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Discuss the stepwise determination of the (Lewis) structure of nitric acid . |
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Answer» Solution :Let us draw the Lewis structure for nitric acid . 1. Skeletal structure `{:(H,O,N,O),(,,O,):}` 2. Total number of valence electrons in `HNO_(3)` ` = [1 xx 1 "(HYDROGEN )"] + [1 + 5 "(nitrogen)"]` `+ [3 xx 6 "(OXYGEN)"] = 1 + 5 + 18 = 24` 3. Draw single bonds between atoms. Four bonds can be drawn as shown in the figure for `HNO_(3)` which account for eight electrons (4 bond pairs ). `H- O - UNDERSET(O)underset(|)N-O` 4. Distribute the remaining sixteen `(24 - 8 = 16)` electrons as eight lone pairs startingfrom most electrongative atom, the oxygen. Six lone pairs are distributed to the two terminal oxygens ( THREE each ) to satisfy their octet and two pairs are distributed to the oxygen that is connected to hydrogen to satisfy its octet . H- underset(* *) overset(* *)O-underset( :underset(* *)(O: ))underset(|) N -underset(* *)overset(* *) (O):` 5. VERIFY wheather all the atoms have octet conguration . In the above distribution , the nitrogen has one pair short for octet . Therefore, move one of the lone pair from the terminal oxygen to form another bond with nitrogen. The Lewis structure of nitric acid is given as `H - underset(* *) overset(* *)O - underset( : underset(* *)O: )underset(|)N = underset(* *) overset(* *)O` |
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