Text for Accessibility: 2. What is the oxidation number and coordination number

Question

Text for Accessibility: 2. What is the oxidation number and coordination number of rhodium in the coordination compound K[13hCI(OH)(C204)2]

(b) Calculate the number of moles of ethylenediamine in 4.0 g of H2NCH2CH2NH2. (c) Since iron(III) and ethylenediamine react in a 1:3 mole ratio to form [Fe(H2NCH2CH2NH2)3]^3+ (aq), what is the limiting reactant in a mixture of 8.0 mL of 0.40 g/mL of FeCI3.6H2O and 4.0 g of ethylenediamine. (d) How many moles of [Fe(H2NCH2CH2NH2)3]^3+ product do you expect to form, in theory?

Explanation / Answer

2) the givne compound is

K[RhCl(OH)(C204)2].6H20

the common oxidation states are

K is +1

OH is -1

Cl is -1

C204 is -2

let Rh be y

now

the overall charge on the compound is 0

so

equating the charges on all the ions to 0

1 + y -1-1 -4 = 0

y = +5

so the oxidation number is +5

now the coordintaion number is the number of ligands attached to the central metal atom

Cl- , OH- are mono ligands

C2042- is biligand .

total ligand sites = 1 + 1 + ( 2 x 2) = 6

so the coordination number of Rh is 6

3)

mass of FeCl3 = conc x volume

= 0.4 x 8

= 3.2 g

moles = mass / molar mass

moles of FeCl3.6H20 = 3.2 / 270

moles of FeCl3.6H20 =0.01185

FeCl3 ----> Fe+3 + 3Cl-

from the above reaction

moles of Fe+3 = moles of FeCl3 = 0.01185

b)

moles = mass / molar mass

moles of ehtylenediamine = 4 / 60 = 0.06667

c)

given   1:3 ratio

so

moles of ethylenediamine required = 3 x moles of Fe+3

= 3 x 0.01185

= 0.0355

but

0.06667 moles of ethylenediamine is present

so
ethylenediamine   is in excess

FeCl3.6H20 is limting reagent

d)

from the given reaction

moles of [Fe(H2NCH2CH2NH2)3]3+ formed = moles of Fe+3 reacted = 0.01185

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