Chemistry Class 11 NCERT Solutions: Chapter 4 Chemical Bonding and Molecular Structure Part 8

Q: 34. Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.

Answer:

The given conditions should be satisfied by atomic orbitals to form molecular orbitals:

(A) The combining atomic orbitals must have the same or nearly the same energy. This means that in a homonuclear molecule, the 1s-atomic orbital of an atom can combine with the 1s-atomic orbital of another atom, and not with the 2s-orbital.

(B) The combining atomic orbitals must have proper orientations to ensure that the overlap is maximum.

(C) The extent of overlapping should be large.

Q: 35. Use molecular orbital theory to explain why the Equation molecule does not exist.

Answer: The electronic configuration of Beryllium is Equation

The molecular orbital electronic configuration for Equation molecule can be written as:

Equation

Hence, the bond order for Equation is Equation

Where,

Equation

Equation Number of electrons in anti-bonding orbitals

Equation

A negative or zero bond order means that the molecule is unstable. Hence, Equation molecule does not exist.

Q: 36. Compare the relative stability of the following species and indicate their magnetic properties;

Equation

Answer: There are 16 electrons in a molecule of Dioxygen, 8 from each oxygen atom. The electronic configuration of oxygen molecule can be written as:

Equation Since the 1s orbital of each oxygen atom is not involved in boding, the number of bonding electrons Equation and the number of anti-bonding orbitals Equation

Bond order Equation

Equation

Equation

Similarly, the electronic configuration of Equation can be written as:

Equation

Equation

Equation

Bond order of Equation

=2.5

Electronic configuration of Equation

Equation

Electronic configuration of Equation

Equation

Equation

Equation

Bond order of Equation

=1

Bond dissociation energy is directly proportional to bond order. Thus, the higher the bond order, the greater will be the stability. On this basis, the order of stability is Equation

Q: 37. write the significance of a plus and a minus sign shown in representing the orbitals.

Answer: Molecular orbitals are represented by wave functions. A plus sign in an orbital indicates a positive wave function while a minus sign in an orbital represents a negative wave Function.

Q: 38. Describe the hybridisation in case of Equation Why are the axial bonds longer as compared to equatorial bonds?

Answer: The ground state and excited state outer electronic configurations of phosphorus Equation are:

Ground state:

Image of Ground state for Chemical Bonding and Molecular Structure

Image of Ground State

Image of Ground state for Chemical Bonding and Molecular Structure

Excited state:

Image of Excited state Chemical Bonding and Molecular Structure

Image of Excited State

Image of Excited state Chemical Bonding and Molecular Structure

Phosphorus atom is Equation hybridized in the excited state. These orbitals are filled by the electron pairs donated by five Cl atoms as:

Image of Polycaprolactone for Chemical Bonding and Molecular Structure

Image of PCl 5

Image of Polycaprolactone for Chemical Bonding and Molecular Structure

The five Equation d hybrid orbitals are directed towards the five corners of the trigonal bipyramidals. Hence, the geometry of PCl_5 can be represented as:

Image:

Quadrilateral poly1 Quadrilateral poly1: Polygon A, B, C, D Triangle poly2 Triangle poly2: Polygon D, B, E Segment a Segment a: Segment [A, B] of Quadrilateral poly1 Segment b Segment b: Segment [B, C] of Quadrilateral poly1 Segment c Segment c: Segment [C, D] of Quadrilateral poly1 Segment d Segment d: Segment [D, A] of Quadrilateral poly1 Segment e Segment e: Segment [D, B] of Triangle poly2 Segment d_1 Segment d_1: Segment [B, E] of Triangle poly2 Segment b_1 Segment b_1: Segment [E, D] of Triangle poly2 Segment g Segment g: Segment [D, F] Segment f Segment f: Segment [G, B] Segment h Segment h: Segment [H, E] Segment i Segment i: Segment [A, I] Segment j Segment j: Segment [J, C] Segment k Segment k: Segment [K, L] Segment l Segment l: Segment [E, C] Cl text4 = "Cl" Cl text4_1 = "Cl" Cl text4_2 = "Cl" Cl text4_3 = "Cl" Cl text4_4 = "Cl" P text5 = "P"

Image of Trigonal Bipyramidal

Image of trigonal bipyramidal for Chemical Bonding and Molecular Structure

There are five Equation sigma bonds in Equation . Three Equation bonds lie in one plane and make an angle of Equation with each other. These bonds are called equatorial bonds.

The remaining two Equation bonds lie above and below the equatorial plane and make an angle of Equation with the plane. These bonds are called axial bonds.

As the axial bond pairs suffer more repulsion from the equatorial bond pairs, axial bonds are slightly than equatorial bonds.

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