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s-character: sp (50%) > sp² (33%) > sp³ (25%). CH₃C≡CH has an sp-hybridized carbon in the triple bond.
Ref: NCERT Class 11 Chemistry > Chapter 8: Organic Chemistry - Some Basic Principles and Techniques > Topic: Qualitative and Quantitative Analysis - Lassaigne Test and Estimation
In HC≡N (hydrogen cyanide), the carbon forms a triple bond with nitrogen and a single bond with hydrogen, indicating sp hybridization.
Ref: NCERT Class 11 Chemistry > Chapter 8: Organic Chemistry - Some Basic Principles and Techniques > Topic: Isomerism - Stereoisomerism - Geometrical and Optical Isomerism
Electronegativity increases with s-character: sp (50%) > sp² (33%) > sp³ (25%). Thus, sp hybridized carbon is the most electronegative.
Ref: NCERT Class 11 Chemistry > Chapter 8: Organic Chemistry - Some Basic Principles and Techniques > Topic: Isomerism - Stereoisomerism - Geometrical and Optical Isomerism
In CS₂ , carbon has 2 double bonds (2 bonding domains), requiring sp hybridization for a linear shape.
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: Valence Bond Theory and Hybridisation - sp sp2 sp3 dsp2
In C₂H₂ , each carbon is sp hybridized, forming a triple bond and a linear structure with a 180° bond angle.
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: Valence Bond Theory and Hybridisation - sp sp2 sp3 dsp2
In BeCl₂ , beryllium uses one s and one p orbital to form two sp hybrid orbitals, resulting in a linear geometry.
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: Ionic Bonding - Lattice Enthalpy and Properties of Ionic Compounds
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