Class 11th

Get insights from 8k questions on Class 11th, answered by students, alumni, and experts. You may also ask and answer any question you like about Class 11th

Follow Ask Question
8k

Questions

0

Discussions

37

Active Users

0

Followers

New answer posted

6 months ago

0 Follower 114 Views

V
Vishal Baghel

Contributor-Level 10

Single bond contains only one sigma bond and double bond contains one sigma and one pi bond.

  1. In this structure, nine single bonds and three double bonds are present. So, there are 12σ and 3π bonds present.
  2. In this structure, eighteen single bonds are present. So, there are only 18σ bonds present.
  3. In this structure, four single bonds are present. So, there are only 4σ bonds present.
  4. In this structure, four single bonds and two double bonds are present. So, there are 6σ and 2π bonds present.
  5. In this structure, five single bonds and one double bond are present. So, there are 6σ and 1π bonds present.
  6. In this structure, seven single bonds

...more

New answer posted

6 months ago

0 Follower 117 Views

P
Payal Gupta

Contributor-Level 10

4.26. In BF3, B atom is sp2 hybridised. In NH3, N is sphybridised.
After the reaction, hybridisation of B changes from sp2 to sp3, while the hybridisation of N remains the same.

New question posted

6 months ago

0 Follower 20 Views

New answer posted

6 months ago

0 Follower 43 Views

P
Payal Gupta

Contributor-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 AlCl4, the empty 3porbital is also involved. So, the hybridisation is sp3 and the shape is tetrahedral

New answer posted

6 months ago

0 Follower 20 Views

P
Payal Gupta

Contributor-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°.

sphybridisation: 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

...more

New answer posted

6 months ago

0 Follower 174 Views

P
Payal Gupta

Contributor-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 ago

0 Follower 11 Views

P
Payal Gupta

Contributor-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 ago

0 Follower 9 Views

P
Payal Gupta

Contributor-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 ago

0 Follower 23 Views

New answer posted

6 months ago

0 Follower 132 Views

P
Payal Gupta

Contributor-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.

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 679k Reviews
  • 1800k Answers

Share Your College Life Experience

×
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.