There are two ways of balancing Redox reactions:

• Oxidation number method
• Half equation method

Oxidation method: The steps to be followed-

1. Write the skeletal equation of reactants and products.
2. Indicate the oxidation number of all the elements involved in the reaction.
3. Calculate the increase or decrease in oxidation number per atom. Also, identify the oxidizing and reducing agents.

1. Multiply the formula of oxidizing agent and reducing agent by suitable integers, so as to equalize the total increase or decrease in oxidation number as calculated in step c.
2. Balance all atoms other than H and O.
3. Finally balance H and O atoms by adding water molecules using hit and trial method.
4. In case of Ionic reactions:

• For acidic medium
• First balance O atoms by adding water molecules to the deficient side.
• Balance H+ ions to the side deficient in H atoms.
• For Basic medium
• First balance oxygen atom by adding water molecules to the deficient side.
• Then to balance hydrogen, add water molecules equal to the number of deficiency of H atoms.
• Also add equal number of OH^- ions to opposite side of the equation.

Example: Permagnate ion reacts with bromide ion in basic medium to give manganese dioxide and Bromate ion .

Step1: the skeletal ionic equation is :

MnO₄^- (aq) +Br^- (aq) ---> MnO₂ +BrO₃^-

Step 2: assign oxidation numbers for Mn and Br

Step3: calculate the increase and decrease in oxidation number and make the change equal :

Step: 4 as the reaction occurs in basic medium, and the ionic charges are not equal on both sides, add 2OH^- ions on the right to make it equal.

Step5: finally count the hydrogen atoms and add appropriate number of water molecules on the left side to achieve balanced Redox reaction.

Half reaction method or Ion electron method

1. Write the skeletal equation and indicate the oxidation number of all the elements which occur in skeletal equation
2. Find out the species that are oxidized and reduced.
3. Split the skeletal equation into two half reactions: oxidation half reaction and reduction half reaction
4. Balance the two-half equation separately by rules described below:

• In each half reaction first balance the atoms of element that has undergone a change in oxidation number.
• Add electrons to whatever side is necessary to make up the difference in oxidation number in each half reaction.
• Balance the charge by adding H⁺ ions, if the reaction occurs in acidic medium .For basic medium, add OH^- ions if the reaction occurs in basic medium.
• Balance oxygen atoms by adding required number of water molecules to the side deficient in oxygen atoms
• In the acidic medium, H atoms are balanced by adding H ⁺ ions to the side deficient in H atoms.
• However, in the basic medium H atoms are balanced by adding water molecules equal to number to H atoms deficient.
• Add equal number of OH^- ions to opposite side of equation.
• The two half reactions are then multiplied by suitable integers .so that the total number of electrons gained in half reaction becomes equal to total number of electrons lost in another half reaction.
• Then the two half reactions are added up.
• To verify the balancing, check whether the total charge on either is equal or not.

Example: Let us consider the skeletal equation:

Fe²⁺ + Cr₂O₇^2- -->  Fe³⁺ +Cr³⁺

Step 1: Separate the equation in to two halves:

Oxidation half reaction: Fe² --> Fe³⁺

 Reduction half reaction: Cr₂O₇^2-  -->    Cr³⁺

Step 2: Balance the atoms other than hydrogen and oxygen in each half reaction individually. Here the oxidation half reaction is already balanced with respect to Fe atoms .For the reduction half reaction, we multiply the Cr³⁺ by 2 to balance Cr atoms.

Step 3: For reactions occurring in acidic medium, add water molecules to balance oxygen atoms and hydrogen ions are balanced by adding H atoms. Thus, we get:

Cr₂O₇^2- + 14 H⁺  +  6e^- --> 2 Cr³⁺ + 7H₂O

Step 4: Add electrons to one side of the half reaction to balance the charges .if needed make the number of electrons equal in two half reactions by multiplying one or both half reaction by suitable coefficient.

The oxidation half reaction is thus written again to balance the charge .Now in the reduction half reaction there are 12 positive charges on the left hand side and only 6 positive charge on right hand side .Therefore, we add six electrons to left hand side .

Cr₂O₇^2- + 14 H⁺  +  6e^- --> 2 Cr³⁺ + 7H₂O

To equalize the number of electrons in both reactions, we multiply oxidation half reaction by 6 and write as:

6Fe²⁺  -->  6Fe³⁺ +6e^-

Step 5: We add the two half reactions to achieve the overall reaction and cancel the electrons on each side .This give us net ionic equation:

6Fe²⁺  + Cr₂O₇^2- + 14 H_⁺ -->    2Cr³⁺+6Fe³⁺ +7H₂O

Step6: Verify that the equation contains the same type and number of atoms and the same charges on both sides of the equation. This last check reveals that the equation is fully balanced with respect to number atoms and the charges.

It can be defined as the electrons which are there in degenerate orbitals have a parallel spin and tend to exchange their position.  The exchange energy is the energy released when two or more electrons with the same spin-exchange their positions in the degenerate orbitals of a subshell. Exchange energy is nothing but the energy released during this process. When the orbitals are half-filled or completely filled then the number of exchanges is maximum. Therefore, its stability is maximum.

Ionisation enthalpy: It is defined as the minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom.
M(g) →M⁺(g) + e^–I₁  = First ionisation enthalpy
Similarly, second and third electrons are also removed by providing successive ionisation enthalpies.
Factors on which Ionisation Enthalpy Depends:
(i) Size of the atom: The larger the atomic size, smaller is the value of ionisation enthalpy. In a larger atom, the outer electrons are far away from the nucleus and thus force of attraction with which they are attracted by the nucleus is less and hence can be easily removed.
Ionisation enthalpy ∝ 1/ Atomic size
(ii) Screening effect: Higher the screening effect, the lesser is the value of ionisation enthalpy as the screening effect reduces the force of attraction towards nucleus and hence the outer electrons can be easily removed.
Ionisation enthalpy ∝  1/ Screening effect
(iii) Nuclear charge: As the nuclear charge increases among atoms having same number of energy shells, the ionisation enthalpy increases because the force of attraction towards nucleus increases.
Ionisation enthalpy ∝ Nuclear charge
(iv) Half filled and fully filled orbitals: The atoms having half filled and fully filled orbitals are comparatively more stable, hence more energy is required to remove the electron from such atoms. The ionization enthalpy is rather higher than the expected value in case of such an atom.
Ionisation enthalpy ∝ Stable electronic configuration
(v) Shape of orbital: The s-orbital is more close to nucleus than the p-orbital of the same orbit. Thus, it is easier to remove electron from a p-orbital in comparison to s-orbital. In general, the ionisation enthalpy follows the following order
(s>p> d>f) orbitals of the same orbit.
Variation of ionisation enthalpy in the periodic table
ln general, the ionisation energy decreases down the group due to increase in atomic size. On the other hand, the ionisation energy increases across the period from left to right, again due to decrease in atomic size from left to right.

The energy of one photon is

E= hc/λ

 = (6.626×10⁻³⁴Js×3×10⁸m/s​)/ 0.57×10⁻⁶

 =3.487×10⁻¹⁹J .

25 watt corresponds to 25 J per second.

The number of photons emitted per second is 25​ / 3.487×10⁻¹⁹=7.17×10¹⁹/s. 

Air contains 20% oxygen by volume. Calculate the theoretical volume of air which will be required for burning completely 500 m³ of acetylene gas. All volumes are measured under the same conditions of temperature and pressure.

A n s w e r

(d) pentagonal bipyramid ?

 Iodine heptafluoride, also known as iodine(VII) fluoride or iodine fluoride, is an interhalogen compound with the chemical formula IF7. It has an unusual pentagonal bipyramidal structure, as predicted by VSEPR theory.

ANSWER
In Lyman series, the series of lines are in the Ultraviolet region 

In Balmer series, the series of lines are in the Visible region 

The series of lines in Paschen and Brackett series lies in an Infrared region

Hence, the correct option is Ad

According to law of conservation of mass,

Mass of reactants = Mass of products.

AgNO3 + NaCl = AgCl + NaNO3

xg 5.85g = 14.35g 8.5g

x 5.85 = 14.35 8.5

x. = 17.0g a n s w e r

According to law of conservation of mass,

Mass of reactants = Mass of products.

AgNO3 + NaCl = AgCl + NaNO3

xg 5.85g = 14.35g 8.5g

x 5.85 = 14.35 8.5

x. = 17.0g answer

A process which under some conditions may take place by itself or by initiation independent of the rate is called spontaneous process.

A process which can take place by itself or has an urge or tendency to take place is called spontaneous process.

A spontaneous process is simply a process which is feasible.

The rate of the process may vary from extremely slow to extremely fast.

Examples of processes which take place by themselves :

1) Dissolution of common salt in water

2) Evaporation of water in an open vessel

A process which can neither take place by itself nor by initiation is called a non spontaneous process.

For Example

1) Flow of water up a hill

2) Flow of heat from cold body to a hot body

3) Diffusion of gas from low pressure to a high pressure

4) Dissolution of sand in water

To designate an orbital n,l and m quantum numbers are required and s represents the (spin ) of the electron present in the orbital.

Composition of Calcium Hydrogen Carbonate - C₂H₂CaO₆

ElementS symbol Mass Percent Calcium Ca 24.7225%

Hydrogen H 1.2435%

Carbon C 14.8178%

Oxygen O 59.2162%

The Valence Shell Electron Pair Repulsion Theory abbreviated as VSEPR theory is based on the premise that there is a repulsion between the pairs of valence electrons in all atoms, and the atoms will always tend to arrange themselves in a manner in which this electron pair repulsion is minimalized. This arrangement of the atom determines the geometry of the resulting molecule.

The different geometries that molecules can assume keeping with VSEPR theory can be seen in the illustration provided below.

VSEPR Theory – Different Geometries that Molecules can Assume