Q14.148CP - Chemistry: Molecular Nature Of Matter And Change

Question

Two substances with empirical formula HNO are hyponitrous acid  $\left( \mu =62.04g/\mathrm{mol}\right)$ and nitroxyl $\left( \mu =31.02g/\mathrm{mol}\right)$ . 

(a) What is the molecular formula of each species? 

(b) For each species, draw the Lewis structure having the lowest formal charges. (Hint: Hyponitrous acid has an N N bond.) 

(c) Predict the shape around the N atoms of each species.

(d) When hyponitrous acid loses two protons, it forms the hyponitrite ion. Draw cis and trans forms of this ion.

Step-by-Step Solution

Answer

     a. Hyponitrous acid's molecular formula is H₂N₂O₂  . And nitroxyl's molecular formula is HNO.

     b. The Lewis structure of   H₂N₂O₂   , with lowest formal charges, is as follows:

   
            The Lewis structure of HNO, with lowest formal charges, is as follows:

         

      c. HNO and   H₂N₂O₂  both have a bent shape.

      d. The following are the cis and trans versions of hyponitrite:

Step 1: (a)

The following is the relationship between the molecular formula and the empirical formula:

$\text{ molecular formula = }{\left(\text{empirical formula}\right)}_{\text{n}}$ 

The molar mass is divided by the empirical mass to get n.

HNO has an empirical mass of 31.02 g/mol.

Determine the value of n in the case of hyponitrous acid.

$\begin{array}{rcl}\mathrm{n}& =& \frac{\text{62.04 g/mol}}{\text{31.02 g/mol}}\\ & =& 2\\ \mathrm{Molecular} \mathrm{formula}& =& {\left(\text{HNO}\right)}_{\text{2}}\\ & =& {\text{ H}}_{\text{2}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\\ & & \end{array}$ 

As a result, hyponitrous acid's molecular formula is H₂N₂O₂  .

Calculate the nitroxyl value of n.

$\begin{array}{rcl}\mathrm{n}& =& \frac{\text{31.02 g/mol}}{\text{31.02 g/mol}}\\ & =& \text{1}\\ \mathrm{Molecular} \mathrm{formula}& =& {\left(\text{HNO}\right)}_{\text{1}}\\ & =& \text{ HNO}\\ & & \end{array}$ 

As a result, nitroxyl's molecular formula is HNO.

Step 2: (b)

Draw the Lewis structure of  H₂N₂O₂   as follows:

In a Lewis structure, the center atom is the least electronegative. As a result, nitrogen is the center atom in this example since it is the least electronegative of the three. Always keep in mind that hydrogen atoms are terminal atoms.

There are 24 valence electrons in all (1 electron from each H, 5 electrons from each N, and 6 electrons from each O).

Create the skeletal structure as follows:

Subtract 10 electrons from total valence electrons to account for five bonds in the skeletal structure. Distribute the remaining 14 electrons to the terminal atoms, then the center atom, to complete the octet of each atom.

One nitrogen atom is still missing an octet. To form a double bond between nitrogen atoms, move one of the lone pairs on nitrogen.

All of the atoms in the aforementioned structure have full octets.

Calculate each atom's formal charge as follows:

$$\begin{gathered} \text{Formal} \text{charge}=\text{valance} \text{electrons}-\left(\text{unshared} \text{electrons}-\frac{1}{2}\text{bonding} \text{electrons}\right) \\ \text{Formal} \text{charge} \text{on} \text{H}=1-\left(0+\frac{1}{2}\times 2\right)=0 \\ \text{Formal} \text{charge} \text{on} \text{O}=6-\left(4+\frac{1}{2}\times 4\right)=0 \\ \text{Formal} \text{charge} \text{on} \text{N}=5-\left(2+\frac{1}{2}\times 6\right)=0 \end{gathered}$$

Therefore, the Lewis structure of H₂N₂O₂, with the lowest formal charges, is as follows:

Draw the Lewis structure of HNO as follows:

The total number of valence electrons in HNO is 12.

Draw the skeleton structure as follows:

Subtract 4 electrons from the total valence electrons to account for two bonds in the skeletal structure. Distribute the remaining 8 electrons to the terminal atoms, then the center atom, to complete the octet of each atom.

One nitrogen atom is still missing an octet. To create a double bond between nitrogen and oxygen, move one of the lone pairs on oxygen.

In the above structure, all the atoms have complete octets.

Calculate the formal charge on each atom as follows:

$$\begin{gathered} \text{Formal charge on H}=1-\left(0+\frac{1}{2}\times 2\right)=0 \\ \text{Formal charge on O}=6-\left(4+\frac{1}{2}\times 4\right)=0 \\ \text{Formal charge on N}=5-\left(2+\frac{1}{2}\times 6\right)=0 \end{gathered}$$

Therefore, the Lewis structure of HNO, with the lowest formal charges, is as follows:

Step 3: (c)

H₂N₂O₂  has the following shape around the N atoms:

Both nitrogen atoms have one lone pair of electrons and are connected to two oxygen atoms in H₂N₂O₂. As a result, there are three electron groups around the nitrogen atoms. As a result,  H₂N₂O₂ belongs to the VSEPR class ${\text{AX}}_{\text{2}}\text{E}$. A stands for the center atom, X for the bound atoms, and E for the lone pairs.

A molecule of class ${\text{AX}}_{\text{2}}\text{E}$ has a trigonal planar electron arrangement and a bent molecular shape, according to VSEPR theory.

As a result,  H₂N₂O₂  has a bent form.

 

HNO has the following shape around the N atoms:

The nitrogen atom in HNO is linked to oxygen and hydrogen atoms and possesses one lone pair of electrons. As a result, there are three electron groups surrounding the nitrogen atom. As a result, HNO's VSEPR class is ${\text{AX}}_{\text{2}}\text{E}$. A stands for the center atom, X for the bound atoms, and E for the lone pairs.

A molecule of class ${\text{AX}}_{\text{2}}\text{E}$ has a trigonal planar electron arrangement and a bent molecular shape, according to VSEPR theory.

As a result, HNO has a bent form.

Step 4: (d)

The following is the Lewis structure of the hyponitrite ion:

A cis-form molecule is one in which the same sorts of groups are on the same side of a double bond. The shape is called trans-form when the two groups are on opposing sides.

The following are the cis and trans versions of hyponitrite: