Hybridisation of atomic orbitals:
Hybridisation is the process in which atomic orbitals of nearly equal energy within the same atom mix together to form an equal number of new, equivalent orbitals called hybrid orbitals. These hybrid orbitals have definite shapes and orientations in space, which help explain the geometry of molecules.
Shapes of hybrid orbitals:
1. sp hybrid orbitals
• Formed by the mixing of one s and one p orbital.
• Two equivalent sp hybrid orbitals are formed.
• The orbitals are oriented in a linear geometry at an angle of 180°.
• Each sp hybrid orbital has one large lobe and one small lobe.
Shape: Linear
2. sp2 hybrid orbitals
• Formed by the mixing of one s and two p orbitals.
• Three equivalent sp2 hybrid orbitals are formed.
• The orbitals lie in the same plane and are oriented at an angle of 120°.
• One unhybridised p orbital remains perpendicular to the plane.
Shape: Trigonal planar
3. sp3 hybrid orbitals
• Formed by the mixing of one s and three p orbitals.
• Four equivalent sp3 hybrid orbitals are formed.
• The orbitals are directed towards the corners of a tetrahedron.
• The angle between the orbitals is 109.5°.
Shape: Tetrahedral
Conclusion:
Hybridisation explains the shapes and bond angles of molecules by forming hybrid orbitals with specific geometrical orientations such as linear (sp), trigonal planar (sp2), and tetrahedral (sp3).
Match List - I with List - II.
| List - I (Complex) | List - II (Hybridisation) |
|---|---|
| (A) \([\text{CoF}_6]^{3-}\) | (I) \( d^2 sp^3 \) |
| (B) \([\text{NiCl}_4]^{2-}\) | (II) \( sp^3 \) |
| (C) \([\text{Co(NH}_3)_6]^{3+}\) | (III) \( sp^3 d^2 \) |
| (D) \([\text{Ni(CN}_4]^{2-}\) | (IV) \( dsp^2 \) |
Choose the correct answer from the options given below:
Arrange the following in increasing order of solubility product:
\[ {Ca(OH)}_2, {AgBr}, {PbS}, {HgS} \]
Concentrated nitric acid is labelled as 75% by mass. The volume in mL of the solution which contains 30 g of nitric acid is:
Given: Density of nitric acid solution is 1.25 g/mL.
Match List - I with List - II.
List - I (Saccharides) List - II (Glycosidic linkages found)
(A) Sucrose (I) \( \alpha 1 - 4 \)
(B) Maltose (II) \( \alpha 1 - 4 \) and \( \alpha 1 - 6 \)
(C) Lactose (III) \( \alpha 1 - \beta 2 \)
(D) Amylopectin (IV) \( \beta 1 - 4 \)
Choose the correct answer from the options given below: