De broglie atomic model

Introduction :
Bohr assumed that an electron is a particle and postulated that it revolves around the nucleus in an orbit in which the angular momentum ( mvr ) of the electron is an integral multiple of   `(h)/(2pi)` .
                      m v r  =  n`(h)/(2pi)`   .
But according to de-Broglie an electron in motion with high speed is associated with wave nature. For an electron moving around the nucleus in the circular path, two cases of electron waves of different wavelengths are possible .

In one case the circumference of the electron orbit ( 2`pi`r ) is an integral multiple of the wavelength ( `lambda` ) of the electron. This is shown in the figure (1). In the second case, the circumference of the electron orbit is not an integral multiple of the wavelength . This is shown in figure (2) .



In the first case of atomic model above, the two ends of the electron wave meet to give a regular series of crests and troughs. In this case the electron waves are said to be in-phase. This means that there is constructive interference of the electron - waves. In such a case, the electron is considered to behave as a standing wave or non-energy radiating wave. This extends round the nucleus in a circular orbit.
In the second case of atomic model above, here the circumference of the Bohr's orbit ( 2`pi`r ) is bigger or smaller than n`lambda` , the electron waves are said to be out of phase. This is shown in the figure (2). In this case, a destructive interference of waves occur causing radiation of energy.
Therefore the necessary condition to get an electron - wave in phase is that the circumference of the Bohr's orbit ( 2`pi`r ) is equal to integral ( whole number ) multiple of the wave length (`lambda`) of the electron - wave  
                       n`lambda`   =   2`pi`r    
                        `lambda`    =    `(2pir)/(n)`     ;   n   =   integer or whole number
According to de Broglie theory   `lambda`   =   `(h)/(mv)`    
               `:.`    `(2pir)/(n)`   =   `(h)/(mv)`
        or           m v r   =   n  *  `h/(2pi)`  
This is Bohr's postulate that stipulates the angular momentum of an electron moving round the nucleus is an integral multiple of `h/(2pi)` . In other words, de Broglie theory and the Bohr's theory are in agreement with each other. In case the circumference of the Bohr's orbit (2`pi`) is bigger or smaller than n`lambda`  , destructive interference of waves takes place and the orbit cannot exist.

Solved Problem on De broglie atomic model:

Q ) Find the momentum of a particle whose de Broglie wavelength is 1A.
Solution :  de Broglie Wavelength , `lambda`  =  1 A   =   1  *   10^-10  m
As per de Broglie's equation   `lambda`  =   `h/p`    
`:.`  Momentum of particle ,  p   =    `h/lambda`     
                                                          =    `(6.6256*10^-34kgm^2s^-1)/(1*10^-10m)`
Momentum of particle, p  =  6.6256*10^-24 kgms^-1 .

Another problem on De broglie atomic model:

Q ) A 8 kg bowling ball rolls with a velocity of 9 metre per second. a ) What is de Broglie's wavelength of the bowling ball. b) Why does the bowling ball exhibit no observable wave behaviour ?
Solution :  Given mass of the ball m = 8 kg   and velocity  v  =  9 m / s

a ) de Broglie's Wavelength is given by   `lambda` =  `h/p`   
where   p  =  m v     and    h  is Planck's constant .
In this case    `lambda` =  `(6.6256*10^-34)/(8*9)`
                          `lambda` =  0.092 * 10^-34

b ) Wave  cannot be seen for macroscopic objects like bowling balls. The de Broglie wavelength for this object is smaller than the atomic nucleus and impossible to detect .

Ionisation energy and electron affinity

Introduction
Ionisation Energy Ionisation energy is one of the important properties of elements . If energy is supplied to an atom , electrons are promoted to higher energy levels . If sufficient energy is supplied , an element in the outer most  shell can be completely removed from the atom , resulting in the formation of a positive ion .

The minimum energy required to remove the most loosely bound electron from an isolated gaseous atom is called ionisation energy . It is also called the first ionisation energy . ( I₁ )
      M _(g)    +   I₁    `|->`   M⁺ _(g)   +  e^-

The minimum energy required to remove another electron affinity from the uni positive ion is called second ionisation energy ( I₂ ) .
     M _(g)    +   I₂    `|->`   M²⁺ _(g)   +  e^-

The second ionisation energy ( I₂ ) is greater than the first ionisation energy  . On removing an electron from an atom , the uni positive ion formed will have more effective nuclear charge than the number of electrons .

This decreases the repulsions between the electrons and increases the nuclear attraction on the electrons . As a result , more energy is required to remove an electron from the uni positive ion . Hence the second ionisation energy ( I₂ ) is greater than the ionisation energy ( I₁ )  .

Similarly the third ionisation energy ( I₃ ) is greater than the second ionisation energy . An atom has as many ionisation energies as the number of electrons present in it . The order of ionization energies
     I₁  <   I₂   <   I₃   <  ........I_n
where n is the number of electrons in the atoms . Ionisation energies are determined from spectral studies as well as from discharge tube experiments . They are measured in electron volts (eV) atom^-1 .
    1 eV   =  1.602 * 10^-19 J ,  hence 1 mol of eV has energy  1.602 *10^-19 * 6.023*10²³
               = 96.45 k j mol^-1 .

The discharge tube is filled with gas whose ionisation energy is required . At low voltages , there is no flow of electricity . But , on increasing the voltage between anode and cathode , the gas ionizes at a particular voltage , which is indicated by sudden increase in the flow of electricity . That particular voltage is called ionization energy  .

The magnitude of ionisation of an atom depends on the following factors

• Atomic radius
• Nuclear charge
• Screening or shielding effect on the outer most electrons from the attraction of the nucleus .
• Completely filled or half filled nature of sub shells .

Ionisation potentials of some elements

Electron Affinity

Electron Affinity
Electron affinity is another important property of elements like ionisation energy is required to remove an electron from an atom . Conversely , when electron is added to an atom , energy is released .

Electron affinity of an element is the energy released when an electron is added to a neutral gaseous atom of that element .
Energy is released when only one electron is added to an  atom forming a uni negative ion . The negative ion prevents entry of further electrons due to repulsive forces between negative charges . Thus energy is needed to overcome these repulsive forces between uni-negative ion ( X^- ) and electron ( e^- ) and to add one more electron energy is required . Hence , second electron affinity value usually has a positive value .

Electron affinity depend on the size and the effective nuclear charge of atom . They cannot be determined directly , but are obtained indirectly using the Born-Haber cycle . Electron affinity are measured in kj mol^-1 .
Halogens have high values for electron affinity due to their small atomic sizes and requirement of only one electron to get the nearest inert gas configuration . Depending on thermodynamic notation that energy liberated is shown by negative sign , the electron affinity values are mentioned with numerals carrying negative sign before them . If energy is absored during the addition of an electron to an atom , then it is shown by  positive sign .

Variation of electron affinity in a group and in a period .

Variation of Electron affinity in a group

From top to bottom as the atomic size increases , the electron affinity decreases , in a period . But the electron affinity of the second element in the group is greater than the first one . For example , in halogens , the electron affinity of fluorine is -333 kj mol^-1 while the electron affinity of chlorine is -348 kj mol^-1 . It is because fluorine atom is smaller in size than chlorine atom and has strong inter repulsions . During the addition of electrons of fluorine atom the electronic repulsions are overcome at the expense of some liberated energy and hence , the overall energy liberated is less than that of chlorine atom .
Similarly phosphorus has bigger electron affinity value than nitrogen and sulphur has greater electron affinity than oxygen in V and VI groups .

Variation of electron affinity in a period

In a period , as we move from left to right , the atomic size decreases and the nature of the element changes from metallic to non metallic and this results in an increase in the electron affinity values . In a given period halogen has the highest electron affinity . Since the outer shells of zero group elements are filled with electrons to attain octet structure sp²sp⁶ , they do not accept electron and their electron affinities are treated as zero .

Net Ionic Equation Example

On the basis of type of chemical bonds in compounds, they can be mainly two types; ionic and covalent compounds. In ionic compounds, cations and anions are bonded by electrostatic force of attraction while in covalent compounds; elements are bonded with covalent bonds. Due to different types of bonds, these compounds show different type of reactions.

In covalent compounds, molecules involve in reactions while in ionic compounds, ions take part in reaction. Therefore chemical reactions of ionic compounds can be represents in three different types; molecular equation, ionic equation and net-ionic equation.

Molecular equation represents reacting substance in its molecular form while ionic and net-ionic equation shows in ionic form.

Let’s take a Net Ionic Equation Examples of reaction of silver nitrate with potassium chloride to form white precipitate of silver chloride and potassium nitrate. The molecular equation for this reaction would be;

AgNO_3(aq) + KCl_(aq)   AgCl () + KNO_3(aq) 

While the ionic equation will show all ions of reacting substances excluding precipitate as it exists in solid state.

Ag⁺_(aq)  + NO₃^-_(aq)  + K⁺_(aq)  + Cl^-_(aq)   AgCl()+ K⁺_(aq)  + NO₃^-_(aq) ..........(1)

You can observer in equation (1) only silver ion and the chloride ion takes part in the reaction to form AgCl precipitate while potassium ion and nitrate ion remains unchanged. These ions are known as spectator icons. Therefore we can eliminate from the ionic equation. Hence net ionic equation helps to understand the involvement of reacting substances in given reaction and net ionic equations examples show only those ions which really take part in reaction or get some change.

Ag⁺ + Cl^-  AgCl () .......(2)

Let’s take some more examples of Net Ionic Equations and practice Net Ionic Equations from their molecular equation.

• Silver nitrate and sodium bromide to form silver bromide :

1. Molecular equation:   

AgNO₃ (aq)   +   NaBr (aq)         AgBr (s)   +   NaNO₃ (aq)

2. Ionic equation:    

Ag⁺ (aq)+ NO₃^¯ (aq) +  Na⁺ (aq)+ Br¯ (aq)      AgBr (s)    +   Na⁺ (aq)  +  NO₃¯ (aq)

3. Net-Ionic equation:    Ag⁺ (aq)   +   Br¯ (aq)       AgBr (s)

• Potassium carbonate and calcium nitrate to form calcium carbonate :

1.  Molecular equation:  Ca(NO₃)₂ (aq)    +   K₂CO₃ (aq)       CaCO₃ (s)   +    2 KNO₃ (aq)

2. Ionic equation:  

Ca⁺ (aq)  +  2 NO₃¯ (aq)  +  2 K⁺ (aq)  +  CO₃^2- (aq)    CaCO₃ (s)  +  2 K⁺ (aq) +  2 NO₃¯ (aq)

3. Net-Ionic equation:  Ca²⁺ (aq)    +   CO₃^2- (aq)      CaCO₃ (s)

• Barium bromide and potassium sulphate to form barium sulphate :

1. Molecular equation:   BaBr₂ (aq)  +  K₂SO₄ (aq)         BaSO₄ (s)   +   2 KBr (aq)

2. Ionic equation:  

Ba²⁺ (aq) +  2 Br¯ (aq) +  2 K⁺ (aq) + SO₄^2- (aq)    BaSO₄ (s) +  2 K⁺ (aq) +  2 Br¯ (aq)

3. Net Ionic equation:  Ba²⁺ (aq)  +  SO₄^2- (aq)       BaSO₄ (s)

Newtons law of heating and cooling

Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature (i.e. the temperature of its surroundings).

Newton's Law makes a statement about an instantaneous rate of change of the temperature. We will see that when we translate this verbal statement into a differential equation, we arrive at a differential equation. The solution to this equation will then be a function that tracks the complete record of the temperature over time. Newton's Law would enable us to solve the following problem.

 Example 1: The Big Pot of Soup As part of his summer job at a restaurant, Jim learned to cook up a big pot of soup late at night, just before closing time, so that there would be plenty of soup to feed customers the next day. He also found out that, while refrigeration was essential to preserve the soup overnight, the soup was too hot to be put directly into the fridge when it was ready. (The soup had just boiled at 100 degrees C, and the fridge was not powerful enough to accommodate a big pot of soup if it was any warmer than 20 degrees C). Jim discovered that by cooling the pot in a sink full of cold water, (kept running, so that its temperature was roughly constant at 5 degrees C) and stirring occasionally, he could bring that temperature of the soup to 60 degrees C in ten minutes. How long before closing time should the soup be ready so that Jim could put it in the fridge and leave on time ?


Solution: Let us summarize the information briefly and define notation for this problem.
Let

= Temperature of the soup at time t (in min).

= Initial Temperature of the soup =100 deg.

= Ambient temperature (temp of water in sink) = 5 deg .

Given: The rate of change of the temperature , is (by Newton's Law of Cooling equation) proportional to the difference between the temperature of the soup and the ambient temperature This means that:



Here a bit of care is needed: Clearly if the soup is hotter than the water in the sink , then the soup is cooling down which means that the derivative should be negative. (Remember the connection between a decreasing function and the sign of the derivative ?). This means that the equation we need has to have the following sign pattern:


where is a positive constant.
This equation is another example of a differential equation. The independent variable is for time, the function we want to find is , and the quantities are constants. In fact, from Jim's measurements, we know that , but we still don't know what value to put in for the constant . 

Net Ionic Equations

An ionic chemical reaction involves the interactions of ionic species to form a new compound. For example, if we add a solution of barium chloride (BaCl2) with a solution of sodium sulphate (Na2SO4), it forms an insoluble solid (precipitate) of barium sulphate (BaSO4). The molecular and ionic equation for the given reaction would be;

Na₂SO₄ + BaCl₂  2NaCl + BaSO₄

2Na⁺_(aq) + SO₄^2-_(aq) + Ba²⁺_(aq) + 2Cl^-_(aq)  2Na⁺_(aq) + 2Cl^-_(aq) + BaSO_4(s)

A total ionic equation indicates all soluble ionic materials like ions with subscript (aq). For insoluble ionic solids, (s) stands for their solid state. Those ions which are remain same before and after reaction are called as spectator ions. For writing a net ionic equation, subtract these spectator ions from ionic equation. Let’s see what is a net ionic equation for is given reaction. Here sodium and chloride ions are common and have to eliminate from net-ionic-equation;

2Na⁺_(aq) + SO₄^2-_(aq) + Ba²⁺_(aq) + 2Cl^-_(aq)  2Na⁺_(aq) + 2Cl^-_(aq) + BaSO_4(s)

Ba²⁺_(aq) + SO4^2-_(aq)  BaSO_4(s)

Let’s see how to writing net ionic equations of reaction of sodium phosphate and calcium chloride to form an insoluble white solid form of calcium phosphate. The molecular equation for this reaction is as follow;

2Na₃PO₄(aq) + 3 CaCl₂(aq)  6 NaCl(aq) + Ca₃(PO₄)₂(s)

First write complete ionic equation with the use of chemical reaction. It shows all ions involve in reaction in their physical state and also indicate the correct formula, charge and number of each ion.

Hence ionic equation would be;

6 Na⁺ (aq) + 2 PO₄^3- (aq) + 3 Ca²⁺ (aq) + 6 Cl^- (aq) 6 Na⁺ (aq) + 6 Cl^- (aq) + Ca₃(PO₄)₂ (s)

Now subtract spectator ions (Na+, Cl-) from reaction to get net ionic equation.

6 Na⁺ (aq) + 2 PO₄^3- (aq) + 3 Ca²⁺ (aq) + 6 Cl^- (aq) 6 Na⁺ (aq) + 6 Cl^- (aq) + Ca₃(PO₄)₂ (s)

2 PO₄^3- (aq) + 3 Ca²⁺ (aq)  Ca₃(PO₄)₂(s)

Overall net ionic equations show soluble, strong electrolytes reacting in the form of ions. Now days many net ionic equation Solver are available to solve equation from molecular equation. For writing ionic equations, you must remember that most alkali metal compounds, ammonium compounds, halide, nitrates, chlorate compounds are soluble in nature. Carbonate, phosphate, oxalates, sulphide, chromate compounds are insoluble in nature. Except group-1, oxides are insoluble in water and form acids or bases. Only HCl, HBr, HI, HNO3, H2SO4, HClO4 are strong acids and ionize completely while all other are weak acids.

Oxidation State of Carbon

An element forms a molecule by sharing or donating/accepting electrons.

If it is sharing, it is covalent bond, if it is by donating/accepting electrons, it is an ionic bond.
The degree of oxidation of an element in a molecule is called oxidation state.This is particularly significant where an element can exhibit variable valency.
Oxidation states of free atoms are zero.

The oxidation state of free ion is the charge carried by it.

The sum of all oxidation states in a neutral molecule is zero.

The oxidation states of hydrogen is +1 and oxygen -2 with a few exceptions.

Introduction :

Carbon has configuration 2s² 2p²
as shown under

The valency and therefore the oxidation state is expected to be 4 or -4 as it would donate or accept 4 electrons to attain stability.

The oxidation state is therefore variable in carbon from -4 to +4
Concider a compound like CO₂

oxygen has -2 oxidation state,
the sum of oxidation state of carbon and oxygen is zero since the molecule is neutral.
-2 x 2 + carbon oxidation state=0

oxidation state of carbon = +4
Similarly for a compound like CH4,
the oxidation state would be

+1 x 4 + oxidation state of carbon = 0
oxidation state of carbon = -4

Now consider another reaction where in oxidation number of carbon undergoes change and has variable oxidation state CH₂=CH₂ + H₂O  CH₃-CH₂OH

oxidation states of carbon in ethene: -2 , -2 ; and in ethanol they are : -3 -1

Other Compounds.

Let us consider a molecule of CO.

The oxygen has -2 oxidation state.

The oxidation state of carbon is therefore found out as under
-2 + oxidation state of carbon = 0
oxidation state = -2

Let us consider a molecule of CF4.

The fluorine has -1 oxidation state.

The oxidation state of carbon is therefore found out as under
-1x 4 + oxidation state of carbon = 0
oxidation state=+4

Let us consider a molecule of CH3Cl.

The chlorine has -1 oxidation state. The hydrogen has +1 oxidation state.

The oxidation state of carbon is therefore found out as under
-1  + 1 x 3 + oxidation state of carbon = 0
oxidation state= +2

Liquid oxygen temperature

Introduction :

Oxygen gas is in the air around us which also includes many other gases. Air is cooled to a certain temperature will change into a liquid or a solid.Among the elements in the universe, oxygen is the third most abundant and makes up to 21% of the earth's atmosphere.Carl Wilhelm and Joseph Priestley independently discovered oxygen, by heating mercuric oxide (HgO).

How we Obtain Liquid Oxygen?

Oxygen turns into a liquid at -183 degrees Celsius. To bring these gases to such low temperatures, the air is cooled and compressed. When allowed to expand again, their temperature drops further. To extract the oxygen, the liquid is so warmed that just nitrogen turns back into gas - leaving behind liquid oxygen.

Liquid form of oxygen depends on two things:

• Pressure
• Temperature

Storing of liquid oxygen:
Oxygen is kept in cryogenic, insulated containers called dewars which keep the oxygen in liquid form at -170 degrees Celsius. Temperature outside the storage tank is much higher than the liquid temperature. Due to the heat leak, the liquid tends to warm up thus producing gaseous oxygen that contributes to a rise in pressure inside the tank.

Image on liquid oxygen  

Temperatures of Liquid Oxygen

Oxygen turns into a pale blue color liquid at temperatures below its boiling point of -183�C and weighs 1.14 times the weight of water. At temperatures more than -118.6�C, the liquid oxygen will become a gas in spite of the pressure exerted on it. This is the critical temperature for liquid oxygen. 1 Liter of liquid oxygen can turn into 860 Litres of gas.
Boiling Point at 1 atm: -183.0°C, 90.15K, -297.4°F
Oxygen is in liquid phase from -297.4°F @ 1 atm. to -181°F and 731psi.

Conclusion to Temperatures of Liquid Oxygen

We can conclude that, liquid oxygen, in combination with liquid hydrogen, makes an excellent rocket fuel. Liquid oxygen is a versatile and efficient method of supplying oxygen to patients at home.

Triple Point Temperature

Introduction:

Definition Temperature:

The temperature means the degree of hotness and coldness of any particular substance.There are so many different types of the temperature such as the critical temperature, Curie temperature, triple point temperature etc. Critical temperature for the conductors is the temperature at which it loses its resistance completely. Curie temperature is the temperature at which the magnet loses its magnetism completely on heating. Here we discuss about the triple point temperature of any substance. The image below shows the different phases of water.

Examples of Triple Point Temperature

The triple point temperature of the substance is defined as the temperature at which that particular substance exists in all the three phases of matter, i.e., it exists in the form of solid, liquid and gas at that particular triple point temperature in the thermodynamic equilibrium. For example, the triple point temperature of the mercury is – 38.84°C, the triple point temperature of hydrogen is 13.8033 K and the triple point temperature of water is 0°C or 273.16 K.
Generally, the combination of pressure and the temperature at which the substance can exists in the stable equilibrium. In case of water, the triple point temperature is 273.16 K that means water exists in the form of ice, water and vapour at temperature 273.16K. In case of the triple point temperature, the surfaces separating the different phases should be perfectly flat so that there is no effect of the surface tension.

Conclusion for Triple Point Temperature

From the above discussion and the concept of the triple point temperature, the phase diagram of water is too complex. At the high temperature the pressure increases results first in liquid and then solid water. At the lower temperature, the liquid state disappears very slowly and then water converted into gas to solid. When the temperature is above that of the triple point at the constant pressure, the ice convert solid into liquid (water) and then steam (water vapour).
At the temperature below the triple point at the constant pressure almost zero that means in the outer space the liquid form of the water does not exists. In the process of the sublimation, the ice directly converts in the form of steam when it is heated.