Please give me solutions for TM1.3.4 ... and give me the details how to solve...
ID: 303826 • Letter: P
Question
Please give me solutions for TM1.3.4 ... and give me the details how to solve... Answer is on the bottom.Please give me solutions for TM1.3.4 ... and give me the details how to solve... Answer is on the bottom.
Please give me solutions for TM1.3.4 ... and give me the details how to solve... Answer is on the bottom.
TRANSITION METALS TM1. Consider aqueous solutions of the following coordination compounds: Co(NHs)dla, Pt(NHs)l Na Ptls, and Cr(NHa)da. If aqueous AgNO, is added to separate beakers containing solutions of each coordination compound, how many moles of Agl will precipitate per mole of transition metal present? Assume that each transition metal ion forms an octahedral complex. 12. Draw structures of each of the following a. cis-dichloroethylenediamineplatinum(l) mi Co cis-tetraamminechloronitrocobalt(Ill) ion TM3. The complex ion NiCl2 has two unpaired electrons, whereas Ni(CN)42 is diamagnetic. Propose structures for these two complex ions TM4. Ammonia and potassium iodide solutions are added to an aqueous solution of chromium(lI) nitrate. A solid is isolated (compound A), and the following data are collected: when 0.105g of compound A was strongly heated in excess 0:0.0203g Cro, was formed. b. a. In a second experiment, it took 32.93 mL of 0.100 M HCI to titrate completely the ammonia present in 0.341g compound A. c. Compound A was found to contain 73.53% iodine by mass. The freezing point of water was lowered by 0.64°C when 0.601g compound A was dissolved in 10.00g water (K d. 1.86 °C kg/mol) What does each piece of data tell you about the compound? What is the formula of the compound? What is the structure of the complex ion present? (Hints: Cr+ is expected to be six-coordinate, with NH and possibly I' as ligands. The I' ions will be the counterions, if needed.) (Hint: This is related to calculations you will do for Project Werner!) Abbreviated Solutions: TM1. 3, 2, 0,1 TM3. tetrahedral, square planar TM4. [Cr(NHs)sl))2
Explanation / Answer
In general, Werner,s proposed two valencies for CMI in coordination complexes
Primary/ ionizable: which are present outer from coordination sphere and we can detect by chemical test.
Secondary/Non- ionizable: which are present in coordination sphere and we can not detect by chemical test.
TM 1:
1. In octahedral complex [Co(NH3)6] I3, When AgNo3 is added, 3 mol of silver iodide is formed, because 3 iodide is primary valency which is ionizable.
[Co(NH3)6] I3 + AgNO3 = 3 AgI + Co(NH3)6]3+
2. In octahedral complex [Pt(NH3)4I2 ]I, When AgNo3 is added, 1mol of silver iodide is formed, because 1 iodide is primary valency which is ionizable.
3. In octahedral complex Na2[PtI6 ], When AgNo3 is added, there is not present Iodide as ionic valency, 6 I- present as secondary/ non ionisable vaency, therefore 0 is answer.
4. In [Cr(NH3)4I2] I, When AgNo3 is added, 1 mol of silver iodide is formed, because 1 iodide is primary valency which is ionizable.
TM3.
In [NiCl4]2- X - 4(Cl-) = -2 therefore X = +2
Coordination no. 4 of Central metal ion and Ni has ox. state is +2.
Valence el. configuration is : 3d8 4S0
In d- orbital has 2-unpaired electrons and Cl- ligand is weak field therefore electrons are not paired up and for hybridization outer orbital complex (SP3) hybrization takes place thus geometry is tetrahedral and nickel has 2 unpaired electron in this complex.
In [NiCN4]2- X - 4(CN-) = -2 therefore X = +2
Coordination no. 4 of Central metal ion and Ni has ox. state is +2.
Valence el. configuration is : 3d8 4S0
While CN- is a strong field ligand, Electrons of d- orbitals are paired up (CFSE>PE),
one d-orbital will be vacant thus for accomodation of 4- cynide ion, 1 d, 1S and 2P are hybridized, dsp2
Hybridization takesplace therefore square planar geometry is formed and complex will be diamagnetic.
TM4.
When ammonia and potassium iodide solution is added in cromium (III) nitrate,penta ammine iodocromium(III) iodidie is formed.
Octahedral complex will be formed, and73.53% iodine by mass was observed it means 2 iodide ion present as ionizable or primary valences.
And From a, we also concluded that cromium is in (III) ion
therefore 5 ammine and 1 Iodo reqiured as secondary valencied which are non ionizable.
thus formula will be
[Cr(NH3)5I] I2