Question

1. name the following compounds:

a. $\ce{n_{2}o}$
b. $\ce{no_{2}}$
c. $\ce{n_{2}o_{3}}$
d. $\ce{no}$
e. $\ce{s_{2}cl_{2}}$
f. $\ce{so_{2}}$
g. $\ce{s_{2}o_{3}}$
h. $\ce{s_{2}o}$

Explanation:

To name these compounds, we use the rules for naming covalent compounds. We use prefixes to indicate the number of each atom. The prefixes are: mono - 1, di - 2, tri - 3, tetra - 4, penta - 5, hexa - 6, hepta - 7, octa - 8, nona - 9, deca - 10. For the second element, we change the ending to - ide.

a. \(N_2O\): There are 2 nitrogen (\(N\)) atoms and 1 oxygen (\(O\)) atom. The prefix for 2 is di - and for 1 is mono -. So it is dinitrogen monoxide.

b. \(NO_2\): 1 nitrogen (\(N\)) atom and 2 oxygen (\(O\)) atoms. Prefix for 1 is mono - (but we usually omit mono - for the first element if there is only one), and for 2 is di -. So it is nitrogen dioxide.

c. \(N_2O_3\): 2 nitrogen (\(N\)) atoms and 3 oxygen (\(O\)) atoms. Prefixes di - and tri -. So it is dinitrogen trioxide.

d. \(NO\): 1 nitrogen (\(N\)) and 1 oxygen (\(O\)). Prefixes mono - (omitted for first element) and mono -. So it is nitrogen monoxide.

e. \(S_2Cl_2\): 2 sulfur (\(S\)) and 2 chlorine (\(Cl\)) atoms. Prefixes di - for both. So it is disulfur dichloride.

f. \(SO_2\): 1 sulfur (\(S\)) and 2 oxygen (\(O\)) atoms. Prefixes mono - (omitted) and di -. So it is sulfur dioxide.

g. \(S_2O_7\): 2 sulfur (\(S\)) and 7 oxygen (\(O\)) atoms. Prefixes di - and hepta -. So it is disulfur heptoxide.

h. \(S_2O\): 2 sulfur (\(S\)) and 1 oxygen (\(O\)) atoms. Prefixes di - and mono -. So it is disulfur monoxide.

Answer:

a. dinitrogen monoxide
b. nitrogen dioxide
c. dinitrogen trioxide
d. nitrogen monoxide
e. disulfur dichloride
f. sulfur dioxide
g. disulfur heptoxide
h. disulfur monoxide