Extra Credit 23

S19.1.2

Initially, find the atom’s atomic number. Then, find the charge of the atom. Utilizing the order of orbitals, fill it in according to the number of electrons in the atom. Hint: utilize the periodic table as a visual.

S19.1.4

Lanthanoids are an active metal (relatively) which are easily oxidized. Additionally, they can only stay in combined form, like the alkaline earth metals, to exist.

S19.1.6

MnO−4 because the ions contain metal which must be reduced. Both Vanadium (which is harder to reduce than Chromium) and Chromium are hard to reduce in comparison to Manganese. This leaves the answer to be MnO-4 as the strongest oxidizing agent out of the three options.

S19.1.8

a, b, c, d, e

S19.1.10

In relation to the Lewis acid-base theory, the Lewis acid accepts lone pair electrons; thus, it is also known as the electron pair acceptor. This may be any chemical species. Acids are substances that must be lower than 7. Therefore, oxides of manganese is most likely going to become more acidic in (aq) solutions if the oxidation number increases.

Q19.1.12

Firstly, the overall reaction must be determined. Fe2O3(s) + 3CO(g) -> 2Fe(l) + 3CO2(g) is where CO is created though the breakdown of CO2.

Now, the calculation of CO is: Mass CO= 1 ton of Fe2O3 x 1 ton mol Fe2O3/159.692 ton Fe2O3 x 3 ton mol CO/1 ton mol Fe2O3 x 28.010ton CO/1 ton mol CO= .53 ton CO.

Deriving from this, we are now able to find O2: Mass O2= .53 ton CO x 1 ton mol CO/28.010 ton CO x 1 ton mol O2/2 ton mol CO x 31.998 ton O2/1 ton mol O2 = .3 ton O2/tonFe2O3 which is .3 ton of O2 that needs to be divided by .19 since the assumed air is 19 percent oxygen. Thus, it is 1.58 ton air.

Now, lastly, we can get (air)= 4.58 tone air/ton Fe2O3 x 2000lb/ton air x 453.59 g/lb x 1 mol air/29.9g x 22.4 L/mol which give us the Fe2O3 conversion to .3 ton O2/tonFe2O3.

Q19.1.14

Finding the mol of AgCl is 3.03707 g x 1mol/143.321g= .0211906 mol Cl which is now multiplied by 35.45 g/mol that gives us .75127 g Cl

To get percent mass, we get .75127g/2.5624g (the sample) x 100= 29.32% Cl.

As for the identity of the salt, we have to subtract 2.5624 g by the given Cl g which is .75127 g. This gives us 1.8111 g. 1.8111g/.0211906 mol Cl is 85.47 g mol-1 which gives us with Rubidium (RbCl to be exact).

Q19.1.16

1. MnCO3(s)+HI(aq)->MnI2(aq)+H2O(l)+CO2(g)
2. 6CoO(s)+O2(g)>2Co3O4−>6CoO(s)+O2(g)
3. 4 La(s)+3O2(g)->2La2O3(s)
4. V(s)+VCl4(s)>2VCl2(s)
5. 2Co(s)+3F2(g)->2CoF3(s)
6. CrO3(s)+2Cs+(aq)+2OH−(aq)->2Cs+(aq)+CrO42−(aq)+H2O(l)

S19.1.18

The electrolytic process for refining copper is to extract the copper through the process of electrolysis.

Q19.1.20

H2S Gas is formed due to its nonoxidizing acid nature.

Q19.1.22

Co2+->Co3+ +e-

N5+ +e- -> N4+

2O2- -> O2 + 4e-

The balancing aspect of the Co atoms show that we have even bumbers of Co(No3)2. Thus, the production of 4CO(No3)2->2Co2O3 + 8NO2 +O2 through the doubled of 2Co(NO3)2 and eight N atoms reduced by eight electrons and the production of 8 electrons from oxidation.

S19.1.24

[CrO₄]^2- is more stable because Chromium is in the 3d orbital while Tungsten is in the 4d orbital, which has a higher energy level and makes it less stable.

Q19.2.2

Give the coordination numbers and write the formulas for each of the following, including all isomers where appropriate:

1. tetrahydroxozincate(II) ion (tetrahedral)
2. hexacyanopalladate(IV) ion
3. dichloroaurate ion (note that aurum is Latin for "gold")
4. diaminedichloroplatinum(II)
5. potassium diaminetetrachlorochromate(III)
6. hexaaminecobalt(III) hexacyanochromate(III)
7. dibromobis(ethylenediamine) cobalt(III) nitrate

1. 4, [Zn(OH)₄]²⁻;
2. 6, [Pd(CN)₆]²⁻;
3. 2, [AuCl₂]⁻;
4. 4, [Pt(NH₃)₂Cl₂];
5. 6, K[Cr(NH₃)₂Cl₄];
6. 6, [Co(NH₃)₆][Cr(CN)₆];
7. 6, [Co(en)₂Br₂]NO₃

Q19.2.4

Geometric isomers are formed by transition metals. Cis configuration would have the same element adjacent to each other; however, trans configuration would the two elements directly across from each other.

S19.2.6

Q: Name the elements to the right:

1. tricarbonatocobaltate(III) ion;
2. tetraaminecopper(II) ion;
3. tetraaminedibromocobalt(III) sulfate;
4. tetraamineplatinum(II) tetrachloroplatinate(II);
5. tris-(ethylenediamine)chromium(III) nitrate;
6. diaminedibromopalladium(II);
7. potassium pentachlorocuprate(II);
8. diaminedichlorozinc(II)

Q19.2.8

1. They would have no isomers, the coordination complexes that has two different ligands in the cis or trans will have a position from a ligand of interest form isomers, which in this case does not.
2. They would have no isomers. Same explanation as a.
3. The Cl ligands consist of two which can form cis or trans transfiguration. With the Cis, there will be an optical isomer.

Q19.3.4

There are three unpaired electrons leading to [Co(H2O)6]Cl2

Q19.3.6

1. 4
2. 2
3. 1
4. 5
5. 0

The reason being, when an electron is unpaired (in both atom or ion), the magnetic moment due to the spin makes the ion/atom paramagnetic. The greater the unpaired electrons, the larger the magnetic moment. The observed magnetic moment can help determine the number of unpaired electrons that are present.

Q19.3.8

1. [Fe9BN]6]4-
2. [Co(NH3)6]3+
3. [Mn(CN)6]4-

The complexes can be predicted of their transition with the crystal field theory. The different bonding model shows the stability, structures, colors, and magnetic properties. CFT focus on nonbonding electrons of the central metal ion in coordination complexes not on the metal-ligand bonds.