Chemical Bonding and Atomic Structure MCQs for NEET — Chemistry Questions with Answers

Theorectical dipole momentum $ H = q \times d [ 1 D = 1 \times 10 ^ {-18} esa cm ] $ $ = 4.8 \times 10 ^ {-10} \times 1.275 \times 10 ^ {-8} $ $ = 6.12 \times 10 ^ {-10} e.s.u cm $ $ = 6.12 D $ $ % of ionic character = { Practical ( \mu ) \over Theoretical (\mu) } $ $ = { 1.03 \over 6.12} \times 100 $ $ = 16.83 $ $ \approx 17 \% $

Bond enthalpy is affected by the number of nonbonding electron pairs on bonded atoms. Nonbonding electron pairs can cause repulsion, which weakens the bond and thus lowers the bond enthalpy.

KO₂ (Potassium superoxide) is paramagnetic due to the presence of unpaired electrons in its molecular orbital configuration. The presence of a single unpaired electron in the antibonding Ï€* orbital makes KO₂ paramagnetic.

The bond order of CO is 3. The bond order of NO⁺ is also 3, NO has a bond order of 2.5, N₂ has a bond order of 3, and CN^- has a bond order of 3. Therefore, NO (neutral) does not have the same bond order as CO.

In VSEPR (Valence Shell Electron Pair Repulsion) theory, the repulsion between lone pairs (Lp) of electrons is greater than the repulsion between lone pair and bonding pair (Bp) of electrons, which in turn is greater than the repulsion between bonding pairs of electrons. Hence, the correct order is Lp-Lp > Lp-Bp > Bp-Bp.

The VSEPR (Valence Shell Electron Pair Repulsion) theory is primarily used to determine the geometrical structure of molecules. It predicts the shape of individual molecules based on the extent of electron-pair electrostatic repulsion.

The bond order of $O_2$ is 2, CN is 3, and $NO^+$ is also 3. Therefore, CN and $NO^+$ have equal bond orders.