Write the formulas for the following compounds: (a) copper(l) cyanide, (b) strontium chlorite, (c) perbromic acid,(d) hydroiodic acid, (e) disodium ammonium phosphate, (f) potassium dihydrogen phosphate, (g) iodine heptafluoride, (h) tetraphosphorus decasulfide, (i) mercury(ll) oxide, (j) mercury(l) iodide, (k) selenium hexafluoride.
The Correct Answer and Explanation is :
Here are the chemical formulas for the compounds you’ve listed, along with a brief explanation of how each formula is derived:
- Copper(I) cyanide
Formula: ( \text{CuCN} )
Copper(I) has a +1 charge, and the cyanide ion (CN(^-)) has a -1 charge. To balance the charges, one copper ion combines with one cyanide ion to form copper(I) cyanide. - Strontium chlorite
Formula: ( \text{Sr(ClO}_2\text{)}_2 )
Strontium (Sr) has a +2 charge, while chlorite (ClO(_2)^-) has a -1 charge. To balance the charges, two chlorite ions are needed for each strontium ion, resulting in ( \text{Sr(ClO}_2\text{)}_2 ). - Perbromic acid
Formula: ( \text{HBrO}_4 )
Perbromic acid contains the perbromate ion (BrO(_4)^-) with a -1 charge. The hydrogen ion (H(^+)) balances this charge, so the formula is ( \text{HBrO}_4 ). - Hydroiodic acid
Formula: ( \text{HI} )
Hydroiodic acid consists of hydrogen (H(^+)) and iodide (I(^-)). The charges balance in a 1:1 ratio, so the formula is ( \text{HI} ). - Disodium ammonium phosphate
Formula: ( \text{Na}_2\text{NH}_4\text{PO}_4 )
The ammonium ion (NH(_4)^+) is paired with phosphate (PO(_4)^{3-}). To balance charges, two sodium ions (Na(^+)) are needed for each ammonium ion, giving the formula ( \text{Na}_2\text{NH}_4\text{PO}_4 ). - Potassium dihydrogen phosphate
Formula: ( \text{KH}_2\text{PO}_4 )
Potassium (K(^+)) combines with the dihydrogen phosphate ion (H(_2)PO(_4)^-) to form the neutral compound ( \text{KH}_2\text{PO}_4 ). - Iodine heptafluoride
Formula: ( \text{IF}_7 )
Iodine forms a +7 oxidation state in iodine heptafluoride, while fluoride (F(^-)) has a -1 charge. Seven fluoride ions balance the iodine ion, so the formula is ( \text{IF}_7 ). - Tetraphosphorus decasulfide
Formula: ( \text{P}4\text{S}{10} )
This compound consists of four phosphorus atoms and ten sulfur atoms, so its formula is ( \text{P}4\text{S}{10} ). - Mercury(II) oxide
Formula: ( \text{HgO} )
Mercury(II) has a +2 charge, and oxygen has a -2 charge. The charges balance in a 1:1 ratio, so the formula is ( \text{HgO} ). - Mercury(I) iodide
Formula: ( \text{Hg}_2\text{I}_2 )
Mercury(I) exists as a diatomic ion, ( \text{Hg}_2^{2+} ). Iodide has a -1 charge, so two iodide ions are needed to balance the mercury(I) cation, resulting in ( \text{Hg}_2\text{I}_2 ). - Selenium hexafluoride
Formula: ( \text{SeF}_6 )
Selenium hexafluoride consists of one selenium atom and six fluoride ions. The formula is ( \text{SeF}_6 ), where the selenium atom is in the +6 oxidation state.
Summary of Charges and Formula Construction:
- Copper(I) cyanide (CuCN): Copper(I) has +1 charge; cyanide has -1.
- Strontium chlorite (Sr(ClO₂)₂): Strontium has +2; chlorite has -1.
- Perbromic acid (HBrO₄): Hydrogen +1; perbromate has -1.
- Hydroiodic acid (HI): Hydrogen +1; iodide has -1.
- Disodium ammonium phosphate (Na₂NH₄PO₄): Sodium +1; ammonium +1; phosphate -3.
- Potassium dihydrogen phosphate (KH₂PO₄): Potassium +1; dihydrogen phosphate -1.
- Iodine heptafluoride (IF₇): Iodine +7; fluoride -1.
- Tetraphosphorus decasulfide (P₄S₁₀): Phosphorus and sulfur in fixed ratios.
- Mercury(II) oxide (HgO): Mercury(II) +2; oxide -2.
- Mercury(I) iodide (Hg₂I₂): Mercury(I) exists as Hg₂²⁺; iodide -1.
- Selenium hexafluoride (SeF₆): Selenium +6; fluoride -1.
Understanding oxidation states and charge balance is essential when deriving these formulas.