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.

The post SP Hybridization appeared first on The Fact Factor.

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.

In this article, we shall study the concept of empirical formula and a molecular formula of a compound.

Empirical Formula:

The empirical formula of a substance represents the simplest relative whole number ratio of the atoms of each element contained in the molecule of the substance.

Example: the empirical formula for glucose is CH₂O. Thus the molecule of glucose contains atoms of carbon, hydrogen and oxygen in the ratio 1:2:1.

Molecular Formula:

The formula which gives an actual number of atoms of different elements present in the molecule of the compound is called molecular formula.

Example: A molecular formula for glucose is C₆H₁₂O₆. Thus the molecule of glucose contains 6 atoms of carbon, 12 atoms of hydrogen and 6 atoms oxygen.

The molecular formula is a simple multiple of the empirical formula. The relation between Empirical formula and the molecular formula is given by

Molecular formula = n x Empirical formula

Where n is a whole number and is a ratio of molecular mass to empirical formula mass.

e.g. The empirical formula for glucose is CH₂O.

Thus the empirical formula mass of glucose is (12 x 1 + 1 x 2 + 16 x 1  = 30 )

The molecular formula for glucose is C₆H₁₂O₆

Thus the molecular mass of glucose is (12 x 6 + 1 x 12 + 16 x 6  = 180 )

Now, Molecular formula mass = n x Empirical formula mass

∴ 180 = n x 30

∴ n = 6

Determination of Empirical Formulae/ Molecular Formulae:

Following steps are involved in the determination of empirical formula of organic compound.

1. Determination of atomic ratio: The percentage concentration of each element in the compound is divided by its respective atomic weight to get the atomic ratio. It is to be noted that if the sum of the percentage of constituent elements is not 100%, then the rest of the percentage is assumed to be that of oxygen.
2. Determination of simplest ratio: The ratio so obtained is divided by the smallest ratio in order to obtain the least ratio or the simplest ratio.
3. Determination of whole-number ratio: If the ratios are fractional, reduce them to the smallest possible whole numbers by multiplying throughout by suitable integer. This gives the simplest whole-number ratio.
4. Writing the empirical formula: Write down the symbols present side by side with the above numbers as a subscript to the lower right corner of each. This gives the empirical or the simplest formula.
5. Find the value of n: Find the value of the ratio of molecular mass to empirical formula mass and denote it as n
6. Find molecular formula using relation

Molecular Formula = n x Empirical Formula.

Numerical Problems:

Example – 01:

An organic substance contains 65% carbon, 3.5 % hydrogen and 9.59%  nitrogen. Find its empirical formula.

Solution:

The sum of the percentage of carbon, hydrogen and nitrogen is not 100%.

Hence the rest of the part of the compound is oxygen.

Percentage of oxygen = 100 – (65 + 3.5 + 9.59) = 21.91%

The simplest ratio of C to H to N to O is 8:5:1:2

Hence empirical formula = Ans. C₈H₅NO₂.

Example – 02:

An organic compound contains C, H, and O and is found to have 32% carbon, 4% of hydrogen, the remaining being oxygen. What is the molecular formula? If it contains six atoms of oxygen per molecule?

Solution:

The sum of the percentage of carbon and hydrogen s not 100%.

Hence the rest of the part of the compound is oxygen.

Percentage of oxygen = 100 – (32+4) = 64%

The simplest ratio is 1 :1.5:1.5 i.e. 2:3:3

Hence emperical formula of compound is C₂H₃O₃.

The compound contains six atoms of oxygen

n = Number of oxygen in molecular formula/Number of oxygen in molecular formula = 6/3 =2

Therefore, molecular formula = 2 x (Empirical formula)

Molecular formula = 2 x (C₂H₃O₃) = C₄H₆O₆

The molecular formula for the compound is C₄H₆O₆

Example – 03:

An organic compound contains 40% carbon, 6.66% of hydrogen.  What is the molecular formula of the compound? If its molecular weight is 180.

Solution:

The sum of the percentage of carbon and hydrogen is not 100%.

Hence the rest of the part of the compound is oxygen.

Percentage of oxygen = 100 – (40 + 6.66) = 53.34%

Simplest ratio = 1:2:1

Hence empirical formula = CH₂O.

Empirical formula weight = 12 x 1 + 1 x 2 + 16 x 1 = 30

n = Molecular mass / empirical formula mass = 180/30 = 6

Therefore, molecular formula = n x (Empirical formula)

Molecular formula= 6 x (CH₂O)

Molecular formula = C₆H₁₂O₆.

Therefore molecular formula of the organic compound is C₆H₁₂O₆.

Example – 04:

An organic monobasic acid contains 18.6% carbon, 1.55% of hydrogen, 55.04% chlorine, and 24.81% oxygen.  What is the molecular formula of the acid, If its molecular weight is 129?

Solution:

The sum of the percentage of carbon, hydrogen, and chlorine is not 100%.

Hence the rest of the part of the compound is oxygen.

Percentage of oxygen = 100 – (18.6 + 1.55 + 55.04) =  24.81%

Simplest ratio = 1:1:1:1

Hence empirical formula = CHClO.

Empirical formula weight = 12 x 1 + 1 x 1 + 35.5 x 1 + 16 x 1 = 64.5

n = Molecular mass / empirical formula mass = 129/64.5 = 2

Therefore, molecular formula = n x (Empirical formula)

Molecular formula = 2 x (CHClO)

Molecular formula = C₂H₂Cl₂O₂.

Therefore molecular formula of the organic acid is C₂H₂Cl₂O₂.

But the acid is monobasic acid, hence it should contain one – COOH group

i.e. the molecular formula of organic monobasic acid is CHCl₂COOH.

Example – 05:

Analysis of Vitamin C shows that it contains 40.92% carbon by mass, 4.58% hydrogen, and 54.50% oxygen. Determine the simplest formula of vitamin C.

Solution:

The simplest ratio is 1 :1.3:1 i.e. 3:4:3

Hence the simplest formula of vitamin C is C₃H₄O₃.

Science > Chemistry > Introduction to Organic Chemistry > Empirical and Molecular Formulae of Organic Compounds

The post Empirical and Molecular Formulae of Organic Compounds appeared first on The Fact Factor.

In the last artivle we have studied what is organic chemistry? Why it is termed as chemistry of carbon compounds nowaday? In this article we shall study classification of organic compounds. Depending upon the structure organic compounds are classified into two types. viz. open-chain organic compounds or aliphatic organic compounds or Acyclic organic compounds and closed chain organic compounds or cyclic organic compounds

Open Chain Organic Compounds:

Open chain organic compounds are organic compounds that contain an open chain of carbon atoms which may be straight-chain or branched-chain.

Closed Chain Organic Compounds:

Closed chain organic compounds are organic compounds that contain one or more closed chains or rings of carbon atoms. Closed chain organic compounds are further classified into two types. viz. homocyclic organic compounds or carbocyclic organic compounds. and heterocyclic organic compounds

Homocyclic Carbon Compounds: 

Homocyclic organic compounds are organic compounds that contain one or more closed rings of carbon atoms only. Homocyclic organic compounds are further classified into two types. viz. alicyclic organic compounds and aromatic organic compounds

Alicyclic Organic Compounds: 

Alicyclic organic compounds are homocyclic organic compounds that have properties similar to that of aliphatic compounds.

Aromatic Organic Compounds:

Aromatic organic compounds are homocyclic organic compounds that contain at least one benzene ring.

Heterocyclic Organic Compounds: 

Heterocyclic organic compounds are organic compounds that contain at least one atom other than a carbon atom in the ring.

Homologous Series:

A series of organic compounds which have a common general formula and in which the two successive members of the series differ by – CH₂ – is known as homologous series. The individual member of the homologous series is called homologue.

Alkane Series:

Alcohol Series:

Characteristics of Homologous Series:

• Members of the same homologous series are represented by the same general formula. E.g. all alkanes are represented by the same general formula C_nH_{2n + 2}.
• They can be prepared by similar methods of preparation.
• They have the same functional group hence have a number of chemical properties in common which are called the general properties.
• They show a regular gradation in physical properties such as melting and boiling points.
• Each member of the homologous series is known as the homologue of the other elements and differs from its next higher or next lower homologue by a common difference –CH₂–
• Each member of the homologous series differs from its next higher or next lower homologue in molecular weight by 14.

Science > Chemistry > Introduction to Organic Chemistry > Classification of Organic Compounds

The post Classification of Organic Compounds 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

The post Pressure-Temperature Relation (Gay-Lussac’s law) appeared first on The Fact Factor.

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

The post Properties of Substance appeared first on The Fact Factor.

In the last few articles, we have studied the methods of preparations of alkyl halides. In this article, we shall study the physical properties of alkyl halides. Some physical properties of alkyl halides are as follows:

State:

Lower members (methyl chloride, methyl bromide, methyl fluoride, ethyl bromide, ethyl chloride and ethyl bromide) are gases and higher members are liquids (Up to C18) and solids (Greater than C18).

Odour:

In the pure state, the haloalkanes up to C18 possess pleasant sweet odour. All higher haloalkanes are odourless.

Colour:

Pure haloalkanes are colourless. However, bromoalkanes and iodoalkanes on storing for long period, when exposed to light develop colour.

Boiling Points:

Haloalkanes have higher boiling points as compared to those compared to corresponding alkanes. This is due to their polarity and strong dipole-dipole attractive interaction between haloalkane molecules and greater magnitude of van der Wall’s forces.

• For the same alkyl group the boiling points of haloalkanes are in the order RCl < RBr < RI, because with the increase in the size of halogen atom the magnitude of van der Wall forces of attraction increases.
• Among isomeric alkyl halides, the boiling point decreases with an increase in branching in the alkyl group, because with branching the molecule attains a spherical shape with less surface area. As a result, interparticle forces become weaker. Hence the boiling point decreases. The order of boiling point is Primary  > Secondary >= iso > Tertiary
• For the same halogen, the boiling point increases with the increase in the molecular mass because with the increase in the size of the alkyl group the magnitude of van der Wall forces of attraction increases. i.e R – X < R -CH2-X  < R -CH2-CH2-X
• As the number of halogen in a molecule increases the boiling point of the compound increases because of the increase in the number of halogen atoms the magnitude of van der Wall forces of attraction increases. i.e. CH₃Cl < CH₂Cl₂ < CHCl₃ < CCl₄

Solubility:

Alkyl halides are polar in nature (dipole moment 2.05 to 2.15 D) but they are not able to form hydrogen bonds with water molecules. Hence they are sparingly soluble in water. But they are soluble in organic solvents like alcohols, ethers and benzene.

Density:

Alkyl chlorides are generally lighter than water, while alkyl bromides and alkyl iodides are heavier than water. The order of density is RI > RBr > RCl. Poly chlorides are heavier than water. Thus the density of alkyl halides increases with the increase in the number and atomic mass of the halogen atoms. Methyl iodide is the heaviest of all the haloalkanes.

Scientific Reasons:

Density:

Arrange the following in the order of decreasing density. 1-Chloropropane, 1-Iodopropane, 1-Bromopropane

• For the same alkyl group, the density of alkyl halides increases with the increase in the number and atomic mass of the halogen atoms. The atomic mass of I > Atomic mass of Br >Atomic mass of Cl.
• Hence boiling point of 1-Iodopropane > 1-Bromopropane > 1-Chloropropane.

Arrange each of the following set of compounds in the order of increasing densities:

CHCl₃, CH₂Cl₂, CCl₄, CH₃Cl:

• The order of density is RI > RBr > RCl. Poly chlorides are heavier than water. Thus the density of alkyl halides increases with the increase in the number and atomic mass of the halogen atoms.
• Hence, the order of densities is CH₃Cl. < CH₂Cl₂ < CHCl₃ < CCl₄.

C₂H₅Cl, C₂H₅I, C₂H₅Br:

• The order of density is RI > RBr > RCl. Thus the density of alkyl halides increases with the increase in the number and atomic mass of the halogen atoms.
• Hence, the order of densities is C₂H₅Cl < C₂H₅Br < C₂H₅I.

Which alkyl halide has the highest density and why?

• For the same alkyl group, the order of density is R-I > R-Br > R-Cl. Thus R-I will have the highest density.
• For the same halogen group, with the increase in the branching, the molecule acquires the spherical shape with less surface area. Thus the tertiary butyl group will have the smallest size.
• From the above two points, we can say that tertiary butyl iodide should have the highest density.

Boiling Points:

Which isomer of C₅H₁₁Cl has the highest boiling point? Why?

• Consider the following two isomers of C₅H₁₁Cl 

• Among isomeric alkyl halides, the boiling point decreases with an increase in branching in the alkyl group.
• 1-Chloropentane is a straight-chain isomer. It has the strongest interparticle forces. Hence it has the highest boiling point among all the isomers. While 1-Chloro-2,2-dimethylpropane has the highest number of branches in all possible isomers, hence it has the weakest interparticle forces. Hence it has the lowest boiling point among all the isomers.

Which isomer of C₄H₉Cl has the highest boiling point? Why?

• Among isomeric alkyl halides, the boiling point decreases with an increase in branching in the alkyl group.
• 1-Chlorobutane (n-Butyl chloride) is a straight-chain isomer. It has the strongest interparticle forces. Hence it has the highest boiling point among all the isomers. While 2-Chloro-2-methylpropane (tert-Butyl chloride) has the highest number of branches in all possible isomers, hence it has the weakest interparticle forces. Hence it has the lowest boiling point among all the isomers.

Arrange in the order of increasing boiling points. 

Bromobenzene. chlorobenzene, iodobenzene:

• The boiling points of mono halogen derivatives of benzene follow the order Iodo > Bromo > Chloro.
• Hence boiling point of Chlorobenzene <  Bromobenzene < Iodobenzene .

n-pentyl chloride, iso-pentyl chloride, neo-pentyl chloride:

• Among isomeric alkyl halides, the boiling point decreases with the increase in branching in the alkyl group, because with branching the molecule attains spherical shape with less surface area. As a result, interparticle forces become weaker. Hence the boiling point decreases. The order of boiling point is Primary  > Secondary >= iso > Tertiary
• Hence boiling point of neo-pentyl chloride < iso-pentyl chloride < n-pentyl chloride.

Bromomethane, Bromoform, Chloromethane, Dibromomethane:

• For the same alkyl group the boiling points of haloalkanes are in the order RCl < RBr< RI, because with the increase in the size of halogen atom the magnitude of van der Wall forces of attraction increases. For the same halogen, the boiling point increases with the increase in the molecular mass. As the number of halogen in a molecule increases the boiling point of the compound increases.
• Hence boiling point of Chloromethane < Bromomethane <  Dibromomethane < Bromoform.

1-Chloropropane, isopropyl chloride, 1-Chlorobutane:

• As the number of halogen in a molecule increases the boiling point of the compound increases. Among isomeric alkyl halides, the boiling point decreases with the increase in branching in the alkyl group, because with branching the molecule attains a spherical shape with less surface area. As a result, interparticle forces become weaker. Hence the boiling point decreases. The order of boiling point is Primary  > Secondary >= iso > Tertiary.
• Hence boiling point of isopropyl chloride < 1-Chloropropane < 1-Chlorobutane.

Methyl chloride, methyl bromide, methyl iodide:

• For the same alkyl group the boiling points of haloalkanes are in the order RCl < RBr < RI, because with the increase in the size of halogen atom the magnitude of van der Wall forces of attraction increases.
• Hence the order of boiling points is Methyl chloride (CH₃Cl) < methyl bromide (CH₃Br) < methyl iodide (CH₃I).

Methyl bromide, methylene bromide, bromoform:

• As the number of halogen in a molecule increases the boiling point of the compound increases.
• Hence the order of boiling points is methyl bromide (CH₃Br) < methylene bromide (CH₂Br₂) < Bromoform (CHBr₃).

Propane, n-propyl bromide, isopropyl bromide:

• Haloalkanes have higher boiling points as compared to those compared to corresponding alkanes. This is due to their polarity and strong dipole-dipole attractive interaction between haloalkane molecules.
• Among isomeric alkyl halides, the boiling point decreases with the increase in branching in the alkyl group, because with branching the molecule attains a spherical shape with less surface area. As a result, interparticle forces become weaker. Hence the boiling point decreases. The order of boiling point is Primary  > Secondary >= iso > Tertiary.
• Hence the order of boiling points is propane (alkane) < isopropyl bromide (isoalkyl halide) < n-propyl bromide (primary alkyl halide).

n-butyl chloride, iso-butyl chloride, tert-butyl chloride:

• Among isomeric alkyl halides, the boiling point decreases with the increase in branching in the alkyl group, because with branching the molecule attains a spherical shape with less surface area. As a result, interparticle forces become weaker. Hence the boiling point decreases. The order of boiling point is Primary  > Secondary >= iso > Tertiary.
• Hence the order of boiling points is tert-Butyl chloride < iso-butyl chloride < n-butyl chloride.

1-Bromopropane, isopropyl bromide, 1- Bromobutane:

• Among isomeric alkyl halides, the boiling point decreases with the increase in branching in the alkyl group, because with branching the molecule attains spherical shape with less surface area. As a result, interparticle forces become weaker. Hence the boiling point decreases. The order of boiling point is Primary  > Secondary >= iso > Tertiary. or the same halogen, the boiling point increases with the increase in the molecular mass.
• Hence the order of boiling points is isopropyl bromide  <  1-Bromopropane <  1- Bromobutane

Explain why 1-Chlorobutane has higher B.P. than 2-Chlorobutane?

Among isomeric alkyl halides, the boiling point decreases with the increase in branching in the alkyl group, because with branching the molecule attains spherical shape with less surface area. As a result, interparticle forces become weaker. Hence the boiling point decreases.

The order of boiling point is Primary  > Secondary >= iso > Tertiary. Hence 1-Chlorobutane (primary alkyl halide) has higher B.P. than 2-Chlorobutane (secondary alkyl halide).

Explain why Bromoethane has a higher boiling point than Chloroethane. OR Out of ethyl bromide and ethyl chloride which has a higher boiling point and why?

For the same alkyl group the boiling points of haloalkanes are in the order RCl < RBr < RI, because with the increase in the size of halogen atom the magnitude of van der Wall forces of attraction increases. Hence Bromoethane has a higher boiling point than Chloroethane.

Solubility:

Explain why choroform is not soluble in water although it is polar. OR alkyl halides though polar, are immiscible with water.

• A substance is soluble in water if its molecules are capable of forming hydrogen bonds with water. Chloroform molecules do not form the hydrogen bond with water.
• The energy required to break the bonds between haloalkane molecules is much larger than the energy released during the formation of the bond between haloalkane molecules and water molecules. Hence chloroform is not soluble in water although it is polar.

Alkyl halides are insoluble in water though they contain polar C-X bond. Explain.

• Alkyl halides are polar in nature but they are not able to form hydrogen bonds with water molecules. Hence they are sparingly soluble in water. But they are soluble in organic solvents like alcohols, ethers and benzene.

Dipole Moment:

Which one of the following has the highest dipole moment?

CH₂Cl₂, CHCl₃, CCl₄.

• The dipole moment of CH₂Cl₂ is the highest while that of CCl₄ is zero. Dipole moment of CH₂Cl₂ is greater than  CHCl₂ because the dipole moment of the third C-Cl bond of CHCl₃ opposes the dipole moment of the remaining two C-Cl bonds.

Science > Chemistry > Organic Chemistry > Halogen Derivatives of Alkanes > Physical Properties of Alkyl Halides

The post Physical Properties of Alkyl Halides appeared first on The Fact Factor.

In the last three articles, we have studied the methods of preparations of alkyl halides from alkanes, alkenes, alcohols. In this article, we shall study the method of the preparation of alkyl halids by halide exchange method. This method is typical for preparations of alkyl iodides/fluorides/bromides.

Preparation of Alkyl iodides by Halide Exchange Method(Finkelstein Reaction)

General Reaction:

When alkyl chlorides or bromides when treated with NaI in presence of dry acetone give alkyl iodides.

R–Cl or R–Br  +  NaI → RI + NaCl  or NaBr

Alkyl chloride / bromide   sodium iodide   →    Alkyl iodide sodium chloride / bromide

Example – 1: Preparation of Ethyl iodide (Iodoethane) from ethyl chloride (Chloroethane):

C₂H₅-Cl    +      NaI   C₂H₅I      +   NaCl

Ethyl chloride    sodium iodide    Ethyl iodide  sodium chloride

Example – 2: Preparation of Ethyl iodide (Iodoethane) from ethyl bromide (Bromoethane):

C₂H₅-Br    +      NaI   C₂H₅I      +   NaBr

Ethyl bromide    sodium iodide   Ethyl iodide       sodium bromide

Preparation of Alkyl fluorides by Halide Exchange Method(Swart’s Reaction):

The direct reaction of alkanes with fluorine is highly explosive in nature, hence it can’t be produced by direct fluorination of alkanes.

General Reaction:

When alkyl halides are treated with salts like AgF, Hg₂F₂, CoF₃, SbF₃ fluoroalkanes can be obtained.

Example – 1: Preparation of Ethyl fluoride (Fluoroethane) from ethyl chloride (Chloroethane):

C₂H₅Cl          +     AgF    →  C₂H₅F           +   AgCl

Ethyl chloride   silver fluoride   Ethyl fluoride       silver chloride

Example – 2: Preparation of Methyl fluoride (Fluoromethane) from methyl bromide (Bromoethane):

2  CH₃Br          +         Hg₂F₂     →       2 CH₃F           +   Hg₂Br₂

Methyl bromide Mercurous fluoride  Methyl fluoride  Mercurous bromide

For replacement of two or three halogens CoF₃, SbF₃ are used

Example – 3

Preparation of Alkyl bromides from silver salt of fatty acid (Borodine Hunsdiecker Reaction)

General Reaction:

When silver salt of fatty acid is refluxed with bromine in CCl4, alkyl bromide is obtained.

RCOOAg   +    Br₂        RBr  +  CO₂ +  AgBr

Silver salt of fatty acid   bromine    alkyl bromide   silver bromide

Example -1: Preparation of Methyl Bromide (Bromomethane) from silver acetate:

CH₃COOAg   +    Br₂        CH₃Br  +  CO₂ +  AgBr

Silver acetate bromine    methyl bromide  silver bromide

Example -2: Preparation of Ethyl Bromide (Brommethane) from silver propionate:

C₂H₅COOAg   +    Br₂        C₂H₅Br  +  CO₂ +  AgBr

Silver propionate   bromine          ethyl bromide    silver bromide

Note:

Chloroalkanes can be obtained by this method but yield is very low. Iodoalkanes can not be obtained by this method because iodine forms easter with  silver salt of fatty acid.

Science > Chemistry > Organic Chemistry > Halogen Derivatives of Alkanes > Preparation of Alkyl Halides by Halide Exchange Method

The post Preparation of Alkyl Halides by Halide Exchange Method appeared first on The Fact Factor.