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write formulas for the following type i compounds: 5. calcium iodide 6.…

Question

write formulas for the following type i compounds:

  1. calcium iodide
  2. sodium sulfide
  3. aluminum oxide
  4. beryllium bromide

name the following type ii compounds:

  1. fef₃
  2. cro₃
  3. vcl₃
  4. tio₂

write formulas for the following type ii compounds:

  1. molybdenum (iv) oxide
  2. iron (iii) sulfide
  3. nickel (iv) chloride
  4. chromium (vi) fluoride

name the following type iii compounds:

  1. n₂o₅
  2. pcl₅
  3. of₂

Explanation:

Step1: Determine charges for type I compounds

Type I compounds are between metals and non - metals where metal has a fixed charge. Calcium has a +2 charge ($Ca^{2+}$) and iodine has a -1 charge ($I^-$), so calcium iodide is $CaI_2$. Sodium has a +1 charge ($Na^+$) and sulfur has a -2 charge ($S^{2 -}$), so sodium sulfide is $Na_2S$. Aluminum has a +3 charge ($Al^{3+}$) and oxygen has a -2 charge ($O^{2 -}$), so aluminum oxide is $Al_2O_3$. Beryllium has a +2 charge ($Be^{2+}$) and bromine has a -1 charge ($Br^-$), so beryllium bromide is $BeBr_2$.

Step2: Name type II compounds

Type II compounds involve metals with variable charges. For $FeF_3$, iron has a +3 charge and it is iron(III) fluoride. For $CrO_3$, chromium has a +6 charge and it is chromium(VI) oxide. For $VCl_3$, vanadium has a +3 charge and it is vanadium(III) chloride. For $TiO_2$, titanium has a +4 charge and it is titanium(IV) oxide.

Step3: Write formulas for type II compounds

Molybdenum(IV) has a +4 charge ($Mo^{4+}$) and oxygen has a -2 charge ($O^{2 -}$), so molybdenum(IV) oxide is $MoO_2$. Iron(III) has a +3 charge ($Fe^{3+}$) and sulfur has a -2 charge ($S^{2 -}$), so iron(III) sulfide is $Fe_2S_3$. Nickel(IV) has a +4 charge ($Ni^{4+}$) and chlorine has a -1 charge ($Cl^-$), so nickel(IV) chloride is $NiCl_4$. Chromium(VI) has a +6 charge ($Cr^{6+}$) and fluorine has a -1 charge ($F^-$), so chromium(VI) fluoride is $CrF_6$.

Step4: Name type III compounds

Type III compounds are between non - metals. $N_2O_5$ is dinitrogen pentoxide. $PCl_5$ is phosphorus pentachloride. $OF_2$ is oxygen difluoride.

Answer:

  1. $CaI_2$
  2. $Na_2S$
  3. $Al_2O_3$
  4. $BeBr_2$
  5. Iron(III) fluoride
  6. Chromium(VI) oxide
  7. Vanadium(III) chloride
  8. Titanium(IV) oxide
  9. $MoO_2$
  10. $Fe_2S_3$
  11. $NiCl_4$
  12. $CrF_6$
  13. Dinitrogen pentoxide
  14. Phosphorus pentachloride
  15. Oxygen difluoride