NW5.  Solutions to Selected Problems

Problem NW1.1. 

a) graphite is between 1 & 2        b) stishovite is between 9 & 10        c)  opal  is between 6 & 7

d) obsidian is about 5                   e)  ruby is about 9                               f) emerald is between 7 & 8  

Problem NW3.1. 

a) forsterite:  Mg₂SiO₄:  If each nesosilicate unit has a charge of 4^- because of the formal charges on the four oxygens, then the two magnesiums must combine to balance that charge with a 4⁺ charge.  Each magnesium must have a 2⁺ charge of its own.  That seems reasonable, since magnesium is two atoms from the left edge of the periodic table; it would have a 2⁺ charge in a noble gas configuration.

b) fayalite: Fe₂SiO₄: In order for this to work, each iron would have to have a 2⁺ charge.  Transition metal charges are harder to predict than alkali and alkaline earth metals because they are farther from the edge of the periodic table, but it turns out that the two most common charges on an iron ion are 2⁺ and 3⁺, so this charge seems reasonable.

Problem NW3.2. 

a) If each nesosilicate has a charge of 4^-, then the total negative charge is 100 x 4^- = 400^-.  If there are 25 magnesium ions, each with a 2⁺ charge, the total positive charge is 25 x 2⁺ = 50⁺.  That leaves a net negative charge of 400^- + 50⁺ = 350^-.  The irons would balance that charge with an equal positive charge.  The number of iron ions would be 350+/3+ = 150.

b) If we divide these ratios by 100 to arrive at the one nesosilicate in the formula, we get Mg_0.25Fe_1.5SiO₄.  It doesn't mean there is a quarter of a magnesium ion anywhere; this formula is just the overall ratio of atoms.

Problem NW3.3. 

 So far we have Al₂Fe₃.  The positive charge = 2 x 3⁺ + 3 x 2⁺ = 12⁺.  That charge must be balanced by the nesosilicate anions, which are each 4^-.  So the number of silicates is 12^-/4^- = 3.  The formula is Al₂Fe₃(SiO₄)₃.

Problem NW3.4. 

 The anions are nesosilicate (4^-) + sorosilicate (6^-) + oxide (2^-) + hydroxide (1^-).  The total charge is 13^-.

Calcium is an alkaline earth metal, two atoms from the left edge of the periodic table.  The two calciums would partially balance that charge with 2 x 2⁺ = 4⁺ charge; the net charge is now 13^- + 4⁺ = 9^-.

If each Al or Fe has charge 3⁺, then the number needed to balance the charge is 9⁺/3⁺ = 3 Fe/Al ions.

This site was written by Chris P. Schaller, Ph.D., College of Saint Benedict / Saint John's University (retired) with other authors as noted on individual pages.  It is freely available for educational use.

Structure & Reactivity in Organic, Biological and Inorganic Chemistry by Chris Schaller is licensed under a Creative Commons Attribution-NonCommercial 3.0 Unported License. 

Send corrections to cschaller@csbsju.edu

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