There is huge difference in ionisation energy between IE3 and IE4.
This means, lots of energy are required to remove 4th electron.
So, after 3 electrons are removed, the element has acquired noble gas configuration,.
So, there are 3 valence electrons in the element.
The element with 3 valence electrons in period 5 is Indium
Answer: In
11. The successive ionization energies of a certain element in period 5 are 577.9 kl/mol, 1x...
Consider this set of ionization energies. IE1 = 578 kJ/mol, IE2 = 1820 kJ/mol, IE3 = 2750 kJ/mol, IE4 = 11,600 kJ/mol. To which third-period element do these ionization energies belong?
9. A certain 4th row element has the following ionization energies: 12 = 1310 kJ/mol I3 = 2652 kJ/mol 14 = 4175 kJ/mol Is = 9581 kJ/mol 16 = 11533 kJ/mol Now tell me, most peerless predictor of periodic properties, what is the identity of the element? Please provide an explanation as well.
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:
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
15. Below is a list of successive ionization energies (in kJ/mol) for a period 3 element, X. What is the most likely formula for a stable ion of X? 151 = 1000 IE2 = 2250 IE3 = 3360 IE4= 4560 E5= 7010 TE6= 8500 IE2 = 27.100 d. X a. X2 b. X30 c. X²
please explain why
3. The first six ionization energies (kJ/mol) of a certain SECOND-ROW element are as follows: Ij = 1,086 12 = 2350 13 = 4620 14 = 6220 Is = 38,000 16 = 47,261 The most likely identity of this element is a) B b) 0 c) N d) Be e) C
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 23 2 pts What element in period 3 has the following successive ionization energies? 1E1:578 kJ/mol IE2: 1817 kJ/mol 1E3: 2745 kJ/mol IE4: 11578 kJ/mol IES: 14842 kJ/mol IE6: 18379 kJ/mol 1E7:23326 kJ/mol IE8:27464 kJ/mol
Identify the group number of the element that has the following successive ionization energies (in kJ/mol). E1 = 1,060 IE2 = 1,890 IE3 = 2,905 IE4 = 4,950 IE5 = 6,270 IE6 = 21,200 IE7 = 25,400