JEE MAIN 2020 NEWS
NATIONAL LEVEL ONLINE TEST
Both, Hydrogen and S & P block elements are important topics for JEE Main and the candidates need to have a detailed understanding of structures of various acids like P, S, Cl, N, Nitrogen Oxides, Dimers of Boron and Aluminium, oxides and their acidic-basic structure. S-Block Elements are a part of Inorganic Chemistry and hold a significant weightage in the JEE Main. Check JEE Main Chemistry Syllabus
JEE Main is expected to be taken between July 18 and 23, 2020, the candidates appearing for the same have almost 45 days in hand to prepare for the exam. So pull up your socks so that you can clear the exam with flying colors.
Both the elements of Group 1 and Group 2 of the modern periodic table are being referred to as S block elements. S block includes 14 elements namely
|hydrogen (H)||Rubidium (Rb)|
|Lithium (Li)||Calcium (Ca)|
|Helium (He)||Cesium (Cs)|
|Sodium (Na)||Strontium (Sr)|
|Beryllium (Be)||Francium (Fr)|
|Potassium (K)||Barium (Ba)|
|Magnesium (Mg)||Radium (Ra)|
The S block elements with only one electron in their s-orbital are referred to as group one or alkali metals while the S block elements having with electrons filling their s-orbital are called group two or alkaline earth metals. The electrons inside an atom hold different sub-orbitals of available energy levels in increasing order of energy. The last electron of an atom might find a place in either s, p, d, and f subshells. Accordingly, the elements of the atom with the last valence electron present in the s-suborbital are known as S block elements.
Practice with JEE Main Question Paper
Both, alkali and alkaline earth elements show a regular gradation in their properties in their respective group elements. The first member of both S block elements, Lithium and Beryllium are different from the rest of their members; however, at the same time, they resemble more with the diagonal element present in the next column.
The anomaly of these S block elements is mainly due to:
Low atomic and ionic size
Greater charge density (charge/volume of the atom)
Absence of d-orbitals.
Greater polarization of s block elements results in making the first element more covalent and makes it different from the rest which is ionic. Also, their similarity regarding the size and charge density makes them look similar to the element diagonally placed in the next group (diagonal relationship). It is being witnesses that both the physical and chemical properties of these S block elements keep on changing in a particular trend as the atomic number of the elements increases. Changes in the various properties of the group are as mentioned below:
Group 1 elements are being referred to as Alkali Metals, including Lithium (Li), Sodium (Na), Potassium (K), Rubidium (Rb), Caesium (Cs) and Francium (Fr). This entire group lies in the s block of the periodic table.
Alkali Metals are shiny and high reactive metals.
They are included in some solutions like oil to lower reactivity with the air.
They are soft and can be cut using a knife.
Sodium is abundant and francium is rare.
Alkali Metals are soft and can be cut with the help of a knife except lithium.
The atoms of alkali metals are of the maximum size in their respective periods.
The first ionization energy of the alkali metals is low as compared to the elements in the second group.
The alkali metals show more than 1 oxidation state.
The alkali metals have lesser values of reduction potential and thus have a higher tendency to lose electrons and act like good reducing agents.
Both, the melting and boiling points of alkali metals are quite less as the intermetallic bonds in them are quite weak.
All the alkali metals can form ionic compounds.
The alkali metals are excellent conductors of heat and electricity.
Alkali metals (except Li) exhibit the photoelectric effect
The alkali metals and their salts have a characteristic color to a flame
Alkali Metals react with oxygen to produce oxides, peroxides, and superoxides; however, Lithium is only Alkali that forms monoxides. While the remaining other alkali metals can form peroxides and superoxides.
4 Li +O2 → 2Li2 O (oxide)
They form hydroxides when mixed with water- 2Li +2H2O → 2LiOH+H2
They produce hydrides with hydrogen- 2 Li + H2 → 2 LiH
Since they have lesser first ionization energy, so they can react vigorously with halogen to producre halides- 2Na(s) + Cl2(g) → 2NaCl(s)
Alkali metals are solvable in liquid ammonia and form ammoniated ions which impart blue color to the solution- M + ( x + y ) NH3 → M+ ( NH3 )x + e-(NH3)y
The Alkali metals also forms salts when combined with Oxoacids.
|Benefits of Alkali Metals||General Characteristics of Compounds of Alkali Metals|
|They are used to make alloys.||All monoxides of alkali metals are elementary.|
|Sodium is important during nerve impulse transmission.||They react with nitrates and liberate nitrites.|
|Radium is useful in treat cancer cells.||Hydroxides of alkali metals have a strong base.|
|Potassium helps to open and close stomata.||They are miscible in polar solvents.|
|Potassium hydroxide works as a precipitating agent.||Alkali Metals are electropositive and metallic character keeps on increasing from lithium to francium.|
All the alkali metals, their oxides, peroxides and superoxides can dissolve easily in water. They produce corresponding hydroxides which are strong alkalies.
2Na + 2H2O → 2NaOH + H2
Na2O + 2H2O 2NaOH
Na2O2 + 2H2O → 2NaOH + H2O2
2KO2 + 2H2O results in 2KOH + H2O2 + O2
b) The basic strength of these hydroxides increases when we move down the group Li to Cs.
c) All these hydroxides can easily dissolve in water and thermally stable except lithium hydroxide.
d) Alkali metals react with other acids to form salts.
NaOH + HCI → NacI + H2O
2NaOH + H2 SO4 → Na2SO4 + 2H2O
M2O + 2HX → 2MX + H2O
MOH + HX → MX + H2O
M2CO3 + 2HX results in 2MX + CO2 + H2O
a) Standard enthalpies of formation in (kJ/mol-1)
b) Covalent Character-
Small cation and large anion favors covalency.
Order: LiCl > NaCl > KCl > RbCl > CsCl & . LiI > LiBr > LiCl > LiF
Greater the charge on the cation greater is its polarizing power and hence larger is the covalent character: Na+CI- < Mg+2CI2 < AI+3 CI3
If the charge on the anion is higher, it can gets polarized thus imparting more covalent character to the compound formed.
Lattice Energies are the amount of energy needed to separate one mole of solid ionic compound into its gaseous ions. If the lattice energy is greater than it will have a higher melting point of the alkali metals halide and lower is its solubility in water
The amount of energy released when one mole of gaseous ions combine with water to form hydrated ions.
M+ (g) + aq results in M+ (aq) + hydration energy
X- (g) + aq - (aq) + hydration energy
If the hydration energy is higher than the solubility of the compound in water will also be higher.
The solubility of the majority of the alkali metal halides except those of fluorides lowers if the group is descending. Since the decrease in hydration energy is greater than the corresponding decrease in the lattice energy.
Because of the higher hydration energy of Li+ ion, Lithium halides can easily dissolve in water except for LiF which is sparingly soluble due to its high lattice energy.
For the same alkali metal, the melting point will keep on decreasing order-wise
For the same halide ion, the melting point of lithium halides will be less as compared to of the corresponding sodium halides. The halides will decrease as we move down from Na to group Cs.
The low melting point of LiCl (887 K) as compared to NaCl is due to the fact that LiCl is covalent in nature while NaCl is ionic.
Li has anomalous properties mainly because
They are small in size
They have higher polarizing power
Lithium shows a diagonal relationship with magnesium as both the elements have same polarizing power.
The melting point and boiling point of lithium are much higher.
Lithium is quite harder as compared to other alkali metals. Magnesium is also hard metal.
Lithium reacts with oxygen least readily to form normal oxide whereas other alkali metals form peroxides and superoxides.
LiOH like Mg (OH)2 have a weaker base. Hydroxides of other alkali metals are strong bases.
Due to their appreciable covalent nature, the halides and alkyls of lithum and magnesium are soluble in organic solvents.
Unlike other alkali metals, lithium reacts directly with carbon to produce ionic carbide. Magnesium also forms same carbide.
The carbonates, hydroxides, and nitrates of lithium, as well as magnesium, decompose when they are heated:
Li2CO3 → Li2O + CO2
MgCO3 → MgO + CO2
2LiOH → Li2O + H2O
Mg (OH)2 → MgO + H2O
4LiNO3 → 2Li2O + 4NO2 + O2
2Mg ( NO3)2 → 2Mg + 4NO2 +O2
The salts of remaining alkali metals are stable towards heat.
When Lithium is heated, it decomposes to form lithium oxide, Li2O while the other alkali metals nitrate decomposes to offer the corresponding nitrite.
4LiNO3 → 2Li2O + 4NO2 + O2
2NaNO3 → 2NaNO2 + O2
2KNO3 → 2KNO2 + O2
Li2CO3, LiOH, LiF and Li3PO4
They are the only alkali metal salts which are not dissolved in water. The corresponding magnesium compounds are also insoluble in water. Hydrogen carbonates of both lithium and magnesium are not isolated in solid state. Hydrogen carbonates of other alkali metals can be isolated in a solid state.
Sodium Hydroxide is stable towards heat but it reduced to metal when heated with carbon
2NaOH + 2C → 2Na +2CO + H2
FeCl3 + 3NaOH →Fe(OH)3 + 3NaCl
NH4Cl + NaOH results in NaCl + NH3 + H2O
HgCl2 + 2NaOH results in HgO + 2NaCl + H2O
Increase in Zn(OH)2 + 2NaOH → Na2ZnO2 + 2H2Oh
Increase in Al2O3 + 2NaOH → 2NaAlO2 + H2O
SiO2 + 2NaOH → Na2SiO3 + H2O
3P + 3 NaOH +3H2O results in PH3 + 3NaH2PO2
2Al + 2 NaOH + 2H2O results in 3H2 + 2NaAlO2
It is used in the manufacturing of paper, soap, etc.
Also it is useful in in petroleum refining.
Used to make cotton.
Sodium Hydroxide is used to prepare sodium metal and many salts of sodium.
How it is prepared?
Preparation- Solvay process
Carbon dioxide gas is bubbled through a brine solution saturated with ammonia which results in the resulting creation of sodium hydrogen carbonate.
NH3 + H2O + CO2 → NH4HCO3
NaCI + NH4HCO3 → NaHCO3 + NH4CI
Sodium hydrogen carbonate formed precipitates out mainly due to the presence of more NaCl. The precipitated NaHCO3 is filtered off and later ignited to get Na2CO3.
2NaHCO3 → Na2CO3 + CO2 + H2O
Its properties include:
The aqueous solution absorbs CO2 yielding sparingly soluble sodium bicarbonate.
Na2CO3 + H2O + CO2 → 2NaHCO3
It dissolves in acids with an effervescence of carbondioxide and is causticised with lime to produce caustic soda.
Na2CO3 + 2HCl →2NaCl + H2O + CO2
Na2CO3 + Ca(OH)2 → 2NaOH + CaCO3
Fusion with silica, sodium carbonate yields sodium silicate.
Na2CO3 + SiO2 → Na2SiO3 + CO2
Hydrolysis – being a salt of a strong base (NaOH) and weak acid (H2CO3), when dissolved in water sodium carbonate. Undergoes hydrolysis to form an alkaline solution
Na2CO3 + 2H2O→ H2CO3 + 2NaOH
It is used for softening hard water.
A mixture of sodium carbonate & potassium carbonate is used a fusion mixture.
As an important laboratory reagent both in qualitative and quantitative analysis.
It is used in paper, paints and textile industries.
It is used for washing purposes in the laundry.
It is used in the manufacture of glass, borax, soap, and caustic soda.
Ionic hydrides are formed when hydrogen reacts with S-block metals, excluding Be and Mg. These s-block elements are present in Group 1 and Group 2 of the periodic table and among the most active metals. Group 1 metals are referred to as alkali metals and have a charge of +1 Group 2 metals are referred to as alkaline earth metals and have a charge of +2. Both Group 1 and Group 2 metals have low electronegativity values (less than 1.2). A few examples it includes:
Example 1- what will be the reaction between Hydrogen and Group 1 Alkali Metal (M can represent any Group 1 alkali metal)
Example 2- what will be the reaction between Hydrogen and Group 2 Alkaline Earth Metal (here M represent Ca, Sr, or Ba)
In ionic hydrides, hydrogen acts like a "halogen" in its hydride form needing 1 electron to fill its valence shell.
Example 3: what will happen in case Hydrogen reacts with Group 14 Elements (C is interchangeable with Si, Ge, and Sn)
A hydrocarbon is compound that includes only of the two elements namely carbon and hydrogen. There is no fixed limit to how many chains of C-H can be formed. Hydrocarbons are arranged in straight chains, in branched chains, or ring structures.
Chemistry is the most rewarding subject among other subjects that include the questions of JEE Main. Therefore, students must pay special attention to all the topics taught in Chemistry.
The students need to assess all the sections carefully and pay more attention to more important topics holding higher weightage. It is important to allocate quality time on those sections that need more time and devotion.
Prepare a strong strategy and adhering to the same is important while studying for Chemistry for JEE Main. Since Chemistry subject is quite vast and mostly it includes theoretical learning, students need to study regularly so as to keep the flow.
Keeping some time aside for the revisions is also important so that most essential topics are revised for JEE Main Chemistry. Since the concepts are more theoretical, so revising all the topics will help in making it memories fresh.
Studying from different books is not suggested; rather the students need to focus on some good books and refer to them for their preparation. Referring to different books results will make the students confused and will result in the wastage of time as well.
Memorizing some important topics like the Periodic Table, Atomic Weights, Hydrogen, Atomic Numbers, etc. is important while preparing for the Chemistry section.
The syllabus of JEE Mains Chemistry is vast and based on the exam analysis of JEE Main January 2020 and previous years, some of the important topics and chapters with their expected weightage are mentioned below:
|Name of Chapter||Topics to be Covered||Expected Number of Questions||Expected Marks|
|Physical Chemistry - 1||5||20|
|Inorganic Chemistry||S-Block Periodic Table & Periodicity in Properties Chemical Bonding||5||20|
For JEE’ Chemistry preparation, studying from good quality books plays an essential role. The candidates must not get confused between many books as it results in confusion among students. Some of the best books of Chemistry that can help in the preparation of JEE Main is mentioned below:
Modern Approach to Chemical Calculations- R.C. Mukherjee
Organic Chemistry -O. P. Tandon
Concept of Physical Chemistry- P. Bahadur
Physical Chemistry -P. W. Atkins
Concise Inorganic Chemistry- J. D. Lee
Organic Chemistry - Morrison and Boyd
1. Develop the habit of preparing your notes, as it will help a lot while revising the syllabus at the end. Prepare 2 notes, one for the theoretical concept and while other for writing the formulas especially in case you are solving the numerical question.
2. Revision is a must; the more you revise the topics, the better your understanding will be.
3. Managing speed and accuracy in JEE Main is very important to score well. No matter what speed you are preparing, accuracy is more crucial.
4. Focus on more relevant and important topics in the examination context, so that going through the syllabus thoroughly.
5. Revise the last year’s questions, as it offers an insight into the exam and types of questions asked in the exam.
For the students who have already cleared 12th, reading some of the beneath topics from NCERT books are a must:
Oxidizing and Reducing Agents of Organic Chemistry
The division of JEE Main Chemistry syllabus for 45 days is mentioned below:
|Total No. Of Days Left for Exam||45|
|Total No. Of Topics to be Revised in Chemistry||28|
|Total No. Of Topics to be Revised Per Day (Including Sunday)||Almost 2|
|Revision of Chemistry Syllabus to be Completed in||At least 10 days|
Early morning is the best time to prepare for the JEE Main 2020, and the students are suggested to prepare for at least two hours daily. However, cover all the subjects in one session is not possible, therefore students need to divide their accordingly.