Identify Pairs of Species Without Similar Lewis Dot Structures

To determine which pairs of species do not have similar Lewis dot structures, we evaluate the total number of valence electrons, the formal charges, hybridization, and the resulting Lewis structures for each given pair:

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Pair A: $[\text{SO}_3^{2-}, \text{CO}_3^{2-}]$

1. $\text{SO}_3^{2-}$ (Sulfite ion):

   • Total valence electrons = $6 \,(\text{from S}) + 3 \times 6 \,(\text{from O}) + 2 \,(\text{negative charge}) = 26\text{ electrons}$.
   • Central atom: $\text{S}$ forms 3 single bonds with oxygen atoms and retains $1$ lone pair.
   • Geometry / Shape: Trigonal pyramidal ($\text{sp}^3$ hybridized central atom).

2. $\text{CO}_3^{2-}$ (Carbonate ion):

   • Total valence electrons = $4 \,(\text{from C}) + 3 \times 6 \,(\text{from O}) + 2 \,(\text{negative charge}) = 24\text{ electrons}$.
   • Central atom: $\text{C}$ forms $1$ double bond and $2$ single bonds with oxygen atoms (0 lone pairs on $\text{C}$).
   • Geometry / Shape: Trigonal planar ($\text{sp}^2$ hybridized central atom).

Since $\text{SO}_3^{2-}$ and $\text{CO}_3^{2-}$ have a different number of valence electrons ($26$ vs $24$) and different molecular geometries, Pair A does not have similar Lewis dot structures.

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Pair B: $[\text{O}_2^{2-}, \text{F}_2]$

1. $\text{O}_2^{2-}$ (Peroxide ion):

   • Total valence electrons = $2 \times 6 + 2 = 14\text{ electrons}$.
   • Structure: $[:\ddot{\text{O}}-\ddot{\text{O}}:]^{2-}$ (single bond with 3 lone pairs on each oxygen atom).

2. $\text{F}_2$ (Fluorine molecule):

   • Total valence electrons = $2 \times 7 = 14\text{ electrons}$.
   • Structure: $:\ddot{\text{F}}-\ddot{\text{F}}:$ (single bond with 3 lone pairs on each fluorine atom).

Both species are isoelectronic and possess identical Lewis dot structures.

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Pair C: $[\text{CN}^-, \text{CO}]$

1. $\text{CN}^-$ (Cyanide ion):

   • Total valence electrons = $4 \,(\text{from C}) + 5 \,(\text{from N}) + 1 = 10\text{ electrons}$.
   • Structure: $[:\text{C}\equiv\text{N}:]^-$ (triple bond with 1 lone pair on $\text{C}$ and 1 lone pair on $\text{N}$).

2. $\text{CO}$ (Carbon monoxide):

   • Total valence electrons = $4 \,(\text{from C}) + 6 \,(\text{from O}) = 10\text{ electrons}$.
   • Structure: $:\text{C}\equiv\text{O}:$ (triple bond with 1 lone pair on $\text{C}$ and 1 lone pair on $\text{O}$).

Both species are isoelectronic and possess identical Lewis dot structures.

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Pair D: $[\text{NH}_3, \text{H}_3\text{O}^+]$

1. $\text{NH}_3$ (Ammonia):

   • Total valence electrons = $5 \,(\text{from N}) + 3 \times 1 \,(\text{from H}) = 8\text{ electrons}$.
   • Structure: Central $\text{N}$ forms 3 single bonds with $\text{H}$ and has 1 lone pair (Trigonal pyramidal).

2. $\text{H}_3\text{O}^+$ (Hydronium ion):

   • Total valence electrons = $6 \,(\text{from O}) + 3 \times 1 \,(\text{from H}) - 1 = 8\text{ electrons}$.
   • Structure: Central $\text{O}$ forms 3 single bonds with $\text{H}$ and has 1 lone pair (Trigonal pyramidal).

Both species are isoelectronic and possess similar Lewis dot structures.

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Pair E: $[\text{MnO}_4^-, \text{CrO}_4^{2-}]$

1. $\text{MnO}_4^-$ (Permanganate ion):

   • Manganese ($\text{Mn}$) has 7 valence electrons ($3d^5 4s^2$).
   • To minimize formal charge, $\text{Mn}$ forms 3 double bonds ($\text{Mn}=\text{O}$) and 1 single bond ($\text{Mn}-\text{O}^-$).

2. $\text{CrO}_4^{2-}$ (Chromate ion):

   • Chromium ($\text{Cr}$) has 6 valence electrons ($3d^5 4s^1$).
   • To minimize formal charge, $\text{Cr}$ forms 2 double bonds ($\text{Cr}=\text{O}$) and 2 single bonds ($\text{Cr}-\text{O}^-$).

Since the number of double bonds and single bonds around the central metal atom differs between $\text{MnO}_4^-$ and $\text{CrO}_4^{2-}$, Pair E does not have similar Lewis dot structures.

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Conclusion:

The pairs that do not have similar Lewis dot structures are A and E.

Correct Option: B