Class 11th
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New answer posted
6 months agoContributor-Level 10
Single bond contains only one sigma bond and double bond contains one sigma and one pi bond.

- In this structure, nine single bonds and three double bonds are present. So, there are 12σ and 3π bonds present.
- In this structure, eighteen single bonds are present. So, there are only 18σ bonds present.
- In this structure, four single bonds are present. So, there are only 4σ bonds present.
- In this structure, four single bonds and two double bonds are present. So, there are 6σ and 2π bonds present.
- In this structure, five single bonds and one double bond are present. So, there are 6σ and 1π bonds present.
- In this structure, seven single bonds
New answer posted
6 months agoContributor-Level 10
4.26. In BF3, B atom is sp2 hybridised. In NH3, N is sp3 hybridised.
After the reaction, hybridisation of B changes from sp2 to sp3, while the hybridisation of N remains the same.
New question posted
6 months agoNew answer posted
6 months agoContributor-Level 10
4.25. Electronic configuration of 13Al in ground state= 1s2 2s2 2p6 3s2 3p1
Electronic configuration of 13Al in excited state = 1s2 2s2 2p6 3s1 3px13py1
Hence, hybridisation will be sp2 and this makes the geometry to be Trigonal planar (in case of AlCl3).
In AlCl–4, the empty 3pz orbital is also involved. So, the hybridisation is sp3 and the shape is tetrahedral
New answer posted
6 months agoContributor-Level 10
4.24. Hybridisation: It is defined as the process of intermixing of atomic orbitals of slightly different energies to give rise to new hybridized orbitals having equivalent energy and identical shapes.
Shapes of Orbitals:
sp hybridisation: When one s-and one p-orbital intermix then it is called sp-hybridisation. For example, in BeF2, Be atom undergoes sp-hybridisation. It has linear shape. Bond angle is 180°.
sp2 hybridisation: One s-and two p-orbitals get hybridised to form three equivalent hybrid orbitals. The three hybrid orbitals directed towards three corners of an equilateral triangle. It is, therefore, kno
New answer posted
6 months agoContributor-Level 10
4.23. The dipole moment of NH3 (1.47D) is higher than the dipole moment of NF3 (0.24D), even though the molecular geometry is pyramidal for both the molecules. In each molecule, N atom has one lone pair. F is more electronegative than H and N? F bond is more polar than N? H bond. Hence, NF3? is expected to have much larger dipole moment than NH3. However, reverse is true. In case of ammonia, the direction of the lone pair dipole moment and the bond pair dipole moment is same whereas in case of NF3, it is opposite. Thus, in ammonia molecule, individual dipole moment vectors add whereas in NF3, they cancel each other.

New answer posted
6 months agoContributor-Level 10
4.22. BeH2is a linear molecular (H—Be—H), the bond angle = 180°. Be—H bonds are polar due to difference in their electronegativity but the bond polarities cancel each other. Thus, molecule has resultant dipole moment of zero
New answer posted
6 months agoContributor-Level 10
4.21. According to VSEPR theory, if CH4 were square planar, the bond angle would be 90°. For tetrahedral structure, the bond angle is 109°28? Therefore, in square planar structure, repulsion between bond pairs would be more and thus the stability will be less.
Moreover, the orbitals of carbon in ground and excited states look as shown below. In the excited state, the electrons from 4 H atoms occupy one 2s orbital and three 2p orbitals. This results in an sp3 hybridisation, which is the hybridisation of a tetrahedral geometry

New question posted
6 months agoNew answer posted
6 months agoContributor-Level 10
4.19. N2 < SO2 < ClF3 < K2O
The ionic character in a molecule is dependent upon the electronegativity difference between the constituting atoms. The greater the difference, the greater will be the ionic character of the molecule.
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