Question

The formal charges of atoms (1), (2) and (3) in the ion \[ \left[\overset{(1)}{O}=\overset{(2)}{N}=\overset{(3)}{O}\right]^+ \] is

• A. \(0,\;+2,\;-1\)
• B. \(0,\;+1,\;0\)
• C. \(+2,\;0,\;-1\)
• D. \(+1,\;0,\;0\)

Hint:

For formal charge calculations: \[ \text{F.C.}= V-L-\frac{B}{2} \] where \(V\) is valence electrons, \(L\) is lone-pair electrons and \(B\) is bonding electrons.

Answer 1

The Correct Option is B

Step 1: Write the formal charge formula.
The formal charge on an atom is \[ \text{F.C.} = V - L - \tfrac{1}{2} B \] where $V$ is the number of valence electrons, $L$ is the lone pair (non bonding) electrons, and $B$ is the bonding electrons. The ion is $[O=N=O]^+$ with atoms labelled $(1)$, $(2)$, $(3)$.
Step 2: Formal charge on atom $(1)$, oxygen.
This oxygen has a double bond and two lone pairs. So $V=6$, $L=4$, $B=4$. \[ \text{F.C.} = 6 - 4 - \tfrac{1}{2}(4) = 0 \]
Step 3: Formal charge on atom $(2)$, nitrogen.
The central nitrogen forms two double bonds and has no lone pair. So $V=5$, $L=0$, $B=8$. \[ \text{F.C.} = 5 - 0 - \tfrac{1}{2}(8) = 5 - 4 = +1 \]
Step 4: Formal charge on atom $(3)$, oxygen.
This oxygen is equivalent to atom $(1)$, with one double bond and two lone pairs. So $V=6$, $L=4$, $B=4$. \[ \text{F.C.} = 6 - 4 - \tfrac{1}{2}(4) = 0 \]
Step 5: Check the total charge.
Adding the formal charges gives $0 + (+1) + 0 = +1$, which correctly equals the overall charge on the ion. This confirms our assignment.
Step 6: State the answer.
The formal charges of atoms $(1)$, $(2)$, $(3)$ are $0$, $+1$, $0$, matching the key.
\[ \boxed{0,\; +1,\; 0} \]