Question

Ionization energies (in kJ/mol) for two unknown elements are given in the table below.

　　　　　 　　　　1st 　　　　2nd 　　　　3rd 　　　　4th

element A 　　　  735 　　　  1445 　　　 7730 　　　 xxx

element B 　　　  580 　　　  1815 　　　 2740 　　　 11,600

1) What can you determine about the formulas of ions formed from A and B?

2) Write the most likely compound of A and B when chemically combined with chlorine? Explain.

Answer 1

est 2nd 9th 735 7730 XXX 1815 2740 11,600 4 J AGB so for element Difference betwee gst Given ~ Tonisation energy.(in kJ/mol ) 3rd Element 1445 Element B S80 1 ③ - Defermined the fomula ion → If the difference of two successive ionisation potential is approx (960-1450 kJ/mol or less then higher oxidation state of element will be more stable. --> 9H the difference of two successive ionisation potential is more than (1950 k] Imol then lower oxidation state of element will be more stable. and Ionisation ene A energy- OT and ist $ AL = 1445-735 DI = Flo ki imol less than 1450 kJ/mol SI - 3rd- and AT = 7730 - 1995 ors 6285 more than 1450 kg/mol 6. The ionisation of element from and to 3rd is required more than 1450 kJ/mol energy - 8o (and ionisation state more stable for A ist Formula in for A AT 2nd ++ A j.e A ART 2 t more stabe

SI SI = > SI 1 3rd Because A2A for element (B ora 2nd - 784 OI: 1815 - 586 1235 kJlmol less than 1450 kg/mol 3rd - and AT = 2740 -1815 925 Kolmol less than 1450 kJ/mol. DIE 9th Ara 1),(00 - 2790 DI= 8860 kJ/mol than 1450 kJ/mol The difference between 9 th y god jonisation energy is 8860 k) Jmol. so 3rd ionisation estate (ie lower oxidation state] more stable Formula of ion of element B → B B 3t ion more stable A & B react with ela formed compound Atch Ach more stable more 3+ B Bt ca Belz Because it more stable