In this article, we shall solve problems to calculate number of moles and number of molecules and atoms present in given quantity of a substance.

Example 01:

3.49 g of ammonia at STP occupies 4.48 dm³. Calculate molar mass of ammonia.

Given: Mass of gas = m = 3.49 g = 3.49 x 10^-3 kg, Volume of gas = V = 4.48 dm³ = 4.48 x 10^-3 m³, P = 1.01325 x 10⁵ Pa, T = 273.15 K, Universal gas constant = R = 8.314 J K^-1 mol^-1.

To Find: Molar mass of gas = M =?

Solution:

PV = nRT

∴ M = mRT/ PV

∴ M = (3.49 x 10^-3 x 8.314 x 273.15) / (1.01325 x 10⁵ x 4.48 x 10^-3 )

∴ M = 17.46 x 10^-3 kg mol^-1 = 17.46 g mol^-1

Ans: Molecular mass of the gas is 17.46 g mol^-1.

Example 02:

2.8 x 10^-4 kg of a gas at STP occupies 0.224 dm³. Calculate molar mass of the gas.

Given: Mass of gas = m = 2.8 x 10^-3 kg, Volume of gas = V = 0.224 dm³ = 0.224 x 10^-3 m³, P = 1.01325 x 10⁵ Pa, T = 273.15 K, Universal gas constant = R = 8.314 J K^-1 mol^-1.

To Find: Molar mass of gas = M =?

Solution:

PV = nRT

∴ M = mRT/ PV

∴ M = (2.8 x 10^-3 x 8.314 x 273.15) / (1.01325 x 10⁵ x 0.224 x 10^-3 )

∴ M = 28.015 x 10^-3 kg mol^-1 = 28.02 g mol^-1

Ans: Molecular mass of the gas is 28.02 g mol^-1.

Example 03:

Calculate the mass of

a) 2.5 gram-atom of calcium. Ans:100 g.

Given: Given mass = 2.5 gram-atom

Solution:

Atomic mass of calcium = 40 g

1 gram atom of calcium corresponds to 40 g of calcium

2.5 gram atom of calcium corresponds to 40 x 2.5 = 100 g of calcium

Ans: Mass of calcium = 100 g

b) 1.5 gram-molecule of water.

Given: Given mass = 1.5 gram molecule

Solution:

Molecular mass of water (H₂O) = 1 x 2 + 16 x 1

Molecular mass of water (H₂O) = 2 + 16 = 18 g

1 gram molecule of water corresponds to 18 g of water

1.5 gram molecule of water corresponds to 18 x 1.5 = 27 g of water

Ans: Mass of water = 27 g

c) 2 gram-molecule of sulphuric acid.

Given: Given mass = 2 gram molecule

Solution:

Molecular mass of sulphuric acid (H₂SO₄) = 1 x 2 + 32 x 1 + 16 x 4

Molecular mass of sulphuric acid (H₂SO₄) = 2 + 32 + 64 = 98 g

1 gram molecule of sulphuric acid corresponds to 98 g of sulphuric acid

2 gram molecule of sulphuric acid corresponds to 98 x 2 = 196 g of sulphuric acid

Ans: Mass of sulphuric acid = 196 g

d) 0.5 gram-molecule of iodine.

Given: Given mass = 0.5 gram molecule

Solution:

Molecular mass of iodine (I₂) = 127 x 2 = 254 g

1 gram molecule of iodine corresponds to 127 g of iodine

0.5 gram molecule of iodine corresponds to 254 x 0.5 = 127 g of iodine

Ans: Mass of iodine = 127 g

e) 1.5 gram-molecule of sucrose. Ans: 513 g.

Given: Given mass = 1.5 gram sucrose

Solution:

Molecular mass of sucrose (C₁₂H₂₂O₁₁) = 12 x 12 + 1 x 22 + 16 x11

Molecular mass of sucrose = 342 g

1 gram molecule of sucrose corresponds to 342 g of sucrose

1.5 gram molecule of sucrose corresponds to 342 x 1.5 = 513 g of sucrose

Ans: Mass of sucrose = 513 g

Example 04:

Calculate the number of gram-atoms and gram-molecules of 25.4 mg of iodine.

Given: Given mass = 25.4 mg of iodine = 25.4 x 10^-3 g of iodine

Solution:

Molecular mass of iodine (I₂) = 127 x 2 = 254 g

254 g of iodine corresponds to 1 gram molecule of iodine.

25.4 x 10^-3 g of iodine corresponds to (1 x 25.4 x 10^-3)/ 254 gram molecule of iodine.

25.4 x 10^-3 g of iodine corresponds to 1 x 10^-4 gram molecule of iodine.

Iodine is diatomic molecule

25.4 x 10^-3 g of iodine corresponds to 1 x 10^-4 x 2 gram atom of iodine

25.4 x 10^-3 g of iodine corresponds to 2 x 10^-4 gram atom of iodine

Ans:2 x 10^-4 gram-atom and 1 x 10^-4 gram-molecule of molecule.

Example 05:

What is the molar mass of a gas if 1.00 dm³ of the gas weighs 1.50 g at 273 K and 1 atmospheric pressure?

Given: Mass of gas = m = 1.50 g = 1.50 x 10^-3 kg, Volume of gas = V = 1 dm³ = 1 x 10^-3 m³, P = 1 atm = 1.01325 x 10⁵ Pa, T = 273 K, Universal gas constant = R = 8.314 J K^-1 mol^-1.

To Find: Molar mass of gas = M =?

Solution:

PV = nRT

∴ M = mRT/ PV

∴ M = (1.50 x 10^-3 x 8.314 x 273) / (1.01325 x 10⁵ x 1 x 10^-3 )

∴ M = 33.59 x 10^-3 kg mol^-1 = 33.59 g mol^-1

Ans: Molecular mass of the gas is 33.59 g mol^-1.

Example 06:

What is the mass of one mole electrons.

Given: Mass of each electron = 9.1 x 10^-31 kg, Avogadro’s number = N = 6.022 x 10²³.

1mole of electrons contain 6.022 x 10²³ electrons

Mass of 1 mole of electrons = 6.022 x 10²³ x 9.1 x 10^-31

Mass of 1 mole of electrons = 5.48 x 10^-7 kg

Ans: Mass of 1 mole of electrons = 5.48 x 10^-7 kg

Example 07:

What is the molar volume of water at 273 K. Given density of water as 1.00 g cm^-3.

Solution:

Molecular mass of water (H₂O) = 1 x 2 + 16 x 1

Molecular mass of water (H₂O) = 2 + 16 = 18 g

Molar volume = Molar mass/ Density = 18 g /1 g cm^-3 = 18 cm³

Ans: molar volume of water at 273 K is 18 cm³

Example 08:

Calculate the molar mass of glucose and find number of atoms of each kind in 0.18 g of glucose.

Solution:

Molecular mass of glucose (C6H12O6) = 12 x 6 + 1 x 12 + 16 x 6

Molecular mass of sucrose = 72 + 12 + 96 = 180 g mol^-1 or 180 u

Number of moles = Given mass/Molecular mass

Number of moles = 0.18/180 = 10^-3

1 mole of glucose contain 6.022 x 10²³ molecules of glucose.

10^-3 mole of glucose contain 6.022 x 10²³ x 10^-3 molecules of glucose.

10^-3 mole of glucose contain 6.022 x 10²⁰ molecules of glucose.

There are 6 carbon atoms, 12 hydrogen atoms and 6 oxygen atoms in one molecule of glucose.

Number of carbon atoms = 6.022 x 10²⁰ x 6 = 36.13 x 10²⁰

Number of hydrogen atoms = 6.022 x 10²⁰ x 12 = 72.16 x 10²⁰

Number of oxygen atoms = 6.022 x 10²⁰ x 6 = 36.13 x 10²⁰

Example 09:

Calculate the number of atoms of each kind present in 3.42 g of sucrose.

Solution:

Molecular mass of sucrose (C₁₂H₂₂O₁₁) = 12 x 12 + 1 x 22 + 16 x11

Molecular mass of sucrose = 144 + 22 + 176 = 342 g

Number of moles = Given mass/Molecular mass

Number of moles = 3.42/342 = 10^-2

1 mole of sucrose contain 6.022 x 10²³ molecules of sucrose.

10^-2 mole of sucrose contain 6.022 x 10²³ x 10^-2 molecules of sucrose.

10^-2 mole of sucrose contain 6.022 x 10²¹ molecules of sucrose.

There are 12 carbon atoms, 22 hydrogen atoms and 11 oxygen atoms in one molecule of sucrose.

Number of carbon atoms = 6.022 x 10²¹ x 12 = 72.26 x 10²¹

Number of hydrogen atoms = 6.022 x 10²¹ x 22 = 132.5 x 10²¹

Number of oxygen atoms = 6.022 x 10²¹ x 11 = 66.24 x 10²¹

Example 10:

Calculate the number of atoms of each kind present in 5.6 g of urea (CO(NH₂)₂). Molecular mass of urea = 60 g mol^-1.

Solution:

Molecular mass of urea (CO(NH₂)₂) = 60 g

Number of moles = Given mass/Molecular mass

Number of moles = 5.6/60 = 0.0933 mole of urea

1 mole of urea contain 6.022 x 10²³ molecules of urea.

0.0933 mole of urea contain 6.022 x 10²³ x 0.0933 molecules of urea.

0.0933 mole of urea contain 5.618 x 10²² molecules of urea.

There are 1 carbon atom, 4 hydrogen atoms, 1 oxygen atom and 2 nitrogen atoms in one molecule of urea.

Number of carbon atoms = 5.618 x 10²² x 1 = 5.618 x 10²²

Number of hydrogen atoms = 5.618 x 10²² x 4 = 22.47 x 10²²

Number of oxygen atoms = 5.618 x 10²² x 1 = 5.618 x 10²²

Number of nitrogen atoms = 5.618 x 10²² x 2 = 11.24 x 10²²

Example 11:

Calculate the number of atoms of each kind present in 72.5 g of isopropanal (C₃H₇OH).  Molecular mass of isopropanol = 60 g mol^-1.

Solution:

Molecular mass of isopropanal (C₃H₇OH) = 60 g mol^-1

Number of moles = Given mass/Molecular mass

Number of moles = 72.5/60 = 1.208

1 mole of isopropanal contain 6.022 x 10²³ molecules of isopropanal.

1.208 mole of isopropanal contain 6.022 x 10²³ x 1.208 molecules of isopropanal.

1.208 mole of isopropanal contain 7.275 x 10²³ molecules of isopropanal.

There are 3 carbon atoms, 8 hydrogen atoms and 1 oxygen atom in one molecule of isopropanal.

Number of carbon atoms = 7.275 x 10²³ x 3 = 21.83 x 10²³

Number of hydrogen atoms = 7.275 x 10²³ x 8 = 58.2 x 10²³

Number of oxygen atoms = 7.275 x 10²³ x 1 = 7.275 x 10²³

Example 12:

Calculate the number of water molecules in a drop of water weighing 0.05 g. If this drop evaporates in one hour. Calculate the number of molecules evaporating per second.

Solution:

Molecular mass of water (H₂O) = 1 x 2 + 16 x 1

Molecular mass of water (H₂O) = 2 + 16 = 18 g

Number of moles = Given mass/Molecular mass

Number of moles = 0.05/18 = 0.0028

1 mole of water contains 6.022 x 10²³ molecules of water.

0.0028 mole of water contains 6.022 x 10²³ x 0.0028 molecules of water.

0.0028 mole of water contains 1.686 x 10²¹ molecules of water.

Rate of evaporation = 1.686 x 10²¹ / (1 x 60 x 60)

= 4.68 x 10¹⁷ molecules s-1.

Ans: Rate of evaporation is 4.68 x 10¹⁷ molecules s^-1.

Example 13:

How many oxygen are present in 300 g of calcium carbonate?

Solution:

Molecular mass of calcium carbonate (CaCO3) = 40 x 1 + 12 x 1 + 16 x 3

Molecular mass of calcium carbonate (CaCO₃) = 100 g

Number of moles = Given mass/ Molecular mass

Number of moles = 300/100 = 3

1 mole of calcium carbonate contains 6.022 x 10²³ molecules of calcium carbonate.

3 moles of calcium carbonate contain 6.022 x 10²³ x 3 molecules of calcium carbonate.

3 moles of calcium carbonate contain 18.066 x 10²³ molecules of calcium carbonate.

Each molecule of calcium carbonate contains 3 atoms of oxygen

Number of oxygen atom in 3 moles of calcium carbonate

= 18.066 x 10²³ x 3 = 5.42 x 10²⁴

Ans: Number of oxygen atom in 300 g of calcium carbonate is 5.42 x 10²⁴

Example 14:

How many molecules of water of hydration are present in 252 mg of oxalic acid. (H₂C₂O₄.2H₂O)

Solution:

Molecular mass of oxalic acid. (H₂C₂O₄.2H₂O)

= 1 x 2 + 12 x 2 + 16 x 4 + 2(1 x 2 + 16 x 1)

Molecular mass of oxalic acid. (H₂C₂O₄.2H₂O) = 126 g

Number of moles = Given mass/Molecular mass

Number of moles = 252 x 10^-3/126 = 2 x 10^-3

1 mole of oxalic acid contains 6.022 x 10²³ molecules of oxalic acid.

2 x 10^-3 mole of oxalic acid contains 6.022 x 10²³ x 2 x 10^-3 molecules of oxalic acid.

2 x 10^-3 mole of oxalic acid contains 12.044 x 10²⁰ molecules of oxalic acid.

For each molecule of oxalic acid, there are 2 molecules of water.

Number of molecules of water of hydration = 12.044 x 10²⁰ x 2

= 24.088 x 10²⁰

Ans: Number of molecules of water of hydration are 2.408 x 10²¹.

The post Calculation of Number of Moles and Molecules appeared first on The Fact Factor.

In this article, we shall study hydrolysis of different types of salts and we shall derive expression for hydrolysis constant for each type of salt.

Hydrolysis of salt of strong acid and weak base.

These salts on hydrolysis produce strong acids and weak bases. The resulting solution is acidic in nature.

Consider hydrolysis of ammonium chloride (NH₄Cl). It gives strong acid HCl and weak base NH₄OH, when treated with water and equilibrium exist as,

NH₄Cl + H₂O   ⇌   NH₄OH    + HCl

(weak base )  (strong acid)

Applying ionic theory

NH₄⁺  + Cl^– + H₂O  ⇌     NH₄OH    + H⁺  + Cl^–

Cancelling common ions of both the sides

NH₄⁺  + H₂O  ⇌    NH₄OH    + H⁺

This solution contains free H+ ion. It is acidic to litmus and pH < 7.

Note:

Some salts like CuSO₄, FeCl₃ on hydrolysis do not give clear solution but give turbid solution. It is due to formation of insoluble hydroxides.

Expression for  hydrolysis constant for a salt of strong acid and weak base.

Let one mole of a salt BA of strong acid and weak base be dissolved in water and solution is made Vdm3. The hydrolysis of salt takes place as follows

BA + H₂O         ⇌         BOH    + HA

(weak base )  (strong acid)

Applying ionic theory

B⁺  + A- + H₂O   ⇌   BOH    + H⁺  + A-

Cancelling common ions of both the sides

By applying the law of mass action to above equilibrium,

As water is in large excess [H₂O] = constant

Where K_h is constant called hydrolysis constant.

For a salt of strong acid and weak base h is very small. Hence , 1 – h  = h

Thus, the degree of hydrolysis of a salt of strong acid and weak base is inversely proportional to the square root of concentration and directly proportional to the square root of dilution.

Relation between  hydrolysis constant (K_h) of salt of strong acid and weak base and dissociation constant(K_b) of a weak base.

The hydrolysis of a salt of strong acid and weak base, BA leads to an equilibrium.

B⁺  + H₂O   ⇌   BOH    + H⁺ 

Besides above equilibrium there are two more equilibria.

Weak base BOH which ionizes to small extent.

BOH   ⇌ B⁺ +  OH^–

Where K_b is dissociation constant for the weak base.

Water is slightly ionized.

H₂O      ⇌      H⁺   +    OH ^–

k_w = [H⁺][OH^–] ……….. (3)

Where k_w is ionic product of water

Dividing equation (3) by (2), we get

Thus hydrolysis constant of a salt of strong acid and weak base is a ratio of ionic product of water and dissociation constant of a weak base.

Expression of concentration of [H⁺] in aqueous solution of strong acid and weak base.

The hydrolysis of a salt of strong acid and weak base, BA leads to an equilibrium.

Hydrolysis of salt of weak acid and strong base.

These salts on hydrolysis produce weak acids and strong bases. The resulting solution is basic in nature.

Consider hydrolysis of sodium acetate (CH₃COONa). It gives weak acid CH₃COOH and strong base NaOH, when treated with water and equilibrium exist as,

CH₃COONa  + H 2O ⇌  CH₃COOH + NaOH

(weak acid)  (strong base)

Applying ionic theory

CH₃COO^–  + Na⁺ + H₂O  ⇌    CH₃COOH    + Na⁺  + OH^–

Cancelling common ions of both the sides

CH₃COO^–  +  H₂O   ⇌   CH₃COOH    +  OH^–

This solution contains free OH- ions. It is basic to litmus and pH > 7.

Expression for  hydrolysis constant for a salt of weak acid and strong base.

Let one mole of a salt BA of weak acid and strong base be dissolved in water and solution is made Vdm3. The hydrolysis of salt takes place as follows

BA + H₂O  ⇌    BOH    + HA

(strong base )  (weak acid)

Applying ionic theory

B⁺  + A^– + H₂O  ⇌     B⁺  +  OH^– + HA

Cancelling common ions of both the sides

By applying the law of mass action to above equilibrium,

As water is in large excess [H₂O] = constant

Whre K_h is constant called hydrolysis constant.

For a salt of strong acid and weak base h is very small. Hence, 1 – h  = h

Thus, the degree of hydrolysis  of a salt of weak acid and strong base is inversely proportional to the square root of concentration and directly proportional to the square root of dilution.

Relation between  hydrolysis constant (K_h) of salt of weak acid and strong base and dissociation constant (K_a) of a weak acid.

The hydrolysis of a salt of weak acid and strong base, BA leads to an equilibrium.

A^– + H₂O  ⇌      OH^–  +  HA

Besides above equilibrium there are two more equilibria.

Weak acid which ionizes to small extent.

HA   ⇌ H⁺ +  A ^–

Where Ka is dissociation constant for the weak acid.

Water is slightly ionized.

H₂O      ⇌      H⁺   +    OH ^–

k_w = [H⁺][OH^–] ……….. (3)

Where k_w is ionic product of water

Dividing equation (3) by (2), we get

Thus hydrolysis constant of a salt of weak acid and strong base is a ratio of ionic product of water and dissociation constant of a weak acid.

Expression of concentration of [OH^–] in aqueous solution of weak acid and strong base.

The hydrolysis of a salt of strong acid and weak base, BA leads to an equilibrium.

Hydrolysis of salt of weak acid and weak base:

These salts on hydrolysis produce weak acids and weak bases. The nature of resulting solution depends on the relative strengths of the weak acid and the base formed..

Consider hydrolysis of ammonium acetate (CH₃COONH₄). It gives weak acid CH₃COOH and weak base NH₄OH, when treated with water and equilibrium exist as,

CH₃COONH₄ + H₂O  ⇌ CH₃COOH + NH₄OH

(weak acid)  (weak base)

Applying ionic theory

CH₃COO^–  + NH₄⁺ + H₂O  ⇌ CH₃COOH + NH₄OH

As there are no  free OH^– or H⁺ ions. The relative strength of CH₃COOH and NH₄OH are the same. It is neutral to litmus and pH = 7.

Expression for  hydrolysis constant for a salt of weak acid and weak base.

Let one mole of a salt BA of weak acid and strong base be dissolved in water and solution is made V dm³. The hydrolysis of salt takes place as follows

BA + H₂O  ⇌    BOH    + HA

(weak base )  (weak acid)

Applying ionic theory

By applying the law of mass action to above equilibrium,

As water is in large excess [H₂O] = constant

Where K_h is constant called hydrolysis constant.

For a salt of strong acid and weak base h is very small. Hence, 1 – h  = h

Thus, the degree of hydrolysis  of a salt of weak acid and weak base is independent of  concentration and dilution.

Note:

When h to be found from K_h, use formula

When K_h to be found from h, use formula

Relation between  hydrolysis constant (K_h) of salt of weak acid and weak base, dissociation constant(K_a) of a weak acid and dissociation constant(K_b) of a weak base.

The hydrolysis of a salt of weak acid and weak base, BA leads to an equilibrium.

B⁺  + A^– + H₂O ⇌     BOH + HA

Besides above equilibrium there are three more equilibria.

Weak acid which ionizes to small extent.

Where K_a is dissociation constant for the weak acid.

Weak base BOH which ionizes to small extent.

Where k_b is dissociation constant for the weak base.

Water is slightly ionized.

H₂O      ⇌      H⁺   +    OH ^–

k_w = [H⁺][OH^–] ……….. (4)

Where k_w is ionic product of water

Where KW is ionic product of water

Dividing equation (2) by (3) and (4), we get

.

………  (5)

From equation (1) and (5)

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The mixing of one s – orbital and one p – orbital of the same atom of nearly same energy to form a set of diagonally arranged two identical hybrid orbitals of equivalent energy is called sp hybridization. These hybrid orbitals are arranged in a linear manner around central atom and are at an angle of 180^o to one another.

Formation of BeF₂ Molecule:

Ground State of Beryllium Atom:

Atomic number of beryllium is 4. Its configuration in ground state is 1s², 2s², 2p⁰.

Beryllium atom in ground state:

Excited state of Beryllium Atom:

During combination with fluorine, the 2s electron pair is split up and one electron is promoted to empty 2p_x orbital. This condition is called excited state of beryllium. In excited state one electron of 2s migrates to 2p orbital forming 2 – half filled orbitals.
Beryllium atom in excited state:

Hybridization of Beryllium Atom:

One 2s orbital and one 2p orbitals of beryllium mix up forming two hybrid orbitals of equivalent energy. These two new equivalent orbitals are called sp hybrid orbitals. They are identical in all respect.

Beryllium atom in excited state:

Angle and Geometry :

Two sp hybridized orbitals formed, repel each other and These hybrid orbitals are arranged in a linear manner around central beryllium atom and are at an angle of 180^o to one another. Each sp hybrid orbital contain unpaired electron. In each sp hybrid orbital, one of the lobes is bigger because of more concentration of electron density. Only bigger lobe is involved in bond formation.

Thus in BeF₂ molecule has linear or diagonal structure with boron atom at the centre and two fluorine atoms at the either sides of beryllium. F-Be-F bond angle is 180^o.

Bond:

Two sp hybrid orbitals of boron atom having one unpaired electron each overlap separately with 1p orbitals of two fluorine atoms along the axis forming two covalent bonds (sigma bonds). The bonds between beryllium and fluorine are sp- p overlap. Thus F – Be — F bond angles are 180^o. Molecule is diagonal or linear. Both Be-F bonds in beryllium difluoride are of equal strength.

Diagram:

Type and Geometry of Beryllium difluoride Molecule:

BeF₂ Is linear molecule, while H₂O is angular molecule.

In BeF₂ molecule, beryllium undergoes sp hybridization and achieve electronic configuration 1s², 2s¹2p_x¹ in excited state. During formation of beryllium difluoride boron undergoes sp hybridization One 2s orbital and one 2p orbital of beryllium mix up forming two hybrid orbitals of equivalent energy. These two new equivalent orbitals are called sp hybrid orbitals. They are identical in all respect. Two sp hybridized orbitals formed, repel each other and they are directed diagonally opposite in space.  Angle between them is 180⁰. Two sp hybrid orbitals of boron atom having one unpaired electron each overlap separately with 1p orbitals of two fluorine atoms along the axis forming two covalent bonds (sigma bonds). Thus BeF₂ Is linear molecule.

In water, the oxygen atom is sp³ hybridized. Two hybrid orbitals have paired electrons (lone pair) and they are non – bonding orbitals.  Other two orbitals are half – filled (singly occupied) and they are bonding orbitals. These hybridized orbitals are in four directions of four corners of tetrahedron. Four sp³ hybridized orbitals formed, repel each other and they should be directed towards the four corners of a regular  tetrahedron and  Angle between them should be 109.5⁰.  The non bonding electron repel each other strongly and occupy more space than the electron pairs involved in bonding.  The force of repulsion between electron pair decreases in following order. lone pair – lone pair > lone pair- bond pair >  bond pair – bond pair. Hence the bond angle between two lone pairs i.e. H-O-H angle decreases from109.5⁰ to 104.5⁰. Thus water molecule is angular.

Formation of acetylene (C₂H ₂) Molecule:

Ground State of Carbon Atom:
Atomic number of carbon is 6. Its configuration in ground state is 1s², 2s², 2p² i.e. 1s² 2s², 2p_x¹ 2p_y¹ 2p_z⁰

Carbon atom in ground state:

Excited state of Carbon Atom:

During combination with hydrogen, the 2s electron pair is split up and one electron is promoted to empty 2p_z orbital. This condition is called excited state of carbon. In excited state one electron of 2s migrates to 2p orbital forming 4 – half filled orbitals.
Carbon atom in excited state:

Hybridization:

In acetylene there is sp hybridization. One 2s orbital and one 2p orbitals of carbon mix up forming two hybrid orbitals of equivalent energy. These two new equivalent orbitals are called sp hybrid orbitals. They are identical in all respect. Two ‘p’ orbitals of each carbon atom remains unhybridized.

Angle and Geometry:

Two sp hybridized orbitals formed, repel each other and These hybrid orbitals are arranged along x- axis in a linear manner around central carbon atom and are at an angle of 180⁰ to one another. The unhybridized p_y and p_z orbital remain perpendicular to hybrid orbitals along y-axis and z- axis ,
mutually perpendicular. Each sp hybrid orbital and unhybrid orbitals contain unpaired electron. In each sp hybrid orbital, one of the lobes is bigger because of more concentration of electron density. Only bigger lobe is involved in bond formation.

As all the four atoms in C₂H₂ molecule being in the same line, the molecule is diagonal or linear. H-C-C bond angle is 180^o.

Formation of Bonds:

1. Sigma Bond Formation :

A covalent bond formed by collinear or coaxial or in the line of internuclear axis. Overlapping of orbitals is known as sigma bond. One Sp hybrid orbital of one carbon atom overlaps with One hybrid orbital of other carbon atom by head on collision forming sigma bond. One (Sp- Sp ) overlap. Remaining one hybrid orbitals of each carbon atom overlap with ‘s’ orbital of two hydrogen atoms separately forming two sigma bonds. (2 C – H). Two (Sp– s)overlaps. Both C-H bond in Acetylene are of equal strength. Thus there are Three sigma bonds. Sigma bonds are stronger.

2) Formation of pi Bond :
The covalent bond formed by collateral or sidewise overlapping is called pi bond. The unhybridized 2 p_x and 2 p_z orbitals of each carbon atom being perpendicular to each other and to the plane of H-C-C-H axis overlap laterally with one another to form two week pi bond between two
carbon atoms by two p – p overlap. (one 2 p_y -2 p_y )and (one 2p_z-2 p_z) . Thus two (p-p) -overlaps.

In ethylene molecule there are 3 sigma bonds and 2 pi bonds. There is a triple bond between carbon and carbon consisting one sigma and two pi bonds.

Diagram:

Type and Geometry of Acetylene Molecule:

There is only one pi bond in ethylene molecule but there are two pi bonds in the acetylene molecule:

In ethylene molecule carbon atom shows sp² hybridization in excited state. The resulting three hybrid orbitals form two C-H bonds and One pi bond of sigma type. Thus each carbon atom is left with unhybridized p_z orbitals with lobes above and below the plane of hybridized orbitals. These two unhybridized orbitals overlap each other laterally and form a single pi bond between two carbon atoms.

In acetylene molecule carbon atom shows sp hybridization in excited state. The resulting two orbitals are linear and form one C-H bond and one C-C bond of sigma type. Thus each carbon is left with two unhybridized p_y and p_z orbitals, which are mutually perpendicular to H-C-C-H axis. These unhybridized orbitals overlap each other laterally and form two pi bonds between two carbon atoms.

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Mixing of one ‘s’ orbital and two ‘p’ – orbitals of nearly same energy forming set of trigonally arranged three identical orbitals of equal energy is known as sp² hybridization.

Geometry sp² Hybridization:

The hybrid orbitals are arranged around the central atom in a plane at an angle of 120° to one another. When these three orbitals overlap with appropriate orbitals of three other atoms, three bonds are formed and the resulting molecule has a trigonal planar structure. Each bond angle is 120°.

Diagram:

Formation of Boron Trifluoride Molecules:

Ground State of Boron Atom:

Atomic number of boron is 5. Its configuration in ground state is 1s², 2s², 2p¹

Boron atom in ground state:

Excited State of Boron Atom:

During combination with fluorine, the 2s electron pair is split up and one electron is promoted to empty 2p_y orbital. This condition is called excited state of boron. In excited state one electron of 2s migrates to 2p orbital forming 3 – half filled orbitals.

Boron atom in excited state:

Hybridization of Boron Atom:
One 2s orbital and two 2p orbitals of boron mix up forming three hybrid orbitals of equivalent energy. These three new equivalent orbitals are called sp² hybrid orbitals. They are identical in all respect

Boron atom in hybridized state:

Angle and Geometry:

• Three sp² hybridized orbitals formed, repel each other and they are directed towards the three corners of an equilateral triangle. The angle between them is 120°.
• Each sp² hybrid orbital contains an unpaired electron.
• In each sp² hybrid orbital, one of the lobes is bigger because of more concentration of electron density. Only bigger lobe is involved in bond formation.
• Thus BF₃ molecule has a trigonal planar structure with boron atom at the centre and three fluorine atoms at the four corners of equilateral triangle. F-B-F bond angle is 120°.

Bond Formation:

Three sp² hybrid orbitals of boron atom having one unpaired electron each overlap separately with 2p orbitals of three fluorine atoms along the axis forming three covalent bonds (sigma). Thus in BF₃ molecule has a planar structure with a boron atom at the centre and three fluorine atoms at the three corners of an equilateral triangle. F-B-F bond angle is 120°.

Bond:

• There are three sigma bonds..
• The bonds between boron and fluorine are sp²– p.
• Thus F – B — F bond angles are 120°. The molecule is trigonal planar.
• All B-F bonds in boron trifluoride are of equal strength.

Diagram:

Type and Geometry of Boron trifluoride Molecule:


The BF₃ molecule has a planar structure while NH₃ molecule has a pyramidal structure.
During formation of boron trifluoride, boron undergoes sp² hybridization and achieve the electronic configuration 1s², 2s¹2p_x¹2p_y¹. In the excited state, one 2s orbital and two 2p orbitals of boron mix up forming three hybrid orbitals of equivalent energy. These three new equivalent orbitals are called sp² hybrid orbitals. They are identical in all respect. Three sp² hybridized orbitals formed, repel each other and they are directed towards the three corners of an equilateral triangle (in one plane). Angle between them is 120^o. Three sp² hybrid orbitals of boron atom having one unpaired electron each overlap separately with 1p orbitals of three fluorine atoms along the axis forming three covalent bonds (sigma bonds). Thus boron trifluoride has triangular planar structure.

During formation of ammonia nitrogen undergoes sp³ hybridization. One 2s orbital and three 2p orbitals of mix up forming four hybrid orbitals of equivalent energy. These four new equivalent orbitals are called sp³ hybrid orbitals. They are identical in all respect and they should be directed towards the four corners of a regular tetrahedron and Angle between them should be 109.5^o. Three hybridized orbitals contain unpaired electron. The fourth hybridized orbital has lone pair of electron. The three half filled (containing unpaired electron) sp³ hybrid orbitals of nitrogen overlap axially with three half filled 1s orbitals of three hydrogen atoms separately to form three covalent N-H bonds (sigma bonds). The fourth hybrid orbital containing lone pair of electron remains non bonded. The non bonding electron repel each other strongly and occupy more space than the electron pairs involved in bonding. The force of repulsion between lone pair- bond pair is greater than bond pair – bond pair. Hence the
bond angle i.e. H-N-H angle decreases from109.5^o to 107^o.

Formation of Ethylene Molecule:

Ground State of Carbon Atom:
Atomic number of carbon is 6. Its configuration in ground state is 1s², 2s², 2p² i.e. 1s² 2s², 2p_x¹ 2p_y¹ 2p_z⁰

Carbon atom in ground state:

Excited state of Carbon Atom:

During combination with hydrogen, the 2s electron pair is split up and one electron is promoted to empty 2p_z orbital. This condition is called excited state of carbon. In excited state one electron of 2s migrates to 2p orbital forming 4 – half filled orbitals.
Carbon atom in excited state:

Hybridization of Carbon Atom:

In ethylene there is sp² hybridization. One 2s orbital and two 2p orbitals of carbon mix up forming three hybrid orbitals of equivalent energy. These three new equivalent orbitals are called sp² hybrid orbitals. They are identical in all respect. One ‘p’ orbital of each carbon atom remains unhybridized

Carbon atom in hybridized state:

Angle and Geometry:

Three sp² hybridized orbitals formed, repel each other and they are directed in a plane towards the three corners of an equilateral triangle. Angle between them is 120^o. The unhybridized p_z orbital remain perpendicular to this plane. Each sp² hybrid orbital contain unpaired electron. In each sp² hybrid orbital, one of the lobes is bigger because of more concentration of electron density. Only bigger lobe is involved in bond formation.

As all the six atoms in C₂H₆ molecule being in the same plane, the molecule is planar. H-C-H bond angle is 120^o. H-C-C bond angle is 120^o.

Bonds Formation:
1. Sigma Bond Formation :

A covalent bond formed by collinear or coaxial or in the line of internuclear axis. Overlapping of orbitals is known as sigma bond. One Sp² hybrid orbital of one carbon atom overlaps with One hybrid orbital of other carbon atom by head on collision forming sigma bond. One (Sp²– Sp² ) overlap.

Remaining two hybrid orbitals of each carbon atom overlap with ‘s’ orbital of four hydrogen atoms separately forming four sigma bonds. (C – H). Four (Sp²– s) overlaps. All C-H bond in ethylene are of equal strength.

Thus there are five sigma bonds. Sigma bonds are stronger.

2. Formation of pi Bond:

The covalent bond formed by collateral or sidewise overlapping is called pi bond. The unhybridized 2 p_z orbitals of each carbon atom being perpendicular to the plane of four hydrogen atoms and carbon atoms overlap laterally with one another to form a week pi bond between two carbon atoms by p – p overlap. One (p-p) -pi bond. This bond consists of two equal electron cloud one lying above the plane of the atom and other lying below this plane.

Hence, in ethylene molecule there are 5 sigma bonds and 1 pi bond.

Diagram :

Type and Geometry of Ethylene Molecule :

The post SP2 Hybridization appeared first on The Fact Factor.

In this article, we shall study to solve problems on the calculations of the number of electrons, protons, and neutrons in atoms, molecules, and species.

Example 01:

Calculate the charge and mass of 1 mole of electrons.

Solution:

1 mole of electron corresponds to 6.023 x 10²³ electrons

Mass of 1 electron = 9.1 x 10^-31 kg

Mass of one mole of electron = 9.1 x 10^-31 x 6.022 x 10²³ = 5.48 x 10^-7 kg

Charge of 1 electron = 1.602 x 10^-19 C

Charge on one mole of electron = 1.602 x 10^-19 x 6.022 x 10²³ = 9.65 x 10⁴ C

Ans: The mass of one mole of electrons is 5.48 x 10^-7 kg and the charge on one mole of electrons is 9.65 x 10⁴ C

Example 02:

Calculate the total number of electrons in 1 mole of ammonia (NH₃).

Solution:

Nitrogen (N):

Atomic number = Z = 7

Number of electrons = Atomic number = 7

Hydrogen (H):

Atomic number = Z = 1

Number of electrons in 1 atom of hydrogen = Atomic number = 1

Number of electrons in 3 hydrogen atoms = 1 x 3 = 3

Number of electrons in 1 molecule of ammonia = 7 + 3 = 10

1 mole of ammonia contains 6.022 x 10²³ molecules of ammonia

Number of electrons in 1 mole of ammonia = 10 x 6.022 x 10²³ = 6.022 x 10²⁴

Ans: The number of electrons in 1 mole of ammonia = 6.022 x 10²⁴

Example 03:

Calculate the total number of electrons in 1 mole of methane (CH₄).

Solution:

Carbon (C):

Atomic number = Z = 6

Number of electrons = Atomic number = 6

Hydrogen (H):

Atomic number = Z = 1

Number of electrons in 1 atom of hydrogen = Atomic number = 1

Number of electrons in 4 hydrogen atoms = 1 x 4 = 4

Number of electrons in 1 molecule of ammonia = 6 + 4 = 10

1 mole of ammonia contains 6.022 x 10²³ molecules of ammonia

Number of electrons in 1 mole of ammonia = 10 x 6.022 x 10²³ = 6.022 x 10²⁴

Ans: The number of electrons in 1 mole of ammonia = 6.022 x 10²⁴

Example 04:

Find the total number and the total mass of neutrons in 7 m of ¹⁴C. Mass of neutron = 1.675 x 10^-27 kg.

Given mass of carbon = 7 mg = 7 x 10^-3 g

Molecular mass of carbon = 14 g

Number of moles of carbon = (7 x 10^-3)/14 = 5 x 10^-4  mol

Number of carbon atoms in 7 mg of ¹⁴C = 5 x 10^-4 x 6.022 x 10²³ = 3.011 x 10²⁰

Number of neutrons in 1 atom of carbon = A – Z = 14 – 6 = 8

Number of neutrons in 7 mg of ¹⁴C = 8 x 3.011 x 10²⁰ = 2.4088 x 10²¹

Mass of 1 neutron = 1.675 x 10^-27 kg.

Mass of neutron in 7 mg of ¹⁴C = 1.675 x 10^-27 x 2.4088 x 10²¹

Mass of neutron in 7 mg of ¹⁴C = 4.0347 x 10^-6 kg

Example 05:

Find total number and total mass of proton in 34 mg of ammonia at STP. Will the answer change if temperature and pressure are changed?

Given mass of ammonia = 34 mg = 34 x 10^-3 g

Molecular mass of ammonia = 14 + 3 = 17 g

Number of moles of ammonia = (34 x 10^-3)/17 = 2 x 10^-3 mol

Nitrogen (N):

Atomic number = Z = 7

Number of protons = Atomic number = 7

Hydrogen (H):

Atomic number = Z = 1

Number of protons in 1 atom of hydrogen = Atomic number = 1

Number of protons in 3 hydrogen atoms = 1 x 3 = 3

Number of protons in 1 molecule of ammonia = 7 + 3 = 10

Number of molecules in 34 mg of ammonia = 2 x 10^-3 x 6.022 x 10²³ = 1.2044 x 10²¹

Number of protons in 34 mg of ammonia = 10 x 1.2044 x 10²¹= 1.2044 x 10²²

Mass of 1 proton = 1.673 x 10^-27 kg.

Mass of protons in 34 mg of ammonia = 1.673 x 10^-27 x 1.2044 x 10²²

Mass of protons in 34 mg of ammonia = 2.015 x 10^-5 kg

Due to change in temperature and pressure, there is no change in number of moles of the gas. Hence there is no effect o change of temperature and pressure.

Example 06:

Calculate the number of electrons which will together weigh 1 gram.

Mass of 1 electron = 9.1 x 10^-31 kg = 9.1 x 10^-28 g

Number of electrons in 1 gram = 1/(9.1 x 10^-28) = 1.098 x 10²⁷ electrons

Example 07:

2 x 10⁸ atoms of carbon are arranged side by side, calculate the radius of carbon atom, if the length of this arrangement is 2.4 cm

Diameter of carbon atom = 2.4/(2 x 10⁸) = 1.2 x 10^-8  cm

Radius of carbon atom = (1.2 x 10^-8)/2 = 6 x 10^-9 cm = 6 x 10^-11 m = 0.6 x 10^-10  m = 0.6 angstrom

Example 08:

Find the total number and total mass of neutrons in 18 mL of water. The specific gravity of water = 1.

Given volume of water = 18 mL = 18 x 10^-3 L

Mass of water = Volume x density = 18 x 10^-3 L x 1 kg/L = 18 x 10^-3 kg = 18 g

Molecular mass of water= 2 + 16 = 18 g

Number of moles of water = 18/18 = 1 mol

Oxygen (O):

Atomic number = Z = 8

Atomic mass number = A = 16

Number of neutrons = 16 – 8 = 8

Hydrogen (H):

Atomic number = Z = 1

Atomic mass of hydrogen = A = 1

Number of neutrons in 1 atom of hydrogen = 1 – 1 = 0

Number of neutrons in 2 hydrogen atoms = 0 x 2 = 0

Number of neutrons in 1 molecule of water = 8 + 0 = 8

Number of molecules in 18 mL of water = 1  x 6.022 x 10²³ = 6.022 x 10²³

Number of neutrons in 18 mL of water = 8 x 6.022 x 10²³= 4.8176 x 10²⁴

Mass of 1 neutron = 1.675 x 10^-27 kg.

Mass of neutrons in 18 mL of water = 1.675 x 10^-27 x 4.8176 x 10²⁴

Mass of neutrons in 18 mL of water = 8.0695 x 10^-3 kg

Due to change in temperature and pressure, there is no change in number of moles of the gas. Hence there is no effect o change of temperature and pressure.

Science > Chemistry > Atomic Structure > Problems on Calculation of Number of Electrons, Protons, and Neutrons

The post Problems on Calculation of Mass of Electrons, Protons, and Neutrons appeared first on The Fact Factor.

In this article, we shall study to solve problems based on the calculation of atomic number, atomic mass number, and neutron number

Atomic number (Z) :

The number of protons (positive charge) present in the nucleus of an atom of a particular element is called the atomic number of that element. It is denoted by letter ‘Z’.

Neutron number (N): 

The number of neutrons present in the nucleus of an atom is known as neutron number. It is denoted by ‘N’

Mass number (A): 

The total number of protons and neutrons present in the nucleus of an atom of the element is called mass number. The mass number is denoted as ‘A’.

A  =   Z + N

Example 01:

Calculate the number of electrons, protons, and neutrons in the following atoms.

Atomic Number = Z = 15

Atomic Mass Number = A = 31

Number of protons = atomic number = 15

Number of electrons = atomic number = 15

Neutron number = Number of neutrons = A – Z = 31 – 15 = 16

Number of nucleons = Atomic mass number = 31

Atomic Number = Z = 12

Atomic Mass Number = A = 24

Number of protons = atomic number = 12

Number of electrons = atomic number = 12

Neutron number = Number of neutrons = A – Z = 24 – 12 = 12

Number of nucleons = Atomic mass number = 24

Atomic Number = Z = 17

Atomic Mass Number = A = 37

Number of protons = atomic number = 17

Number of electrons = atomic number = 17

Neutron number = Number of neutrons = A – Z = 37 – 17 = 20

Number of nucleons = Atomic mass number = 37

Atomic Number = Z = 18

Atomic Mass Number = A = 40

Number of protons = atomic number = 18

Number of electrons = atomic number = 18

Neutron number = Number of neutrons = A – Z = 40 – 18 = 22

Number of nucleons = Atomic mass number = 40

Atomic Number = Z = 35

Atomic Mass Number = A = 80

Number of protons = atomic number = 35

Number of electrons = atomic number = 35

Neutron number = Number of neutrons = A – Z = 80 – 35 = 45

Number of nucleons = Atomic mass number = 80

Atomic Number = Z = 20

Atomic Mass Number = A = 40

Number of protons = atomic number = 20

Number of electrons = atomic number = 20

Neutron number = Number of neutrons = A – Z = 40 – 20 = 20

Number of nucleons = Atomic mass number = 40

Atomic Number = Z = 6

Atomic Mass Number = A = 13

Number of protons = atomic number = 6

Number of electrons = atomic number = 6

Neutron number = Number of neutrons = A – Z = 13 – 6 = 7

Number of nucleons = Atomic mass number = 13

Atomic Number = Z = 8

Atomic Mass Number = A = 16

Number of protons = atomic number = 8

Number of electrons = atomic number = 8

Neutron number = Number of neutrons = A – Z = 16 – 8 = 8

Number of nucleons = Atomic mass number = 16

Atomic Number = Z = 26

Atomic Mass Number = A = 56

Number of protons = atomic number = 26

Number of electrons = atomic number = 26

Neutron number = Number of neutrons = A – Z = 56 – 26 = 30

Number of nucleons = Atomic mass number = 56

Atomic Number = Z = 38

Atomic Mass Number = A = 88

Number of protons = atomic number = 38

Number of electrons = atomic number = 38

Neutron number = Number of neutrons = A – Z = 88 – 38 = 50

Number of nucleons = Atomic mass number = 88

Example 02:

Calculate the number of electrons, protons, and neutrons in the following species.

Atomic Number = Z = 15

Atomic Mass Number = A = 31

Number of protons = atomic number = 15

P^3- → P + 3 e^–

Number of electrons = 15 + 3 = 18

Number of neutrons = A – Z = 31 – 15 = 16

Number of nucleons = Atomic mass number = 31

Atomic Number = Z = 35

Atomic Mass Number = A = 80

Number of protons = atomic number = 35

Br^– → Br + e^–

Number of electrons = 35 + 1= 36

Number of neutrons = A – Z = 80 – 35 = 45

Number of nucleons = Atomic mass number = 80

Atomic Number = Z = 20

Atomic Mass Number = A = 40

Number of protons = atomic number = 20

Ca²⁺ → Ca – 2e^–

Number of electrons = 20 – 2= 18

Number of neutrons = A – Z = 40 – 20 = 20

Number of nucleons = Atomic mass number = 40

H₃PO₄

Hydrogen (H):

Atomic Number = Z = 1

Atomic Mass Number = A = 1

Number of protons per atom= atomic number = 1

Number of electrons per atom = Z = 1

Number of neutrons per atom = A – Z = 1 – 1 = 0

There are 3 hydrogen atoms in the molecule

Total number of protons in 3 hydrogens = 1 x 3 = 3

Total number of electrons in 3 hydrogens = 1 x 3 = 3

Total number of neutrons in 3 hydrogens = 0 x 3 = 0

Phosphorous (P):

Atomic Number = Z = 15

Atomic Mass Number = A = 31

Number of protons per atom= atomic number = 15

Number of electrons per atom = Z = 15

Number of neutrons per atom = A – Z = 31 – 15 = 16

There is 1 phosphorous atom in the molecule

Total number of protons in 1 phosphorous = 15 x 1 = 15

Total number of electrons in 1 phosphorous = 15 x 1 = 15

Total number of neutrons in 1 phosphorous = 16 x 1 = 16

Oxygen (O):

Atomic Number = Z = 8

Atomic Mass Number = A = 16

Number of protons per atom= atomic number = 8

Number of electrons per atom = Z = 8

Number of neutrons per atom = A – Z = 16 – 8 = 8

There are 4 oxygen atom in the molecule

Total number of protons in 4 oxygen= 8 x 4 = 32

Total number of electrons in 4 oxygen = 8 x 4 = 32

Total number of neutrons in 4 oxygen = 8 x 4 = 32

Ans:

Total number of protons in H₃PO₄ = 3 + 15 + 32 = 50

Total number of electrons in H₃PO₄ = 3 + 15 + 32 = 50

Total number of neutrons in H₃PO₄ = 0 + 16 + 32 = 48

NH₄⁺

Nitrogen (N):

Atomic Number = Z = 7

Atomic Mass Number = A = 14

Number of protons per atom= atomic number = 14

Number of electrons per atom = Z = 14

Number of neutrons per atom = A – Z = 14 – 7 = 7

Hydrogen (H):

Atomic Number = Z = 1

Atomic Mass Number = A = 1

Number of protons per atom= atomic number = 20

Number of electrons per atom = Z = 1

Number of neutrons per atom = A – Z = 1 – 1 = 0

There are 4 hydrogen atoms in the species

Total number of protons in 3 hydrogens = 1 x 4 = 4

Total number of electrons in 3 hydrogens = 1 x 4 = 4

Total number of neutrons in 3 hydrogens = 0 x 4 = 0

Total number of protons in NH₄⁺ = 7 + 4 = 11

NH₄⁺ → NH₄ – e^–

Total number of electrons in NH₄⁺ = 7 + 4 – 1 = 10

Total number of neutrons in NH₄⁺ = 7 + 4 = 11

ClO₃^–

Chlorine (Cl):

Atomic Number = Z = 17

Atomic Mass Number = A = 37

Number of protons per atom= atomic number = 17

Number of electrons per atom = Z = 17

Number of neutrons per atom = A – Z = 37 – 17 = 20

Oxygen (O):

Atomic Number = Z = 8

Atomic Mass Number = A = 16

Number of protons per atom= atomic number = 8

Number of electrons per atom = Z = 8

Number of neutrons per atom = A – Z = 16 – 8 = 8

There are 3 oxygen atom in the species

Total number of protons in 4 oxygen= 8 x 3 = 24

Total number of electrons in 4 oxygen = 8 x 3 = 24

Total number of neutrons in 4 oxygen = 8 x 3 = 24

Total number of protons in ClO₃^– = 17 + 24 = 41

ClO₃^– → ClO₃ + e^–

Total number of electrons in ClO₃^– = 17 + 24 + 1 = 42

Total number of neutrons in ClO₃^– = 20 + 24 = 44

Example 03:

Complete the table

Solution:

Zn

Atomic Number = Z = 30

Atomic Mass Number = A = 64

Number of protons = atomic number = 30

Number of electrons = atomic number = 30

Number of neutrons = A – Z = 64 – 30 = 34

Sr²⁺

Atomic Number = Z = 38

Atomic Mass Number = A = 90

Number of protons = atomic number = 38

Sr²⁺ → Sr – 2e^–

Number of electrons = 38 – 2 = 36

Number of neutrons = A – Z = 90 – 38 = 52

Te

Number of protons = 43

Number of Neutrons = N = 56

Atomic number =Number of protons = Z = 43

Atomic mass number = Z + N = 43 + 56 = 99

Number off electrons = Atomic number = 43

Br^–

Number of neutrons = N = 44

Number of electrons = 36

Br^– → Br + e^–

Number of protons = 36 – 1 = 35

Atomic Number = Number of protons = Z = 35

Atomic mass number = Z + N = 35 + 44 = 79

N

Number of neutrons = N = 7

Number of electrons = 7

Number of protons = 7

Atomic Number = Number of protons = Z = 7

Atomic mass number = Z + N = 7 + 7 = 14

Ca²⁺

Atomic Number = Z = 20

Number of neutrons = N = 20

Atomic Mass Number = Z + N = 20 + 20 = 40

Number of protons = atomic number = 20

Ca²⁺ → Ca – 2e^–

Number of electrons = 20 – 2 = 18

O

Atomic Number = Z = 8

Atomic Mass Number = A = 16

Number of protons = atomic number = 8

Number of electrons = atomic number = 8

Number of neutrons = A – Z = 16 – 8 = 8

The completed table is as follows:

Example 04:

The numbers of electrons, protons, and neutrons in a monoatomic species are equal to 36, 35, and 45 respectively. Assign proper symbol.

Solution:

Number of electrons = 36

Number of protons = Atomic number = Z = 35

Number of neutrons = N = 45

Atomic mass number = A = Z + N = 35 + 45 = 80

Charge on species = Z – Number of electrons = 35 – 36 = -1

The species is

Example 05:

The numbers of electrons, protons, and neutrons in a monoatomic species are equal to 18, 16, and 16 respectively. Assign proper symbol.

Solution:

Number of electrons = 18

Number of protons = Atomic number = Z = 16

Number of neutrons = N = 16

Atomic mass number = A = Z + N = 16 + 16 = 32

Charge on species = Z – Number of electrons = 16 – 18 = -2

The species is

Example 06:

Find the number of electrons in fluorine atom, fluorine molecule, and fluoride ion. The atomic number and mass number of fluorine are 9 and 19 respectively.

Solution:

Atomic number = Z = 9

Atomic mass number A = 19

Fluorine Atom (F):

Number of protons = Atomic number = 9

Number of Protons = Atomic number = 9

Number of electrons = Atomic number = 9

Number of neutrons = a – z = 19 – 9 =10

Fluorine Molecule (F₂):

There are 2 atoms in a molecule of fluorine

Number of protons in fluorine molecule= 9 x 2 = 18

Number of electrons in fluorine molecule= 9 x 2 = 18

Number of neutrons in fluorine molecule= 10 x 2 = 20

Fluoride Ion (F^–):

Number of Protons = Atomic number = 9

F^– → F + e^–

Number of electrons = 9 + 1 = 10

Number of neutrons = a – z = 19 – 9 =10

Example 07:

An isotope of atomic mass 24 had 12 neutrons in its nucleus. What is its atomic number? Represent the isotope in symbolic form.

Solution:

Atomic mass number = A = 24

Number of neutrons = N = 12

Number of protons = A – Z = 24 – 12= 12

Atomic number = Number of protons = Z = 12

The element is

Isotopes, Isotones, and Isobars

• Different atoms of the same element having the same atomic number but having different mass numbers are known as isotopes.
• Atoms of the different elements having a different atomic number but having the same mass numbers are known as isobars.
• Atoms of the different elements having the different atomic number, different mass number but having the same neutron number are known as isotones.

Example 08:

Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A):

• Z = 17 and A = 35

• Z = 92, A = 233

• Z = 4, A = 9

Example 09:

Give an isobar, an isotone, and an isotope of

Isobar
Isotone
Isotope

Science > Chemistry > Atomic Structure > Problems Based on Atomic Number, Mass Number, and Neutron Number

The post Problems Based on Atomic Number, Mass Number, and Neutron Number appeared first on The Fact Factor.

Mathematical relationships between volume, pressure, and temperature of a given mass of gas are referred to as Gas laws. In this article. we shall study Pressure-Temperature relation or Gay-Lussac’s law.

Gay-Lussac’s Law:

Statement:

At constant volume the pressure of a given mass of a gas increases or decreases by 1/273 of its pressure at 0^oC for every degree rise or fall in temperature.

Explanation:

Let P_o be the volume of a gas at 0 °C, Let this gas be heated through t °C, Let P_t be the volume of the gas at t °C. then,

Alternate Statement of Gay-Lussac’s law:

Thus at constant volume, the pressure of the certain mass of enclosed gas is directly proportional to the absolute temperature of the gas.

In general

This relation is called the pressure-temperature relation.

Graphical Representation:

A graph is drawn by taking the absolute temperature on the x-axis and pressure on the y-axis. The graph is as follows. This graph is also known as a P-T  diagram.

Numerical Problems:

Example 01:

A steel tank contains air at a pressure of 15 bar at 20 ^oC. The tank is provided with a safety valve which can withstand a pressure of 35 bar. Calculate the temperature to which the tank can be safely heated.

Solution:

Given: Initial pressure P₁ = 15 bar, Initial temperature = 20 ^oC = 20 + 273 = 293 K, Final pressure P₂ = 35 bar

To Find: Temperature up to which tank can be heated = T₂ =?

By Gay-Lussac’s Law

T₂ = (P₂ x T₁)/P₁ = (35 x 293)/15 = 683.67 K

T₂ = 683.67 – 273.15 = 410.15 ^oC

Ans: The tank can be heated up to 410.15 ^oC

Example 02:

An iron tank contains helium at a pressure of 2.5 atm at 25 ^oC. The tank can withstand a maximum pressure of 10 atm. The building in which the tank has been installed catches fire. Predict whether the tank will blow up first or melt if the melting point of iron is 1535 ^oC.

Solution:

Given: Initial pressure P₁ = 2.5 atm, Initial temperature = 25 ^oC = 25 + 273 = 298 K, Melting point of iron = T₂ = 1535 ^oC = 1535 + 273 = 1808 K

To Find: Final pressure = P₂ =?

By Gay-Lussac’s Law

P₂ = (T₂ x P₁)/T₁ = (1808 x 2.5)/298 = 15.16 atm

The pressure at the melting point is 15.16 atm, which is much more than the maximum pressure that the tank can withstand 10 atm. Hence the tank will blow up before reaching the melting point.

Ans: The tank will blow up before reaching the melting point.

Science > Chemistry > States of Matter > Charle’s Law

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A matter is defined as anything that has mass, which occupies space and may be perceived by senses. There are three states of matter, viz. (a) solid, (b) liquid, and (c) gaseous states. Whatever may be the state of matter, it is composed of particles. In this article, we shall study the particle model of matter.

Particle and Kinetic Model of Matter:

The particle model is also known as a dynamic particle model. On the basis of the particle model, different states of matter can be explained easily. Some assumptions of this model are as follows.

• All matter is made of tiny particles. However, the arrangement and the distribution of particles are different in the three states of matter.
• Empty spaces exist between these particles. These empty spaces are called voids.
• The particles exert force attraction on one another but the magnitude of these interparticle forces is different in the three states of matter.
• The particles are not stationary and have a tendency to acquire motion. In solids, they are fixed at a position and only vibrate about their mean position. In liquids and solids besides vibrational motion, the particles have translatory motion.
• With the increase in the temperature the kinetic energy of the particles hence the thermal energy increases.

Evidence of Particle Nature of Matter:

If we add potassium permanganate in water kept in a glass jar. We can observe the purple coloured particles separate from the crystals of potassium permanganate and spread in water. Ultimately whole water turns purple.

If we add crystals of salt in water, they settle at the bottom. Gradually their size starts reducing and ultimately the crystal disappear but whole water gets a uniform salty taste. Besides the volume of water does not increase. it indicates salt particles occupy inter-particulate spaces.

When a scent bottle is opened at one corner of a room the fragrance can be smelt at any corner of the room. The molecules of scent occupy the inter-particulate space.

Evidence of Kinetic Nature of the Particles of Matter:

The English Botanist Robert Brown, in 1927 observed that colloidal particles exhibit continuous random motion in all directions in a straight line.  He found such movement when pollen grains were suspended in water. 

The phenomenon of continuous zig-zag movement of colloidal particles in straight-line paths in a random direction is known as a Brownian movement. A pollen grain is placed on the surface of water taken in a beaker. It shows the Brownian movement. The pollen grain is surrounded by a large number of water molecules that constantly bombard the pollen grain. On unequal bombardment, the pollen grain gets pushed in certain directions. This experiment proves the kinetic nature of particles of matter.

Characteristics of Particles of the matter:

Particles of matter are very small.

All matter is made up of very small particles that are not visible to naked eye. It can be proved by the following experiment. Take two or three crystals of potassium permanganate and add them in 100ml of water. The solution formed is deep purple in colour. Now take 10ml of this solution and add it to another beaker containing 00ml of fresh water, again you will observe that the colour of the water will change but the solution will be faint compared to that in the first case. Repeat this procedure four or more time. In every step, we observe that the colour of the water changes but it will become fainter and fainter.

The solution remains coloured even at a very high dilution. Which shows that potassium permanganate added is broken into very very small particles exhibiting their characteristic properties. Hence we can conclude that particles of matter are very very small.

Particles have spaces between them.

If we add crystals of salt in water, they settle at the bottom. Gradually their size starts reducing and ultimately the crystal disappear but whole water gets a uniform salty taste. Besides the volume of water does not increase. it indicates salt particles occupy inter-particle spaces present between water particles.

Particles are constantly moving.

A pollen grain is placed on the surface of water taken in a beaker. It shows Brownian movement. The pollen grain is surrounded by a large number of water molecules which constantly bombard the pollen grain. On unequal bombardment, the pollen grain gets pushed in certain directions. This experiment shows that the particles of matter are constantly moving. Thus they possess kinetic energy. As the temperature rises, particles move faster. Hence with the increase in temperature the kinetic energy of the particles also increase.

Particles Attract each other.

Particles of matter have a force acting on them. This force keeps the particles together. The strength of this force of attraction varies from one kind of matter to another.  This force of attraction varies from substance to substance it can be verified by the fact that some forces can be powdered by applying small force, while some break into crystals, while some do not break. This force of attraction between the particles of the same substance is called cohesion. The attractive forces between the particles are maximum in solids and minimum or negligible in case of gases.

States of Matter on the Basis of Particle and Kinetic Model:

Solid State:

At room temperature, the particles of solids occupy definite positions. They can vibrate about their mean positions, but they cannot move from one position to another. The particles are bound to one another strongly. Hence solids have definite shape and definite volume.

Liquid State:

The particles of liquid do not have fixed positions, and they slide over one another within the bulk of the liquid. Thus particles can move from one position to another, but cannot leave the bulk. The particles are bound loosely. Hence liquids have definite volume but don’t have a definite shape. They take shape of the container in which they are kept.

Gaseous State:

In gases, the particles lie far apart and exert a very weak force of attraction on one another. Actually, the particles of gas are in random motion can change position continuously, and can move away from each other. They occupy the whole space available. Only walls of containers restrict their movement. Hence gases have neither definite shape nor definite volume.

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All matter has physical and chemical properties. Extensive properties are those properties of a substance which depend on the amount of substance. They vary with the amount of the substance. Examples: Mass, weight, and volume. Intensive properties are those properties of a substance which do not depend on the amount of substance. Examples: colour, melting point, boiling point, electrical conductivity, and physical state at a given temperature.

Physical Properties of Substance:

Physical properties are characteristics that can be measured or observed without changing the composition of the substance under study. All samples of a pure substance have the same chemical and physical properties. Physical properties can be extensive or intensive. 

Mass

A mass is the amount of matter that is found in a substance. Mass is expressed in terms of kilograms (kg).

Density

Density is the measurement of mass with respect to and in a relationship with volume. The mass of a substance per its unit volume is called density. Density is expressed in kilograms per cubic metre (kg/m³). Mathematically

Density = Mass / Volume

The density depends on the temperature and pressure of the substance. The effect is prominent in cases of gases. The application of increasing temperature decreases its density because its volume increases with increasing temperatures, and the application of increasing pressure increases density because the volume decreases with increasing pressure.

Volume

Volume is the measurement of the quantity or amount of matter in a three dimensional space. It is the space occupied by the substance. Volume is expressed in cubic metres (m³).

Boiling Point:

The temperature at which a liquid changes its state to a gas at atmospheric pressure is called the boiling point of that liquid.  It is defined as the temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure. This is the point at which both liquid and gaseous phase exists at equilibrium. The boiling point of the substance also varies with pressure and is specified at standard pressure.

The boiling point of a liquid is a characteristic property and can be treated as a criterion for the purity of liquid.  It increases with the increase in external pressure. Liquids having greater intermolecular forces have high boiling points.

Melting Point:

The temperature at which a solid changes its state to a liquid at atmospheric pressure is called the melting point of that solid. This is the point at which both liquid and solid phase exists at equilibrium. The melting point of the substance also varies with pressure and is specified at standard pressure.

The melting point of a liquid is a characteristic property and can be treated as a criterion for the purity of a solid.  It increases with the increase in external pressure. Solids having greater intermolecular forces have high melting points.

Conductivity

Conductivity is the measure of a substance’s ability, or lack of ability, to conduct electricity or heat. Some matter has a high level of conductivity and other matter has a high level of resistance to the conduction of electricity.

Heat Capacity

Simply stated, heat capacity is the amount of heat that must be added or taken away from a substance to achieve a certain temperature. Heat capacity is also referred to as thermal capacity and the amount of heat that is added or taken away is measured in terms of joules per kelvin.

Deliquescence:

Deliquescence refers to the property of a substance to absorb water from the air to dissolve itself and form an aqueous solution. Materials showing deliquescence are termed deliquescent. In order to be deliquescent, a substance must both absorb a large amount of water and be sufficiently soluble to dissolve in it. Examples: Sodium hydroxide, potassium hydroxide, anhydrous potassium chloride, anhydrous magnesium chloride, anhydrous ferric chloride show deliquescence.

Hygroscopicity:

Hygroscopicity is the tendency of a solid substance to absorb moisture from the surrounding atmosphere and are converted into hydroxides or hydrates. Anhydrous copper sulphate, quick lime (CaO), anhydrous sodium carbonate, etc. are hygroscopic in nature.

Efflorescence:

Efflorescence is a spontaneous loss of water by a hydrated salt, which occurs when the aqueous vapor pressure of the hydrate is greater than the partial pressure of the water vapour in the air. Washing soda (Na₂CO₃·10H₂O), Glauber’s salt or sodium sulphate (Na₂SO₄·10H₂O), Ferrous sulphate (FeSO₄·7H₂O), potash alum (K₂SO₄· Al₂(SO₄)₃.24H₂O) show efflorescence.

Mechanical Properties of Substance:

Strength:

It is the property of a material which opposes the deformation or breakdown of material in presence of external forces or load. Engineering materials must have the suitable mechanical strength to be capable to work under different mechanical forces or loads. It is shown by solids.

Toughness:

Toughness is the ability of a material to absorb energy and gets plastically deformed without fracturing. For good toughness, materials should have good strength as well as ductility. To be tough, the material should be capable to withstand both high stress and strain. It is shown by solids.

Elasticity:

Within elastic limit, the solid completely regains its original shape, size or volume after removal of deforming force, then the property is called elasticity. Steel, copper, aluminium show elastic behaviour.

Plasticity:

If a body is stressed beyond elastic limit, and it does not regain original shape, size, and volume after removal of deforming force, then the property is called plasticity. These substances can be given required shape very easily. Example: Plaster of paris

Hardness:

It is the ability of a material to resist permanent shape change due to external stress. There are various measures of hardness – Scratch Hardness, Indentation Hardness, and Rebound Hardness. Scratch Hardness is the ability of materials to oppose the scratches to the outer surface layer due to external force. It is shown by solids.

It is measured on Mohs’ scale. The Mohs’ scale of mineral hardness is a qualitative ordinal scale that characterizes the scratch resistance of different minerals through the ability of a harder material to scratch a softer material. It was created by the German geologist and mineralogist Friedrich Mohs in 1812.

On Moh’s scale hardness of a diamond is maximum (10) and that of talk is minimum (1). If a material can scratch topaz but can’t scratch corundum, then it possesses hardness equal to 8.

Brittleness:

The brittleness of a material indicates that how easily it gets fractured when it is subjected to a force or load. The solids of non-metal are generally brittle in nature. The brittleness of the material is temperature-dependent. Some metals which are ductile at normal temperature become brittle at low temperature. Hardness and brittleness are inverse properties. The harder the substance, the more brittle it is. It is shown by solids.

Malleability:

Malleability is a property of solid materials which indicates that how easily a material gets deformed under compressive stress. Malleability is often categorized by the ability of the material to be formed in the form of a thin sheet by hammering or rolling. This mechanical property is an aspect of the plasticity of the material. The malleability of material is temperature-dependent. With the rise in temperature, the malleability of material increases. This is the characteristic property of metals. Copper, aluminium, gold, silver show malleability. Gold is the most malleable metal.

Ductility:

Ductility is a property of a solid material indicates that how easily a material gets deformed under tensile stress. Ductility is often categorized by the ability of a material to get stretched into a wire by pulling or drawing. This mechanical property is also an aspect of the plasticity of material and is temperature-dependent. With the rise in temperature, the ductility of material increases. This is a characteristic property of metals. Copper, aluminium, gold, silver show ductility. Platinum is the most ductile metal.

Creep:

Creep is the property of a material that indicates the tendency of a material to move slowly and deform permanently under the influence of external mechanical stress. It results due to long time exposure to large external mechanical stress within the limit of yielding. Creep is more severe in materials that are subjected to heat for a long time.

Resilience:

Resilience is the ability of material to absorb the energy when it is deformed elastically by applying stress and release the energy when stress is removed. Proof resilience is defined as the maximum energy that can be absorbed without permanent deformation. The modulus of resilience is defined as the maximum energy that can be absorbed per unit volume without permanent deformation.

Fatigue:

Fatigue is the weakening of a material caused by the repeated loading of the material. When a material is subjected to cyclic loading, and loading greater than a certain threshold value but much below the strength of the material (ultimate tensile strength limit or yield stress limit), microscopic cracks begin to form at grain boundaries and interfaces. Eventually, the crack reaches a critical size. This crack propagates suddenly and the structure gets fractured.

Chemical Properties of Substance:

Chemical properties are characteristics that can only be measured or observed as matter transforms into a particular type of matter. The tendency of matter to react chemically with other substances is known as reactivity.

Reactivity:

The tendency of matter to combine chemically with other substances is known as reactivity. Certain materials like chlorine, potassium, sodium, etc. are highly reactive, whereas others like gold, platinum, etc. are extremely inactive.

Flammability:

The tendency of matter to burn is referred to as flammability. As matter burns, it reacts with oxygen and transforms into various substances. Example: wood, paper, etc. are flammable. Petrol, ethyl alcohol are highly flammable.

Toxicity:

Toxicity refers to the extent to which a chemical element or a combination of chemicals may harm an organism. Methyl alcohol, methyl isocyanate are highly toxic.

Reactivity with Acids and Bases:

A substance’s ability to react with an acid or a base is a chemical property.

Science > Chemistry > Introduction to Chemistry > Properties of Substance

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Atomic Radius:

The size of an atom is very small (120 pm). Secondly, since the electron cloud surrounding the atom does not have a sharp boundary, the determination of the atomic size cannot be precise.   One practical approach of finding the size of an atom of a non-metallic element is to measure the distance between two atoms when they are bound together by a single bond in a covalent molecule and from this value, the “Covalent Radius” of the element can be calculated.

The atomic radius (atomic size) may be regarded as the distance from the centre of the atom to the outermost (valence) shell of electrons.

The electron density in an atom is greatly influenced by the presence of other atoms around the bonding atom and the nature (type) of bonding with neighbouring atoms, Depending upon this the terms like crystal radius, covalent radius, van der Walls’ radius, tetrahedral radius, etc. are used. Hence the definition given in above point is arbitrary.

Other Terms Related to Atomic Radius:

Crystal Radius:

It is defined as one half of the distance between the centres of nuclei of two adjacent atoms in a metallic crystal. For e.g. the distance between two sodium atoms in a sodium crystal is 372 pm. Hence crystal radius of sodium = 372 / 2 = 186 pm.

Covalent Radius:

Covalent radius is defined as one half of the distance between the centres of the two similar nuclei of two similar atoms bonded together by a single covalent bond. For e.g. the distance between two oxygen atoms in molecular oxygen is 132 pm. Hence crystal radius of oxygen = 132 / 2 = 66 pm. Covalent radii are additive. This property can be used to find the internuclear distance between two molecules forming a single covalent bond among themselves.

van der Walls’ Radius: 

van der Walls’ radius is defined as one half of the distance between the nuclei of the two atoms of the same substance at their closest approach. For e.g. the closest distance between two hydrogen atoms without forming the bond is 240 pm. Hence van der Walls’ radius of sodium = 240 / 2 = 120 pm. Generally, atomic radii of inert gases are expressed in terms of van der Walls’ radius. van der Wall’s forces are weaker hence the distance between the atoms is larger.

The covalent radius is the smallest of all the radii because covalent bonds are formed due to overlapping of orbitals and there is penetration of one atom in another.

Factors Affecting Atomic Size:

Number of Shells:

Atomic size increases with the increase in the number of electronic shells. Thus atomic radius is directly proportional to the number of electronic shells.

Nuclear Charge:

As the nuclear charge increases the atomic radius decreases due to an increase in the attractive force on the outermost electrons. Thus atomic radius is inversely proportional to the nuclear charge.

Screening Effect:

In an atom having more electrons and particularly more electron shells, it is observed that the inner orbits decrease the attraction between the electrons in the outer orbit and nucleus. Thus they act as a screen or shield between the nucleus and electrons of outer orbit. This effect is known as the screening effect. As the screening effect increases, the atomic radius increases. Thus atomic radius is directly proportional to the screening effect. For a given quantum shell, the shielding ability of inner electrons decreases in the order of s > p > d > f.

Effective nuclear charge: The effective nuclear charge is the difference between the actual nuclear charge and the screening effect constant.charge.  Z_eff = Z – σ. The atomic radius is inversely proportional to the effective nuclear charge.

Periodic Trend in Atomic Radius Along the Period:

The atomic radii of the elements of the second period and the graphical representation of variation for the second period are given below.

The atomic radii of the elements of the third period are given below.

The trend for ‘s’ block and ‘p’ block elements: 

The atomic size generally decreases across a period.

Explanation:

As we move from left to right in a period atomic number increases, hence the nuclear charges increases. The valence electrons are added to the same orbit of all the elements in the same period, hence screening effect and number of shells are the same. The effective nuclear charge increases across the period.  Hence the attractive force on the electrons in the outermost shell increases. Hence the atomic radius decreases.

Thus in a period alkali metals have the largest atomic radius and it gradually decreases across the period and it is minimum for the halogen elements. The inert gases have the largest atomic radii in the period because for them van der Wall’s radii are considered. Radii of aluminium and gallium are equal in spite of the fact they belong to the same group. This is due to the presence of then d – block elements. screening power of d elements is less.

The trend for ‘d’ block elements:

In case of d block elements as we move from left to right across the period, atomic number increases. The nuclear charge increases. But the electrons are added to penultimate i.e. (n-1) shell, hence the electron cloud density of inner shells increases which increases the screening effect. Thus nuclear charge increases and screening effect increases. Hence there is decreases in the atomic radius but the extent of variation is very small compared to s block and p block elements. The extent of variation is so small that all of them can be considered to have almost equal atomic radii.

The trend for ‘f’ block elements:

In case of ‘f’ block elements as we move from left to right across the period, atomic number increases. The nuclear charge increases. But the electrons are added to pre-penultimate i.e. (n-2) shell, hence the electron cloud density of inner shells increases which increases the screening effect strongly. Thus nuclear charge increases and screening effect increases. The effect of nuclear charge is almost balanced by the screening effect. Hence there is a very small decrease in the atomic radius compared to d block elements. Hence the properties of ‘f’ block elements are almost similar.  This type of contraction among the first and second series of ‘f’ block elements is called lanthanide contraction.

Periodic Trend in Atomic Radius Along the Group:

Trend:  The atomic size increases down the group.

Explanation: As we move down the group atomic number increases, hence the nuclear charges increases. The number of shells increases hence screening effect increases. Hence the attractive force on the electrons in the outermost shell decreases. Hence the atomic radius increases.

The atomic radii of the elements of the group 1 are given below.

The atomic radii of the halogens elements of the group 17 are given below.

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