Given below are two statements: Statement I : The number of species among BF _ 4 ^ - , SiF _ 4 , XeF _ 4 and SF _ 4 , that have unequal E - F bond lengths is two. Here, E is the central atom. Statement II : Among O _ 2 ^ - , O _ 2 ^ 2- , F _ 2 and O _ 2 ^ + , O _ 2 ^ - has the highest bond order. In the light of the above statements, choose the correct answer from the options given below
A. Both Statement I and Statement II are false
Statement I: Let's analyze the geometry of the given species. BF_4^-: sp^3 hybridization, tetrahedral geometry. All B-F bond lengths are equal. SiF_4: sp^3 hybridization, tetrahedral geometry. All Si-F bond lengths are equal. XeF_4: sp^3d^2 hybridization, square planar geometry. All Xe-F bond lengths are equal. SF_4: sp^3d hybridization, see-saw geometry. It has axial and equatorial bonds. Axial bonds are longer than equatorial bonds due to greater repulsion. Thus, only SF_4 has unequal bond lengths. The number of species is one, not two. Statement I is false. Statement II: Bond order calculation using Molecular Orbital Theory: Total electrons in O_2^- = 17, Bond Order = 1.5 Total electrons in O_2^ 2- = 18, Bond Order = 1.0 Total electrons in F_2 = 18, Bond Order = 1.0 Total electrons in O_2^+ = 15, Bond Order = 2.5 Among these, O_2^+ has the highest bond order (2.5), not O_2^-. Statement II is false. Both Statement I and Statement II are false.
Related: Chemistry — Chemical Bonding and Molecular Structure · All PYQ Banks