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Question
Draw the most plausible resonance structure for the cyanate ion, $\mathrm{OCN}^{-}$.

Asked By FrostBite26 at

Answered By Expert

Dwight

Expert · 3.7k answers · 3k people helped

To determine the most plausible resonance structure for the cyanate ion,

\mathrm{OCN}^{-}, we need to consider the distribution of electrons and the formal charges on each atom. Here’s a step-by-step guide to drawing the most plausible resonance structure:

Solution By Steps

Step 1: Identify the atoms and their electronegativities

Oxygen (O) is the most electronegative atom, followed by nitrogen (N), and carbon ©.

Step 2: Assign a central atom

Typically, the least electronegative atom is the central atom. In this case, carbon © is the least electronegative among the three.

Step 3: Draw the basic structure

Place C in the center, with O and N bonded to it.

Step 4: Determine the number of valence electrons

O: 6 valence electrons

C: 4 valence electrons

N: 5 valence electrons

An additional electron due to the negative charge: 1

Total valence electrons: 6 + 4 + 5 + 1 = 16

Step 5: Arrange the electrons to form bonds and lone pairs

Form a triple bond between C and N, and a single bond between C and O.

Assign the remaining electrons to satisfy the octet rule for each atom.

Step 6: Calculate formal charges

Formal charge = valence electrons - (lone pair electrons + 1/2 bonding electrons)

For O: 6 - (4 + 1/2 * 2) = 0

For C: 4 - (0 + 1/2 * 8) = 0

For N: 5 - (2 + 1/2 * 6) = 0

Step 7: Consider resonance structures

Move the electrons around to form different bonding patterns, ensuring that the formal charges are minimized.

Final Answer

The most plausible resonance structure for the cyanate ion,

\mathrm{OCN}^{-}, is one where carbon is the central atom, with a triple bond to nitrogen and a single bond to oxygen. All atoms have a formal charge of zero, and the structure is stable.

\begin{array}{ccc}

& :O^{\ominus} & \\

& | & \\

:N &\equiv &C: \\

\end{array}

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