QUESTION IMAGE
Question
- cabr₂ 45. ba₃p₂ name to formula to find the formula you must first determine each ions charge. write the formulas for these binary ionic compounds. 51. k₂se 52. becl₂ 53. rubidium sulfide 54. mercury(ii) oxide 55. calcium nitride 56. zinc acetate 57. uranium(vi) fluoride 58. silver phosphide 59. copper (i) nitride 60. magnesium hydroxide 61. cesium oxalate 62. lead(ii) nitrite 63. calcium phosphate 64. tin(iv) fluoride 65. iron(ii) oxide 66. iron(iii) oxide
Step1: Determine ion charges
For each compound, identify the charges of the cation and anion. For example, in magnesium hydroxide, magnesium ($Mg$) has a +2 charge ($Mg^{2 + }$) and hydroxide ($OH^-$) has a - 1 charge.
Step2: Apply charge - balance rule
The formula of an ionic compound is written such that the total positive charge equals the total negative charge. For magnesium hydroxide, to balance the +2 charge of $Mg^{2+}$, we need 2 $OH^-$ ions.
Step3: Write chemical formulas
- Rubidium has a +1 charge ($Rb^+$) and sulfide has a - 2 charge ($S^{2 - }$). So the formula is $Rb_2S$.
- Mercury(II) has a +2 charge ($Hg^{2+}$) and oxide has a - 2 charge ($O^{2 - }$), formula is $HgO$.
- Calcium has a +2 charge ($Ca^{2+}$) and nitride has a - 3 charge ($N^{3 - }$), formula is $Ca_3N_2$.
- Zinc has a +2 charge ($Zn^{2+}$) and acetate ($C_2H_3O_2^-$) has a - 1 charge, formula is $Zn(C_2H_3O_2)_2$.
- Uranium(VI) has a +6 charge ($U^{6+}$) and fluoride has a - 1 charge ($F^-$), formula is $UF_6$.
- Silver has a +1 charge ($Ag^+$) and phosphide has a - 3 charge ($P^{3 - }$), formula is $Ag_3P$.
- Copper(I) has a +1 charge ($Cu^+$) and nitride has a - 3 charge ($N^{3 - }$), formula is $Cu_3N$.
- Magnesium has a +2 charge ($Mg^{2+}$) and hydroxide ($OH^-$) has a - 1 charge, formula is $Mg(OH)_2$.
- Cesium has a +1 charge ($Cs^+$) and oxalate ($C_2O_4^{2 - }$) has a - 2 charge, formula is $Cs_2C_2O_4$.
- Lead(II) has a +2 charge ($Pb^{2+}$) and nitrite ($NO_2^-$) has a - 1 charge, formula is $Pb(NO_2)_2$.
- Calcium has a +2 charge ($Ca^{2+}$) and phosphate ($PO_4^{3 - }$) has a - 3 charge, formula is $Ca_3(PO_4)_2$.
- Tin(IV) has a +4 charge ($Sn^{4+}$) and fluoride has a - 1 charge ($F^-$), formula is $SnF_4$.
- Iron(II) has a +2 charge ($Fe^{2+}$) and oxide has a - 2 charge ($O^{2 - }$), formula is $FeO$.
- Iron(III) has a +3 charge ($Fe^{3+}$) and oxide has a - 2 charge ($O^{2 - }$), formula is $Fe_2O_3$.
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- $Rb_2S$
- $HgO$
- $Ca_3N_2$
- $Zn(C_2H_3O_2)_2$
- $UF_6$
- $Ag_3P$
- $Cu_3N$
- $Mg(OH)_2$
- $Cs_2C_2O_4$
- $Pb(NO_2)_2$
- $Ca_3(PO_4)_2$
- $SnF_4$
- $FeO$
- $Fe_2O_3$