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Concept explainers
Interpretation: The complex which is diamagnetic is to be determined from the given option.
Concept introduction: The
The properties of the coordination compounds depend upon the primary and secondary valancy of the metal ion in the coordination sphere. The electrical conductivity depends upon the number of ions that are produced by complex.
The strong-field ligands results in pairing of electrons present in the complex and leads to diamagnetic species , while the low-field ligand do not have tendency to pair up the electrons therefore forms paramagnetic species.
The strong field ligands lead to splitting to a higher extent than the weak field ligands and the
The five d orbitals get divided into two sets that is
Electronic configuration: It is defined as the distribution of electrons present in the atom over orbitals following certain rules like electrons starts filling the lower energy orbital to higher energy, pairing of electrons does not occur until all the orbitals are singly filled and finally no electrons present in orbital can have same set of quantum numbers.
Oxidation number: The total number of electrons in an atom after losing or gaining electrons to make a bond with another atom. It indicates the charge of an ion.
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Chapter 22 Solutions
Bundle: Chemistry & Chemical Reactivity, Loose-Leaf Version, 9th + OWLv2, 4 terms (24 Months) Printed Access Card
- we were assigned to dilute 900ppm in to 18ppm by using only 250ml vol flask. firstly we did calc and convert 900ppm to 0.9 ppm to dilute in 1 liter. to begin the experiment we took 0,225g of kmno4 and dissolved in to 250 vol flask. then further we took 10 ml sample sol and dissolved in to 100 ml vol flask and put it in to a spectrometer and got value of 0.145A . upon further calc we got v2 as 50ml . need to find DF, % error (expval and accptVal), molarity, molality. please write the whole report. thank you The format, tables, introduction, procedure and observation, result, calculations, discussion and conclusionarrow_forwardQ5. Predict the organic product(s) for the following transformations. If no reaction will take place (or the reaction is not synthetically useful), write "N.R.". Determine what type of transition state is present for each reaction (think Hammond Postulate). I Br₂ CH3 F2, light CH3 Heat CH3 F₂ Heat Br2, light 12, light CH3 Cl2, light Noarrow_forwardNonearrow_forward
- In the phase diagram of steel (two components Fe and C), region A is the gamma austenite solid and region B contains the gamma solid and liquid. Indicate the degrees of freedom that the fields A and B have,arrow_forwardFor a condensed binary system in equilibrium at constant pressure, indicate the maximum number of phases that can exist.arrow_forwardPart V. Label ad match the carbons in compounds Jane and Diane w/ the corresponding peak no. in the Spectra (Note: use the given peak no. To label the carbons, other peak no are intentionally omitted) 7 4 2 -0.13 -0.12 -0.11 -0.10 -0.08 8 CI Jane 1 -0.09 5 210 200 190 180 170 160 150 140 130 120 110 100 -8 90 f1 (ppm) 11 8 172.4 172.0 f1 (ppr HO CI NH Diane 7 3 11 80 80 -80 -R 70 60 60 2 5 -8 50 40 8. 170 160 150 140 130 120 110 100 90 -0 80 70 20 f1 (ppm) 15 30 -20 20 -60 60 -0.07 -0.06 -0.05 -0.04 -0.03 -0.02 -0.01 -0.00 -0.01 10 -0.17 16 15 56 16 -0.16 -0.15 -0.14 -0.13 -0.12 -0.11 -0.10 -0.09 -0.08 -0.07 -0.06 -0.05 -0.04 17.8 17.6 17.4 17.2 17.0 f1 (ppm) -0.03 -0.02 550 106 40 30 20 20 -0.01 -0.00 F-0.01 10 0arrow_forward
- n Feb 3 A T + 4. (2 pts) Draw the structure of the major component of the Limonene isolated. Explain how you confirmed the structure. 5. (2 pts) Draw the fragment corresponding to the base peak in the Mass spectrum of Limonene. 6. (1 pts) Predict the 1H NMR spectral data of R-Limonene. Proton NMR: 5.3 pon multiplet (H Ringarrow_forwardPart VI. Ca H 10 O is the molecular formula of compound Tom and gives the in the table below. Give a possible structure for compound Tom. 13C Signals summarized C1 C2 C3 C4 C5 C6 C7 13C shift (ppm) 23.5 27.0 33.0 35.8 127 162 205 DEPT-90 + DEPT-135 + +arrow_forward2. Using the following data to calculate the value of AvapH o of water at 298K. AvapH o of water at 373K is 40.7 kJ/mol; molar heat capacity of liquid water at constant pressure is 75.2J mol-1 K-1 and molar heat capacity of water vapor at constant pressure is 33.6 J mol-1 K-1.arrow_forward
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