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GENERAL CHEMISTRY.

Is a branch of chemistry which deals with behaviour and characteristics of electrons.

This subtopic deals with electron occupying space in an atom. The electron occurs in the orbit/ shell at the region called orbital. The orbital is the region where there is a 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:-

· Atomic 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’s atomic model.
                 • Rutherford atomic model.
                 • Wave particles duality nature of matter.
                 • Heisenberg uncertainity principle.


       DALTON’S ATOMIC THEORY

Dalton’s atomic theory includes the following main points:-

i. Matter is made up of small indivisible particles called atoms.

ii. Atom is neither created nor destroyed.

iii. Atoms of the 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 valid. This resulted into discover of modern atomic theory. The following include point of modern 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	₁¹P	1.00	Nucleus
Neutron	0	₀¹n	1.00	Nucleus
Electron	-1	_-1⁰e	1/1840	Shell

ii. Atom is neither created nor destroyed was not valid due to the existence of radioactivity therefore Atom can be created or destroyed by either nuclear fission or fusion.

iii. Atoms of the same elements are similar especially in mass was not valid due to the existence of isotopes. Atoms of the same elements have either same or different mass.

iv. Atoms of elements are different especially in mass was not valid. Atoms of different elements have either same or different mass.

v. Atoms of different elements when combine they do so in small ratio of whole number was not valid because different elements combine by using variable ratio of whole number.

THOMPSON's EXPERIMENT (DISCOVERY OF ELECTRONS)

       Thompson’s conducted an experiment to investigate if air conducts electricity. The following circuit was used during the experiment.

    • The emission tube have electrode at each end which is connected to the external circuit. The emission tube is connected to the vacuum pump in order to maintain the low pressure in        the emission tube.

The circuit is switched on which results into the following observation:-

   • The bulb emitted light which indicate that the gas conduct electricity.

   • There is glowing of emission tube or emission of light.

   • There is fluorescence of emission tube.

   • The stream of rays running from cathode to the anode. Through investigation of properties of cathode rays by using magnetic field, electric field and gold electroscope results into
      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. Is a 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. This state make atom to be stable.

Excited state: Is a state of an atom when electron filled in the highest energy level before filling lowest energy level available. If the electron excited jumps to the extent that the nuclear attractive force 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 to the highest energy level where the nuclear attractive force cannot act up on it. This electron cannot return back to the ground state it result the atom left to positively charged. The convergent limit is a factor which causes some electrons to move from cathode to anode. These were stream of rays called cathode rays which later was electrons.

SIGNIFICANCE OF CONVERGENT LIMIT

These include the following:-

  • It resulted into discover of the ionization energy. This ionization energy used in the inorganic section.

      • It resulted into formation of ion particles. The ionic particle is more reactive when take part during chemical reaction.

      • It 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 are arranged in an atom. The first simple model of the atom was proposed by J.J.Thomson in 1898.The Thomson atomic model is popularly known as the Thomson's "plum-pudding" model of the atom.

Thomson considered an atom to be a sphere (radius = 10^-10m or 10^-8 cm) of uniform positive charge into which the negatively charged electrons were embedded. This model is like plum-pudding dotted with raisins

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's scattering experiment. In this experiment, he bombarded a thin sheet (0.00006 cm thickness) of gold with alpha α particles. The α-particles were obtained from a radioactive substance. The -particles are doubly ionized helium atoms (He²*).

The scattered α -particles produced tiny flashes on striking with the zinc sulphide screen. These tiny flashes were observed with a movable microscope. The experimental set up used in the famous α-scattering experiment is shown in figure below. The following observations were made from the scattering experiment.

Structure of atom

The α -scattering from metal foils. The -particles are produced by a radioactive source.Since lead absorbs α -particles a lead plate with a hole is used to obtain a beam of a particles.The α -particles scattered from the metal foil strike the fluorescent (zinc sulphide) screen and produce tiny flashes. A movable microscope is used to view the flashes.

i. Most of the a-particles passed through the metal foil without any change in their path i.e. they remained undeflected.

ii. Some of the α -particles deflected through small angles.

iii. Only a few of them (1 in 10,000) were actually deflected by as much as 90°, or even larger angles. One in 20,000 particles returned back suffering a deflection of 180°.

Explanation. The results of the scattering experiment could not be explained by the Thomson's atomic. Calculations showed that a charge spread over a sphere of radius 10^-8 cm could deflect α -particles only through small angles. The deflections of onlyα -particles through larger angels as observed would be possible If the positive charge in the atom is spread over a sphere of radius of about 10^-13cm thus, the α-particles scattering result could not be explained by Thomson's atomic model.

RUTHERFORD'S NUCLEAR MODEL OF THE ATOM

On the basis of α-particle scattering experiment, Rutherford put forwarded in 1912, his nuclei model of the atom. According to this assumptions.

i. An atom consists of a positively charged nucleus surrounded by a system of electronics. Electrons are moving around the nucleus. The positive charge of the nucleus is due to the
protons present in it.

ii. Electrons and the nucleus are held together by coulombic force of attraction,

        iii. The effective volume of the nucleus is extremely small as compared to the effective volume of the atom. From the experiments, it was found that, Approximate radius of the
               nucleus of an atom 10^-14– 10^-15 m (10^-12 – 10^-13 cm), Approximate radius of sphere of electrons (or the radius of an atom) 10^-10 m (10^-8 cm).Since, the volume varies as r³,
               hence the volume occupied by the nucleus is about 10¹² times the volume of the atom.

iv. Almost the entire mass of the atom is concentrated in the nucleus.

v. The positive charge on the nuclei of different elements are always integral multiples of the electron charge, but opposite in sign. Since, each atom is electrically neutral, hence in
an atom, the number of positive charges on the nucleus of an atom is equal to the number of electrons in it. The Rutherford's model of an atom is shown in the figure below

Rutherford, in 1912, put forwarded an atomic model. According to this model an atom consists of positively charged nucleus which is surrounded by a system of electrons. The electrons and I nucleus are held together by electrostatic forces. Because of the electrostatic attraction between the nucleus and the electrons, the electrons should ultimately fall into the nucleus. But, it does not happen. In order to explain why electrons not fall into the nucleus, Rutherford postulated that the electrons are not stationary, but are revolving about the nucleus in orbits. But, this explanation did not solve the problem completely. The physicists had observed that a revolving electric charge must emit radiation and thus lose energy. Thus, an electron revolving around a nucleus in an orbit should emit radiation and lose energy

If so, the continuous loss of energy should slow down electrons. As a result, the electron will not be able to stand the attraction of the nucleus and gradually move towards the nucleus. The electron should, therefore, follow a spiral path and ultimately fall into the nucleus within 10^-8s. The atom should thus, collapse. But, it does not happen, as the atoms are stable. Thus, there must be something wrong in the Rutherford's nuclear model of the atom.

RUTHERFORD ATOMIC MODEL

           (i) The atom consists of extreme denser region at center called nucleus which have positive charged particles.
   (ii) The negative particles (electron(s) of atoms revolve around the nucleus in a path called orbit.
                 (iii) The large area of atom is empty space.

SHORT COMING OF RUTHERFORD ATOMIC MODEL.

Rutherford atomic model accounted that electron revolve around the nucleus in the path called orbit. This is an imagination because no any scientist proved occurrence but discovered that when charged particle revolve around the opposite charge reach a point when they collapse due to the attraction force occurs between the nuclear and electron(s) hence electron capture. But this proved the truth of Rutherford because no any atom discovered to collapse.

BOHR’S ATOMIC MODEL The electrons revolve around the nucleus in a path called orbits which have certain energy.
The electron when revolve in stationery state does not radiate energy.
The electrons emit or absorb energy when shift from one energy level to another.
The electronic motions are those which its angular momentum is integral multiple of where n = 1, 2, 3...

SHORT COMING OF BOHR’S ATOMIC MODEL

Bohr’s accounted that electrons revolve in a single plane. But discovered that movement of electrons is not restricted in only on plane.
The Bohr’s model did not explain spectral of multi-electrons atom but accounted only spectral line of uni-electron hydrogen atom.
The Bohr’s model did not account for electrons which were found in a chemical bond.
Bohr viewed an electron as being placed at a certain distance from the nucleus. However, it was proved by Weaver Heisenberg’s Uncertainity Principle:–It is impossible at any
movement to predict the exact position and velocity of an electron in an atom.

MASS SPECTROMETER

Is an instrument which is used to determine the relative atomic mass of an element.

Mass spectrometers have an ability to determine the relative molecular mass of a compound and formula of a compound. In a spectrographic plate.

Mass spectrograph is a plate of mass spectrometers which detect and record the relative atomic mass of an element. The spectrographic plate is a detector or recorder. The following is a diagram of mass spectrometer which has different part.

Vaporization. The instrument is evacuated in order to prevent the interference of air. The solid sample must be heated until it forms gaseous atoms and vaporized before it’s introduced into the mass spectrometer.

Ionization. The vaporized sample of element is introduced in the ionization chamber. This region has electric filament which emits fast moving electrons. These electrons collide with atom of an element as a result the atom split which leave only nucleus of positive charge. The nucleus left is where mass of atom is concentrated.

Acceleration. The ion particles formed in the ionization chamber are accelerated by two plate of negative charge toward the magnetic field. These two plates are connected to negative potential which accelerate ions as a beam of light toward the magnetic field.

Deflection. The beam of ions deflected in the magnetic field. The extent of deflection depends on the mass charge ratio m/e. The light particle deflected more than the heavy particles. The deflected ions strike on the detector. The ions at the same mass charge ratio strike on the same mass spot.

Detection. Once the ions strike the detector in the collector that converts the intensity of the ions into electric signals which is amplified by the amplifier into large electric current.

Recording. The current is used to operate a pen that moves on a paper tracing the peak of the isotope. The ion is recorded in terms of atomic mass and relative abundance. When they
fall on the photographic plate they produce a mass spectrum consisting of a series of line at different point. The mass of the ions detected on the photographic plate. Relative
abundance is the percentage of an isotopic atom in the element. The ions particles when strike the detector produce mass spectrum. The mass spectrum recorded as a peak.

The following include example of mass spectrum.

CALCULATION OF RELATIVE ATOMIC MASS

Relative atomic mass of an element is the average value of all the known isotope atomic weight relative to the proportional abundances. Relative atomic mass is more useful in chemistry than its simple atomic mass of element because simple atomic weight does not consider existence of isotopic forms of elements. The relative abundance of isotopic element is applied to determine relative atomic mass. Example isotopic element X contained Y% of _z^aX and W% of _z^bX, the relative atomic mass of element X is determined as given below.

Since y + w = 100

The length of peaks of each isotopic atom applied to find the relative atomic mass. Each atom have peak due to the different mass charge ratio m/e.

ATOMIC SPECTRUM

Definition: Are light waves which have definite line and colour because they have intermediate wave length which cannot be detected by a human eye. These spectrums have no harmful affect on human. Atomic spectrum produced once the atom gain energy which cause the electron to be excited and jump from lowest energy level. This results into an atom being unstable in order to maintain the stability these electrons return back to its group state which accompanied with release of energy inform of radiation. These radiation have wavelength which detected by a human eye of definite colour definite wave and definite line.

Continuous Spectrum: are spectrums which contain all possible frequency over wide range of energy. Continuous spectrums are colourless and have no definite line because they contain very short wave lengths which are not detected by an eye. The continuous spectrum recorded in a spectrographic plate as given below.

      Line spectrum: is a spectrum which consists of scattered definite line. These spectrums have very long wave length. The spectrographic plate of line spectrum

           Band Spectrum: Band spectrum are spectrums which consists of a group of definite line in small bands. The spectrographic plate of band spectrum includes the following:


BOHR-ATOMIC THEORY

(i) H - SPECTRUM

Definition:

Is a definite line and colour which resulted when electric discharge is passed through hydrogen gas in the emission tube under very low pressure. The H – spectrum recorded in the spectrographic plate. The following is a horizontal diagram of H-spectrum

UV	Violet	Infrared Red	Red
X - ray				Radio electron
- ray				Television wave
UV – ray
Invisible

Note: wave length increases

The explanation of horizontal diagram in terms of atomic structure:-

First band: is a colourless band or invisible band. These are spectrum produced by electrons which excited from the first shell. The electron from the first shell experience stronger nuclear attractive force which use high energy in order to jump toward highest energy level. When these electron returned back release high amount of energy which have shorter wave length. These wave lengths are not detected by a human eye. The radiations formed are continuous spectrum such as X – ray, sun rays etc.

Second band: is a visible band which has definite line and colour. These are line spectrum, produced by electron which excited from the second shell. The electron in the second shell need moderate energy in order to jump towards highest energy level when returned back release a normal energy which its wave length detected by a human eye.

Third band: Is invisible band. These spectrums are colourless and have no definite line.The electrons which produce these spectrum caused the scattered spectrum which appear as a colourless band. This is due to the lowest energy and highest wave length. The following include vertical diagram of H-spectrum.

The vertical diagram of Hydrogen Spectrum

Lyman series: is a series of spectrum which resulted by electron excited from the first shell (n =1). These series correspond to the invisible band or colourless band.

Balmer series: is a series of spectrum which resulted by the electron which excited from the second shell (n = 2) this corresponds with visible or violet band.

Paschen’s series: is a series of spectrum which resulted with electron excited from the third shell (n = 3) etc.

Bracket series: is a series of spectrum which resulted with electron which excited from the forth shell (n = 4) etc.

p-fund series: is a series of spectrum which resulted with the electron which exited from the fifth shell (n=5).

PLANCK’S QUANTUM THEORY

Planck put forward the Planck’s quantum theory. This theory has three main points which include the following;-

i. Any radiation should be association with energy.

ii. The energy is released inform of radiation, occur in small packets called quanta.

iii.The energy is directly proportion to the frequency.

E f

E = hf……………………..(1)

Since h = Planck’s constant

h = 6.63 x 10 ^-34 JS

n= 1

Since f =

E = ………………….(2)

C = 3.0 x 10⁸ ms^-1

REYDBERG EQUATION

Reydberg put forward a principle which applied to find wave length of spectrum. The wave length of H-spectrum determined is applied to find frequency and energy of the H- spectrum. The wave number is inversely proportional to the square of energy level differences ( n²).

1/V = RH n²

V = RH

Since V = 1/λ
= ………………. (3)

Where λ= wave length
n₁= lowest energy level

n₂= highest energy level

RH = Reydberg constant RH is 1.09 x 10⁷m^-1

The value of n₁ and n₂for H-spectrum can be obtained if given number of line and number of series. The value n₁ is equal number of series given. But the value n₂ is equal to the number of line plus the number of series. Example the value of n₁ and n₂of third line of Balmer’s series include the following:-

Number of line = Third line

Number of series = Balmer’s series

n₁ = number of series = 2

n₂ = number of line + Number of series

= 3 + 2

n = 5

But:

       = 1.09 X 10⁷m^-1[1/2²-1/5²]

           = 1.09 X 10⁷m^-1[1/4-1/25]

           = 1.09 X 10⁷m^-1[(25-4)/100]

           = 1.09 X 10⁷m^-1[(21/100]
λ = 4.45 x 10^-7m of third line

E=6.3 x 10^-3 Js x 3.0 x 10⁸ms^-1

4.45 x 10^-7m

E=4.25 x 10^-19J

Also;

f =

f = 6.74 x 10¹⁴s^-1

f = 6.74 x 10¹⁴Hz

TRANSITION ENERGY

Is the energy which is required to shift electron from one shell to another. The energy required to shift electrons from one shell to another should be equal to the energy difference between those two shells. The energy difference between lowest energy level E₁ and highest energy level E₂ is obtained by using the following expression;-

E = E2 – E1……………… (1)

The trend of transition of electrons include the following;

E = E₂ – E₁ - Transition take place from E₁ – E₂ exactly.

E > E₂ – E₁ – Transition take place from E₁ toward above E₂

E < E₂ – E₁ – Transition take place from E1 and hang between E₁ and E₂

But if electrons gain enough energy which is equal to the ionization energy result the electron to jump completely from ground state to infinite. These electrons cannot return back instead leave atom ionized positively. The energy supplied to the atom results to be excited and electron move completely from the ground state to the infinite. The energy supplied to the electrons is used as ionization energy and as kinetic energy. The ionization energy of electrons is equal to the amount of energy associated by electron in the shell or quantum number where it belongs.

The following expression is used to find the energy associated by electrons or energy of that shell.

1/λ = RH

Let n₁ = n

n₂= ∞

since 1/ = 0

1/λ = ………………………..(i)
E = hc/λ
E/hc = 1/λ ……………………..(ii)

Substitute (ii) into (i)

E =

Since R, h and C are constant

1eV = 1.6 x 10^-19J

1MeV = 10⁶eV

1MeV = 1.6 x 10^-13J

But discovered that the energy increase from the nucleus atom toward the high energy level. The energy increases from first shell toward seventh shell. From seventh shell the energy is zero or the energy at infinite is zero. Then the energy from infinite which is zero towards the lowest shell decreases and results to be negative value of energy

n∞ = 0.00ev

n₇ = -0.15ev

n₆ = -0.21ev

n₅ = -0.54ev

n₄ = -0.84ev

n₃ = -1.50ev

n₂ = -3.40ev

n₁ = -13.60ev

E = E₂ – E₁

= E – En

= 0 – En

= 0 –

E =

The negative value occurs because the energy of infinite is zero. Since energy increase from nuclear towards highest energy level result the energy of each shell be negative value. This formula is used to determine energy of electron at each shell.

Example

i. Find the first ionization energy of potassium.

ii. Find the wavelength, frequency and energy of third line of Balmer’s series ( RH = 1.09 x 107m^-1)

iii.If the wavelength of the first number of Balmer’s series is 6563Å. Calculate Reydberg constant and wavelength of the first member of the Lyman series.

iv.Find energy associated with electrons in the quantum number 2.

Solution:

i. E = I.E

k = 2:8:8:1

E =

= -1.36 x 10^-19J

E = E₂ – E₁

E = 0 - -1.36 x 10^-19J

E = 1.39 x 10^-19J

ii. n₁= 2

n₂ = 3 + 1

λ =

                          λ= 4.86 x 10^-7m

iii.    λ = 6563A
From

RH =

RH = 1.09 x 10^-3 A^-1

RH = 1.9 x 10 ^-3 x 1010m^-1

RH = 1.09 x 10⁷m^-1

iv. E =?

n = 2

E =

E = J

E= -5.44 x 10^-19J

Example

The U.V light has a wave length 2950 Å. Calculate its frequency energy 1A = 10^-10M

Given J

If the electron dropped from E₄ to E₂: Find the frequency and wave length energy release

Given line spectrum

Wave length

i. Which line has highest frequency and energy?

ii. Which line has lowest frequency and energy?

iii. Find the energy and frequency of each line in (i) and (ii) above?

	E₅
	E₄
	E₃
e⁰	E₂
	E₁

State either or not the transition of electrons would occur if energy supplied is

i. Greater than E₄ – E₂

ii. Equal to E₄ – E₂

iii. Less than E₄ – E₂

iv. Greater than E₃ – E₂ but less than E₄ – E₂

v. Smaller than E₄ – E₂

Solution:

a) = 2950A

f = c

f =

f = 1.02x10^-2 S^-1

E= hf

E = 6.3 x 10^-34Js x 1.02 x 10^-2 S^-1

E = 6.426 x 10^-36J

b) E₄ = -1.36 x 10^-19J

E₂ = -15.44 x 10^-19J

E = E₂ – E₁

E = E₄ – E₂

E = (-1.36 x 10^-19) – (-5.44 x 10^-19)

E = 4.08 x 10^-1

But:

E = hf

f =

f = 6.15 x 10¹⁴ S^-1

Hence:

F =

λ =

λ = 4.88 x 10^-7m

C. i) A line of wave length 2030A

ii) A line of wave length 8092 A

iii) f =?

E =?

Line of 2030A

Since f =

f =

f = 1.48 x 10¹⁵S^-1

E = hf

E = 6.3 x 10^-34Js x 1.48 x 10¹⁵

E=9.32 X 10^-19 J

D) Solution

i) Greater than E₄ – E₂result transition of electron from E₁ and above E₄

ii) Equal E₄ – E₂ transition take place from E₂ to E₄

iii) Smaller than E₄ – E₂ result transition of electrons from E₂ but cannot reach instead hang between E₂ and E₄

iv) Greater than E₃ – E₂ but smaller than E₄– E₂ result transition of electron from E₂ and above E₃ but cannot reach E₄

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 Broglie’s derived an expression which applied to find the De Broglie’s wave length.

De Broglie’s wave length is in terms of mass and momentum.

Hence from Einstein equation;

E = mc² …………………….(i)

From Planck’s equation

E = ……………… (ii)

Compare equation (i) and (ii)

mc² =

λ =

De Broglie’s wave length in terms of energy

From:

λ = multiply by 1/c in both side

λ = [1/c]

Since E = mc²

Example

Alpha particles emitted from radium have energy of 4.4MeV. What is the de-Broglie’s wave length?

The mass of moving particles is 9.01 x 10^-19g. What is the de-Broglie’s wave length?

The momentum of particles is 2.0 x 10^-10gm/s. What is the de-Broglie’s wave length?

Given that:

h = 6.63 x 10^-34 JS

C = 3.0 x 10⁸m/s

Solution

a. E = 4.4MeV

λ =?

De Broglie’s

λ =

λ = 4.52 x 10^-13m

b. Solution

m = 9.01 x 10-¹⁹g

λ =?

De Broglie’s

λ =

= 2.33 x 10^-24m

c. P = 2.0 x 10^-10gms

λ =? De Broglie’s

λ =

λ = 3.315 x 10^-24Jg^-1m^-1s^-1

HEISENBERG UNCERTAINITY PRINCIPLE

Heisenberg uncertainity principle state that, “It is impossible to determine position and momentum of electrons simultaneously with greater accuracy.” It is impossible to determined position and momentum of electrons because;-

(i)The size of electron is very small and as such radiations of high energy extremely small wavelength are required to detect it.

     (ii) Impact of these high energy photons changes both the direction and speed of the electron.
          Thus; the very act of measurement disturbs the position of electron. The uncertainties in the determination of these two quantities vary inversely. If one is determined fairly
          accurately, the other must be corresponding less accurate. The distance of electron: Is a position of electron from the nucleus and momentum of electrons is product of mass of
          electrons and velocity of the electron, Heisenberg put forward an expression which is used to determine uncertainity position and momentum of electrons.

Where: P – uncertainity momentum

X – Uncertainity position

P = x

P = …………………. (1)

Since = proportionality constant

Also

mc = …………………. (2)

Example

The uncertainity in the momentum of particles is 3.3 x 10^-16gms^-1. Find accuracy with which its position can be determined

X =

= 1.59 x 10^-29m

WAVE MECHANICS

Wave mechanics deal with electron occupying space. The electrons revolve the nucleus through orbit and occur in the orbital.

Orbital:Is a region within an atomic sublevel that can be occupied by maximum two electrons that have opposite spin. Any orbital can accommodate a maximum of two electrons.

The energy level of atoms is a specific region around the nucleus electrons can occupy in atoms. A shell is defined as a complete group of orbital possessing the quantum number. The

electrons arranged in the atom starting from orbit from nucleus toward the orbit of highest energy level. The following include position of locating electrons in the atom.

i. Principal quantum number (n)

The atom shells called principal quantum number (main energy level). The electrons filled in atom in order of increasing the energy from the nucleus. Each shell have specific constant number or electrons obtained taking

e = 2n²

Old name	1	2	3	4
New name	K	L	M	N

(ii)Subsidiary quantum number (L)

Subsidiary quantum number is a sub energy level. The sub energy level known as azimuthal quantum number, sub shell or sub energy level. The shell divides to form sub shell. The total number of sub shells in each shell is equal to the number of shell from the nucleus. This describes shapes of orbitals (sub-energy level).Values of subsidiary quantum number start from 0-------n-1

The names of sub shell include the following;

iii. Magnetic quantum number (m)

This is a quantum number which describes the orbital of each sub shell. This sub shell divided and form orbital. Each sub shells have specific number of orbital.

(iv) Spin Quantum number (s)

These quantum numbers describe the spinning of electrons in the orbital. Maximum number of electrons in the orbital in only two which spines in opposite direction. Since each orbital has two maximum electrons result the sub shell to have constant total number of electrons which accommodated in all shells S², p⁶, d¹⁰, f¹⁴, g¹⁸, h²² and i²⁶. In order to write electrons of element consider the knowledge of all four quantum numbers.

The following series of sub shell used to write the electrons structure.

This arrangement used to write the electronic configuration.

Arrangement used to write the electronic configuration has some irregularities or abnormality. This irregularity occurs because there is overlapping of orbital. The orbital of low energy jump toward highest energy level and orbital of high energy drop down towards lowest energy level. Example 4s arranged first before 3d orbital.

RULES WHICH GOVERN ARRANGEMENT OF ELECTRONS IN ORBITALS

1. Aufbau’s principle – State that "in the ground state of orbital the lowest energy is filled by electrons before filling orbital of the highest energy".

This means that electron filled in order of increasing energy level. The orbital of lowest energy when is full of electrons then the electron filled on the next orbital of highest energy level. Consider the following arrangement

2. Hund’s rule of maximum multiplicity

State that, “The electrons pair in the orbital of the same sub shell is allowed if all orbital of the same energy are occupied by electrons of maximum multiplicity. This means that the degenerate orbital are not allowed to pair up unless each orbital is singly occupied. The incoming pairing is done in an opposite spin. Consider a case with 5 electrons in valency shell.

3. Paul’s exclusive principle

States that, “Two electrons in an atom can never have exactly similar set of all quantum numbers.” This means that the maximum number of electrons which is occupied in the orbital is only two. Two electrons may occupy the same orbital only on condition that they have their spin in the opposite direction. Two electrons may have three quantum numbers the same; n, l and ml but the fourth, ms must be different.

Sometimes the electrons structure or electronic diagram of ions is written according to what kind of ions given.

There are two kinds of ions which include the following;-

Cation or metallic ion is formed through losing of electrons. The electrons which are lost occur in the orbital out the of the noble gas structure. The electron removed starting from orbital of low energy toward the orbital of highest energy.

Anion or non-metallic ion – is formed by gaining electrons. Electrons which gain is added in the orbital having unpaired electrons starting from orbital of lowest energy to the orbital of highest energy level

QUANTUM NUMBER

Is a number which describe the characteristics of electrons or is a number which is used to characterize electrons as they occupy orbitals in different energy levels. The quantum number is a number which describe main energy level, orientation of orbitals and spinning of electrons.

There are four type of quantum number which includes the following:-

1.Principle quantum number (n): Is a number which specifies the location and the energy of electron. This is a number which describe the main energy level of electrons. The principle quantum number called shell orbit, main energy level and stationary state. The principle quantum number have specific name from the nucleus. K, L, M, N etc.

2. Subsidiary principle quantum number (L): Is a number which specifies the shape of orbitals. Subsidiary principle quantum number is a sub energy level. The sub energy level is known as azimuthal number sub shell or sub energy level. The shell is divided to form sub shell. The total number of sub shells in each shell is equal to the number of shells from the nucleus. The names of sub shells include the following; s, p, d, f, g, h, i. The value of sub energy level i.e. 0, 1, 2, …L = n - 1.The value of L must be smaller that n because L = n – 1

n	Value of L	Sub shell
1	0	1s
2	0, 1	2s2p
3	0, 1, 2	3sp3d
4	0, 1, 2, 3	4s4p4d4f
5	0, 1, 2, 3, 4	5s5p5d5f5f

3. Magnetic quantum number: Is a number which specifies the number of orbital present in a given value of subsidiary quantum number. This is a quantum number which describes the orientation of orbital. The magnetic quantum number is a number of orbital. When the sub shell divides form orbital, these orbitals are called Magnetic quantum number.

1s – Orbital S = X

4. Spin quantum number (s) – This is a quantum number which describes the direction in which the electrons spin. There are two maximum electrons, these electrons are moving in opposite direction. They revolve clockwise + ½ and another revolve anti – clock wise – ½

APPLICATION OF QUANTUM NUMBER

Quantum number is applied to find the number of sub shell (L), number of orbital (m) and total number of electrons (s) of the given quantum number (n). There are two methods which are applied to find total sub shells, orbital and electrons.

Find total number of electrons in the principle quantum number 2.

(By using tree diagram)

Total shells = 2 (s, p)

Orbitals = 4 (0, 0, +1, -1)

Electrons = 8

Also: by using tabular form

(b) Find the total number of electrons in the principle quantum number 3

Soln: (by using tree diagram)

Example

(a) The modern theory of electron behavior is based on two assumptions. State them

(b) Define the following terms

i. Orbital.

ii. Energy levels of atoms.

iii A shell.

iv. Quantum.

Solution:

(a) The modern theory of electron behavior is based on two assumptions.

(i) An electron has a dual nature i.e. it behaves both as a particle and wave.

(ii) It is practically impossible to determine simultaneously both the position and momentum of an electron with any degree of precision.

(b) (i) An Orbital is a region within an atomic sublevel that can be occupied by a maximum of two electrons that have opposite spin. There are s,p,d, and f orbitals is a volume of space surrounding the nucleus within which there is a more than 95% chances of finding electrons. Any orbital can accommodate a maximum of two electrons.

(ii) Energy level of atoms in specific region around the nucleus that electrons can occupy in atoms.

(iii) A shell is defined as a complete group of orbital possessing the same quantum number.

(iv) Quantum is the smallest countable, discrete packet or increment of radiant energy that can be absorbed or emitted.

Example

Explain the meaning of the following terms:

(a) Quantization of energy and radiation.

(b) Wave particles duality of matter.

Solution

(a)Light and other forms of electromagnetic radiations are not emitted continuously but in discrete “packets” called photons. Energy of an electromagnetic radiation is also not emitted continuously but in “packets” called quanta. A photon of radiation carries a quantum of energy. Energy and radiation are quantized in that electron shall fall from one energy level to another and not anywhere else in between.

(b) According to de Broglie, every sub – atomic particles has both wave and particle properties. Photons of light and other electromagnetic radiations, are regarded and wavelengths. They are regarded as waves because they have specific frequencies. Although electrons have extremely small mass, they have both velocity and momentum as particles.

(c) Quantum numbers are used to characterize electrons as they occupy orbitals in different energy levels. Each electron is characterized by four quantum numbers:-

The principal quantum number, n, which specifies the location and energy of the electron.

The azimuthal, subsidiary or angular momentum quantum number, l, which specifies the shapes of the orbitals.

The magnetic quantum number, m, which specifies the orbital orientation in space.

The spin quantum number, s, which specifies the direction in which the electron is spinning. These quantum numbers completely describe the stationary stage of the electron.

Example

(a) Distinguish between the following terms:-

(i) An orbital and orbit.

(ii) S – Orbital and P – orbital.

(iii) A quantum of light and quantization of energy.

(iv) Quantum shell and quantum number.

(b) Explain briefly the meaning of the following quantum numbers:-

(i) n

(ii) l

(iii) m

(iv) ms

(c) In a tabular form specify all the four quantum number for each electron in an atom whose n value is 2. Given all the orbitals are full of electrons.

(d) Given the value of the quantum numbers n, l and m for the electron with the highest energy in sodium atom in the ground state.

(e) Write down all the quantum number of all the electrons in the ground state of nitrogen atom.

(f) Give the values of all the four quantum number for 2p electrons in Nitrogen.

(g) Briefly explain why the following quantum numbers are not allowed.

i) n = 1, l = 1, m = 0

ii) n = 1, l = 0, m = 2

iii) n = 4, l = 3, m = 4

iv) n = 0, l = -1, m = 1

v) n = 2, l = -1, m = 1

Solution

(a) (i) An orbit is a well define circular path in which electrons were assumed to revolve around the nucleus. Whereas an orbital is a three dimensional region of space around the nucleus whereby there is high probability of finding an electron.

(ii) S–Orbital is spherical and hence non directional whereas p-orbital is dumbbell in shape and it is directional.

(iii) A quantum of light is a photon of radiation emitted when an electron jumps from one energy level to another whereas quantization of energy means energy emitted or absorbed following electron transition between one energy level and another is not continuous rather it is in form of small packet called quanta.

(iv) A quantum shell is the energy level of an electron in a given atom whereas quantum number refers to specifies way of defining an electron in a given atom whereas quantum number refers to specified way of defining an electron in a given orbital of an atom using n, l, m and ms values.

(b) (i) n – represents the principle quantum number;

It specifies the location and the energy of an electron.

It is a measure of the volume of the electrons cloud.

As the value of n increases or becomes less negative it means the energy levels of the electrons get further away from the nucleus n can have only integers 1, 2, 3, 4 to infinite represented by K,L,M,N etc. Also called azimuthal quantum number or subsidiary quantum number, specifies the shapes of the orbital, when n = 1 there’s only l s orbital where n = 2, there’s 2s and 2p orbital, thus for a given value of n, l, are all from 0, 1, 2…up to l = n - 1. Thus when n = 4 values are 0, 1, 2 and 3. Which represents 4s, 4p, 4d and 4, â„“ sub–levels m Magnetic quantum number specifies the number of orbital present in a given value of â„“.

It specifies the orientating i.e. the direction of the orbital to magnetic field in which it’s placed.

It accounts for the splitting of the spectral lines observed when an atom emitting radiation is placed in a magnetic field, ms – Refers to spin quantum number.

(i) It indicates the direction in which the electron spins.

(ii) Only two values are allowed for an electrons i.e. electrons may spin clockwise, shown as +1/2 or may spin anticlockwise as shown as ½ or

(c)

Principle (n)	Azimuthal	Magnetic quantum	Spins ms
n = 1	â„“ = 0	m = 0	⁺¹/₂, ^–1/₂
n = 2	â„“ = 0 â„“ = 1	m = 0 m = -1 m = 0 m = 1	⁺¹/₂ , ^–1/₂ ⁺¹/₂ , ^–1/₂ ⁺¹/₂ , ^–1/₂ ⁺¹/₂ , ^–1/₂ Total of 10 electrons

(d) When n = 3 in Na atom that electron is found in s – orbital thus â„“ = 0 and also ml = 0

(e) 1 electrons:

(1^st) n = 1, l = 0, m = 0, s = +¹/₂ (2^nd) n = 1, l = 0, m = 0, s = ^-1/₂

Electrons:

1^st n = 1, l = 0, m = 0, s = ⁺¹/₂ (2^nd) n = 1, l = 0, m = 0, s = ^-1/₂

2 Px; n = 2, l = 1, m = ⁺1, s = ⁺1/2 2Py; n = 2, l = ^-1, m = ^-1, s = ⁺1/2
2Pz; n = 2, l = 1, m = 0, s = ⁺1/2

(f) 2P electrons, there’s

2Px; n = 2, l = +1, m = +1, s = +1/2 2Py: n = 2 , l= ^-1, m = ^-1, s = ⁺1/2

2Pz: n = 2 l = 1 m = 0, s = +1/2

(g) (i) l value must be smaller than n value since l = n-1

(ii) When l = 0 value of m is only 0

(iii) For l= 3, m can range from^-3 to ⁺3. Thus m=⁺4 is not allowed

(iv) n value cannot be equal to zero.

(v) l value cannot be a negative number.

Example

(a) briefly explain the meaning of the following quantum numbers:-

(i) ml

(ii) n

(iii) l

(iv) ms

(b) If n = 2, tabulate the values of 1, _ml and ms.

(d) Briefly explain why an electron cannot have the quantum number n = 2, l = 2 and ml = 2

Solution

(a) (i) ml – Magnetic quantum number with values from -l to +l. These are numbers showing the number of sub – level in each energy level.

– Describes orientation

(ii) n – Principal (shell) Quantum number value of 1, 2, 3 etc theses are numbers used to represent the main energy level in which an electrons is found.

(iii) l – Azimuthal or subsidiary or Angular or orbital Quantum number. They specify the shape of orbitals.

(iv) ms – Magnetic spin Quantum number with values ⁺1/2 and ^-1/2 shows the spinning of electrons in orbitals (clockwise and anticlockwise direction).

(a) For n = 2, the scheme below illustrates the value of 1, ml and ms.
         l=0, 1
         ml=-1, 0, +1, 0
         ms= ±1/2, ±1/2, ±1/2, ±1/2

Therefore the total number of electrons = 8

(b) If n = 3, possible values of 1 and m are;

l = 0, 1 and 2

m = 0 for l = 0

m =^-1, 0 and ⁺1 for l = 1

m = ^-2, ^-1, 0, ⁺1 and 2 for l = 2

Note that: l=n-1

(c) An electrons cannot have quantum numbers, n = 2, l = 2 and m = 2. This is because when n = 2, l can be 0 and 1; then ml can be 0, ^-1, 0, ⁺1 different values which can occupy in orbitals (i.e. they have n, l and m values the same but they must spin in opposite directions and hence have different ms values).

RULES FOR FILLING ELECTRONS IN THE ORBITALS OF AN ATOM

The orbitals of an atom can be filled with electrons by applying the following rules.

i)Aufbau Principle

This is also known as the building up principle. According to this principle "The electrons in an atom are so arranged that they occupy orbitals in the order of their increasing energy."

Thus, the orbital with the lowest energy will be filled first, then the next higher in energy, and so on. Since, the energy of an orbital in the absence of any magnetic field depends upon the principal quantum number (n) and the azimuthal quantum number (l), hence the order of filling orbitals with electrons may be obtained from the following generalizations.

a) The orbitals for which n + I is the lowest is filled first.

b) When two orbitals have the same values of n+l, the orbital having the lower value of n is filled first.

The order of filling of various orbitals with electrons obtained by this rule is given below:

1s, 2s. 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f,5d, 6p, 7s, 5f…

To remember this sequence may be a difficult task. Given a long side is a simple way of working out this order. In this method a series of arrows running from upper right to the lower left gives the order of orbitals with increasing energy.

ii) Pauli’s Exclusion principle

In 1925 Wolfgang Pauli discovered what is known as the exclusion principle. This principle is very useful in constructing the electronic configuration of atoms. According to this principle “No two electrons in an atom can have the same values for all the four quantum numbers”.

For example in 1s orbital of helium atom there are two electrons. According to the concept of quantum number and Pauli’s rule, their quantum numbers are:

The + and - sign before j refers to the clockwise and anticlockwise spins of the electrons.

Thus, the two electrons having the same values of n, I and m could have different values

of s, i.e., their spins are in the opposite directions. This leads to a very significant observation that

"Each orbital can accommodate at the maximum two electrons with opposite spins."

Applications of the Pauli's exclusion principles

The Pauli's exclusion principle leads to the following conclusions:

(a) An orbital cannot have more than two electrons.

(b) In any main energy level (shell), the maximum number of electrons is twice the in number of orbitals, or is equal to 2n², where n is the principal quantum number.

iii) Hund's Rule of Maximum Multiplicity

The rules discussed above do not give any idea for filling electrons into the orbitals having equal energies (such states are called degenerate states). For example, three p-orbitals, i.e. px, py and pz, have equal energy. How should electrons be filled into these orbitals? Let us take an example in which three electrons are to be filled into three p-orbitals. The three electrons can be filled into three p orbitals in two different ways as shown below.

Now, which of the two is correct? The answer to this question is given by Hund's rule, Hund's rule states that,

"When more than one orbitals of equal energies are available, then the electron*,] first occupy these orbitals separately with parallel spins. The pairing of electrons will | only after all the orbitals of a given sub-level are singly occupied."

According to the Hund's rule, the correct way of filling three electrons in three p orbitals that in which each orbital is singly occupied, (arrangement II above).

Hund's rule is also known as the Hund's rule of maximum multiplicity.

Explanation

Two electrons with parallel spins, tend to be as far apart as possible to minimize the electrostatic repulsion. Therefore, the electrons prefer to occupy the orbitals singly
as far as possible. When all the orbitals get singly occupied, then the incoming electron has two choices either to pair up with the other electron or to go to the next higher orbital.

When vacant orbital of suitable energy is not available, then the incoming electron will have no choice except to pair up with another electron.

Example

(a) State the;

(i) Heisenberg uncertainty principle.

(ii) Hund’s rule.

(iii) Paul’s Principle.

(iv) Aufbau’s Principle.

(b) Brief explain why the following sets of quantum number are NOT allowed in hydrogen atom:

(i) n = 1, l = 1, ml = 0

(ii) n = 1, 1 = 0, ml = 2

(iii) n = 4,1 = 3, ml = 4

(iv) n = 0, 1 = 0, ml = 0 and

(v) n = 2, 1 = ^-1, ml = 1

(i) 1s

(ii) 2p

(iii) 3d

(iv) 4f

(d) How many sublevels are there in each of the following shells?

(i) K

(ii) N

(iii) L

(e) Which of the following electronic configuration is correct for 14p in its ground state?

State the principle violated in each case

Solution

(a) (i) The Heisenberg’s uncertainly principle states that; “it is impossible to measure simultaneously the exact position and exact velocity (momentum) of an electron.” Any attempt to measure one quality will distract the measurements of the other quantity

(ii) Hund’s rule states that, “electron pairing is not allowed until all orbitals of the particular energy sub – level are occupied by at least one electron.”

(iii) Paul’s Exclusion Principle, “states that no two electrons may have all the four quantum number the same.” Two electrons may have three quantum numbers of the same; the paired electrons are always in opposite direction at the same time.

(E.g. and ).

(iv)Aufbau Principle states that, “In electrons configuration, electrons are filled in order of increasing energy levels.” Thus the lowest energy available must be filled up first. The orbital energy increase in the following order; is 2s, 2p, 3s, 4p, 5p etc.

(b) (i) n = 1, 1 = 1, ml = 0 this is not allowed because l must be smaller than n, in the case l = 1 which is not allowed.

(ii) n = 1, 1 = 0, ml = 2. This is not allowed because 1 = 0 and ml = 0

(iii) n = 4, 1 = 3, ml = 2. This is not allowed because for 1 = 3 can range from ^-3, ⁺3, thus, ⁺4 is not allowed.

(iv) n = 0, l = 0, ml = 0; this is not allowed because n cannot be zero.

(v) n = 2, l = -1, ml = 1. This is not allowed because l cannot be a negative number and n can never be = 0

(ii) 2 p sub energy has three p – orbital (which is 1 s – orbital).

(iii) 3 d sub energy level has five d – orbital [which are 3 d (x² – y²), 3dz² 3dxy, 3 and 3dyz].

(iv) 4 f sub energy level has seven f – orbital [which are 4xy, 4fyz, 4fz, 4f(x² – y²),

4f(x²-z²) and 4f (y² – z²)].

(d) (i) K has one sub level (which is l s²).

(ii) N has four sub levels (which are 4s², 4p⁶, 4d¹⁰ and 4f¹⁴).

(iii) L has two sub levels (which are 2s2 and 2p6).

(e) Is the correct electronic configuration of 15p

This is correct since it obeys principle governing electron in an atom distribution

(i) [(Ne)]

(ii) [(Ne)] aufbau (building up) principle is violated as it states that electrons available as the case here.

(iii) [(Ne)]

(iv) [(Ne)] Hund’s Rule of maximum that multiplicity is violated as it states electrons occupy orbital as singly as possible

Example

(a) State the rules used in filling electrons in various orbitals of an atom.

(b) By using 3 different ways for each of the following atoms write their electron configuration

(i) C

(ii) N

(iii) Ag

(c)Write the electronic configuration of Na⁺ and F^- then show the other element whose electronic configuration resembles these ions. Show that electronic configurations of Cu and Cr are unusually written. Give reasons.

Solution

(a)(i)Paul’s exclusion principle states that; “no two electrons may have all the four quantum number the same i.e. electrons may occupy the same orbital only on conditions that they have their spins in the opposite direction.”

(ii) Hund’s rule of maximum multiplicity states that, “when more than one orbital with equal energies are available, electrons tend occupy those orbital separately first with parallel spins and separately first with parallel spin and pairing of electrons will start only after all the orbitals of a given sub level are singly occupied.”

(iii) Aufbau principle states that, “Electrons in an atom are so arranged that they occupy the orbitals in the order of their increasing energy.”

(b) (i) C Atomic number 6

Using Box method:

1s² 2s² 2p²

Using noble gas structure;

[He] 2p²

(ii) N atomic number 7

Using box method

Using sub levels only;

1s² 2s² 2p³

Using noble gas structure;

[He] 2p³

(iii) Ag atomic number 47

Using box method;

Note: 4d is filled first to maintain stability of full filled d – orbital before 5s is filled.

Using sub levels only;

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 4d¹⁰5s¹

Using noble gas structure;

[Kr] 4d¹⁰ 5s¹

(c)

(i) Na⁺ number of electrons also 10

1s² 2s² 2p⁶ resembles Ne

(ii) F – number of electrons also 10

1s² 2s² 2p⁶ resembling Ne

(d) Cu Atomic number 29

e.g. [Ar] 4s¹ 3d¹⁰

Cr Atomic number 24

e.g. [Ar] 4s¹ 3d⁵ full filled and half filled orbitals are very stable electronic structures. The stability gained is the cause of 4s electrons to be unpaired and make 3d orbitals either full filled or half filled.

Note that: In writing the electronic configuration the atoms or orbitals of the same value are usually written together irrespective of their relation energy levels.

E.g. Cu 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹ not 1s² 2s²2p⁶ 3s² 2p⁶ 4s¹ 3d¹⁰. To save time and space noble gases are often utilized e.g. sodium [Ne] 3s¹ and Iron [Ar] 3d⁶ 4s²

CHEMICAL BONDING AND MOLECULAR STRUCTURE

Definition:-chemical bond is the chemical force /force of attraction which keeps the atoms in any molecule together.

Types of the chemical bond

i) Ionic/Electrovalent bond

ii) Covalent bond

iii) Coordinate bond

iv) Polar covalent bond

v) Non-polar covalent bond

vi) Hydrogen-bonding

i) IONIC/ ELECTROVALENT BOND.

An ionic or electrovalent bond is formed by complete transfer of one or more electrons from the atom of the metal to that of a non-metal.

Note: - As a result of electron transfer the following changes occur in the reacting atom:-

(a) Both the atoms acquire stable noble gas configurations.

(b) The atoms that loses its electrons becomes positively charged ion called cation, the atom which gain this electrons becomes negatively charged ions called anion.

(c) The two oppositely charged ions .i.e. the cation and the anion are then held together by the coulombic force of attraction to form an ionic bond .

Note:- During the formation of ionic bond, a certain amount of energy is released.

e.g Na + Cl → Na⁺cl^-

This ionic bond may be defined as the coulombic force of attraction which holds the oppositely charged ions together.

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

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)

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

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

(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

(iii) Formation of molecules having triple bond :-
(a) Formation of nitrogen (N₂) molecule.

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.

(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

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.

(ii) POLAR COVALENT BOND
When a covalent bond is formed between two atoms of different elements, the bonding pair of electrons does not lie exactly midway between the two atoms. It lies more towards the atom which has more affinity for electrons.
The atom with higher affinity for electrons thus develops a slight negative charge and the atom with lesser affinity for electrons a slight positive charge, such molecules are called polar molecules.The covalent bond between two unlike atoms which differ in the affinities for electrons is said to be a polar covalent bond.
Example of polar covalent bond are hydrogen chloride (H Cl) , water (H₂O)

CAUSES OF POLARITY IN BONDS
The cause of polarity in bond is due to the electronegativity difference.
Example of the electronegativity difference between hydrogen and halogen atoms is as follows:-
halogen atoms: (F,Cl,Br,I)

         electronegativity differences:-

H – F    > H – Cl       > H – Br      > H – I
      (4- 2.1)          (3.0- 2.1)              (2.8- 2.1)        (2.5- 2.1)
         1.9                 0.9                         0.7             0.4

VALENCE SHELL ELECTRON PAIR REPULSION (VSEPR) THEORY.

The VSEPR theory was proposed by R.J Gillespie and R.S Nyholmn in 1957. This theory was developed to predict the shapes of the molecule in which the atom are bonded together with single bond only. This theory is based on the repulsions between the electron pairs in the valence shell of the atoms in the molecule. The main postulate of the vsepr theory are:-

     (i) The geometry of a molecule is determined by the total number of electron pairs (bonding and non-bonding) around the central atom of the molecule.
          The shape of the molecule depends upon the orientation of these electron pairs in the space around the central atom.
     (ii) The electron pairs (shared or lone pairs) around the central atom in a molecule tend to stay as far away from each other as possible so as to minimize the repulsion forces
           between them.

(iii) The strength of repulsion between different electron pairs follows the order.

Lone pair-lone pair > lone pair-shared pair > shared pair-shared pair
The shared pair of electrons are also called bond pair of electrons.The presence of lone pair(s) of electrons on the central atom causes some distortions in the expected regular shape of the molecules.

PREDICTING THE SHAPE OF MOLECULES ON THE BASIS OF VESPR THEORY

(i) Molecules with two bond pairs
In a molecule having two bonds pairs of electrons around its central atom, the bond pairs are located in the opposite side (at an angle of 180º) of the central atom so that the repulsion between them is minimum.

(ii) Molecules with three bond pair

In a molecule having three bond pairs of electrons around its central atom, the electron pair form an equilateral triangular arrangement around central atom. Thus the three bond pairs at 120º with respect of each other. Thus the molecule having three bond pairs around its central atom have trigonal planar (triangular planar shape).

(iii) Molecules with four bond pairs
In a molecule having four bond pairs of electrons the four bond pairs are arranged tetrahedral around the central atom. Thus the four band pairs are at an angle pf 129º 28’ with respect to each other. Therefore the molecule having four bond pairs around its central atom has a tetrahedral shape.

(iv) Molecules with five bond pairs
Five bond pair orient themselves around the central atom in a trigonal bipyramidal way. Three bond pairs are in a plane called equatorial plane, and oriented at an angle of 120º with respect to each other.
Thus a molecule having five bond pairs around its central atom has a triangular bipyramidal shape.

Therefore the molecule of the type shape AB₅ are trigonal bipyramidal in shape e.g PCl₅, PF₅, SbCl₅.

(v) Molecules with six bond pairs
Six bond pairs in a molecule are distributed octahedral around the central atom. Thus a molecule having six bond pairs around its central atom has an octahedral shape. Thus the molecules of the type AB₆ are octahedral. Thus molecules SF₆ has an octahedral

SHAPES OF THE MOLECULES HAVING BOND PAIRS AND LONE PAIR ELECTRONS.

The pair of electrons in the valence shell of an atom which is not involved in bonding is called lone pair of electrons.

For example the nitrogen atom in ammonia molecules (NH₃) has one lone pair of electrons, the oxygen atom in water molecule (H₂O) has two lone pairs of electrons.

(i) Molecules having three bond pairs and one lone pair.
A molecule having three bond pairs and one lone pair of electrons, thus has in all four pairs of electrons around its central atom.
Therefore these four pairs of electrons are distributed tetrahedral around the central atom.

(ii) A molecule of AB₃ type has a triangular pyramidal shape
Typical molecules of this type are NH₃, NF₃,PCl₃ and H₃O⁺.

(iii) Molecules with two bond pairs and two lone pairs
The four electron pairs (two bond pair + two lone pair ) are distributed tetrahedral around the central atom

The two lone pairs are the central atom repel the bond pairs slightly inward due to greater lone pair-bond pair repulsion. As a result the bond angle in such a molecule is less than the tetrahedral vale of 109º 28’
The presence of only two bond pairs is the molecules gives a bent shape (inverted V- shaped) to the molecule. Example include H₂O, H₂S, F₂O and SCl₂.

(iv) Molecules with four band pair and two lone pairs
The four bond pair are distributed in a square planar distribution. The two lone pairs are in a direction at right angles to this plane.

EXERCISE:-
(i) predict the shapes of the following molecules following the VSEPR theory.

         (a)Ammonia (NH₃) molecules
         (b)beryllium chloride(BeCl₃)
         (c)water (H₂O) molecule
         (d)silicon tetrachloride (SiCl₄)

HYBRIDIZATION
Definition:- Is the process of mixing of the atomic orbitals to form new hybrid orbital.
Note:- All hybrid orbitals of a particular kind have equal energy, identical shape and are symmetrically oriented in space.

CHARACTERISTICS OF HYBRID ORBITALS

(i) The number of hybrid orbitals formed is equal to the number of the atomic orbitals participating in hybridization.

(ii) All hybrid orbitals are equivalent in shape and energy but different from the participating atomic orbitals.

(iii) A hybrid orbital which takes part in the bond formation must contain only one electron in it.

(iv) A hybrid orbital, like atomic orbital cannot have more than two electrons. The two electrons should have heir spins paired.

(v) Due to the electronic repulsions between the hybrid orbitals they tend to remain at the maximum distance from each other.

TYPES OF HYBRIDIZATION
The types of hybridization shown by an atom depends upon the requirement of the reaction.

(a) SP³(ess-pee three) hybridization (or tetrahedral hybridization)

(b) sp²(ess-pee two) hybridization (or trigonal hybridization)

(d) s,p & d) hybridization

(a) SP³ HYBRIDIZATION
→ S + 3-P orbitals under mixing to provide new hybrid orbitals.
e.g    S + (P_x+ P_y+ P₂)  →      SP^{3

one s- orbital three p- orbital
   four hybrid orbital}
Each of these hybrid orbitals has 25% S-character and 75%   P-character. These four sp3 hybridized orbitals are directed along the four corner of a tetrahedron and are inclined to each other at an angle 109º 28’
Examples of sp³ hybridization is that of methane (CH₄), the process of hybridization involves the following

During exited state of carbon one electron is promoted from 2S to 2P orbitals .The process of promotion of an electron from 2s to 2p orbital of the carbon atom and subsequent hybridization is illustrated below.

During hybridization/ During excited state ,S is promoted to P orbitals as shown below:

Shape of S is spherical and shape of p is damp bell, then S hybridized with p shape.

Hybrid between S + P

(b) SP² HYBRIDIZATION (trigonal hybridization)
Example of SP² hybrid orbital are the formation of ethane molecule(C₂H₄)

Note:- The three SP² hybrid orbitals are oriented in a plane along the three corners of an equilateral triangle .i.e. they are inclined to each other at an angle of 120º the third p - orbital (say P²) remains unchanged each hybrid orbital has 33.3% s - character and 66.7% p - character

SP² hybridization of Ethene (C₂H₄)/Ethylene

(i) Structure of ethane.

(ii) Electronic configuration of carbon.

(iii) During excitation hybridization takes place. Then;

(c) SP hybridization(or linear hybridization)

Example of SP hybridization is the formation of ethyne(acetylene) molecules(c2h2)

(i) Structure of ethyne (acetylene)molecule

E.g (i)

(ii) Electronic configuration of carbon from ground state.

(iii) During excited state , S is promoted to P - orbital to form hybrid orbital.

(a)

Hybrid shape for carbon 1

(b)

Hybrid shape for carbon 2

Then C₁ combine with C₂

HYDROGEN BONDING

If a hydrogen atom is bonded to a highly electronegative element such as fluorine , oxygen, nitrogen, then the shared pair of electrons lies more towards the electronegative element. This leads to a polarity in the bond in such a way that w slight positive charge get developed on H- atom.

The bond between the hydrogen atom of one molecule and more electronegative atom of the same or another molecule is called hydrogen bond

CONDITION NECESSARY FOR THE FORMATION OF HYDROGEN BOND

(i) Only the molecule in which hydrogen atom is linked to an atom of highly electronegative element are capable of forming hydrogen bond.

(ii) The atom of highly electronegative element should be small.

SOME TYPICAL COMPOUND SHOWING HYDROGEN BONDING

(i) Hydrogen fluoride( HF)

Structure of hydrogen fluoride:-

(ii) Water (H₂O).
Water is a typical polar compound which exhibit hydrogen bonding as shown below:-

Structure of water molecule.

Hydrogen bonding that water exist in the associated form

TYPE S OF HYDROGEN BONDING
There are two types of hydrogen bonding

(i) Intermolecular hydrogen bonding

(ii) Intramolecular hydrogen bonding

(i) INTERMOLECULAR HYDROGEN BONDING.

When the hydrogen bonding is between the h-atom of one molecule and an atom of the electronegative element of another molecule.
For example , hydrogen bonding in water, ammonia etc

(ii) THE INTRAMOLECULAR HYDROGEN BONDING

Is between the hydrogen of one functional group and the electronegative atom of the adjacent functional group in the same molecule. For example the molecule of o-nitrophenol show intramolecular hydrogen bonding. The p-nitrophenol shows intermolecular hydrogen bonding.

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.

(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.

(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.

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 2s²2p⁶ 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)

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

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

(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 1s²2s² 2p²x 2p⁰y 2p⁰z not correct for the ground state of nitrogen?

(e) Mention the disobeyed law from the following arrangements;-

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.

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: = R_H

(i) What do symbols λ, R_H, 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 (R_H = 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.

(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] 4s²3d³
₁₆S = [Ar] 3s²p⁴
₉F = [He] 2s²p⁵
₃₄Cs = [Ar] 4s²d¹⁰ 4p⁴
₁₆In = [Kr] 5s²4d⁸
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 10⁵m/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 5800A^o. 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

(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 3^rd line of visible spectrum.

(RH=1.097 x 10⁷m, C=3 x 10⁸m/s)

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

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