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[ CrCl 2 ( H 2 O ) 4 ] Cl ( H 2 O ) 2 [ CrCl 2 ( H 2 O ) 4 ] + + Cl + water size 12{ \[ ital "CrCl" rSub { size 8{2} } \( H rSub { size 8{2} } O \) rSub { size 8{4} } \] ital "Cl" cdot \( H rSub { size 8{2} } O \) rSub { size 8{2} } rightarrow \[ ital "CrCl" rSub { size 8{2} } \( H rSub { size 8{2} } O \) rSub { size 8{4} } \]rSup { size 8{+{}} } + ital "Cl" rSup { size 8{ - {}} } + ital "water"} {}

The light green compound with two reactive chlorines is apparently [ CrCl ( H 2 O ) 5 ] Cl 2 H 2 O size 12{ \[ ital "CrCl" \( H rSub { size 8{2} } O \) rSub { size 8{5} } \] ital "Cl" rSub { size 8{2} } cdot H rSub { size 8{2} } O} {} , while the violet compound with three reactive chlorines is Cr ( H 2 O ) 6 Cl 3 size 12{ ital "Cr" \( H rSub { size 8{2} } O \) rSub { size 8{6} } ital "Cl" rSub { size 8{3} } } {} .

Closely related to hydrate isomerism is ionization isomerism, where an ion takes the place of water. Consider two different compounds with the formula Co ( NH 3 ) 5 SO 4 Br size 12{ ital "Co" \( ital "NH" rSub { size 8{3} } \) rSub { size 8{5} } ital "SO" rSub { size 8{4} } ital "Br"} {} . One of these, [ Co ( NH 3 ) 5 ( SO 4 ) ] Br size 12{ \[ ital "Co" \( ital "NH" rSub { size 8{3} } \) rSub { size 8{5} } \( ital "SO" rSub { size 8{4} } \) \] ital "Br"} {} , appears red, whereas the other, [ Co ( NH 3 ) 5 Br ] SO 4 size 12{ \[ ital "Co" \( ital "NH" rSub { size 8{3} } \) rSub { size 8{5} } ital "Br" \] ital "SO" rSub { size 8{4} } } {} , appears violet.

In addition to these coordination sphere isomers there are geometrical isomers, which have coordination spheres of the same composition but different geometrical arrangement. Geometrical isomers are distinct compounds and can have different physical properties (although often not too different) such as color, crystal structure, melting point, and so on. For example, dichlorodiamine platinum (II) occurs in the square planar geometry (B) so the chlorine ligands can be either next to one another (cis) or opposite from one another (trans). The compound you will synthesize has an octahedral geometry with two (bidentate) "en" ligands, and two nitro ( NO 2 ) size 12{ \( ital "NO" rSub { size 8{2} } \) } {} ligands. The geometrical isomer you will make is the trans form, in which the NO 2 size 12{ ital "NO" rSub { size 8{2} } } {} ligands are not adjacent to one another. This difference between cis and trans octahedral isomers is shown in Fig 2.

Fig 2. The trans and cis geometrical isomers for octahedral complexes with two bidentate (“en”) and monodentate ( NO 2 ) size 12{ \( ital "NO" rSub { size 8{2} } \) } {} ligands specifically dinitrobis(ethylenediamine)Co(III). The two black balls represent the NO 2 size 12{ ital "NO" rSub { size 8{2} } } {} ligands and the two pairs of linked white balls represent the two ethylenediamine ligands. Cis and trans describe the relationship (relative position) between the two NO 2 size 12{ ital "NO" rSub { size 8{2} } } {} ligands. 

In the procedure that follows we start with a cobalt solution made from the salt hexaquacobalt(II) nitrate, [ Co ( H 2 O ) 6 ] ( NO 3 ) 2 size 12{ \[ ital "Co" \( H rSub { size 8{2} } O \) rSub { size 8{6} } \] \( ital "NO" rSub { size 8{3} } \) rSub { size 8{2} } } {} . When this salt dissolves it ionizes to form two ions of NO 3 size 12{ ital "NO" rSub { size 8{3} } rSup { size 8{ - {}} } } {} and one of Co ( H 2 O ) 6 2 + size 12{ ital "Co" \( H rSub { size 8{2} } O \) rSub { size 8{6} } rSup { size 8{2+{}} } } {} . We wish to prepare a Co(III) compound of ethylenediamine, so we must add ethylenediamine (en) and oxidize the Co(II) to Co(III). Because Co(II) is more reactive than Co(III), we allow it to react with (en) first, and then oxidize the resulting complex ion. In aqueous solution (en) reacts with water to produce OH size 12{ ital "OH" rSup { size 8{ - {}} } } {} ions which can also bind to Co(II), so the pH is adjusted close to 7 first by adding HNO 3 size 12{ ital "HNO" rSub { size 8{3} } } {} . (Other acids would introduce new ligands to compete for the Co.) NaNO 2 size 12{ ital "NaNO" rSub { size 8{2} } } {} is added to provide the ligands that will be trans in the final compound. Lastly, Co(II) is oxidized to Co(III) by bubbling oxygen through the solution.

Experimental procedure

  • Use your 10 mL graduated cylinder to measure out 20 mL of the 20% by weight solution of ethylenediamine in dilute HNO 3 size 12{ ital "HNO" rSub { size 8{3} } } {} .
  • Pour it into a clean 125 mL Erlenmeyer flask. Rinse the graduated cylinder with about 5mL of deionised water (DI water from white handle faucet) and add the rinse water to the flask. Set this aside for a moment and prepare the second set of reactants as described below.
  • Weigh out 9.0 g of hexaquacobalt(II) nitrate and 6.0 g sodium nitrite ( NaNO 2 size 12{ ital "NaNO" rSub { size 8{2} } } {} ) using a rough balance (Record mass on report form). Add these reactants to approximately 15 mL of DI water in an Erlenmeyer flask. After they have dissolved, add the neutralized ethylenediamine solution prepared in steps 1-2. Record your observations.
  • For the next set of instructions, refer to the diagram below. Fit a piece of rubber tubing over an inert gas "IG" tap (on benchtop) and open the valve slowly to obtain a gentle flow of oxygen. Then insert a Pasteur pipet into the other end of the rubber tubing. CAUTION: Too high a gas flow might blow the pipet out of the tubing and cause serious injury. Always adjust the valve carefully while pointing your pipet in a safe direction. Test the flow by immersing the pipet tip in a beaker of water--it should bubble vigorously, but not enough to cause much splashing. When the flow is set to your satisfaction, immerse the tip of the pipet in the Erlenmeyer flask containing the reaction mixture. Secure the flask to a stand with a clamp because the reaction mixture may need about 10 minutes of moderately vigorous bubbling to reach completion. Record your observations.

Questions & Answers

show that the set of all natural number form semi group under the composition of addition
Nikhil Reply
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_3_2_1
felecia
⅗ ⅔½
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The denominator of a certain fraction is 9 more than the numerator. If 6 is added to both terms of the fraction, the value of the fraction becomes 2/3. Find the original fraction. 2. The sum of the least and greatest of 3 consecutive integers is 60. What are the valu
SABAL Reply
1. x + 6 2 -------------- = _ x + 9 + 6 3 x + 6 3 ----------- x -- (cross multiply) x + 15 2 3(x + 6) = 2(x + 15) 3x + 18 = 2x + 30 (-2x from both) x + 18 = 30 (-18 from both) x = 12 Test: 12 + 6 18 2 -------------- = --- = --- 12 + 9 + 6 27 3
Pawel
2. (x) + (x + 2) = 60 2x + 2 = 60 2x = 58 x = 29 29, 30, & 31
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Ifeanyi
on number 2 question How did you got 2x +2
Ifeanyi
combine like terms. x + x + 2 is same as 2x + 2
Pawel
x*x=2
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2+2x=
felecia
Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113?
mariel Reply
Mark = x,. Don = 3x + 1 x + 3x + 1 = 113 4x = 112, x = 28 Mark = 28, Don = 85, 28 + 85 = 113
Pawel
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Harshika Reply
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find the subring of gaussian integers?
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Mark
I need quadratic equation link to Alpa Beta
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find the value of 2x=32
Felix Reply
divide by 2 on each side of the equal sign to solve for x
corri
X=16
Michael
Want to review on complex number 1.What are complex number 2.How to solve complex number problems.
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use the y -intercept and slope to sketch the graph of the equation y=6x
Only Reply
how do we prove the quadratic formular
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Opoku
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Tric Reply
4
Trista
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Sidiki Reply
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Mark
Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
Brenna
(61/11,41/11,−4/11)
Brenna
x=61/11 y=41/11 z=−4/11 x=61/11 y=41/11 z=-4/11
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Need help solving this problem (2/7)^-2
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x+2y-z=7
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what is the coefficient of -4×
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-1
Shedrak
A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place.
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Jeannette has $5 and $10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
August Reply
What is the expressiin for seven less than four times the number of nickels
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How do i figure this problem out.
how do you translate this in Algebraic Expressions
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why surface tension is zero at critical temperature
Shanjida
I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
s.
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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how did you get the value of 2000N.What calculations are needed to arrive at it
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Source:  OpenStax, Honors chemistry lab fall. OpenStax CNX. Nov 15, 2007 Download for free at http://cnx.org/content/col10456/1.16
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