# Homework 10

Q1 1.69

Question: Determine the number of protons and electrons in each of the following ions.

(1) $\small N^{3-}$

(2)$\small Be^{2+}$

(3)$\small K^{+}$

(4)$\small Se^{2-}$

Solution:

Strategy

A. First locate each element on the periodic table. Identify the atomic number associated with each element, or the number in between the element's symbol and scientific name. The atomic number is the number of protons as well as the number of electrons an element possesses in it's neutral state.

B. Next, identify the superscript of each element, or the number to the top right of the element's symbol. This number should include a "+" or "-" sign next to it, indicating the element's charge. A "+" indicates a positively charged ion and a "-" indicates a negatively charged ion. An element becomes positively charged when it loses electrons, and negative when it gains electrons.

C. To solve for each element's total number of electrons, take that element's atomic number, and inversely add or subtract the number in the superscript according to the element's charge. If an element has a "+" sign, it is losing the amount of electrons indicated by the superscript number, so subtract this number from the element's atomic number. The same goes if an element has a "-" sign, instead adding the number in the superscript to the element's atomic number.

(1) N3-

A. Nitrogen's atomic number is 7, so in a neutral state, Nitrogen has 7 protons and electrons.

B. Nitrogen's superscript is "3-" indicating that it is gaining 3 electrons

C. Because it gains 3 electrons, N3- now has a total of 10 electrons.

(2) Be2+

A. Beryllium's atomic number is 4, so Beryllium possesses 4 protons and electrons

B. The superscript "2+" indicates beryllium is losing 2 electrons

C. 2 electrons are subtracted from Beryllium's total 4, so it now has 2 electrons.

(3) K+

A. Potassium as a neutral element possesses 19 protons and electrons, indicated by it's atomic number of 19.

B. The superscript "+" indicates Potassium will lose 1 electron from it's total.

C. 19 total electrons minus 1 makes K+ number of electrons to 18.

(4) Se2-

A. Selenium's atomic number is 34, indicating 34 protons and electrons at ground state.

B. It' superscript "2-" indicates that it will be gaining 2 electrons to it's total.

C. 34 electrons plus 2 makes selenium's total number of electrons 34.

Q2 5.60

Question: Write the name from the formula of the formula from the name for each hydrated ionic compound

(1) Iron(III) phosphate tetrahydrate

(2) CsBr * 4H2O

(3) Cobalt(II) chloride hexahydrate

(4) LiI * 3H2O

Solution:

Strategy

A. Name the cation and anion combination in the first part of the equation, or if the name is given, name the element combination with the anion first, then the cation. The anion is capitalized, while the cation remains lowercase.

B. If the compound involves a metal that forms more than one cation, the number of cations it forms goes in between parentheses in roman numeral form. If the compound does not involve a metal, than list cation charges as normal.

C. For each hydrate at the end of the compound, the prefix at the beginning indicates the number of hydrates. Name the appropriate prefix or number needed for each formula or compound name. Make sure the hydrate is lowercase.

(1) Iron(III) phosphate tetrahydrate

A. Iron phosphate as an element is written as FePO4, since both parts have a charge of 3, they are not needed in the equation.

B. The prefix tetra stands for "4", so therefore tetrahydrate as an element is written as 4H2O

C. The two parts combined make the final formula FePO4*4H2O

(2) CsBr * 4H2O

A. Cs and Br with 1 charge stands for Cesium Bromide

B. 4 as a prefix is Tetra, so the end of the formula is tetrahydrate

C. The final name of the compound is Cesium bromide tetrahydrate

(3) Cobalt(II) chloride hexahydrate ​

A. Cobalt Chloride has a compound formula of CoCl, and with Cobalt's charge of 2 involved, a subscript of 2 is added to the end to make CoCl2

B. The "hexa" prefix stands for 6, meaning there are 6 hydrates, or 6H2O

C. The two parts put together makes the final compound, CoCl2*6H2O

(4) LiI * 3H2O

A. Li and I both with single charges make up the compound Lithium iodide

B. 3 as a prefix i labled as "tri", so with 3 hydrates, the compound is named as trihydrate

C. The final compound put together makes Lithium iodide trihydrate

Help with naming compounds: http://chemwiki.ucdavis.edu/Wikitext...onic_Compounds