Consider the following species: O₂, B₂, N₂, PCl₃, F₂, SF 4 , ClF₃, solid PCl₅, ICl₃. Evaluate: ( C + D ) − ( A + B ) where: A = Number of species that are paramagnetic in their ground state. B = Numbe

Question

Consider the following species:

O₂, B₂, N₂, PCl₃, F₂, SF₄, ClF₃, solid PCl₅, ICl₃.

Evaluate: $(C + D) - (A + B)$

where:

$A$ = Number of species that are paramagnetic in their ground state.

$B$ = Number of species having non-zero dipole moment in their standard molecular (or monomeric) form.

$C$ = Number of species in which the central atom involves d-orbital(s) in hybridization.

$D$ = Number of species containing at least one bond angle less than $109. 5^{\circ}$ .

TestHub · Accepted Answer

Paramagnetic species are O 2 ​ and B 2 ​ , so A = 2 Polar species are PCl 3 ​ , SF 4 ​ , CIF 3 ​ , SnCl 2 ​ , ICl 3 ​ , hence B = 5 d-orbital involvement occurs in SF 4 ​ , CIF 3 ​ , ICl 3 ​ , and PCl 6 − ​ (from solid PCl 5 ​ ), so C = 4 Species having at least one bond angle < 109. 5 ∘ and PCl 3 ​ , SF 4 ​ , CIF 3 ​ , ICl 3 ​ and solid PCl 5 ​ ( PCl 6 ​ − has 9 0 ∘ ) , So D = 5 Therefore ( C + D ) − ( A + B ) = ( 4 + 5 ) − ( 2 + 5 ) = 2

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