NEET Practice Questions (MCQs) with Answers & Solutions

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

$NH_3$ has a higher boiling point than expected due to the presence of intermolecular hydrogen bonds. In ammonia, the nitrogen atom is highly electronegative and forms hydrogen bonds with hydrogen atoms of neighboring ammonia molecules. These hydrogen bonds require more energy to break, resulting in a higher boiling point.

Carbon tetrachloride (CClâ‚„) has a tetrahedral geometry with four chlorine atoms symmetrically arranged around the central carbon atom. This symmetry causes the dipole moments of the C-Cl bonds to cancel each other out, resulting in a net dipole moment of zero.

The molecular shapes and lone pairs of electrons for the given molecules are as follows:

• SFâ‚„ has a 'see-saw' shape with 1 lone pair of electrons.
• CFâ‚„ has a tetrahedral shape with 0 lone pairs of electrons.
• XeFâ‚„ has a square planar shape with 2 lone pairs of electrons. Thus, they have different shapes with 1, 0, and 2 lone pairs of electrons respectively.

The [BF₄]⁻ ion has a regular tetrahedral structure. Boron is surrounded by four fluorine atoms with no lone pairs on the boron atom, giving it a perfect tetrahedral shape. The other options do not have a regular tetrahedral geometry.

Bond strength is directly related to bond order. Bond order refers to the number of chemical bonds between a pair of atoms. Higher bond order means more electrons are shared between the atoms, leading to a stronger bond. Therefore, bond strength increases with bond order.