O2 is paramagnetic due to the presence of two unpaired electrons in its molecular orbitals.

N2 has a triple bond, with a bond order of 3 according to molecular orbital theory.

O2 is paramagnetic due to the presence of two unpaired electrons in its molecular orbitals.

O2 has unpaired electrons in its molecular orbital configuration, making it paramagnetic.

In methane, the bond angle is approximately 109.5 degrees due to its tetrahedral geometry.

In a tetrahedral geometry, the bond angles are approximately 109.5 degrees.

In a trigonal planar geometry, the bond angles are 120 degrees.

The bond angle in H2O is approximately 104.5 degrees due to the two lone pairs on oxygen.

The bond angle in H2O is approximately 104.5 degrees due to the two lone pairs on oxygen.

CO has a bond order of 3, calculated as (10 bonding - 3 antibonding)/2.

O2- has a bond order of 1, calculated as (10 bonding electrons - 7 antibonding electrons)/2 = 1.

B2 has a bond order of 1, calculated as (2 bonding - 0 antibonding)/2.

B2 has a bond order of 1, calculated from its molecular orbital configuration.

CO has a bond order of 3, calculated as (10 bonding electrons - 2 antibonding electrons)/2 = 3.

He2 has a bond order of 0, as it has no net bonding electrons.

N2 has a bond order of 3, calculated as (10 bonding - 0 antibonding)/2 = 3.

O2 has 12 total electrons, leading to a bond order of (10 bonding - 6 antibonding)/2 = 1.

Increasing electronegativity difference between two atoms leads to an increase in ionic character of the bond.

NH3 has four electron pairs (three bonding pairs and one lone pair), giving it a tetrahedral electron pair geometry.

SF6 has six bonding pairs and no lone pairs, resulting in an octahedral electron pair geometry.

Molecules with sp3d hybridization have a trigonal bipyramidal geometry.

A tetrahedral shape corresponds to sp3 hybridization.

In acetylene, each carbon atom is sp hybridized, forming a triple bond between them.

In CO2, the carbon atom forms two double bonds with oxygen atoms, indicating sp hybridization.

A tetrahedral shape corresponds to sp3 hybridization.

In CH4, the central carbon atom forms four equivalent bonds with hydrogen atoms, indicating sp3 hybridization.

The carbon atom in CO2 is sp hybridized, forming two double bonds with oxygen atoms.

Oxygen in H2O is sp3 hybridized, forming two bonds and two lone pairs.

The central nitrogen atom in N2O is sp hybridized, leading to a linear arrangement.

The central nitrogen atom in NH3 is sp3 hybridized due to three bonding pairs and one lone pair.