Ncert Solutions Chemistry Class 11th
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4.46. A co-ordinate bond is formed by:
(a) Sharing of electrons contributed by both the atoms
(b) Complete transfer of electrons
(c) Sharing of electrons contributed by one atom only
(d) None of these
4.46. A co-ordinate bond is formed by:
(a) Sharing of electrons contributed by both the atoms
(b) Complete transfer of electrons
(c) Sharing of electrons contributed by one atom only
(d) None of these
4.46. (c) sharing of electrons contributed by one atom only
Upvote 
<p><strong>4.46. A co-ordinate bond is formed by:</strong></p>
<p><strong>(a)</strong> Sharing of electrons contributed by both the atoms</p>
<p><strong>(b)</strong> Complete transfer of electrons</p>
<p><strong>(c)</strong> Sharing of electrons contributed by one atom only</p>
<p><strong>(d)</strong> None of these</p>
4.45. Assertion: In PCl5, the axial bonds are slightly longer and hence slightly weaker than the equatorial bonds.
Reason: The axial bond pairs suffer more repulsive interaction from the equatorial bond pairs.
4.45. Assertion: In PCl5, the axial bonds are slightly longer and hence slightly weaker than the equatorial bonds.
Reason: The axial bond pairs suffer more repulsive interaction from the equatorial bond pairs.
4.45. (a) It should be noted that all the bond angles in trigonal bipyramidal geometry are not equivalent. In PCl5 the five sp3d orbitals of phosphorus overlap with the singly occupied
p orbitals of chlorine atoms to form five P–Cl sigma bonds. Three P–Cl bond lie in one plane and make an angle
Upvote 4.44. Assertion: A bonding molecular orbital always possesses lower energy than either of the atomic orbitals that have combined to form it.
Reason: Electrons placed in a bonding molecular orbital tend to destabilise a molecule.
4.44. Assertion: A bonding molecular orbital always possesses lower energy than either of the atomic orbitals that have combined to form it.
Reason: Electrons placed in a bonding molecular orbital tend to destabilise a molecule.
4.44. (c) R is an incorrect statement. Electrons placed in a bonding molecular orbital tend to hold the nuclei together and stabilise a molecule. Therefore, a bonding molecular orbital alwayspossesses lower energy than either of the atomic orbitals that have combined to form it.In contrast, th
Upvote 4.43. Assertion: Sigma bond is weaker as compared to the pi bond.
Reason: The overlapping of orbitals takes place to a larger extent in sigma bond while the extent of overlapping occurs to a smaller extent in case of pi bond.
4.43. Assertion: Sigma bond is weaker as compared to the pi bond.
Reason: The overlapping of orbitals takes place to a larger extent in sigma bond while the extent of overlapping occurs to a smaller extent in case of pi bond.
4.43. (d) Basically the strength of a bond depends uponthe extent of overlapping. In case of sigma bond, the overlapping of orbitals takes place to alarger extent. Hence, it is stronger as comparedto the pi bond where the extent of overlappingoccurs to a smaller extent. Further, it isimportant
Upvote 4.42. Assertion: In BF3, the dipole moment is zero
Reason: The three bond moments give anet sum of zero as the resultant of any two is equal and opposite to the third.
4.42. Assertion: In BF3, the dipole moment is zero
Reason: The three bond moments give anet sum of zero as the resultant of any two is equal and opposite to the third.
4.42. (a) In tetra-atomic molecule like BF3, the dipole moment is zero although the B – F bonds are oriented at an angle of 120o to one another, the three bond moments give a net sum of zero as the resultant of any two is equal and opposite to the third
Upvote 4.41. Assertion: The dipole moment in case of BeF2 is zero.
Reason: There are two F atoms in a molecule of BeF2.
4.41. Assertion: The dipole moment in case of BeF2 is zero.
Reason: There are two F atoms in a molecule of BeF2.
4.41. (b) Reason do not explain the assertion, although both are correct statements independently. The dipole moment in case of BeF2 is zero because the two equal bond dipolespoint in opposite directions and cancel theeffect of each other.
Upvote 4.40. What is meant by the term bond order? Calculate the bond order of N2, O2, O2+ and O2.
4.40. What is meant by the term bond order? Calculate the bond order of N2, O2, O2+ and O2.
4.40. Bond order is defined as one half the difference between the number of electrons present in the bonding and the antibonding orbitals i.e.
Bond order (B.O.) = ½ (Nb–Na)
Upvote