Chemical Bonding

i. Electronegativity difference between the atoms.

ii. Number of valence electrons present in the valence shell of the atom.

iii. The force of attraction and repulsion between electrons and protons.

iv. Electron affinity

v. Ionization potential

vi. Atomic size

(a) Six ( all the crosses represent the valence electrons)

(b) Two (two more electrons are required to obtain octet )

(c) One (one more electron needed to obtain stable duplet structure)

(d) Two bonds

(e) Element X can be Hydrogen and Y can be oxygen

Electrons in the inner shells are strongly attracted by the nucleus and are stable thereby they don’t involve in bond formation. The electrons in the outermost shells of atoms are called valence electrons which are highly active an are weakly attracted by the nucleus of atom. So they are easily transferred or migrated, so undergo chemical reactions.

Sodium chloride

i. Sodium chloride is formed between sodium and chlorine.

ii. Sodium (₁₁Na²³) configuration is 1s²2s²2p⁶3s¹

iii. Chlorine (₁₇Cl³⁵) configuration is 1s²2s²2p⁶3s²3p⁵

iv. So, sodium tends to lose a electron i.e. 3s¹ to obtain octet in its 2^nd shell.

v. Chlorine tends to gain a electron to become 3p⁶ to obtain octet in its 3^rd shell.

vi. In this way, both sodium and chlorine become stable.

vii. So, sodium loses electron and gains positive charge and is called as sodium ion which is a cation - Na⁺¹

viii. chlorine gains electron and gains negative charge and is called as chloride ion which is a anion – Cl^-1

ix. ionic bond is formed by the transfer of electrons and the compound formed is sodium chloride.

x.

Calcium oxide

i. Calcium oxide if formed by combination of calcium and oxygen.

ii. Calcium (₂₀Ca⁴⁰) configuration is 1s²2s²2p⁶3s²3p⁶4s².

_iii. Oxygen (₈O¹⁶) configuration is 1s²2s²2p⁴_.

iv. Outer shell of calcium (4^th) contains 2 electrons. So it tends to lose these

2 electrons to obtain octet in its 3^rd shell which then becomes its outermost shell. Oxygen tends to gain 2 electrons to obtain octet in its outermost 2^nd shell.

v. In this way, both calcium and oxygen become stable.

vi. So, calcium loses 2 electrons and gains positive charge and is called as calcium ion which is a cation – Ca⁺²

vii. oxygen gains 2 electrons and gains negative charge and is called as oxide ion which is an anion – O^-2

viii. ionic bond is formed by the transfer of electrons and the compound formed is calcium oxide.

ix.

we know that

A is Carbon – ₆C¹²

B is Sodium – ₁₁Na²³

C is Chlorine – ₁₇Cl³⁵

(i) The configuration of carbon is 1s²2s²2p². It has exactly 4 electrons

in its outermost shell. So, it neither loses nor gains electrons to obtain octet. Rather it shares electrons with other atoms and gets stable by obtaining octet structure. So, it forms covalent compounds but not ionic compounds.

(ii) The configuration of sodium is 1s²2s²2p⁶3s¹. So, to obtain octet

and become stable, it can easily lose an electron which is 3s¹. Therefore, sodium forms ionic compounds.

(iii) The Chlorine (₁₇Cl³⁵) configuration is 1s²2s²2p⁶3s²3p⁵. So, it can

easily gain an electron to obtain octet , thereby forming ionic compounds. But it can also share electrons and form covalent compounds like in the case of HCl. So, chlorine forms both ionic and covalent compounds.

Bond length is defined as the distance between the nuclei of 2 atoms which are involved in bonding. Bond energy is defined as the energy required to break the bond between 2 atoms of a diatomic covalent compound in its gaseous state. Bond energy is generally inversely proportional to bond length. For ex. Chlorine has smaller sized atoms, lesser bond length so high bond energy. Iodine has bigger atoms, more bond length, so less bond energy.

The lesser the bond energy, more is the instability and more is the chemical reactivity. The higher the bond energy, more is the stability and less is the chemical reactivity.

Melting point depends on the force of attraction between molecules and atoms of compounds or elements.

In case of covalent compounds there exists weak vander-waal forces

between molecules. So, lower temperatures are sufficient to break these bonds. But ionic compounds contain strong electrostatic forces which need high energy to break, demanding high melting temperatures. Some exceptions in covalent materials exist like diamond, graphite which demand high melting temperature because of their giant complex structure.

Properties –

i. These compounds contain directional covalent bonds which involve sharing of electrons.

ii. They exhibit isomerism.

iii. They are usually liquids or gases in nature, but some are solids.

iv. They are freely soluble in non-polar solvents like benzene and generally insoluble in polar solvents like water.

v. They have lower melting and boiling points.

Uses-

i. Majority constituents in our body are covalent compounds.

ii. Water is a covalent compound and it has many uses.

iii. Most of the organic compounds are covalent compounds.

iv. Sugar, vinegar, cooking gas are all covalent compounds.

v. Important gases like CO₂, NH₃ are all covalent compounds.

(a) Calcium oxide

(b) Water Molecule:

One oxygen and 2 hydrogen atoms combine to form water molecule.

(c) Chlorine

One atom of oxygen and 2 atoms of hydrogen form 1 molecule of water.

(a) ₄Be⁹ – 1s²2s² – lewis notation is

(b) ₂₀Ca⁴⁰ – 1s²2s²2p⁶3s²3p⁶4s² – lewis notation is

(c) ₃Li⁶ – 1s²2s¹ – lewis notation is

(a)

Bromine is a halogen element. It has 7 electrons in its outermost orbit. It needs one more electron to obtain octet and become stable. Hence, in the case of bromine gas, 2 bromine atoms combine to form a molecule (Br₂). Each bromine atom contributes one electron for covalent bond. So, the 2 electrons involved in the bond are shared by both bromine atoms, hence octet is obtained.

(b)

Calcium has 2 electrons in outermost shell. It tends to lose them to get octet structure. Chlorine being halogen element has 7 electrons in the outermost orbit. It needs one more electron to obtain octet and become stable. So, calcium donates its 2 electrons to 2 different chlorine atoms. Hence, calcium cation, chlorine anions are formed and bond forms.

Thus, CaCl₂ is formed.

(c)

Oxygen has 6 electrons in its outermost orbit. Carbon has 4 electrons in its outermost orbit. So, oxygen tends to gain 2 electrons to obtain octet. Carbon shares electrons to obtain octet. So, in this case, one carbon and 2 oxygen atoms share electrons. Each oxygen contributes its 2 electrons and carbon contributes 4 electrons. 2 pairs of electrons between each oxygen and carbon form bonds. These electrons are shared by both oxygen and carbon. Both carbon and oxygen obtain octet. Thus, 2 double bonds are formed and CO₂ is formed.

(d) Carbon dioxide –

● Nitrogen and hydrogen react to form ammonia.

(a) The electronic configuration of nitrogen is 1s²2s²2p³. Outermost orbit

contains 5 electrons. So, it needs 3 more electrons to obtain octet. Valency of nitrogen is 3. Electronic configuration of hydrogen is 1s¹. Valency of hydrogen is 1. It need to gain one more electron to obtain stable duplet structure. so, one nitrogen contributes its 3 electrons to 3 hydrogen atoms and each hydrogen contributes its one electron for bond formation. The electrons are shared by both nitrogen and hydrogen. Both become stable and form ammonia.

(b)

● Carbon and hydrogen react to form methane.

C + 2H₂→ CH₄

(a) Carbon has 4 electrons in its outermost orbit. Carbon shares

electrons to obtain octet. It needs 4 more electrons to obtain octet. So, valency of carbon is 4. Electronic configuration of hydrogen is 1s¹. Valency of hydrogen is 1. It need to gain one more electron to obtain stable duplet structure. so, carbon contributes its 4 electrons in bond formation with 4 hydrogen atoms. Hydrogen atoms contribute their one electron. So, these electrons are shared by both carbon and hydrogen to form bonds.

(b)

In lewis dot structure of atoms of an element, we represent the nucleus and inner shell electrons by the symbol of element and electrons of outer shell by dots.

Based on this we can understand the valence of an element based on which we can predict the type of bond formed between atom. For ex- calcium has electronic configuration -

₂₀Ca⁴⁰ – 1s²2s²2p⁶3s²3p⁶4s² – lewis notation is

The outermost shell ‘4’ has 2 electrons and they are represented by 2 dots. The inner electrons and nucleus are represented by the symbol Ca.

Octet rule – it states that atoms of elements are stable with 8 electrons in their outermost shell and they tend to undergo chemical changes ( losing or gaining electrons ) to get this structure.

Elements after these chemical changes attain noble gas configuration. Chemically active elements don’t have 8 electrons in their outermost shell and that’s why they lose or gain electrons to obtain stable octet structure.

Atoms having more than 4 electrons and less than 8 electrons, already in their outermost shell, tend to gain electrons thereby becoming negatively charged (anions). Atoms having less than 4 electrons and more than 0 electrons, already in their outermost shell, tend to lose electrons thereby becoming positively charged (cations). The ultimate result of this is obtaining octet in outer most shell.

(a) Formation of N₂ molecule

Nitrogen (₇N¹⁴) has configuration 1s²2s²2p³. Its valence shell contains 5 electrons, i.e. 2 in s and 3 in p orbitals - 2p_x¹2p_y¹2p_z¹

if the p_x orbital of a N atom overlaps with other p_x orbital of another N atom giving (σp_x – p_x) bond, along the internuclear axis. Similarly p_y and p_z orbitals of one N atom overlap with p_y and p_z orbitals of another N atom laterally and perpendicularly to internuclear axis given by(ϖp_y – p_y) (ϖp_z – p_z) bonds. So, N₂ molecule has a triple bond between 2 nitrogen atoms.

(b) Formation of O₂ molecule.

oxygen (₈O¹⁶) has configuration 1s²2s²2p⁴. Its valence shell contains 6 electrons, i.e. 2 in s and 4 in p orbitals - 2p_x²2p_y¹2p_z¹

the p_y orbital of one oxygen atom overlaps with p_y orbital of another oxygen atom along the inter nucleus axis, a sigma p_y – p_y bond is formed.

Similarly, p_z orbital of O atom overlaps with p_z orbital of another O atom laterally, perpendicular to the inter nuclear axis giving pi p_z-p_z bond.

O₂ molecule has a double bond between 2 oxygen atom.

Hybridisation is the concept of mixing atomic orbitals into new hybrid orbitals (with different energies, shapes, etc., than the component atomic orbitals) suitable for the pairing of electrons to form chemical bonds in valence bond theory.

(a) BeCl₂

The electronic configuration of 'Be' in ground state is 1s² 2s². Since there are no unpaired electrons, it undergoes excitation by promoting one of its 2s electron into empty 2p orbital.

Thus in the excited state, the electronic configuration of Be is 1s² 2s¹ 2p¹.

In the excited state, the beryllium atom undergoes 'sp' hybridization by mixing a 2s and one 2p orbitals. Thus, two half filled 'sp' hybrid orbitals are formed, which are arranged linearly.

These half filled sp-orbitals form two σ bonds with two 'Cl' atoms.

Thus BeCl₂ is linear in shape with the bond angle of 180^o.

(b) BF₃

The electronic configuration of 'B' in ground state is 1s² 2s² 2p¹.

In excited state one electron from s orbital gets excited to p orbital and it becomes

1s² 2s¹ 2p_x¹ 2p_y¹. This 1 half filled s orbital and 2 half filled p orbitals intermix to give 3 sp² hybridised orbitals which overlap with empty p orbitals of 3 fluorine atoms to give BF₃. The shape is trigonal planar.