31 Year NEET Previous Year Questions: Chemical Bonding & Molecular Structure - 3 - NEET MCQ

The overlap between s- and p-orbitals occurs along internuclear axis and hence the angle is 180°.

Equilateral or triangular planar shape involves sp² hybridization.

For linear arrangement of atoms the hybridisation should be sp(linear shape, 180° angle). Only H₂S has sp³-hybridization and hence has angular shape while C₂H₂, BeH₂ and CO₂ all involve sp - hybridization and hence have linear arrangement of atoms.

In metallic bonds each ion is surrounded by equal no. of oppositely charged ions hence have electrostatic attraction on all sides and hence do not have directional characteristics.

BeF₂ is linear and hence has zero dipole moment while H₂O, being a bent molecule, has a finite or non-zero dipole moment.

As we move in period from Li → Be → B → C, the electronegativity (EN)  increases and hence the EN difference between the element and Cl decreases and accordingly the covalent character increases Thus option (c) i.e., LiCl < BeCl₂ < BCl₃< CCl₄ is correct.

A σ - bond is stronger than a σ -bond hence option (a) is not correct. Sigma (σ) bonds are formed by head on overlap of  unhybridised s–s, p–p or s–p orbitals and hybridised orbitals (sp, sp², sp³, sp³d and sp³d²) hence σ bonds  are strong bonds. Where as Pi (π)-bonds are formed by side ways overlap of unhybridised p- and d orbitals-hence σ bonds are weak bonds.

Among the F, O, and N, Fluorine has a higher value of electronegativity, this is the reason that fluorine can form the strongest hydrogen bond with the H atom of another molecule.

Just take an idea from the value of the energy of hydrogen bond that is formed with HF, H₂O, and NH₃

_​

means to break the hydrogen of one mole HF molecules, 10Kcal energy is required.

Linear combination of two hybridized orbitals leads to the formation of sigma bond.

The  star marked carbon has a valency of 5 and hence this formula is not correct.

CCl₄ has sp³ hybridisation , tetrahedral geometry and all bond angles of 109° 28'.

Polarity of the bond depends upon the electronegativity difference of the two atoms forming the bond. Greater the electronegativity difference, more is the polarity of the bond.

As the electronegativity difference between N and F is maximum hence this bond is most polar.

The bond length decreases in the order sp³ > sp² > sp.
Because of the triple bond, the carbon-carbon bond distance in ethyne is shortest.

CO2 has sp-hybridization and is linear. SO2 and CO2-3 are planar (sp2) while SO2-4 is tetrahedral (sp3).

H – F shows strongest H-bonds.  Linear combination of two hybridized orbitals leads to the formation of sigma bond.

Sigma bond is stronger than π-bond. The electrons in the π bond are loosely held. The bond is easily broken and is more reactive than σ-bond. Energy released during sigma bond formation is always more than π bond because of greater  extent of overlapping.

For compounds containing ion s of same charge, lattice energy increases as the size the ions decrease. Thus, NaF has highest lattice energy. The size of cation is in the order Na⁺ < K⁺ < Rb⁺ < Cs⁺

The strength of the interactions follows the order vander Waal’s < hydrogen – bonding < dipole-dipole < covalent.

According to Fajan rules, ionic character increases with increase in size of the cation and decrease in size of the anion. Thus, CsF has higher ionic character than NaCl and hence bond in CsF is stronger than in NaCl.

The removal of an electron from a diatomic molecule may increase the bond order as in the conversion  or decrease the bond order as in the conversion,  As a result, the bond energy may increase or decrease. thus, statement (b) is incorrect.

H-bond is the weakest.

BH₃ has sp² hybridization and hence does not have tetrahedral structure while all others have tetrahedral structures.

The stability of the ionic bond depends upon the lattice energy which is expected to be more between Mg and F due to +2 charge on Mg atom.

The angle between the bonds formed by p_x and p_y orbitals is the minimum i.e. 90°

Chemical bonds.

The b.p. of p-nitrophenol is higher than that of o-nitrophenol because in p-nitrophenol there is intermolecular H-bonding but in onitrophenol it is intramolecular H-bonding.

The C–C bond distance decreases as the multiplicity of the bond increases. Thus, bond distance decreases in the order: butane (1.54 Å) > benzene (1.39 Å) > ethene (1.34) Å > ethyne (1.20 Å). Thus in butane, C – C bond distance is the largest.

Paramagnetism is caused by the presence of atoms, ions or molecules with unpaired electrons. In NO the presence of unpaired electron is clear. Therefore it is paramagnetic.

As there is no lone pair on boron in BCl₃ therefore no repulsion takes place. But there is a lone pair on nitrogen in NCl₃. Therefore repulsion takes place. Thus BCl₃ is planar molecule but NCl₃ is a pyramidal molecule.