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Interpretation: Similarities among elements of group 7 is to be stated.
Concept introduction: Periodic table is an arrangement where elements are arranged on the basis of their increasing
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Answer to Problem 50A
In Group 7 of periodic table, all elements are non metals, have high electro negativity among elements of their respective period. Each element has 5 electrons in their valence shell and requires only 1 electron to attain noble configuration.
They all show similarity in physical properties too like have low melting and boiling point. All are colored.
Explanation of Solution
Group 7 of periodic table is known as halogen family. Its family members are F, Cl, Br, I. In halogen family, each and every element has highest electro negativity across their respective period, hence are non metals. Each element has 5 electrons in their valence shell and requires only 1 electron to attain noble configuration. On coming down the group, due to addition of shell, in each successive atom of group 7, atoms gain tendency to lose electron due to reduction in nuclear hold on valence electrons.
They all show similarity in physical properties too like have low melting and boiling point. All are colored.
In Group 7 of periodic table, all elements are non metals, have high electro negativity among elements of their respective period. Each element has 5 electrons in their valence shell and requires only 1 electron to attain noble configuration.
They all show similarity in physical properties too like have low melting and boiling point. All are colored.
Chapter 11 Solutions
World of Chemistry
- This deals with synthetic organic chemistry. Please fill in the blanks appropriately.arrow_forwardUse the References to access important values if needed for this question. What is the IUPAC name of each of the the following? 0 CH3CHCNH₂ CH3 CH3CHCNHCH2CH3 CH3arrow_forwardYou have now performed a liquid-liquid extraction protocol in Experiment 4. In doing so, you manipulated and exploited the acid-base chemistry of one or more of the compounds in your mixture to facilitate their separation into different phases. The key to understanding how liquid- liquid extractions work is by knowing which layer a compound is in, and in what protonation state. The following liquid-liquid extraction is different from the one you performed in Experiment 4, but it uses the same type of logic. Your task is to show how to separate apart Compound A and Compound B. . Complete the following flowchart of a liquid-liquid extraction. Handwritten work is encouraged. • Draw by hand (neatly) only the appropriate organic compound(s) in the boxes. . Specify the reagent(s)/chemicals (name is fine) and concentration as required in Boxes 4 and 5. • Box 7a requires the solvent (name is fine). • Box 7b requires one inorganic compound. • You can neatly complete this assignment by hand and…arrow_forward
- b) Elucidate compound D w) mt at 170 nd shows c-1 stretch at 550cm;' The compound has the ff electronic transitions: 0%o* and no a* 1H NMR Spectrum (CDCl3, 400 MHz) 3.5 3.0 2.5 2.0 1.5 1.0 0.5 ppm 13C{H} NMR Spectrum (CDCl3, 100 MHz) Solvent 80 70 60 50 40 30 20 10 0 ppm ppm ¹H-13C me-HSQC Spectrum ppm (CDCl3, 400 MHz) 5 ¹H-¹H COSY Spectrum (CDCl3, 400 MHz) 0.5 10 3.5 3.0 2.5 2.0 1.5 1.0 10 15 20 20 25 30 30 -35 -1.0 1.5 -2.0 -2.5 3.0 -3.5 0.5 ppm 3.5 3.0 2.5 2.0 1.5 1.0 0.5 ppmarrow_forwardShow work with explanation. don't give Ai generated solutionarrow_forwardRedraw the flowchartarrow_forward
- redraw the flowchart with boxes and molecules written in themarrow_forwardPart I. a) Elucidate the structure of compound A using the following information. • mass spectrum: m+ = 102, m/2=57 312=29 • IR spectrum: 1002.5 % TRANSMITTANCE Ngg 50 40 30 20 90 80 70 60 MICRONS 5 8 9 10 12 13 14 15 16 19 1740 cm M 10 0 4000 3600 3200 2800 2400 2000 1800 1600 13 • CNMR 'H -NMR Peak 8 ppm (H) Integration multiplicity a 1.5 (3H) triplet b 1.3 1.5 (3H) triplet C 2.3 1 (2H) quartet d 4.1 1 (2H) quartet & ppm (c) 10 15 28 60 177 (C=0) b) Elucidate the structure of compound B using the following information 13C/DEPT NMR 150.9 MHz IIL 1400 WAVENUMBERS (CM-1) DEPT-90 DEPT-135 85 80 75 70 65 60 55 50 45 40 35 30 25 20 ppm 1200 1000 800 600 400arrow_forward• Part II. a) Elucidate The structure of compound c w/ molecular formula C10 11202 and the following data below: • IR spectra % TRANSMITTANCE 1002.5 90 80 70 60 50 40 30 20 10 0 4000 3600 3200 2800 2400 2000 1800 1600 • Information from 'HAMR MICRONS 8 9 10 11 14 15 16 19 25 1400 WAVENUMBERS (CM-1) 1200 1000 800 600 400 peak 8 ppm Integration multiplicity a 2.1 1.5 (3H) Singlet b 3.6 1 (2H) singlet с 3.8 1.5 (3H) Singlet d 6.8 1(2H) doublet 7.1 1(2H) doublet Information from 13C-nmR Normal carbon 29ppm Dept 135 Dept -90 + NO peak NO peak 50 ppm 55 ppm + NO peak 114 ppm t 126 ppm No peak NO peak 130 ppm t + 159 ppm No peak NO peak 207 ppm по реак NO peakarrow_forward
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