an ionic compound is formed between\nd. a non - metallic cation and a metallic anion\nb. a metallic cation…
an ionic compound is formed between\nd. a non - metallic cation and a metallic anion\nb. a metallic cation and a non - metallic anion\na. a non - metal and a non - metal\nc. a metalloid and a non - metal\nquestion 2\nwhich of the following best describe the formula for the following compounds lithium nitrate, lithium nitrite\nb. lino₃, lino₂\nc. li₃n, lino₂\ne. li₃n, lino₃\na. lino₂, lino₃\nd. lino₃, lin₃\nquestion 3\nwhile naming ionic compounds, roman numerals are used\nb. to show the number of non - metal atoms present\na. to show the charge of the metal cation\nd. only in binary ionic compounds\nc. to show the number of metal atoms present\nquestion 4\nif the compound formed by radium and bromine has the formula rabr₂, what is the formula of the compound formed by strontium (sr) and iodine (i)?
Answer
Brief Explanations:
- Question 1: Ionic compounds are formed by the transfer of electrons from a metal (which forms a cation) to a non - metal (which forms an anion). So, a metallic cation and a non - metallic anion form an ionic compound.
- Question 2: The formula for lithium nitrate is $LiNO_3$ and for lithium nitrite is $LiNO_2$. Nitrate has the polyatomic ion $NO_3^-$ and nitrite has $NO_2^-$.
- Question 3: Roman numerals are used in naming ionic compounds to show the charge of the metal cation, especially when the metal can have multiple oxidation states.
- Question 4: Radium (Ra) and strontium (Sr) are in the same group (Group 2, alkaline earth metals) and have a + 2 charge. Bromine (Br) and iodine (I) are in Group 17 and have a - 1 charge. So, the formula of the compound formed by Sr and I is $SrI_2$.
Answer:
- Question 1: b. a metallic cation and a non - metallic anion
- Question 2: b. $LiNO_3$, $LiNO_2$
- Question 3: a. to show the charge of the metal cation
- Question 4: $SrI_2$