posted by MELISSA .
THIS IS THE 3RD TIME POSTING THIS. PLEASE HELP ME ANSWER THIS. THE HOMEWORK IS DUE TONIGHT.
Write formulas for the compounds that form from Sr and each of the following polyatomic ion: NO−3, SO2−4, PO3−4.
THE -3, -4, -4 ARE THE CHARGES, CANT TYPE SUBSCRIPT ON HERE.
WHAT I TRIED IS SrNO6,SrSO4,Sr3(PO4)2 AND WRONG
OK... When writing formulas of neutral compounds the sum of the charges of the ions (monatomic or polyatomic) must equal zero. NaCl is 1:1 ratio b/c Na => +1 and Cl => -1. (+1)+(-1)=0.
Now, Strontium is a Group II cation which always has a +2 charge in Ionic Compounds;i.e., metal cation + non-metal anion. So,
1. (Sr+2) + (NO3-1) requires 2 NO3- ions so sum of charges = 0.
(Sr+2) + 2(NO3-1)=> Sr(NO3)2 => (+2)+2(-1) = (+2)+(-2) = 0 Strontium Nitrate is a 1 to 2 ion ratio.
2. First, the sulfate ion only carries a -2 charge not a -4. This might be confusing b/c when writing the ionic compound formula to get Sum Chgs = 0, some people write 2(Sr+2) + 2(SO2-2) in order to => sum = 0. This would lead one to assume that the SO4 ion would be -4, but for ionic compounds, the final form of the formula needs to have subscripts reduced to the lowest whole number ratios. It is much simpler to just recognizs (Sr+2)+(SO4-2) => (+2)+(-2)=0 and use a 1 to 1 combination ratio => SrSO4
3.The phosphate ion does not carry a -4 charge but a -3 charge. Combining the ions so as to have sum = 0 => (Sr+2)+(PO4-3) requires 3(Sr+2)+2(PO4-3) => (Sr)3(PO4)2 Strontium Phosphate.
Also, when writing formulas of ionic compounds, the 'polyions' such as the nitrate ion (NO3-), sulfate ion (SO4)-2 and phosphate ion (PO4-3) need to be in parenthesis so the subscripts outside the parenthesis indicate that you only have '2' ions of the kind as in using PO-3 => Calcium Phosphate => 3(Ca+2)+2(PO4-3)=> Ca3(PO4)2 ... the symbology indicates there are only two Phosphate ions in a formula molecule of the compound. Leaving 'off' the parenthesis => Ca2PO8 which says there are 2Ca, 1P and 8Oxys ... Such a compound does not exist.