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Chapter 22: Problem 64

Both nitramide and hyponitrous acid have the formula \(\mathrm{H}_{2} \mathrm{N}_{2} \mathrm{O}_{2} .\) Hyponitrous acid is a weak diprotic acid; nitramide contains the amide group \(\left(-\mathrm{NH}_{2}\right) .\) Draw plausible Lewis structures for these two substances.

Short Answer

Expert verified
The Lewis structure for Nitramide is: Nitrogen (N) at the center, two Hydrogen (H) and one Nitrogen (N) atom attached to form the amide group (-NH2). An Oxygen (O) atom is attached to the other Nitrogen atom with a double bond. The Lewis structure for Hyponitrous Acid is: Two Nitrogen atoms (N) at the center, each attached to one Hydrogen atom (H) and one Oxygen atom (O) with a double bond.

Step by step solution

01

Lewis Structure for Nitramide

Start by drawing the Nitrogen atom (N) at the center, then connect the two Hydrogen atoms (H) to one of the Nitrogen atoms forming an amide group (-NH2). Connect an Oxygen atom (O) to the other Nitrogen atom. Finally, place a double bond between the Nitrogen and Oxygen atoms, and add the remaining two pairs of electrons to the Oxygen atom.
02

Verify the Nitramide Lewis Structure

The Lewis structure must fit the molecular formula. Therefore, check the count of each atom type: there should be two Hydrogen atoms (H), two Nitrogen atoms (N), and two Oxygen atoms (O). Each atom should be surrounded by a total of eight electrons (except Hydrogen, which should be surrounded by two electrons) - all of these factors should be true for the molecule to abide by the octet rule.
03

Lewis Structure for Hyponitrous Acid

Draw two Nitrogen atoms (N) connected by a single bond in the center. Connect one Hydrogen atom (H) to each Nitrogen atom. Add a double bond between each Nitrogen atom and a separate Oxygen atom (O), and add the remaining two pairs of electrons to the Oxygen atoms.
04

Verify the Hyponitrous Acid Lewis Structure

Revisit the structure and check to make sure all conditions have been fulfilled as in Step 2. Each atom should be surrounded by a total of eight electrons (except Hydrogen, which should be surrounded by two electrons) abiding by the octet rule.

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Most popular questions from this chapter

What is the acid anhydride of \((a) \mathrm{H}_{2} \mathrm{SO}_{4} ;(\mathrm{b}) \mathrm{H}_{2} \mathrm{SO}_{3}\) (c) \(\mathrm{HClO}_{4} ;\) (d) \(\mathrm{HIO}_{3}\) ?

Figure \(15-1\) (page 656 ) shows that \(I_{2}\) is considerably more soluble in \(\mathrm{CCl}_{4}(1)\) than it is in \(\mathrm{H}_{2} \mathrm{O}(1) .\) The concentration of \(I_{2}\) in its saturated aqueous solution is \(1.33 \times 10^{-3} \mathrm{M},\) and the equilibrium achieved when \(\bar{I}_{2}\) distributes itself between \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CCl}_{4}\) is $$\mathrm{I}_{2}(\mathrm{aq}) \rightleftharpoons \mathrm{I}_{2}\left(\mathrm{CCl}_{4}\right) \quad K_{\mathrm{c}}=85.5$$ (a) \(\mathrm{A} 10.0 \mathrm{mL}\) sample of saturated \(\mathrm{I}_{2}(\mathrm{aq})\) is shaken with \(10.0 \mathrm{mL} \mathrm{CCl}_{4} .\) After equilibrium is established, the two liquid layers are separated. How many milligrams of \(I_{2}\) will be in the aqueous layer? (b) If the \(10.0 \mathrm{mL}\) of aqueous layer from part (a) is extracted with a second \(10.0 \mathrm{mL}\) portion of \(\mathrm{CCl}_{4}\) how many milligrams of \(\mathrm{I}_{2}\) will remain in the aqueous layer when equilibrium is reestablished? (c) If the 10.0 mL sample of saturated \(I_{2}(\) aq) in part (a) had originally been extracted with \(20.0 \mathrm{mL} \mathrm{CCl}_{4}\) would the mass of \(I_{2}\) remaining in the aqueous layer have been less than, equal to, or greater than that in part (b)? Explain.

The oxides of the phosphorus(III), antimony(III), and bismuth(III) are \(\mathrm{P}_{4} \mathrm{O}_{6}, \mathrm{Sb}_{4} \mathrm{O}_{6},\) and \(\mathrm{Bi}_{2} \mathrm{O}_{3} .\) Only one of these oxides is amphoteric. Which one? Which of these oxides is most acidic? Which is most basic?

The following bond energies are given for \(298 \mathrm{K}: \mathrm{O}_{2}\) \(498 ; \mathrm{N}_{2}, 946 ; \mathrm{F}_{2}, 159 ; \mathrm{Cl}_{2}, 243 ;\) ClF, \(251 ; \mathrm{OF}\left(\text { in } \mathrm{OF}_{2}\right)\) \(213 ; \mathrm{ClO}\left(\operatorname{in} \mathrm{Cl}_{2} \mathrm{O}\right), 205 ;\) and \(\mathrm{NF}\left(\mathrm{in} \mathrm{NF}_{3}\right), 280 \mathrm{kJmol}^{-14}\) Calculate \(\Delta H_{f}\) at \(298 \mathrm{K}\) for \(1 \mathrm{mol}\) of \((\mathrm{a}) \mathrm{ClF}(\mathrm{g})\) (b) \(\mathrm{OF}_{2}(\mathrm{g}) ;(\mathrm{c}) \mathrm{Cl}_{2} \mathrm{O}(\mathrm{g}) ;(\mathrm{d}) \mathrm{NF}_{3}(\mathrm{g})\).

Make a general prediction about which of the halogen elements, \(\mathrm{F}_{2}, \mathrm{Cl}_{2}, \mathrm{Br}_{2},\) or \(\mathrm{I}_{2},\) displaces other halogens from a solution of halide ions. Which of the halogens is able to displace \(\mathrm{O}_{2}(\mathrm{g})\) from water? Which is able to displace \(\mathrm{H}_{2}(\mathrm{g})\) from water?
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