Write the chemical formulas for the following ionic compounds.
tantalum cyanide
potassium iodide
radium selenide
manganese (IV) nitride
calcium fluoride
zinc oxide
osmium (VIII) carbide
cesium silicide
cadmium cyanide
zinc hydroxide
tungsten (IV) oxalate
thallium (III) arsenide
chromium (VI) sulfide
lithium nitrite
The correct answer and explanation is :
Here are the correct chemical formulas for the given ionic compounds:
- Tantalum cyanide – Ta(CN)₅ (assuming tantalum has a +5 oxidation state)
- Potassium iodide – KI
- Radium selenide – RaSe
- Manganese (IV) nitride – Mn₃N₄
- Calcium fluoride – CaF₂
- Zinc oxide – ZnO
- Osmium (VIII) carbide – OsC₄ (assuming osmium has a +8 oxidation state and carbide as C⁴⁻)
- Cesium silicide – Cs₄Si (assuming silicide as Si⁴⁻)
- Cadmium cyanide – Cd(CN)₂
- Zinc hydroxide – Zn(OH)₂
- Tungsten (IV) oxalate – W(C₂O₄)₂
- Thallium (III) arsenide – TlAs
- Chromium (VI) sulfide – CrS₃
- Lithium nitrite – LiNO₂
Explanation (300 words):
Ionic compounds are formed when metals transfer electrons to nonmetals, creating positively charged cations and negatively charged anions. The formula of an ionic compound must be electrically neutral, meaning the total positive and negative charges must balance.
For binary ionic compounds (e.g., potassium iodide, calcium fluoride, zinc oxide), the metal donates electrons to the nonmetal. The charges on the individual ions determine the formula. For example, CaF₂ consists of Ca²⁺ and F⁻ ions. Since calcium has a +2 charge and fluoride has a -1 charge, two fluoride ions are required to balance the charge.
For compounds with transition metals (e.g., manganese (IV) nitride, chromium (VI) sulfide), the Roman numeral indicates the oxidation state of the metal. In Mn₃N₄, manganese (IV) has a +4 charge, and nitride (N³⁻) has a -3 charge. The least common multiple of 4 and 3 is 12, requiring three Mn⁴⁺ and four N³⁻ to balance.
For polyatomic ionic compounds (e.g., tungsten (IV) oxalate, lithium nitrite), polyatomic ions such as oxalate (C₂O₄²⁻) and nitrite (NO₂⁻) must be included in the correct ratio. W(C₂O₄)₂ has tungsten (IV) with a +4 charge and oxalate with a -2 charge, requiring two oxalate ions.
Understanding the charge interactions and oxidation states helps determine the correct formula, ensuring electrical neutrality.