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directions: name the following ionic compounds. 11. ba(so₄) 12. mgcl₂ 1…

Question

directions: name the following ionic compounds.

  1. ba(so₄)
  2. mgcl₂
  3. ca(co₃)
  4. be₃p₂
  5. al(no₃)₃
  6. cu(no₃)
  7. pb₃n₄
  8. crcl₃
  9. hg(c₂o₄)
  10. sni₄

Explanation:

Brief Explanations

To name ionic compounds, we first identify the cation (positive ion) and the anion (negative ion). For metals that can have multiple charges (like transition metals), we use Roman numerals to indicate the charge. For polyatomic ions (like \(SO_4^{2 -}\), \(NO_3^-\), \(CO_3^{2 -}\), \(C_2O_4^{2 -}\)), we use their standard names.

  1. For \(Ba(SO_4)\):
  • Cation: \(Ba^{2+}\) (barium ion)
  • Anion: \(SO_4^{2 -}\) (sulfate ion)
  • Name: Barium sulfate
  1. For \(MgCl_2\):
  • Cation: \(Mg^{2+}\) (magnesium ion)
  • Anion: \(Cl^-\) (chloride ion)
  • Name: Magnesium chloride
  1. For \(Ca(CO_3)\):
  • Cation: \(Ca^{2+}\) (calcium ion)
  • Anion: \(CO_3^{2 -}\) (carbonate ion)
  • Name: Calcium carbonate
  1. For \(Be_3P_2\):
  • Cation: \(Be^{2+}\) (beryllium ion)
  • Anion: \(P^{3 -}\) (phosphide ion)
  • Name: Beryllium phosphide
  1. For \(Al(NO_3)_3\):
  • Cation: \(Al^{3+}\) (aluminum ion)
  • Anion: \(NO_3^-\) (nitrate ion)
  • Name: Aluminum nitrate
  1. For \(Cu(NO_3)\):
  • Assume \(Cu^{+}\) (copper(I) ion, since nitrate is \(NO_3^-\))
  • Anion: \(NO_3^-\) (nitrate ion)
  • Name: Copper(I) nitrate
  1. For \(Pb_3N_4\):
  • Let charge of \(Pb\) be \(x\). Using charge balance: \(3x+4(- 3)=0\) → \(3x = 12\) → \(x = + 4\)
  • Cation: \(Pb^{4+}\) (lead(IV) ion)
  • Anion: \(N^{3 -}\) (nitride ion)
  • Name: Lead(IV) nitride
  1. For \(CrCl_3\):
  • Let charge of \(Cr\) be \(y\). Using charge balance: \(y+3(-1)=0\) → \(y = + 3\)
  • Cation: \(Cr^{3+}\) (chromium(III) ion)
  • Anion: \(Cl^-\) (chloride ion)
  • Name: Chromium(III) chloride
  1. For \(Hg(C_2O_4)\):
  • Assume \(Hg^{2+}\) (mercury(II) ion, since oxalate is \(C_2O_4^{2 -}\))
  • Anion: \(C_2O_4^{2 -}\) (oxalate ion)
  • Name: Mercury(II) oxalate
  1. For \(SnI_4\):
  • Let charge of \(Sn\) be \(z\). Using charge balance: \(z+4(-1)=0\) → \(z = + 4\)
  • Cation: \(Sn^{4+}\) (tin(IV) ion)
  • Anion: \(I^-\) (iodide ion)
  • Name: Tin(IV) iodide

Answer:

  1. Barium sulfate
  2. Magnesium chloride
  3. Calcium carbonate
  4. Beryllium phosphide
  5. Aluminum nitrate
  6. Copper(I) nitrate
  7. Lead(IV) nitride
  8. Chromium(III) chloride
  9. Mercury(II) oxalate
  10. Tin(IV) iodide