12 public questions tagged with this topic.
In BH₃ , boron is sp² hybridized with 3 bonding pairs and no lone pairs, forming a trigonal planar shape with 120° bond angles.
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: Valence Bond Theory and Hybridisation - sp sp2 sp3 dsp2
In XeF₂ , xenon has 8 valence electrons and forms 2 bonds (4 electrons used), leaving 4 electrons as 3 lone pairs. With sp³d hybridization, the 5 electron pairs (2 bonding, 3 lone) result in a trigonal bipyramidal electron geometry and a linear molecular shape.
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: Valence Bond Theory and Hybridisation - sp sp2 sp3 dsp2
In BrF₃ , bromine is sp³d hybridized with 5 electron pairs (3 bonding, 2 lone), forming a T-shaped geometry with ~87° bond angles.
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: VSEPR Theory and Shapes of Molecules
In BrF₃, bromine has 7 valence electrons and forms 3 single bonds (6 electrons used), leaving 4 electrons as 2 lone pairs. With sp³d hybridization, the 5 electron pairs (3 bonding, 2 lone) result in a T-shaped molecular geometry.
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: VSEPR Theory and Shapes of Molecules
In XeF₆ , xenon has 8 valence electrons, forms 6 bonds (12 electrons used), leaving 2 electrons as 1 lone pair in a sp³d³ hybridized state.
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: Kossel-Lewis Approach and Octet Rule and Lewis Structures
In ClF₅ , chlorine has 7 valence electrons, forms 5 bonds (10 electrons used), leaving 2 electrons as 1 lone pair in a sp³d² hybridized state.
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: Kossel-Lewis Approach and Octet Rule and Lewis Structures
In OF₂ , oxygen is sp³ hybridized with two bonding pairs and two lone pairs, resulting in a bent (V-shaped) geometry.
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: Ionic Bonding - Lattice Enthalpy and Properties of Ionic Compounds
In IF₇ , iodine is sp³d³ hybridized with 7 bonding pairs and no lone pairs, forming a pentagonal bipyramidal shape.
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: Ionic Bonding - Lattice Enthalpy and Properties of Ionic Compounds
In IF₅ , iodine has 7 valence electrons, forms 5 bonds (10 electrons used), leaving 2 electrons as 1 lone pair in a sp³d² hybridized state.
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: Ionic Bonding - Lattice Enthalpy and Properties of Ionic Compounds
In XeF₂ , xenon has 8 valence electrons, forms 2 bonds (4 electrons used), leaving 6 electrons as 3 lone pairs in a sp³d hybridized state.
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: Hydrogen Bonding and Resonance and Dipole Moment
In H₂O , oxygen has two lone pairs and two bonding pairs, resulting in a bent shape with a bond angle of 104.5°.
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: Covalent Bonding - Bond Parameters and Polarity and Fajan's Rule
In SF₂ , sulfur is sp³ hybridized with 4 electron pairs (2 bonding, 2 lone), forming a bent shape. The F-S-F bond angle is ~98°, reduced from 104.5° in H₂O due to increased lone pair repulsion from electronegative fluorine atoms.
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: Covalent Bonding - Bond Parameters and Polarity and Fajan's Rule
