
Since a shorter bond means greater bond strength hence the energy required to separate the bonded atoms( called bond energy) follows the order
Triple bond > Double bond > Single bond

FACTORS FAVOURING THE FORMATION OF COVALENT BOND
The following are the factors that favour the formation of a covalent bond:-
(i) High ionization enthalpy (energy).
The element having higher ionisation enthalpy (energy) cannot lose electrons easily.

(ii) Nearly equal electron gain enthalpy or electron Affinity.
The atoms of the two elements which have equal or nearly equal electron gain enthalpies or electrons affinities tend to complete their outer shells by mutual sharing of electrons.

(iii) Nearly equal electronegativity.
Equal or nearly equal electronegativity of the two combining elements does not permit the transfer of electron(s) from one atom to another.

(iv) High nuclear charge and small atomic size.
High nuclear charge, and smaller atomic size of the combining elements favour covalent bond formation, because the transfer of electrons in such case will not be possible.
COORDINATE COVALENT BOND
Coordinate bond is formed when the shared electron pair is provided by one of the combining atoms.The atom which provides the electron pair is termed as the donor atom ,while the other atoms which accept is termed as the acceptor atom
The bond formed when one sided sharing of electrons take place is called a coordinate or Dative bond. A coordinate bond is presented by an arrow ( → ) pointing towards the acceptor atom.

(i) Formation of coordinate bond during the formation of a molecule or molecular ion:-
(a) Formation of Ammonium (NH4+) ion.
During the formation of ammonium ion ,nitrogen is the donor atom while H+ is the acceptor ion.
EcoleBooks | CHEMISTRY As LEVEL(FORM FIVE) NOTES - GENERAL CHEMISTRY (2)
(b) Formation of coordinate bond between two molecules:-
Two or more stable molecules combine to form a molecular complex,is such a complex molecule, the constituent molecules are held together by coordinate bond
EcoleBooks | CHEMISTRY As LEVEL(FORM FIVE) NOTES - GENERAL CHEMISTRY (2)
The product above is ammonia-borontrifluoride complex
POLARITY IN COVALENT BONDS :-
Depend upon chemical nature of the combining elements, the following two types of covalent bond are formed

(i) Non-polar covalent bond
(ii) Polar covalent bond
(i) NON-POLAR COVALENT BOND
When a covalent bond is formed between two atoms of the same element, the electrons are shared equally between the two atoms .The resulting molecule will be electrically symmetrical.
EcoleBooks | CHEMISTRY As LEVEL(FORM FIVE) NOTES - GENERAL CHEMISTRY (2)

EcoleBooks | CHEMISTRY As LEVEL(FORM FIVE) NOTES - GENERAL CHEMISTRY (2)

EcoleBooks | CHEMISTRY As LEVEL(FORM FIVE) NOTES - GENERAL CHEMISTRY (2)
EcoleBooks | CHEMISTRY As LEVEL(FORM FIVE) NOTES - GENERAL CHEMISTRY (2)

EcoleBooks | CHEMISTRY As LEVEL(FORM FIVE) NOTES - GENERAL CHEMISTRY (2)
EcoleBooks | CHEMISTRY As LEVEL(FORM FIVE) NOTES - GENERAL CHEMISTRY (2)

EFFECT OF HYDROGEN BONDING
Hydrogen bonding affects the physical properties of the compound appreciable .
Some effect of hydrogen bonding are described below:-
(i) Molecular association.
Formation of aggregates containing two or more molecules due to weak electrostatic interactions such as hydrogen bonding for example ,water molecule undergoes molecular association due to hydrogen bonding.
EcoleBooks | CHEMISTRY As LEVEL(FORM FIVE) NOTES - GENERAL CHEMISTRY (2)

(ii) Increase in the melting and the boiling points.
The interaction which affected the melting and boiling points of NH3, H2O and HF is identified to be hydrogen bonding.
EcoleBooks | CHEMISTRY As LEVEL(FORM FIVE) NOTES - GENERAL CHEMISTRY (2)
(iii) Influence on the physical state.
Hydrogen bonding affects the physical state of a substance also for example H2O is liquid while H2S is a gas under room temperature condition

(iv) Solubility of covalent compound in water
Covalent compounds generally do not dissolve in water. However the covalent compound which can form hydrogen bonds with water readily dissolved in it. For example ethanol, Ammonia, Ammine lower aldehyde and Ketones dissolve in water due to their tendency to form hydrogen bond with water.

EcoleBooks | CHEMISTRY As LEVEL(FORM FIVE) NOTES - GENERAL CHEMISTRY (2)
REVIEW QUESTIONS
Question 31
(a) Using the electronic configuration and periodic table, give the name of the element and the number of valence electrons.
(i) 1s2 2s2 2p4
(ii) 1s2 2s2 2p6 3s2 3p3
(b) Use Hund’s rule and other information to write out electronic configuration and orbital diagram of;-
(i) Cobalt
(ii) Nickel
(iii) Zinc
Solution
(a) (i) Oxygen: It has 6 valence electrons in the 2s and 2p sub levels.
(ii) Phosphorus: It has 5 valence electrons (in the 3s and 3p sub level)
EcoleBooks | CHEMISTRY As LEVEL(FORM FIVE) NOTES - GENERAL CHEMISTRY (2)
Question 32
(a) Provide the number of orbital in each of the following shells;-
(i) 1s
(ii) 2p
(iii) 3d
(iv) 4f
(b) Howe many sub – shells are there in each of the following sub – shells? K, N and L
(c) Explain if it is possible for an electron to have the following set of quantum numbers.
n = 3, â„“ = 3; m; m = 3 and s = +1/2
(d) An electron is in a 4f orbital. What possible value for the quantum number can it have?
(e) What sub shell is found in the shell with n value = 4
(f) Write the possible value of the 4 quantum number for the outermost 2 electrons of Calcium.
(g) Write all the four quantum number for an electron added to Cl atom in forming Cl
(h) Write all the four quantum numbers for the outmost 2 electrons in Na
Question 33
(a) Write the total number of electrons possible for an atom whose n = 3. Assume that all the orbitals are full of electrons.
(b) Write electronic configuration of the following elements/ions
(i) O
(ii) K
(iii) Ni at number 28
(iv) Cu+ (at number of Cu = 29)
(v) Mo at number 42
(vi) Cl (at number of Cl = 17
(c) What is the number of unpaired electrons present in the ground state of;-
(i) Fe+3 (At NO. of Fe = 26)
(ii) Cr3+ (At No. of Cr = 24)
(iii) Ni2+ (At No. of Ni = 28
(d) Why is the electronic configuration 1s2 2s2 2p2x 2p0y 2p0z not correct for the ground state of nitrogen?
(e) Mention the disobeyed law from the following arrangements;-
EcoleBooks | CHEMISTRY As LEVEL(FORM FIVE) NOTES - GENERAL CHEMISTRY (2)

Question 1
(a) State two major postulates and five short coming of Bohr’s atomic models.
(b) An electromagnetic radiation was emitted in the Balmer’s series as a result of electron.
Transitions between n= 2 and n= 5 calculate the;
(i) Energy of the radiation in kJmol-1
(ii) Frequency of the radiation in hertz
(iii) Wavelength of the radiation in metres.
(c) Calculate the wavelength of a line in the Balmer Series that is associated with energy transition
E4 E2 (E4 = -1.362 x 10-19J; E2 = -5.448 x 10-19J; h = 6.63 x 10-34J; C = 3 x 10-8 m/s)

Question 9
(a) The atomic spectrum in the visible is given by the following relationship: = RH
(i) What do symbols λ, RH, n1 and n2 represent? Show their SI units
(ii) Calculate the frequency of the third line of the visible spectrum i.e. a transit form n = 5 to n = 2 (RH = 1.097 X 107 M-1)
(b) By means of series of horizontal lines, draw an energy level diagram for visible atomic spectrum of hydrogen and indicate:
(i) The value of n for each line
(ii) By means of an arrow, show the transition which corresponds to the fourth line of emission spectrum in visible region.
(iii) By means of an arrow, indicate the transition which corresponds to the fourth ionization energy of hydrogen atom.
(c) (i) What do you understand by the term convergence limit and what is its significance?
(ii) Why are the discrete lines observed and not a continuous spectrum in Hydrogen spectrum?
Question 25
(a) Define the term electron configuration.
(b) Write down the electron configuration of these elements in their ground;
(i) Hydrogen 1p.
(ii) Beryllium.
(iii) Neon 10p.
(iv) Aluminium 13p.
(v) Calcium 20p.
(vi) Manganese 25p.
(vii) Cobalt 27p.
(viii) Zinc 30p.
(ix) Krypton 36p.
(x) Silicon 14p.

Question 18
Write electronic configuration diagram of the following;-
(a) S-2 (b) Na+ (c) Cr+3 (d) Cr+6 (e) Zn+2
Question 15
Write the electronic configuration of the following;
25V = [Ar] 4s2 3d3
16S = [Ar] 3s2 p4
9F = [He] 2s2 p5
34Cs = [Ar] 4s2 d10 4p4
16In = [Kr] 5s2 4d8
26Fe = [Ar] 4s2 d6
56Ba = [Xe] 6s2
Question 16
Provide electronic configuration of the following;
75Pt
80Br
68I
48Ga
92Cf
58Te
102Ra
17Cl
15P
30Zn
Question10
(a) All radiations are associated with the wave nature and differ from one another terms of their wavelength. Frequency, velocity and energy. Give the relationship between;
(i) Frequency and wavelength.
(ii) Energy and frequency.
(iii) Energy and wavelength.
(b) Ozone (O) protects the earth’s inhabitants from the harmful effects whose wavelength is 2950A. Calculate
(i) Frequency.
(ii) Energy.
for this UV light
(c) If the wavelength of the first line of the Balmer series in a hydrogen spectrum is 6563, calculate the wavelength of the first line of Lyman series in the same spectrum. The uncertainty of the momentum of particle is 3.3 x 10-16gms-1. Find the accuracy in which its position can be determined.
Question 11
a) (i) A small object of mass 10g is thrown with velocity of 200 m/s given h = 6.626 x 10-34J.S. Calculate its wavelength.
(ii) Kinetic energy of a sub – atomic particles is 5.65 x 10-23J. Calculate the frequency of the particle wave (h = 6.626 x 10-34)
(iii) Calculate the momentum of a particle which has a de-Broglie wavelength of 0.1nm (h = 6.6 x 10-34 JS).
(iv) Two particles A and B are in motion, if wavelength λ of A is 5 x 10-8m. Calculate the wavelength λ of B. if its momentum is half that of A.
b) State and describe briefly Heisenberg uncertainty principle.
c) (i) Describe the concept that as the particle size gets smaller determination of its momentum and position simultaneous becomes difficult
(ii) Use the mathematical expression of the uncertainty principle to prove the validity of part c (i) above
d) (i) Calculate the uncertainty in position of an electron if uncertainty in velocity is 5.7 x 105 m/s
(ii) The uncertainty in position and velocity of a particle are 1010m and 5.27 x 1024ms-1 respectively. Calculate the mass of the particle.

Question 12
(a) Outline the postulates and weaknesses of Bohr’s atomic model.
(b) Calculate the mass of a photon of sodium light having wavelength of 5894A and velocity 3 x 108m/s given h = 6.6 x 10-34kgm2s-1 and 3.7 x 10-36kg.
(c) What’s the wavelength of a particle mass 1 gram moving with a velocity of 200ms-1 given h = 6.626 x 10-34J. (Answer 3.31 x 10-29m)
(d) The mass of an oxygen molecule is 5.3 x 10-26kg. Calculate its de Broglie wavelength if it is moving at a speed of 500m/s (Planks constant (h) = 6.626 x 10-34Js (Answer 2.5 x 10-11m)

Question 13
(a) Briefly explain the meaning of these terms, with the help of a diagram
(i) Wavelength
(ii) Frequency
(iii) Velocity
(iv) Amplitude
(v) Wave number
(b) Yellow light from sodium lamp has a wavelength of 5800Ao. Calculate the frequency and wave number of the radiations. Determine the range of frequency of visible light. The wavelength of radiations lying in the visible regions are between (3800 – 7600A0).
Question 14
(a) (i) What do you understand by dual nature of matter?
(ii) 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 400nm
(ii) What is the amount of radiant energy associated with atomic 6.662 x 10-34Js, velocity of light 3 x 108m/s
(c) If the wavelength of a beam of light is 2.8 x 10-7m. Calculate it’s;-
(i) Wavelength in cm.
(ii) Frequency.
(iii) Energy of one of its photons.
(d) (i) Referring to Bohr’s atomic model, what’s ionization energy?
(ii) Calculate ionization energy of hydrogen. Reydberg’s constant RH = 1.097 x 107m-1
(e) (i) Calculate the frequency of the 3rd line of visible spectrum.
(RH=1.097 x 107m, C=3 x 108m/s)

(ii) Given that the wavelength of the 1st line of Balmer series is 6563A. Calculate the wavelength of the 2nd line in the spectrum.




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EcoleBooks | CHEMISTRY As LEVEL(FORM FIVE) NOTES - GENERAL CHEMISTRY (2)

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1 Comment

  • EcoleBooks | CHEMISTRY As LEVEL(FORM FIVE) NOTES - GENERAL CHEMISTRY (2)

    Loth Ellies, June 3, 2024 @ 9:24 pm Reply

    Keep it up but my phone don’t support Ecole app direct

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