Question

2 what is the charge on the cation in each compound below?
a pdcl₂
b al(oh)₃
c feco₃
d pb(so₄)₂
e bi₂o₃

Explanation:

Step1: Recall chloride ion charge

Chloride ion ($Cl^-$) has a charge of - 1. In $PdCl_2$, let the charge of $Pd$ be $x$. Since the compound is neutral, $x + 2\times(-1)=0$, so $x = + 2$.

Step2: Recall hydroxide ion charge

Hydroxide ion ($OH^-$) has a charge of - 1. In $Al(OH)_3$, let the charge of $Al$ be $y$. Since the compound is neutral, $y+3\times(-1) = 0$, so $y=+3$.

Step3: Recall carbonate ion charge

Carbonate ion ($CO_3^{2 -}$) has a charge of - 2. In $FeCO_3$, let the charge of $Fe$ be $z$. Since the compound is neutral, $z+( - 2)=0$, so $z = + 2$.

Step4: Recall sulfate ion charge

Sulfate ion ($SO_4^{2 -}$) has a charge of - 2. In $Pb(SO_4)_2$, let the charge of $Pb$ be $w$. Since the compound is neutral, $w + 2\times(-2)=0$, so $w=+4$.

Step5: Recall oxide ion charge

Oxide ion ($O^{2 -}$) has a charge of - 2. In $Bi_2O_3$, let the charge of $Bi$ be $u$. Since the compound is neutral, $2u+3\times(-2)=0$, solving for $u$ gives $u = + 3$.

Answer:

a. + 2
b. + 3
c. + 2
d. + 4
e. + 3