Question

ionic compound has the formula mnbr₂. complete the table to determine the charge of the transition metal ion in this compound. enter a negative sign before negative numbers. positive numbers do not need a positive sign.
element | charge of one ion | total charge of ion(s) in compound | number of ion(s) in compound
manganese | blank | blank | blank
bromine | blank | blank | blank
*in one formula unit
what is the name of this ionic compound? blanks with dropdowns

Explanation:

Step1: Analyze Bromine ion

Bromine (Br) is a halogen, so its ion has a charge of $-1$. In $\ce{MnBr_{2}}$, the number of Br ions is 2. So total charge of Br ions: $(-1) \times 2 = -2$.

Step2: Determine Manganese ion charge

For the compound to be neutral, total charge of Mn ion must balance Br's total charge. Let charge of Mn ion be $x$. Number of Mn ions is 1. So $x\times1 + (-2) = 0 \implies x = +2$.

Step3: Name the compound

The metal is Manganese (Mn) with charge +2, and the non - metal is Bromine, which forms bromide ions. So the name is Manganese(II) bromide.

Filling the table:

• Manganese:
• Charge of One Ion: $2$
• Total Charge of Ion(s) in Compound: $2$ (since $1\times2 = 2$)
• Number of Ion(s) in Compound: $1$
• Bromine:
• Charge of One Ion: $-1$
• Total Charge of Ion(s) in Compound: $-2$ (since $2\times(-1)= - 2$)
• Number of Ion(s) in Compound: $2$

Answer:

Table Filling:

• Manganese: Charge of One Ion = $2$, Total Charge = $2$, Number of Ions = $1$
• Bromine: Charge of One Ion = $-1$, Total Charge = $-2$, Number of Ions = $2$

Name of the compound: Manganese(II) bromide