naming ionic compounds practice worksheet\nname the following ionic compounds:\n) nh₄cl\n)…

naming ionic compounds practice worksheet\nname the following ionic compounds:\n) nh₄cl\n) fe(no₃)₃\ntibr₃\ncu₃p\nsnse₂\ngaas\npb(so₄)₂\nbe(hco₃)₂\nwrite the formulas for the following compounds:\nchromium (vi) phosphate\nvanadium (iv) carbonate\ncobalt (iii) oxide\ntitanium (ii) acetate\nvanadium (v) sulfide\nchromium (iii) hydroxide\nthallium iodide\nlead (ii) nitride\nsilver bromide\nlots of ionic naming practice problems\nname the following ionic compounds:\n1) nabr\n2) sc(oh)₃\n3) v₂(so₄)₃\n4) nh₄f\n5) caco₃\n6) nipo₄\n7)\n8) zn₃p₂\n9) sr(c₂h₃o₂)₂\n10) cu₂o\n11) ag₃po₄\n12) yclo₃\n13) sns₂\n14)\n15)\n16) pb₃n₂\n17) coco₃\n18)\n19)\n20) fe(hco₃)₂
Answer
Explanation:
Step1: Recall naming rules for ionic compounds
For naming ionic compounds, we name the cation first followed by the anion. For poly - atomic anions, we use their specific names. For metals with variable oxidation states, we indicate the oxidation state in Roman numerals in parentheses.
Step2: Name $NH_4Cl$
The cation is ammonium ($NH_4^+$) and the anion is chloride ($Cl^-$), so the name is ammonium chloride.
Step3: Name $Fe(NO_3)_3$
The cation is iron(III) as the oxidation state of iron is + 3 (since nitrate is $NO_3^-$ and there are 3 of them, so iron must have a + 3 charge to balance the - 3 charge from the nitrates). The anion is nitrate. So the name is iron(III) nitrate.
Step4: Name $TiBr_3$
The cation is titanium(III) (as titanium has an oxidation state of + 3 since there are 3 bromide ions, $Br^-$) and the anion is bromide. So the name is titanium(III) bromide.
Step5: Name $Cu_3P$
The cation is copper(I) (as there are 3 copper atoms and 1 phosphide ion $P^{3 - }$, so each copper has a + 1 charge). The anion is phosphide. So the name is copper(I) phosphide.
Step6: Name $SnSe_2$
The cation is tin(IV) (since there are 2 selenide ions $Se^{2 - }$, so tin has a + 4 charge to balance the - 4 charge from the selenides). The anion is selenide. So the name is tin(IV) selenide.
Step7: Name $GaAs$
The cation is gallium and the anion is arsenide. So the name is gallium arsenide.
Step8: Name $Pb(SO_4)_2$
The cation is lead(IV) (as sulfate is $SO_4^{2 - }$ and there are 2 of them, so lead has a + 4 charge to balance the - 4 charge from the sulfates). The anion is sulfate. So the name is lead(IV) sulfate.
Step9: Name $Be(HCO_3)_2$
The cation is beryllium and the anion is bicarbonate. So the name is beryllium bicarbonate.
Step10: Name $NaBr$
The cation is sodium and the anion is bromide. So the name is sodium bromide.
Step11: Name $Sc(OH)_3$
The cation is scandium(III) (as there are 3 hydroxide ions $OH^-$) and the anion is hydroxide. So the name is scandium(III) hydroxide.
Step12: Name $V_2(SO_4)_3$
The cation is vanadium(III) (since sulfate is $SO_4^{2 - }$ and there are 3 of them with a total charge of - 6, and there are 2 vanadium atoms, so each vanadium has a + 3 charge). The anion is sulfate. So the name is vanadium(III) sulfate.
Step13: Name $NH_4F$
The cation is ammonium and the anion is fluoride. So the name is ammonium fluoride.
Step14: Name $CaCO_3$
The cation is calcium and the anion is carbonate. So the name is calcium carbonate.
Step15: Name $NiPO_4$
The cation is nickel(III) (as phosphate is $PO_4^{3 - }$, so nickel has a + 3 charge to balance the - 3 charge from the phosphate). The anion is phosphate. So the name is nickel(III) phosphate.
Step16: Name $Zn_3P_2$
The cation is zinc and the anion is phosphide. So the name is zinc phosphide.
Step17: Name $Sr(C_2H_3O_2)_2$
The cation is strontium and the anion is acetate. So the name is strontium acetate.
Step18: Name $Cu_2O$
The cation is copper(I) (as there are 2 copper atoms and 1 oxide ion $O^{2 - }$, so each copper has a + 1 charge). The anion is oxide. So the name is copper(I) oxide.
Step19: Name $Ag_3PO_4$
The cation is silver and the anion is phosphate. So the name is silver phosphate.
Step20: Name $YClO_3$
The cation is yttrium(III) (as chlorate is $ClO_3^-$ and yttrium has a + 3 charge to balance the - 1 charge from the chlorate in a neutral compound). The anion is chlorate. So the name is yttrium(III) chlorate.
Step21: Name $SnS_2$
The cation is tin(IV) (since there are 2 sulfide ions $S^{2 - }$, so tin has a + 4 charge to balance the - 4 charge from the sulfides). The anion is sulfide. So the name is tin(IV) sulfide.
Step22: Name $Pb_3N_2$
The cation is lead(II) (as there are 3 lead atoms and 2 nitride ions $N^{3 - }$, so each lead has a + 2 charge to balance the - 6 charge from the nitrides). The anion is nitride. So the name is lead(II) nitride.
Step23: Name $CoCO_3$
The cation is cobalt(II) (as carbonate is $CO_3^{2 - }$, so cobalt has a + 2 charge to balance the - 2 charge from the carbonate). The anion is carbonate. So the name is cobalt(II) carbonate.
Step24: Name $Fe(HCO_3)_2$
The cation is iron(II) (as there are 2 bicarbonate ions $HCO_3^-$ and iron has a + 2 charge to balance the - 2 charge from the bicarbonates). The anion is bicarbonate. So the name is iron(II) bicarbonate.
For writing formulas:
Step1: Chromium(VI) phosphate
Chromium has a + 6 charge ($Cr^{6+}$) and phosphate is $PO_4^{3 - }$. To balance the charges, we need 2 phosphate ions for 1 chromium ion. The formula is $Cr(PO_4)_2$.
Step2: Vanadium(IV) carbonate
Vanadium has a + 4 charge ($V^{4+}$) and carbonate is $CO_3^{2 - }$. To balance the charges, we need 2 carbonate ions for 1 vanadium ion. The formula is $V(CO_3)_2$.
Step3: Cobalt(III) oxide
Cobalt has a + 3 charge ($Co^{3+}$) and oxide is $O^{2 - }$. To balance the charges, we need 2 cobalt ions and 3 oxide ions. The formula is $Co_2O_3$.
Step4: Titanium(II) acetate
Titanium has a + 2 charge ($Ti^{2+}$) and acetate is $C_2H_3O_2^-$. To balance the charges, we need 2 acetate ions for 1 titanium ion. The formula is $Ti(C_2H_3O_2)_2$.
Step5: Vanadium(V) sulfide
Vanadium has a + 5 charge ($V^{5+}$) and sulfide is $S^{2 - }$. To balance the charges, we need 2 vanadium ions and 5 sulfide ions. The formula is $V_2S_5$.
Step6: Chromium(III) hydroxide
Chromium has a + 3 charge ($Cr^{3+}$) and hydroxide is $OH^-$. To balance the charges, we need 3 hydroxide ions for 1 chromium ion. The formula is $Cr(OH)_3$.
Step7: Thallium iodide
Assuming thallium in its common + 1 oxidation state ($Tl^+$) and iodide is $I^-$, the formula is $TlI$.
Step8: Lead(II) nitride
Lead has a + 2 charge ($Pb^{2+}$) and nitride is $N^{3 - }$. To balance the charges, we need 3 lead ions and 2 nitride ions. The formula is $Pb_3N_2$.
Step9: Silver bromide
Silver has a + 1 charge ($Ag^+$) and bromide is $Br^-$. The formula is $AgBr$.
Answer:
- ammonium chloride
- iron(III) nitrate
- titanium(III) bromide
- copper(I) phosphide
- tin(IV) selenide
- gallium arsenide
- lead(IV) sulfate
- beryllium bicarbonate
- sodium bromide
- scandium(III) hydroxide
- vanadium(III) sulfate
- ammonium fluoride
- calcium carbonate
- nickel(III) phosphate
- zinc phosphide
- strontium acetate
- copper(I) oxide
- silver phosphate
- yttrium(III) chlorate
- tin(IV) sulfide
- lead(II) nitride
- cobalt(II) carbonate
- iron(II) bicarbonate Formulas:
- $Cr(PO_4)_2$
- $V(CO_3)_2$
- $Co_2O_3$
- $Ti(C_2H_3O_2)_2$
- $V_2S_5$
- $Cr(OH)_3$
- $TlI$
- $Pb_3N_2$
- $AgBr$