34 public questions tagged with this topic.
For O₂^- : 17 electrons, configuration ends with (π^* 2p_x)² (π^* 2p_y)¹ . Bonding electrons = 10, antibonding = 7. Bond order = 1/2 (10 - 7) = 1.5 .
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.
Beryllium (Be), due to its small size and high charge density, forms compounds with significant covalent character (e.g., BeCl_2). This follows from NCERT principle where the relation explains the outcome clearly for students in simple steps.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.
In CH₃NH₃+, the conjugate acid of methylamine, nitrogen has four bonds (three to H, one to C) and no lone pair, resulting in a tetrahedral geometry.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.
CH₂=CH-C≡CH has 3 π bonds: 1 from CH₂=CH, 2 from C≡CH, more than the others.
Ref: NCERT Class 11 Chemistry > Chapter 8: Organic Chemistry - Some Basic Principles and Techniques > Topic: Isomerism - Structural Isomerism - Chain Position Functional Metamerism
In H₃NBF₃ , nitrogen in NH₃ donates a lone pair to boron in BF₃ , forming a coordinate bond.
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: Valence Bond Theory and Hybridisation - sp sp2 sp3 dsp2
For He₂ : (sigma 1s)² (sigma^* 1s)² . Bonding = 2, antibonding = 2. Bond order = (2 - 2/2) = 0 , so it does not exist.
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: Valence Bond Theory and Hybridisation - sp sp2 sp3 dsp2
N₂ has a triple bond with a bond enthalpy of ~946 kJ/mol, higher than O₂ (498 kJ/mol), F₂ (159 kJ/mol), or Cl₂ (243 kJ/mol).
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: Valence Bond Theory and Hybridisation - sp sp2 sp3 dsp2
In CO₂ , each C=O double bond has a bond length of ~123 pm, shorter than C=C in ethene (134 pm) due to higher bond strength.
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: Valence Bond Theory and Hybridisation - sp sp2 sp3 dsp2
For B₂ : (sigma 1s)² (sigma^* 1s)² (sigma 2s)² (sigma^* 2s)² (pi 2px)¹ (pi 2py)¹ . Bonding = 6, antibonding = 4. Bond order = (6 - 4/2) = 1 , with 2 unpaired electrons (paramagnetic).
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: Molecular Orbital Theory - MOT and Bond Order and Magnetic Properties
For H₂ : (sigma 1s)² . Bonding = 2, antibonding = 0. Bond order = (2 - 0/2) = 1 , with all electrons paired (diamagnetic).
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: Molecular Orbital Theory - MOT and Bond Order and Magnetic Properties
For NO : (sigma 1s)² (sigma^* 1s)² (sigma 2s)² (sigma^* 2s)² (sigma 2pz)² (pi 2px)² (pi 2py)² (pi^* 2px)¹ . Bonding = 10, antibonding = 5. Bond order = (10 - 5/2) = 2.5 , with 1 unpaired electron (paramagnetic).
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: Molecular Orbital Theory - MOT and Bond Order and Magnetic Properties
For B₂ : (sigma 1s)² (sigma^* 1s)² (sigma 2s)² (sigma^* 2s)² (pi 2px)¹ (pi 2py)¹ . Bonding = 6, antibonding = 4. Bond order = (6 - 4/2) = 1 , with 2 unpaired electrons (paramagnetic).
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: Molecular Orbital Theory - MOT and Bond Order and Magnetic Properties
