QQuestionChemistry
QuestionChemistry
What is the correct Lewis structure for CN^- (cyanide ion)?
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Answer
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Step 1:: Determine the total number of valence electrons in the CN$^-$ ion.
5(mathrm{N~valence~electrons}) + 4(mathrm{C~valence~electrons}) + 1(mathrm{extra~electron}) = 10~electrons
The nitrogen atom (N) is in the 15th group of the periodic table, so it has 5 valence electrons. The carbon atom (C) is in the 14th group, so it has 4 valence electrons. Since there is one negative charge on the ion, there is an extra electron associated with the ion.
Step 2:: Draw a skeleton structure for the CN$^-$ ion.
Place the carbon atom and the nitrogen atom in a straight line with a triple bond between them. Since carbon is less electronegative than nitrogen, it will be the central atom.
Step 3:: Distribute the valence electrons around the atoms.
First, place a pair of electrons between the carbon and nitrogen atoms to form the triple bond. This uses up 4 electrons (2 electrons for each bond in the triple bond). Next, place three lone pairs of electrons around the nitrogen atom to complete its octet (5 electrons from the nitrogen valence electrons + 3 electrons from the triple bond = 8 electrons). Now, place the remaining electron as a lone pair on the carbon atom. This will give the carbon atom a formal charge of - 1, which is consistent with the overall charge of the ion.
Step 4:: Check the formal charges on each atom.
mathrm{N~formal~charge} = 5 - 4 - 2 = 0
The formal charge on the carbon atom is: The formal charge on the nitrogen atom is: Since the formal charges add up to the overall charge of the ion, the Lewis structure is valid.
Final Answer
The correct Lewis structure for the CN$^-$ ion is: \begin{array}{c} :\mathrm{N}:::\ | | \ : : \ | | \ :\mathrm{C}: \ \end{array} with a triple bond between C and N, a lone pair on the carbon atom, and three lone pairs on the nitrogen atom.
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