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Chapter 4- Chemical Bonding and Molecular Structure Interview Questions Answers

Question 1 : Explain the formation of a chemical bond.

Answer 1 :

A chemical bond is defined as an attractive force that holds the constituents (atoms, ions etc.) together in a chemical species.
Various theories have been suggested for the formation of chemical bonds such as the electronic theory, valence shell electron pair repulsion theory, valence bond theory, and molecular orbital theory.
A chemical bond formation is attributed to the tendency of a system to attain stability. It was observed that the inertness of noble gases was because of their fully filled outermost orbitals. Hence, it was postulated that the elements having incomplete outermost shells are unstable (reactive). Atoms, therefore, combine with each other and complete their respective octets or duplets to attain the stable configuration of the nearest noble gases. This combination can occur either by sharing of electrons or by transferring one or more electrons from one atom to another. The chemical bond formed as a result of sharing of electrons between atoms is called a covalent bond. An ionic bond is formed as a result of the transference of electrons from one atom to another.

Question 2 : Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br.

Answer 2 :

Mg: There are two valence electrons in Mg atom. Hence, the Lewis dot symbol for Mg is:  
Na: There is only one valence electron in an atom of sodium. Hence, the Lewis dot structure is:  
B: There are 3 valence electrons in Boron atom. Hence, the Lewis dot structure is:  
O: There are six valence electrons in an atom of oxygen. Hence, the Lewis dot structure is:  
N: There are five valence electrons in an atom of nitrogen. Hence, the Lewis dot structure is:  
Br: There are seven valence electrons in bromine. Hence, the Lewis dot structure is:  

Question 3 :

Write Lewis symbols forthe following atoms and ions:

S and S2–; Al and Al3+;H and H

Answer 3 :

(i) S and S2–

The number of valenceelectrons in sulphur is 6.

TheLewis dot symbol of sulphur (S) is 
The dinegative charge infers that there will betwo electrons more in addition to the six valence electrons. Hence, the Lewisdot symbol of S2– is 

(ii) Al and Al3+

The number of valenceelectrons in aluminium is 3.

TheLewis dot symbol of aluminium (Al) is
The tripositive charge on a species infers thatit has donated its three electrons. Hence, the Lewis dot symbol is .

(iii) H and H

The number of valenceelectrons in hydrogen is 1.

TheLewis dot symbol of hydrogen (H) is .
The uninegative charge infers that there will beone electron more in addition to the one valence electron. Hence, the Lewis dotsymbol is .

Question 4 :

Draw the Lewis structuresfor the following molecules and ions:

H2S,SiCl4, BeF2,, HCOOH

Answer 4 :

Question 5 :

Defineoctet rule. Write its significance and limitations.

Answer 5 :

The octet rule or theelectronic theory of chemical bonding was developed by Kossel and Lewis.According to this rule, atoms can combine either by transfer of valenceelectrons from one atom to another or by sharing their valence electrons inorder to attain the nearest noble gas configuration by having an octet in theirvalence shell.

The octet rulesuccessfully explained the formation of chemical bonds depending upon thenature of the element.

Limitations of the octet theory:

The following are thelimitations of the octet rule:

(a) The rule failed topredict the shape and relative stability of molecules.

(b) It is based upon the inert nature ofnoble gases. However, some noble gases like xenon and krypton form compoundssuch as XeF2, KrF2 etc.

(c) The octet rule cannot be applied to theelements in and beyond the third period of the periodic table. The elementspresent in these periods have more than eight valence electrons around thecentral atom. For example: PF5, SF6, etc.

(d) The octet rule is not satisfied for allatoms in a molecule having an odd number of electrons. For example, NO and NOdonot satisfy the octet rule.

(e) This rule cannot be applied to thosecompounds in which the number of electrons surrounding the central atom is lessthan eight. For example, LiCl, BeH2, AlCletc. donot obey the octet rule.

Question 6 :

Writethe favourable factors for the formation of ionic bond.

Answer 6 :

An ionic bond is formed bythe transfer of one or more electrons from one atom to another. Hence, theformation of ionic bonds depends upon the ease with which neutral atoms canlose or gain electrons. Bond formation also depends upon the lattice energy ofthe compound formed.

Hence, favourable factorsfor ionic bond formation are as follows:

(i) Low ionizationenthalpy of metal atom.

(ii) High electron gain enthalpy (Δeg H)of a non-metal atom.

(iii)High lattice energy of the compound formed.

Question 7 :

Discuss the shape of thefollowing molecules using the VSEPR model:

BeCl2, BCl3, SiCl4,AsF5, H2S, PH3

Answer 7 :

BeCl2:

The central atom has no lone pair and thereare two bond pairs. i.e., BeCl2 is of the type AB2.Hence, it has a linear shape.

BCl3:

The central atom has no lone pair and thereare three bond pairs. Hence, it is of the type AB3. Hence, it istrigonal planar.

SiCl4:

The central atom has no lone pair and thereare four bond pairs. Hence, the shape of SiCl4 is tetrahedralbeing the AB4 type molecule.

AsF5:

The central atom has no lone pair and thereare five bond pairs. Hence, AsF5 is of the type AB5.Therefore, the shape is trigonal bipyramidal.

H2S:

The central atom has one lone pair and thereare two bond pairs. Hence, H2S is of the type AB2E. Theshape is Bent.

PH3:

Thecentral atom has one lone pair and there are three bond pairs. Hence, PH3 isof the AB3E type. Therefore, the shape is trigonal pyramidal.

Question 8 :

Although geometries of NH3 andH2O molecules are distorted tetrahedral, bond angle in water is lessthan that of ammonia. Discuss.

Answer 8 :

The molecular geometry of NH3 andH2O can be shown as:

The central atom (N) in NHhasone lone pair and there are three bond pairs. In H2O, there are twolone pairs and two bond pairs.

The two lone pairs present in the oxygenatom of H2O molecule repels the two bond pairs. This repulsion isstronger than the repulsion between the lone pair and the three bond pairs onthe nitrogen atom.

Sincethe repulsions on the bond pairs in H2O molecule are greater thanthat in NH3, the bond angle in water is less than that of ammonia.

Question 9 :

Howdo you express the bond strength in terms of bond order?

Answer 9 :

Bondstrength represents the extent of bonding between two atoms forming a molecule.The larger the bond energy, the stronger is the bond and the greater is thebond order.

Question 10 :

Definethe bond length.

Answer 10 :

Bond length is defined asthe equilibrium distance between the nuclei of two bonded atoms in a molecule.

Bond lengths are expressed in terms ofAngstrom (10–10 m) or picometer

(10–12 m) and are measuredby spectroscopic X-ray diffractions and electron-diffraction techniques.

In an ionic compound, the bond length is thesum of the ionic radii of the constituting atoms (d = r+ + r).In a covalent compound, it is the sum of their covalent radii (d = rA + rB).


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Chapter 4- Chemical Bonding and Molecular Structure Contributors

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