According to the given pattern of ionization energies , first 6 electrons are removed easily with less ionization energies. Ionization energy increases drastically for 7th electron i.e. 7th electron must have been removed from a highly stable Noble gas configuration.
This element has 6 electrons in valence shell.
Only oxygen atom in period 2 has 6 electrons in valence shell.
Hence this element is Oxygen.
Answer -- oxygen
HQ7.45 Homework . Unanswered Identify the period 2 element from the following ionization energies. IE, 13.6...
The first five ionization energies (IE through IEs) of a Period 2 element have the following pattern IE1 IE2 IE3 IE4 IE5 Make a reasonable guess about which element this is. Enter its chemical symbol below
Element J was found to have the following ionization energies (in kJ/mol) 1st IE 2nd IE 3rd IE 4th IE 5th IE 6th IE 631 1235 2389 7099 8844 10720 How many valence electrons does element J have? Explain. (2) If element J is found in the fourth period, identify element J. (1) Write the electronic configuration for element J. (1) Write complete sets of quantum numbers for each of the valence electrons (i.e. n, l, and ml) that you...
10.) Below is a list of successive ionization energies (in kJ/mol) for a period 3 element. Identify the element and explain how you came to that conclusion. IE2 = 2250 IE3 - 3360 TE 4 = 4560 IE5 = 7010 IE6 - 8500 IE 7 = 27,100
Identify the element of Period 2 (name) which has the following successive ionization energies, in kJ/mol. IE1 = 1314 IE2 = 3388 1E3 = 5301 JE4 = 7469 IE5 = 10989 IE6 = 13327 1E7 = 71330 IE8 = 84078 Answer:
To UMOWw26urt twowilgoce Objective Knowledge Check Question 17 The first five ionization energies (IE, through IE,) of a Period 2 element have the following pattern: kJ/mol IE, IE, IE; IE4 IES Make a reasonable guess about which element this is. Enter its chemical symbol below. X 5 ? I Don't Know Submit 2010 Mo dica Type here to search
31. What period 3 element has the following ionization energies (all in l/moly? TE, - 1012 IE, - 1900 IE, - 2910 IE,- 4960 TE,- 6270 IE, -22,200 a. CI b. P c. S d. Mg e. Si 32. Give the set of four quantum numbers that could represent the last electron added (using the Authors principle) to the Clatom. a. -3,1-1, -1,- d. n=2,1-1, m, l, m,- +1/3 b. -3,1-0, m = 1, m,- + c. =3,1=2,m=1, m,-- 33....
Ionization. 25. (3) Which of the following most likely represents the successive ionization energies (IE, through IE) of a noble gas in Period 4, in kJ/mol? (select the best one) IE IE2 IE3 IE4 IES a) 1351 2350 3565 5070 6240 941 2045 2973 4144 6590 c) 419 3052 4420 5877 7975 d) 590 1145 4912 6491 8153
4. Consider an element of Period 2 which has the following successive ionization energies, in kJ/mol. What is the charge of the common ion? IE1, 1314 IE2, 3389 IE3, 5298 IE4, 7471 IEs, 10992 IE6, 13329 IE7, 71345 IE8, 84087 (1) -1 (2) –2 (3) –3 (4) +6 (5) +7
Question 2 1 pts Identify the group number of the element that has the following successive ionization energies (in kJ/mol). IE2 - 577 1 E2 - 1,816 IE3 - 2,881 1E4 - 11,600 IES - 14,800 1E6 - 18,400 IE2 = 23,300 Group 3A Group 5A Group 2A Group 6A Group 7A
Which period 3 element has the following ionization energies (kJ/mol)? IE1 = 738 IE2 = 1451 IE3= 7733 Group of answer choices Mg S P Al Si