Aufbau Rule and filling of electrons in orbitals and sub shells

Aufbau Rule and filling of electrons in orbitals and sub shells

Aufbau rule is not on the name of any scientist as usually happened in sciences. Aufbau means building up.

Electrons are build up in orbitals of different subshells of different energy levels on the basis of energy.

Electrons first fill lower energy level and then move to next higher level.

So electrons start filling from 1s orbital on wards in the following sequence:

1s,2s,2p , 3s,3p,4s,3d,4p,5s,4d,5p,6s,4f,5d,6p,7s,5f,6d,7p,8s

Simply it can be learned as ssp sps dps dps fdps fdps

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How do you find oxidation number of an atom?

How do you find oxidation number of an atom?

If you want to learn about oxidation numbers and the method to assign the oxidation number, you should read the following points.

1.    The oxidation state of group 1 elements is always +1. 

Li, Na, K, Rb, Cs, Fr have +1 oxidation state in all the compounds.

                  For example: Oxidation state of Na in Na₂O  is +1.

2.    The oxidation state of group 2 elements is always +2. 

Mg, Ca, Ba etc will always have +2 Oxidation state.

For Example: Oxidation number of Mg in MgO is +2.

3.    The free element or a free molecule will always have oxidation state equal to zero.

For example: Fe and O₂ will have oxidation state equal to zero.

4.    Fluorine is the most electronegative element so it always has -1 oxidation state.

5.    Other halogens generally have -1 oxidation state.

For example: Cl has -1 oxidation state in NaCl.

In Interhalogen compounds like IF_7,  Iodine has +7 Oxidation state because 

Fluorine always have -1 oxidation state.

6.    Oxygen generally have -2 oxidation state in all oxides. 

For example: Oxidation state of O in CO₂  is -2.

In peroxides ,oxidation state of O is -1.

For example: Oxidation state of O in H₂O₂( Hydrogen peroxide) is -1.

In superoxides, Oxidation state of O is -1/2. 

For example; oxidation state of O in NaO₂ ( Sodium superoxide) is -1/2.

Exceptionally oxygen has +2 oxidation state in OF₂( Oxygen fluoride) because F 

has -1 oxidation state always.

7.    The oxidation state of all the ions ( simple ions or polyatomic ion) is always equal 

to the charge on them. 

For example: Oxidation states of Aluminium and phosphate ions is equal to the 

charge on them.   Al³⁺and (PO4)^3-  

8.    The sum of all the oxidation states of all atoms in compounds is always zero.

For example: The sum of all Oxidation states in H₂O=0

H=+1 so +1x2 = +2

O= -2  so the sum is +2 -2= 0

9.    The oxidation states of transition elements isto be determined by using above 

rules because they have variable oxidation states.

The Oxidation state of Mn in KMnO₄  is to be calculated.

K= +1 because it is in group 1.

Mn=?

O=-2 x4= -8

+1+Mn -8= 0

Mn = +7

Worksheet:

Determine the following oxidation states.

1.    N in NO₃^-       

2.    S in H₂SO₄

3.    Cr in K₂Cr₂O₇

4.    Fe in Fe₂O₃

5.    Fe in Fe₃O₄

6.    Mn in MnO₂      

7.    S in SO₃^2-      

8.    C in CH₄      

9.    S in HSO₄^-    

10. N in NO₂^-        

IB Diploma chemistry guide 2019-2020

IB Diploma chemistry guide 2019-2020

You can download IB diploma chemistry guide from the above link.

This is for the students of IB diploma from the 2016 first exams on wards till now.

It is highly recommended for IB Diploma chemistry students to regularly go through the chemistry guide so that you are aware about the syllabus content which is coming in your exams. Students and teachers sometimes ignore the guide which lend them in a time wasting exercise. For example, earlier the working and structure of mass spectrometer was in IB diploma syllabus but now it is not in syllabus so if you are still preparing about the structure and working of mass spectrometer, you are wasting your time. Only the % abundance of isotopes is in syllabus now. 
So to avoid this situation, you should go through the chemistry guide.

Noble prize in chemistry- 2019 for lithium ion batteries

2019 Nobel Prize in chemistry is awarded jointly to three scientists Goodenough, Whittingham and Yoshino for the development of Lithium Ion battery which revolutionized the fueling , working and safety  of the gazettes like mobiles , tablets and computers.

 A lithium-ion battery or Li-ion battery is a secondary cell or rechargeable battery in terms of chemistry.

Primary cell or battery: The cell which cannot be reused once the reactants are finished or the reaction is completed.

Secondary cell or battery: The cell which can be recharged and reused by reversing the chemical reactions.  Once the chemical reactions are completed, the chemical reaction is reversed to get back the reactants by giving electricity to it which we call recharging.

What is a Chemical bond? How is Ionic bond formed?

Powerpoint presentation on Ionic bond

What is a Chemical bond? How is Ionic bond formed? 

The force of attraction between two or more bonded atoms is 

called chemical bond.

There are three types of chemical bonds:

Octet or duplet rule:

The atoms try to become stable by acquiring the electronic structure of nearest noble gas. Except Helium all the noble gases have eight electrons in their outer most shell so atoms try to get eight electrons in the outermost shell. This is called Octet rule.

Atoms become stable by : Lose of outermost electrons( Metals become stable by lose of outermost electrons)

      Gain of electrons( Non-Metals become stable by gain of electrons) or 
       Sharing of electrons(Non-Metals become stable by sharing of electrons)

The atoms near to He try to acquire 2 electrons in their outermost shell to become stable. It is called duplet rule.

 Definition:

The electrostatic force of attraction between two or more oppositely charged ions is called ionic bond.

Example: The electrostatic force of attraction between Sodium ions (Na⁺) and chloride ions( Cl^-) forms an ionic bond.

Metals loose valence electrons to become positive ions

( cations) because cations have more protons than electrons. Na atom has 11 p and 11 e but after lose of 1 e it left with 10 e but 11 p so 1 p is extra means 1 positive charge is extra so it becomes Na.⁺

Na         – 1e àNa⁺

2,8,1                  2,8

Non-Metals gain electrons to become negative ions( anions) because anions have less protons than electrons. Cl atom has 17 p and 17 e but after gain of 1 e it has with 18 e but 17 p so 1 p is less means 1 positive charge is less so it becomes Cl^- .

Cl         +1e à Cl^- 

2,8,7                 2,8,8

Name of the ionic compounds

The cation is the first name

The anion is the second name. The anions generally end with ide like chloride, oxide, nitride etc

The compound ions like phosphate, carbonate, nitrate etc end with ate.

Sulphate ion, SO₄^2-

Phosphate ion, PO₄^3-

Hydrogen carbonate or bicarbonate ionHCO₃^-

Ethanoate (acetate ion), CH₃CO₂^-

Hydroxide ion, OH^-

Nitrate, NO₃^-

Carbonate, CO₃^2-

^{Oxalate ion C}₂^O₄^-

Ammonium ion is a positive ion , NH₄⁺ 

^{Write the names of the following compounds}

Na₂O , NaOH, MgO, Mg(OH)₂, H₂O, Li₂O

LiCl , MgCl₂, P₄O₁₀, SO₂, SiO₂, Al₂O₃ , Al₂(SO₄)₃

Physical properties of ionic compounds

1. High Melting and boiling point- large amount of energy is needed to break the strong electrostatic force of attraction between oppositely charged ions.

2. Crystalline in name – They form crystal lattice ( Large 3D network of ions)

3. Brittle- They breaks easily because of disturbance in the arrangement of ions.

4. Good conductor of electricity in aqueous solutions because of free movement of ions but bad conductor in solid state.

5. Soluble in water because the ions are separated and surrounded by water molecules.

Hydrogen ion concentration in a solution and pH scale

pH is the degree of acidity or basicity of a chemical substance in aqueous systems.

In simple terms pH is the power of hydrogen ion or hydronium ion concentration.

It can be calculated by :
pH = - log₁₀[H⁺] or 

pH = - log₁₀[H₃O⁺] 

[ ] square bracket represents the concentration in mol/dm³

Example: If a sample of an acid is found to have[ H⁺]=1X10^-4

Than pH = - log₁₀[1X10^-4]

pH = -[-4 log 10]

pH= 4
If you want to find out [ H⁺] than you can use the following formula: pH = 10^-pH

Chemistry data booklet for IB latest

Chemistry data booklet for IB latest 

Latest Chemistry data booklet for IB diploma chemistry students is of fourth edition and it is for the 2019 and 2020 final IB Diploma students. It is in use since 2016 examinations. Students often uses old data booklets which is not suitable for the latest IB questions as they write to reffer on a particular table number which is not similar in old and new data booklet.
Students should also focus on the content of data booklet because if you are familiar about what is already given in the data booklet, you need not to mug it up.

Chemistry Data booklet

How to calculate moles from given mass of a substance?

How to calculate moles from given mass of a substance?Practice questions/worksheet for IB and IGCSE students

Mole(n)= Mass of a substance(m)/ Molar mass(Mr)

so moles of a substance , n= m / Mr

The first question is solved for you.

For calculating Molar mass, you need to add the relative atomic masses of all the atoms in a 

compound or ion. For example Mr of Na₂O = (23x2)+16 = 62 ,

because  Ar( relative atomic mass) of Na=23,

Ar( relative atomic mass) of O= 16

so  moles of 100 g Na₂O= 100/ 62= 1.61 moles

Calculate number of moles in 50 g of Na₂O

Calculate number of moles in 100 g of H₂O

Calculate number of moles in 25 g of NaOH

Calculate number of moles in 75 g of MgO

Calculate number of moles in 65 g of Mg(OH)₂

Calculate number of moles in 15 g of Li₂O

Calculate number of moles in 85 g of HCl

Calculate number of moles in 33 g of LiCl

Calculate number of moles in 48 g of MgCl₂

Calculate number of moles in 77 g of P₄O₁₀

Calculate number of moles in 43 g of H₃PO₄

Calculate number of moles in 94 g of Cl₂O₇

Calculate number of moles in 69 g of HClO₄

Calculate number of moles in 86 g of SO₂

Calculate number of moles in 39 g of H₂SO₃

Calculate number of moles in 68 g of SiO₂

Calculate number of moles in 74 g of Al₂O₃

Calculate number of moles in 116 g of H₂SO₄

Calculate number of moles in 5k g of Al₂(SO₄)₃

Calculate number of moles in 2.5 g of CO

Calculate number of moles in 10 g of NO

 Calculate number of moles in 88 g of CO₂

Calculate number of moles in 56 g of N₂

Calculate number of moles in 235 g of 𝐶𝐻₄

Calculate number of moles in 64 g of 𝑂₂

Calculate number of moles in 45 g of 𝐶₆𝐻₁₂𝑂₆

Calculate number of moles in 99 g of 𝑆𝑂𝐶𝑙₂

Calculate number of moles in 235 g of 𝐹𝑒𝐶𝑙_3.

How to calculate mass from given moles of a substance or compound?

How to calculate mass from given moles of a substance or compound? Practice questions/worksheet on mass to mole

Mole(n)= Mass of a substance(m)/ Molar mass(Mr)

so mass of a substance , m= nx Mr

The first question is solved for you.

For calculating Molar mass, you need to add the relative atomic masses of all the atoms in a 

compound or ion. For example Mr of Na₂O = (23x2)+16 = 62 ,

because  Ar( relative atomic mass) of Na=23,

Ar( relative atomic mass) of O= 16

so  mass of 1.5  moles of Na₂O= 1.5x62= 93 g

Calculate mass  in  1.5 moles of Na₂O

Calculate mass  in  2.5 moles of H₂O

 Calculate mass  in  3.5 moles of NaOH

Calculate mass  in  0.5 moles of MgO

 Calculate mass  in  4.5 moles of Mg(OH)₂

Calculate mass  in  0.5 moles of Li₂O

Calculate mass  in  2.5 moles of HCl

Calculate mass  in  6.5 moles of LiCl

Calculate mass  in  1.5 moles  of MgCl₂

Calculate mass  in  5.5 moles of P₄O₁₀

Calculate mass  in  0.5 moles of H₃PO₄

Calculate mass  in  7.5 moles of Cl₂O₇

Calculate mass  in  1.15 moles of HClO₄

Calculate mass  in  1.25 moles of SO₂

Calculate mass  in  1.35 moles of H₂SO₃

Calculate mass  in  1.45 moles of SiO₂

Calculate mass  in  1.55 moles of Al₂O₃

Calculate mass  in  10.5 moles of H₂SO₄

Calculate mass  in  12.5 moles of Al₂(SO₄)₃

Calculate mass  in  1.35 moles of CO

Calculate mass  in  1.25 moles of NO

 Calculate mass  in  1.45 moles of CO₂

Calculate mass  in  1.55 moles of N₂

Calculate mass  in  1.55 moles of 𝐶𝐻₄

Calculate mass  in  1.25 moles  of 𝑂₂

Calculate mass  in  1.5 moles  of 𝐶₆𝐻₁₂𝑂₆

Calculate mass  in  1.58 moles  of 𝑆𝑂𝐶𝑙₂

 Calculate mass  in  1.50 moles  of 𝐹𝑒𝐶𝑙_3.

What is Empirical formula? How to determine it?

What is Empirical formula? How to determine it?and Determining Molecular formula:

Empirical formula:  Chemical Formula with simplest ratio of atoms of a compound.

Molecular formula: Chemical Formula with definite number of atoms of a compound.

Example:  

C₆H₆ is the molecular formula of benzene.

CH is the simplest ratio of atoms of C₆H₆

Molecular formula = n X Empirical formula 

where n=Molecular formula mass/Empirical formula mass

  molecular formula mass of C₆H₆ is 78 g/mol

Empirical formula mass of CH is 13 g/mol

n= 78/13= 6

Molecular formula = n X Empirical formula = 6 X CH= C₆H₆

Step 1: Write the element symbols along with % given. Consider % as mass in gram.

Step 2: Divide the given mass of each element by relative atomic mass of that element.

( mole=mass/molar mass) so you get moles of each element here.

Step 3: Divide the moles of each element by the smallest moles obtained in step 2.

Step 4: Use the obtained values as number of atoms per element in empirical formula.

Note : if you did not get the whole number values, do not simply round off the values but multiply by suitable integers to make the value whole number. Example if you get 2.5, do not round simply to 3 but multiply 2.5 by 2 to get 5 which is a whole number.

Worked Problem:   A compound has 44.09 % iron and rest chlorine. Determine its empirical formula.  If the molar mass of the compound is 127 g/mol, determine its Molecular formula as well.

Solution: 

Step1	Iron(Fe)=44.09	Chlorine(Cl)=100-44.09=55.91
Step2	44.09/56	55.91/35.5
Step3	0.79/0.79=1	1.57/0.79= 1.98=2
Step4	1	2

So the empirical formula is FeCl₂  so now to find out molecular formula, 

use Molecular formula = n X Empirical formula where n=1

hence Empirical formula = Molecular formula

Practice questions:  

1.       A binary oxide has 74.19 % sodium and rest oxygen. Determine its empirical formula.  If the molar mass of the compound is 62 g/mol, determine its Molecular formula as well.

2.       A compound has 43.39 % sodium, 11.32% carbon  and rest oxygen. Determine its empirical formula.  If the molar mass of the compound is 106 g/mol, determine its Molecular formula as well.

3.       A compound has 34.46 % iron and rest chlorine. Determine its empirical formula.  If the molar mass of the compound is 162.5 g/mol, determine its Molecular formula as well.

4.       A binary oxide has 43.66 % phosphorus and rest oxygen. Determine its empirical formula.  If the molar mass of the compound is 284 g/mol, determine its Molecular formula as well.

5.       A compound has 40 % carbon, 6.6% hydrogen  and rest oxygen. Determine its empirical formula.  If the molar mass of the compound is 180 g/mol, determine its Molecular formula as well.