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Sp 2016 9. 5.59 M&F; 7th edn 10. Q. 5113, M&F; 7th edn 11.5.133 M&F; 7th edn 12.

ID: 1045034 • Letter: S

Question

Sp 2016 9. 5.59 M&F; 7th edn 10. Q. 5113, M&F; 7th edn 11.5.133 M&F; 7th edn 12. Answer both parts (a) Give orbit al notations for electrons in orbitals with the following quantum numbers: (ii) n-41P (iii) n-4.1.2; (b) Give the possible combinations of quantum numbers for the following orbitals: (i) a 3d orbital () a 5p orbital (ii) a 3s orbital 13. Write down the full and shorthand electronic configuration for each of the following (a) F (b) K d) Ca (c) Cr 14. Draw properly labeled energy level diagrams for each the following showing how the electrons are distributed among the orbitals: a) O b) Na c) c KIO, required 23.55 ml of a solution for Na,S,0, for titration using starch as the indicator. What is the molarity of the Na,S O, solution? 16. Balance the following equation then calculate the calculate the molarity of a hydrochloric solution requiring 0.1500 g of Na CO, for titration to the methyl orange end point: Nact)(aq) + HCI(aq)-NaCl(aq) + CO2(g) + H20) 17. A 0.1050 g sample of a divalent metal oxide, MO, was dissolved in 2500 mL of a 0.3000 M hydrochloric acid (an excess), and the unreacted acid required 13.00 mL of a 0.1000 M NaOH for titration to the phenolphthalein end point. What is the molar mass and formula of the oxide? MOS) + 2 HCI(aq)-MC12(aq) + H20()

Explanation / Answer

I am solving the Q12(complete) as per Chegg guidelines, post multiple questions to get the remaining answers

Q12)

(a)

(i) n=2,l=0

l = 0 means it is an s-orbital, therefore the answer is 2s

(ii) n=4,l=1

l = 1 means it is an p-orbital, therefore the answer is 4p

(iii) n=4,l=2

l = 2 means it is an d-orbital, therefore the answer is 4d

b)

i) 3d orbital

n = 3

l = 2

ml = -2,-1,0,1,2

ms = -1/2,1/2

Hence there are 10 possible (5 possible values of ml * 2 possible value of ms)

ii) 5p orbital

n = 5

l = 1

ml = -1,0,1

ms = -1/2,1/2

Hence there are 6 possible (3 possible values of ml * 2 possible value of ms)

iii) 3s orbital

n = 3

l = 0

ml = 0

ms = -1/2,1/2

Hence there are 2possible (1 possible values of ml * 2 possible value of ms)

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