molecular (covalent) compounds\nwhen non - metal atoms share electrons with other non - metal atoms, they…

molecular (covalent) compounds\nwhen non - metal atoms share electrons with other non - metal atoms, they form basic units called molecules. compound composed of molecules are thus molecular (covalent) compounds. we shall learn the nomenclature of binary covalent compounds.\nexercise: define a binary compound\ncase 4. binary covalent compounds\nas with ionic compounds, the second atom in the formula gets an - ide ending.\nunlike ionic compounds, the number of each atom is specified with a prefix (as there are no ions in a molecule)\nthese are the prefixes you need to know:\n1 is mono-\n2 is di-\n3 is tri-\n4 is tetra-\n5 is penta-\n6 is hexa-\n7 is hepta-\n8 is octa-\n9 is nona-\n10 is deca-\nnote: when there is only one atom of the first element, the mono prefix is omitted.\nexamples: nf3 nitrogen trifluoride\np4o10 tetraphosphorus decaoxide\ni2s diiodine monosulfide\nb2cl6 diboron hexachloride\nexercise: name the following:\nn2o4\npcl5\nso2\nso3\ncs2\nbr2o7\nco\np2o3\ncl2o\nsf6\nexercise: give formulas for the following:\niodine pentabromide\nchlorine dibromide\noxygen difluoride\ncarbon tetrachloride\nsulfur hexafluoride\nsilicon dioxide\niodine heptafluoride\nnitrogen monoxide\ndinitrogen trioxide\nalso: you must remember these three very important common names:\nh2o (water)\nnh3 (ammonia)\nch4 (methane)

molecular (covalent) compounds\nwhen non - metal atoms share electrons with other non - metal atoms, they form basic units called molecules. compound composed of molecules are thus molecular (covalent) compounds. we shall learn the nomenclature of binary covalent compounds.\nexercise: define a binary compound\ncase 4. binary covalent compounds\nas with ionic compounds, the second atom in the formula gets an - ide ending.\nunlike ionic compounds, the number of each atom is specified with a prefix (as there are no ions in a molecule)\nthese are the prefixes you need to know:\n1 is mono-\n2 is di-\n3 is tri-\n4 is tetra-\n5 is penta-\n6 is hexa-\n7 is hepta-\n8 is octa-\n9 is nona-\n10 is deca-\nnote: when there is only one atom of the first element, the mono prefix is omitted.\nexamples: nf3 nitrogen trifluoride\np4o10 tetraphosphorus decaoxide\ni2s diiodine monosulfide\nb2cl6 diboron hexachloride\nexercise: name the following:\nn2o4\npcl5\nso2\nso3\ncs2\nbr2o7\nco\np2o3\ncl2o\nsf6\nexercise: give formulas for the following:\niodine pentabromide\nchlorine dibromide\noxygen difluoride\ncarbon tetrachloride\nsulfur hexafluoride\nsilicon dioxide\niodine heptafluoride\nnitrogen monoxide\ndinitrogen trioxide\nalso: you must remember these three very important common names:\nh2o (water)\nnh3 (ammonia)\nch4 (methane)

Answer

Explanation:

Step1: Define binary compound

A binary compound is a compound composed of two different elements.

Step2: Name $N_2O_4$

There are 2 nitrogen atoms (di - prefix) and 4 oxygen atoms (tetra - prefix), so it is dinitrogen tetroxide.

Step3: Name $PCl_5$

There is 1 phosphorus atom (prefix omitted) and 5 chlorine atoms (penta - prefix), so it is phosphorus pentachloride.

Step4: Name $SO_2$

There is 1 sulfur atom (prefix omitted) and 2 oxygen atoms (di - prefix), so it is sulfur dioxide.

Step5: Name $SO_3$

There is 1 sulfur atom (prefix omitted) and 3 oxygen atoms (tri - prefix), so it is sulfur trioxide.

Step6: Name $CS_2$

There is 1 carbon atom (prefix omitted) and 2 sulfur atoms (di - prefix), so it is carbon disulfide.

Step7: Name $Br_2O_7$

There are 2 bromine atoms (di - prefix) and 7 oxygen atoms (hepta - prefix), so it is dibromine heptoxide.

Step8: Name $CO$

There is 1 carbon atom (prefix omitted) and 1 oxygen atom (mono - prefix for oxygen), so it is carbon monoxide.

Step9: Name $P_2O_3$

There are 2 phosphorus atoms (di - prefix) and 3 oxygen atoms (tri - prefix), so it is diphosphorus trioxide.

Step10: Name $Cl_2O$

There are 2 chlorine atoms (di - prefix) and 1 oxygen atom (mono - prefix for oxygen), so it is dichlorine monoxide.

Step11: Name $SF_6$

There is 1 sulfur atom (prefix omitted) and 6 fluorine atoms (hexa - prefix), so it is sulfur hexafluoride.

Step12: Give formula for iodine pentabromide

Iodine is $I$, pentabromide means 5 bromine atoms, so the formula is $IBr_5$.

Step13: Give formula for chlorine dibromide

Chlorine is $Cl$, dibromide means 2 bromine atoms, so the formula is $ClBr_2$.

Step14: Give formula for oxygen difluoride

Oxygen is $O$, difluoride means 2 fluorine atoms, so the formula is $OF_2$.

Step15: Give formula for carbon tetrachloride

Carbon is $C$, tetrachloride means 4 chlorine atoms, so the formula is $CCl_4$.

Step16: Give formula for sulfur hexafluoride

Sulfur is $S$, hexafluoride means 6 fluorine atoms, so the formula is $SF_6$.

Step17: Give formula for silicon dioxide

Silicon is $Si$, dioxide means 2 oxygen atoms, so the formula is $SiO_2$.

Step18: Give formula for iodine heptafluoride

Iodine is $I$, heptafluoride means 7 fluorine atoms, so the formula is $IF_7$.

Step19: Give formula for nitrogen monoxide

Nitrogen is $N$, monoxide means 1 oxygen atom, so the formula is $NO$.

Step20: Give formula for dinitrogen trioxide

Dinitrogen means 2 nitrogen atoms ($N_2$), trioxide means 3 oxygen atoms, so the formula is $N_2O_3$.

Answer:

Define a binary compound: A compound composed of two different elements. $N_2O_4$: Dinitrogen tetroxide $PCl_5$: Phosphorus pentachloride $SO_2$: Sulfur dioxide $SO_3$: Sulfur trioxide $CS_2$: Carbon disulfide $Br_2O_7$: Dibromine heptoxide $CO$: Carbon monoxide $P_2O_3$: Diphosphorus trioxide $Cl_2O$: Dichlorine monoxide $SF_6$: Sulfur hexafluoride Iodine pentabromide: $IBr_5$ Chlorine dibromide: $ClBr_2$ Oxygen difluoride: $OF_2$ Carbon tetrachloride: $CCl_4$ Sulfur hexafluoride: $SF_6$ Silicon dioxide: $SiO_2$ Iodine heptafluoride: $IF_7$ Nitrogen monoxide: $NO$ Dinitrogen trioxide: $N_2O_3$