Tricks in inorganic chemistry
TRICKS FOR MAKING INORGANIC CHEMISTRY LOVABLE
1. Just learn the word LERCMBS DORI .
2. You just remember the place of dotes because they are separated according to the different properties of groups as there is no dot present between CM you can see in 4 point the property CATENATION and METAL METAL BOND ENERGY both properties for 4
Similarly is the case with DORI as this word represent Dipole moment, Oxidizing strength, Reactivity toward hydrogen, Ionic nature
L .E.R.CM.B.S. DORI
1. L - Lattice energy of metal halides = 1 group
2. Enthalpy of hydration = 2 group
3. Relative stability of +3 oxidation state in 3 group
4. Catenation property , metal metal bond energy = 4 group
5. Basicity of hydride = 5 group
6. Stability of halide = 6 group
7. Dipole moment Oxidizing strength Reactivity toward hydrogen Ionic nature
All properties increase from top to bottom except these
2. TRICKS TO ESTIMATE THE PRODUCTS FORM WHEN COMPOUND IS HEATED Maximum time we see that we generally get confuse about what product will form if certain compound is heated or in completing any chemical reaction so there are some easy tricks to find products in just seconds.
METHOD 1: if there are atoms of hydrogen and oxygen remove these in the form of water: EXAMPLE1 : If NH4NO3 is heated what product will form we generally got confuse ? So there are some tips to find product easily:
Write in Molecular Form. WRITE Nitrogen N2 , Oxygen O2 , and write Hydrogen H . Remove water if there is hydrogen and oxygen atom2
N2H4O3 – H2O → N2H2O2
If there is chances of removal of water from compound just remove it again. We get:
N2H2O2 – H20 →N2O + 2H2O (end product )
EXAMPLE 2. IF NH4NO2 is heated what will be the product ? NH4NO2 △→ ? Write nitrogen, Hydrogen and oxygen in its molecular form N2 , H2 , O2 Remove water (if possible remove two molecules of water) -2H2O
Result = N2 + 2H2O
REMEMBER THAT AMMONIUM SALT HAS NO RESIDUE..
N2, NH3 , N2O , NO, NO2 , N2O3 , N2O5 If there is a question that predict which compound will form if NH4NO3 is heated
Write this in the form - NH4NO3 Now oxidation number of nitrogen in NH4 is -3 and in NO3 oxidation number is +5.
NH4+ NO3
-3 +5
Now take the average of these oxidation number like:
-3+5/2
=2/2
=1
and we get oxidation number of 1.
By the rule that oxidation number of products are equal to oxidation number of reactants so, here oxidation number is 1 for reactants therefore on the reactant side the oxidation number should be 1. Choose from the following compounds which having oxidation number 1.
N2, NH3 , N2O , NO, NO2 , N2O3 , N2O5 compounds of nitrogen.
In N2O the Nitrogen has oxidation state of 1 so product will be N2O with removal of water
.
THIS WILL HELPFUL IN :
HEATING OF OXYACID (OXYGEN CONTAINING ACIDS OF NON METALS) DISPROPORTION REACTION WATER ADDITION
3. USE OF HYPO , ORTHO , META , PER , PYRO PREFIXES IN INORGANIC CHEMISTRY :
The acids of Cl are HCLO = O.N OF Cl +1 (hypochlorous acid) HCLO2 = O.N OF Cl +3 (chlorous acid)
HCLO3 = O.N OF Cl +5 (chloric acid ) ……..(common form ) HCLO4= O.N OF Cl +7 (perchloric acid )
1. If O.N decreases from the common form (where ic is use suffix) the ic changes to ous. eg H3PO4 is phosphoric acid and O.N of P is +5 and in H3PO3 O.N is +3 (we use ous) so this will become phosphorous acid .
2. Remember Hypo means to decrease it means that its oxidation state of hypochlorous acid is less than chlorous acid.
3. Remember Per means to increase means oxidation state of perchloric acid is more than chloric acid.
4. boric acid can be written as H3BO3 and is very important acid of boron when boric acid it is also known as orthoboric acid when heated and water is removes we get HBO2 e meta boric acid (meta means after) it means meta compounds are form after heated.
5. Phosphoric acid is generally having formula H3PO4 when two molecules of same compound are heated there is formation of pyro compounds asH3PO4+H3PO4 when 2 units combine and H2O is remove H4P2O7 pyrophosporic acid will form and pyro means heat so if you want to form pyro form just take its two molecules and heat.
4. TRICK TO FIND PARAMAGNEIC OR DIAMAGNETC SPECIES JUST IN SECONDS : By the molecular orbital theory we know by filling of electrons we check that number of unpaired electron if they are not zero paramagnetic if they are zero then they are diamagnetic but it takes very much time to find paramagnetic and diamagnetic species.
1. Total number of electrons if even then diamagnetic species and if odd paramagnetic species.example:
H2 has 2 e- (even) hense diamagnetic H2(+) has 1 e- (odd) hence paramagnetic
2. Remember exception if total number of electrons are:
10 e- or 16 e- according to above rule the both species should be diamagnetic because both are even numbers but remember if total number of e- are 10 e- or 16 e- the species is paramagnetic. Eg N2(2-) has 16 e- hence species is paramagnetic in nature O2 has 16 e- hence species is paramagnetic in nature C2(2+) has 10 e hence species is paramagnetic in nature.
5.STEPS TO FIND BOND ORDER (EASY WAY)
Just count the total number of electrons in the compound
Just for example the compound given is N2,then the number of electrons is 14 in it.
Just remember this trick...The bond order for compounds with total number of electrons 14 will be 3 and for the subsequent one remove 0.5
Eg:
no of - bond
electron order
1. Just learn the word LERCMBS DORI .
2. You just remember the place of dotes because they are separated according to the different properties of groups as there is no dot present between CM you can see in 4 point the property CATENATION and METAL METAL BOND ENERGY both properties for 4
Similarly is the case with DORI as this word represent Dipole moment, Oxidizing strength, Reactivity toward hydrogen, Ionic nature
L .E.R.CM.B.S. DORI
1. L - Lattice energy of metal halides = 1 group
2. Enthalpy of hydration = 2 group
3. Relative stability of +3 oxidation state in 3 group
4. Catenation property , metal metal bond energy = 4 group
5. Basicity of hydride = 5 group
6. Stability of halide = 6 group
7. Dipole moment Oxidizing strength Reactivity toward hydrogen Ionic nature
All properties increase from top to bottom except these
2. TRICKS TO ESTIMATE THE PRODUCTS FORM WHEN COMPOUND IS HEATED Maximum time we see that we generally get confuse about what product will form if certain compound is heated or in completing any chemical reaction so there are some easy tricks to find products in just seconds.
METHOD 1: if there are atoms of hydrogen and oxygen remove these in the form of water: EXAMPLE1 : If NH4NO3 is heated what product will form we generally got confuse ? So there are some tips to find product easily:
Write in Molecular Form. WRITE Nitrogen N2 , Oxygen O2 , and write Hydrogen H . Remove water if there is hydrogen and oxygen atom2
N2H4O3 – H2O → N2H2O2
If there is chances of removal of water from compound just remove it again. We get:
N2H2O2 – H20 →N2O + 2H2O (end product )
EXAMPLE 2. IF NH4NO2 is heated what will be the product ? NH4NO2 △→ ? Write nitrogen, Hydrogen and oxygen in its molecular form N2 , H2 , O2 Remove water (if possible remove two molecules of water) -2H2O
Result = N2 + 2H2O
REMEMBER THAT AMMONIUM SALT HAS NO RESIDUE..
METHOD 2 : USE CONCEPT : AVERAGE O.N OF REACTANT = AVERAGE O.N OF PRODUCT
If Hydrogen and oxygen are heated the product is always H O Remember when we heat nitrogen compounds always one of these compounds will formN2, NH3 , N2O , NO, NO2 , N2O3 , N2O5 If there is a question that predict which compound will form if NH4NO3 is heated
Write this in the form - NH4NO3 Now oxidation number of nitrogen in NH4 is -3 and in NO3 oxidation number is +5.
NH4+ NO3
-3 +5
Now take the average of these oxidation number like:
-3+5/2
=2/2
=1
and we get oxidation number of 1.
By the rule that oxidation number of products are equal to oxidation number of reactants so, here oxidation number is 1 for reactants therefore on the reactant side the oxidation number should be 1. Choose from the following compounds which having oxidation number 1.
N2, NH3 , N2O , NO, NO2 , N2O3 , N2O5 compounds of nitrogen.
In N2O the Nitrogen has oxidation state of 1 so product will be N2O with removal of water
.
THIS WILL HELPFUL IN :
HEATING OF OXYACID (OXYGEN CONTAINING ACIDS OF NON METALS) DISPROPORTION REACTION WATER ADDITION
3. USE OF HYPO , ORTHO , META , PER , PYRO PREFIXES IN INORGANIC CHEMISTRY :
The acids of Cl are HCLO = O.N OF Cl +1 (hypochlorous acid) HCLO2 = O.N OF Cl +3 (chlorous acid)
HCLO3 = O.N OF Cl +5 (chloric acid ) ……..(common form ) HCLO4= O.N OF Cl +7 (perchloric acid )
1. If O.N decreases from the common form (where ic is use suffix) the ic changes to ous. eg H3PO4 is phosphoric acid and O.N of P is +5 and in H3PO3 O.N is +3 (we use ous) so this will become phosphorous acid .
2. Remember Hypo means to decrease it means that its oxidation state of hypochlorous acid is less than chlorous acid.
3. Remember Per means to increase means oxidation state of perchloric acid is more than chloric acid.
4. boric acid can be written as H3BO3 and is very important acid of boron when boric acid it is also known as orthoboric acid when heated and water is removes we get HBO2 e meta boric acid (meta means after) it means meta compounds are form after heated.
5. Phosphoric acid is generally having formula H3PO4 when two molecules of same compound are heated there is formation of pyro compounds asH3PO4+H3PO4 when 2 units combine and H2O is remove H4P2O7 pyrophosporic acid will form and pyro means heat so if you want to form pyro form just take its two molecules and heat.
4. TRICK TO FIND PARAMAGNEIC OR DIAMAGNETC SPECIES JUST IN SECONDS : By the molecular orbital theory we know by filling of electrons we check that number of unpaired electron if they are not zero paramagnetic if they are zero then they are diamagnetic but it takes very much time to find paramagnetic and diamagnetic species.
1. Total number of electrons if even then diamagnetic species and if odd paramagnetic species.example:
H2 has 2 e- (even) hense diamagnetic H2(+) has 1 e- (odd) hence paramagnetic
2. Remember exception if total number of electrons are:
10 e- or 16 e- according to above rule the both species should be diamagnetic because both are even numbers but remember if total number of e- are 10 e- or 16 e- the species is paramagnetic. Eg N2(2-) has 16 e- hence species is paramagnetic in nature O2 has 16 e- hence species is paramagnetic in nature C2(2+) has 10 e hence species is paramagnetic in nature.
5.STEPS TO FIND BOND ORDER (EASY WAY)
Just count the total number of electrons in the compound
Just for example the compound given is N2,then the number of electrons is 14 in it.
Just remember this trick...The bond order for compounds with total number of electrons 14 will be 3 and for the subsequent one remove 0.5
Eg:
no of - bond
electron order
- 14- 3
- 15- 2.5
- 16- 2.0
- 17- 1.5
- 18- 1.0
- 19- 0.5
- 20- 0
- 21- 0.5
- 22- 1.0
- 23- 1.5
- 24- 2.0
- 25- 2.5
- 26- 3.0
- 8- 0
- 9- 0.5
- 10- 1.0
- 11- 1.5
- 12- 2.0
- 13- 2.5
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