Chemical Bonding Terminal Questions with Answer NIOS
National Institute of Open Schooling (NIOS)
Sr. Secondary, Module-1
Lession-4: Chemical Bonding Terminal Questions with Answer
Terminal QUESTIONS with Answer
1. What do you understand by a chemical bond?
Answer:A chemical bond is a force of attraction between atoms, ions and molecules which leads to the formation of compounds. The bond is formed either by the electrostatic attraction between the oppositely charged ions in ionic bonds or sharing of electrons in covalent bonds and stabilizes them through the overall loss of energy. Stronger is the force of attraction (bonding) between constituent atoms more stable will be the resulting compound. There are various types of chemical bonds such as an ionic bond, covalent bond, hydrogen bond, metallic bond etc. The bonding in chemical compounds is explained by the valence shell electron pair repulsion theory and octet theory.
2. Explain that the process of bond formation lead to decrease in energy.
Answer:A chemical bond is a force of attraction between atoms, ions and molecules which leads to the formation of compounds. When a bond is formed, there is an attraction between the space of the nuclei where the force of attraction is greater. As electrons fall to a place of lower potential energy and the total mechanical energy of the system decreases. The energy lost by atoms is released as heat or light. Therefore during the bond formation, there is a decrease in energy.
3. What do you understand by the term bond length?
Answer:Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule. The bigger the size of the atom, the longer the bond length. The bond length decreases in bond multiplicity. It is measured by spectroscopic, X-ray diffraction, rotational spectroscopy, etc. X-ray method is very suitable to crystalline substances and electron diffraction is applicable only to gases.
4. Describe the two possible ways in which the nobel gas electronic configuration is achieved in the process of bond formation.
AnswerThe two possible ways in which the nobel gas electronic configuration is achieved in the process of bond formation are dscribed below-
1. Lewis's Theory of Chemical Bonding
2. Kossel's Theory of Chemical Bonding
1. Lewis's Theory of Chemical Bonding:
According to the Lewis theory, atoms gain, lose, or share electrons in order to achieve a stable electron configuration similar to that of noble gases. Noble gases have complete outer electron shells, making them highly stable and unreactive. Lewis recognized that other atoms tend to react in ways that allow them to acquire a similar electron configuration.
2. Kossel's Theory of Chemical Bonding:
Kossel's theory focuses on achieving a stable electron configuration similar to noble gases. By transferring or sharing electrons, atoms strive to attain a complete outer electron shell with eight valence electrons, except for hydrogen and helium, which can achieve stability with two valence electrons.
Kossel proposed that atoms form bonds by the complete transfer of electrons from one atom to another. This transfer occurs between an atom with low ionization energy (tends to lose electrons) and an atom with a high electron affinity (tends to gain electrons). The result is the formation of positively charged cations and negatively charged anions that are held together by electrostatic forces.
5. What are Lewis electron dot symbol? Show the formation of MgCl2, in terms of Lewis symbol.
Answer:The representation of valence electrons of an atom using dots around the symbol of the element is known as the Lewis electron-dot digram. The arrangement of dots is left, right and above in the symbol of the atom. The dotted electron involved in the bond formation is called bonding electron and the rest dotted electron is called lone pair.
Electron configuration of magnesium is 1s22s22p63s2. So after losing two electrons it forms Mg2+ ion a stable configuration like inert gas Ne.
Similarly, chlorine is having the electronic configuration 1s22s22p63s22p5. So it needs one electron to forms a stable inert gas (like Ar) configuration. Thus, the Lewis dot structure for MgCl2 will be as follows-
6. Define a coordinate bond and give some examples.
Answer:A Co-ordinate bond is a type of covalent bond that is formed by sharing of an electron pair from a single atom. It is also called a dative bond or dipolar bond. the atom that shares an electron pair from itself is termed as the donor and the other atom which accepts these shared pair of electrons is known as a receptor or acceptor. The bond is represented with an arrow →, pointing towards the acceptor from the donor atom. After sharing of electron pairs, each atom gets stability.
Exanmples: NH4Cl, H2SO4, H3O+ etc. shows this type of bond.
7. What is VSEPR Theory? Predict the shape of SF6 using this theory.
Answer:Valence Shell Electron Pair Repulsion Theory (VSEPR Theroy):
This theory was proposed by Sidwick and Powell in 1940 and later on modified by Gillespie and Nyholm in 1957.
This theory predict the shape of simple molecules and ions on the basis of repulsion of electron pairs present in the valence shell of their central atom.
Some Important Postulates of this Theory are given below
1. Electrons involved in the bond formation is called bonding electrons or bond pair (B.P.) and the rest electrons are called lone pairs (L.P.).
2. Electron pairs in valence shell repel one-another as electron clouds are negatively charged.
3. These electron pairs occupy the space at maximum distance for minimum repulsion.
4. The most stable geometrical arrangement of 2,3,4,5,6 electron pairs is linear, triangular, tetrahedral, triangular bipyramidal and octahedral respectively.
5. The central atom in a molecule is surrounded by only B.P. the molecule has regular or symmetrical geometry but in case of B.P. and L.P. the molecule does not have regular geometry.
6. A lone pair occupies more space than a bond pair because lone pair attached with only one atom. Hence the order of repulsion is-
L.P - L.P. > L.P - B.P. > B.P - B.P.
Greater the repulsion, smaller the bond angle.
7. Multiple bonds does not affect the gross geometry of the molecules rather the geometry is exclusively decided by B.P. and L.P.
8. A lone pair and double bond repulsion is much greater than a lone electron and double bond repulsion.
9. A lone pair and a single bond repulsion is larger than a lone pair and double bond repulsion.
According to this theory, the shape of SF6 is normal Octahedral as it has six bond pair and no lone pair on sulphur atom.
8. Why do we need the concept of hybridization? How does it help in explaining the shape of methane?
Answer:9. Give the salient features of molecular orbital theory.
Answer:This theory was proposed by F.Hund and R.S.Mulliken in 1932 and the basic features of this theory are given below-
1. Electrons of the molecule are present in various molecular orbital just like the electrons of an atom are present in atomic orbitals.
2. Molecular orbitals are formed by mixing of atomic orbitals of comparable energies and proper symmetry.
3. An electron in an atomic orbital is influenced by only one nucleus and thus it is monocentric while in a molecular orbital it is influenced by two or more nuclei of the molecule and thus it is polycentric.
4. The total number of molecular orbitals are equal to the number of combining atomic orbitals.
5. Bonding molecular orbital has lower energy and greater satability than the corresponding antibonding molecular orbitals.
6. Molecular orbitals are also filled by electron in accordance with Aufbau principle obeying Pauli's exclusion principle and Hund's rule of maximum multiplicity.
7. Each electron movin in a molecular orbital has a spin od +1/2 and −1/2.
10. Be2 molecule does not exist. Explain on the basis of molecular orbital theory.
Answer:Molecular orbital configuration: σ1s2σ*1s2σ2s2σ*2s2
Based on the molecular orbital configuration, Be2 molecule has zero bond order and there is no overlapping of orbitals due to the absence of filled atomic orbital. That is why Be2 molecule does not exist.
11. Write down the molecular orbital electronic configuration of the following species and compute their bond orders. O2, O2+, O2−, O22−
Answer:Molecular orbital configuration of O2: σ1s2 σ*1s2 σ2s2 σ*2s2 σ2pz2 𝝅2px2 𝝅2py2 𝝅*2px1 𝝅*2py1
Bond Order = 1/2[8 − 4] = 2
Molecular orbital configuration of O2+: σ1s2 σ*1s2 σ2s2 σ*2s2 σ2pz2 𝝅2px2 𝝅2py2 𝝅*2px1
Bond Order = 1/2[7 − 4] = 1.5
Molecular orbital configuration of O2−: σ1s2 σ*1s2 σ2s2 σ*2s2 σ2pz2 𝝅2px2 𝝅2py2 𝝅*2px2𝝅*2py1
Bond Order = 1/2[9 − 4] = 2.5
Molecular orbital configuration of O2−2: σ1s2 σ*1s2 σ2s2 σ*2s2 σ2pz2 𝝅2px2 𝝅2py2 𝝅*2px2𝝅*2py2
Bond Order = 1/2[10 − 4] = 3
12. BF3 is a polar molecule but it does not show dipole moment. Why?
Answer:The structure of BF3 is triangular planar. The three electron pairs located atan angle of 120° and gives the symmetrical planar structure. Due to this symmetrical structure, BF3 does not show dipole moment although B-F bonds are polar.
13. Atom A and B combine to form AB molecule. If the difference in the electronegativity between A and B is 1.7. What type of bond do you expect in AB molecule?
Answer:Atom A and B combine to form AB molecule. If the difference in the electronegativity between A and B is 1.7 then the bond is 50% ionic and 50% covalent.
