JEE Main Chemical Bonding — practice questions
45 free MCQs with worked solutions. Tap any question for the answer + explanation, or practice them all in the app.
Practice JEE Main Chemical Bonding in the app →What is the shape of a molecule whose central atom is $sp^3$ hybridised and has no lone pairs (e.g. $\mathrm{C
Using VSEPR theory, what is the molecular geo
What is the hybridisation of the central P atArrange the bond angles of $\mathrm{H_2O}$, $\mathrm{NH_3}$ and $\mathrm{CH_4}$ in decreasing order.Which of the following molecules has a **non-zero** dipole moment?Using molecular orbital theory, the bond order of the $\mathrm{O_2}$ molecule is:An ELECTROVALENT (ionic) bond is formed when:A COVALENT bond is formed by:The octet rule states atoms tend to:VSEPR stands for:The shape of methane (CH₄) is:Water (H₂O) molecular shape is:Bond order in N₂ molecule is:Dipole moment of CO₂ is:In NH₃, hybridization of N is:BF₃ has shape:In SF₆, central S has hybridization:The order of bond strength: single < double < triple. Bond LENGTHS go in:Bond angle in NH₃ compared to H₂O:Molecule with linear shape (besides CO₂):Number of σ and π bonds in C₂H₄ (ethene):Bond order from MOT (molecular orbital theory):Among NaCl, HCl, H₂O — order of bond polarity:Resonance in molecules:Lattice energy of an ionic crystal depends on:Hybridization in PCl₅ (gaseous phosphorus pentachloride):XeF₄ has shape:In benzene C₆H₆, each C is:Number of π bonds in HCN:Among HF, HCl, HBr, HI: order of bond ENTHALPY (strength):Effective nuclear charge (Z_eff) felt by an electron in 2p orbital of nitrogen, given Z = 7, screening constanAmong NaCl, NaBr, NaI: order of melting points:The bonding in NaCl on the kitchen salt jar is best described as:The shape of the methane (CH$_4$) molecule is:The hybridisation of carbon in ethyne (HC$\equiv$CH) is:Water has a much higher boiling point than H$_2$S (similar molar mass) because:An IONIC bond is formed by:The bond in N2 (nitrogen molecule) is a:In the H3O+ ion, the bond between H+ and H2O is a:The geometry of CH4 (methane) as predicted by VSEPR is:The hybridisation of carbon in C2H4 (ethylene, with a C=C double bond) is:A SIGMA bond differs from a PI bond in that the sigma bond:Molecular Orbital Theory uniquely EXPLAINS why O2 is:The molecule CO2 has zero net DIPOLE MOMENT because:In BENZENE (C6H6), all six C-C bond lengths are equal (1.39 Å) because of: