Question

Formal charge on sulphur atom in the following three Lewis structures I, II and III respectively is \[ \text{I.}\quad \ddot{S}=C=\ddot{N} \] \[ \text{II.}\quad :S-C\equiv N: \] \[ \text{III.}\quad :S\equiv C-N: \]

• A. \(0,+1,-1\)
• B. \(+1,0,-1\)
• C. \(0,-1,+1\)
• D. \(+1,-1,0\)

Hint:

Use \[ \text{Formal Charge} = \text{Valence Electrons} - \text{Lone Pair Electrons} - \frac{\text{Bonding Electrons}}{2}. \] For sulphur: \[ V=6. \] More bonds generally make the formal charge more positive, while more lone pairs make it more negative.

Answer 1

The Correct Option is C

Concept: Formal charge is calculated using \[ \text{F.C.} = V-\left(L+\frac{B}{2}\right), \] where \[ V=\text{valence electrons}, \] \[ L=\text{non-bonding electrons}, \] \[ B=\text{bonding electrons}. \] For sulphur, \[ V=6. \]

Structure I : \[ :S=C=N: \] Sulphur has: \[ L=4 \] (two lone pairs) and a double bond \[ B=4. \] Therefore, \[ \text{F.C.} = 6-\left(4+\frac{4}{2}\right) = 6-(4+2) = 0. \] \[ \boxed{\text{F.C. on S}=0} \]

Structure II : \[ :S-C\equiv N: \] Sulphur has: \[ L=6 \] (three lone pairs) and one single bond \[ B=2. \] Hence, \[ \text{F.C.} = 6-\left(6+\frac{2}{2}\right) = 6-(6+1) = -1. \] \[ \boxed{\text{F.C. on S}=-1} \]

Structure III : \[ :S\equiv C-N: \] Sulphur has: \[ L=2 \] (one lone pair) and one triple bond \[ B=6. \] Thus, \[ \text{F.C.} = 6-\left(2+\frac{6}{2}\right) = 6-(2+3) = +1. \] \[ \boxed{\text{F.C. on S}=+1} \]

Final Result: \[ \boxed{0,\,-1,\,+1} \] \[ \boxed{\text{Answer = (C)}} \]