(a) One mole of atoms of an element 'X' absorb 800 kJ of energy for the formation of unipositive ions. For the conversion of one mole of 'X' atoms into tripositive ione, 6800 kJ of energy is required. Second and third ionisation energies are in 2 : 3 ratio. Calculate the second and third ionisation energies. (b) First and second second ionisation energies of Be_(g) are 900 and 1750 kJ/ mol, respectively. Calculate approximate percentade of Be_((g))^(2+)ions, if 1 g of Be_((g)) absorbs 150 kJ of eneergy (Be atomic weight is 9 ).

(a) One mole of atoms of an element 'X' absorb 800 kJ of energy for th

1.	(a) One mole of atoms of an element 'X' absorb 800 kJ of energy for the formation of unipositive ions. For the conversion of one mole of 'X' atoms into tripositive ione, 6800 kJ of energy is required. Second and third ionisation energies are in 2 : 3 ratio. Calculate the second and third ionisation energies. (b) First and second second ionisation energies of Be_(g) are 900 and 1750 kJ/ mol, respectively. Calculate approximate percentade of Be_((g))^(2+)ions, if 1 g of Be_((g)) absorbs 150 kJ of eneergy (Be atomic weight is 9 ).
Answer» Solution :(a) `x to x^(+) , 800 kJ//mol (IP_1) ltbegt x to x^(+3) , 6800 kJ` ` thereforeIP_(1) + IP_(2) +IP_(3) =6800kJ rArr IP_(2)+IP_(3)=6800-800=6000` ` x^(+) to x^(+2) IP_(2), x^(+2) to x^(+3) to IP_(3)` `because IP_(2): IP_(3) =2:3 rArr(IP_(2))/(IP_(3))=2/3` `rArr IP_(2)=2/3. IP_(3) or IP_(3)=(3IP_2)/2` `rArr IP_(2) + IP_(3) = 6000` `rArr IP_(2) + 3/2. IP_(2) = 6000 rArr 2 IP_(2)+2IP_(3) =6000xx2` `IP_(2) =6000/2 xx2 = 2400 kJ// mol, IP_(3) =6000/5xx3 = 3600 kJ//mol` (B) 1 mole (9g) of beryllium requires 900 kJ to dislodge the first electron. `therefore` 1 g " of Be requires" `=(1xx900)/9=100kJ` `therefore` "Differnce in the amount of energy absorbed " `=150 -100=50kJ` To dislodge the second electron from 1 mole of Beryllium ION, 1750 kJ is required. 50 kJ OS sufficient for `50/1750xx9=0.25=25%` percentage of `Be_((g))^(2+)` ions is 25%.