Question

Table 4.1: Ionization Energies

Complete the ionization energies for each element.

Element		H	He	Li	Be	B	C	N	O	F	Ne	Na	Mg	Al
Atomic Number		1	2	3	4	5	6	7	8	9	10	11	12	13
Ioni zation Energies	1st	1312	2372	520	900	801	1087
	2nd	-	5251	7298	1757		2353	2856	3388	3374	3952	4562	1451
	3rd	-	-	11815	14849		4620	4578	5301	6050	6122	6910	7733
	4th	-	-	-	21007		6223	7475	7469	8408	9371	9543	10543
	5th	-	-	-	-		37831	9445	10990	11023	12177	13354	13630
	6th	-	-	-	-	-	47277	53267	13327	15164	15238	16613	18020
	7th	-	-	-	-	-	-	64360	71330	17868	19999	20117	21711
	8th	-	-	-	-	-	-	-	84078	92038	23070	25496	25661
	9th	-	-	-	-	-	-	-	-	106434	115380	28932	31653
	10th	-	-	-	-	-	-	-	-	-	131432	141362	35458
	11th	-	-	-	-	-	-	-	-	-	-	159075	169988
	12th	-	-	-	-	-	-	-	-	-	-	-	189368

2.    Explain how the successive (first, second, third, etc) ionization potentials change in each atom.

3.     For most atoms, there are one or two huge jumps in the ionization potential.   Explain why these occur. In aluminum, explain these huge increases using the valence electrons.

4.      Explain any trend you observe when looking at only the first ionization energies of each of the 13 elements.

Answer 1

Sol:-

Table 4.1: Ionization Energies

Complete the ionization energies for each element.

Element		H	He	Li	Be	B	C	N	O	F	Ne	Na	Mg	Al
Atomic Number		1	2	3	4	5	6	7	8	9	10	11	12	13
Ioni zation Energies	1st	1312	2372	520	900	801	1087	1402	1314	1681	2081	496	738	578
	2nd	-	5251	7298	1757		2353	2856	3388	3374	3952	4562	1451	1817
	3rd	-	-	11815	14849		4620	4578	5301	6050	6122	6910	7733	2745
	4th	-	-	-	21007		6223	7475	7469	8408	9371	9543	10543	11577
	5th	-	-	-	-		37831	9445	10990	11023	12177	13354	13630	14842
	6th	-	-	-	-	-	47277	53267	13327	15164	15238	16613	18020	18379
	7th	-	-	-	-	-	-	64360	71330	17868	19999	20117	21711	23326
	8th	-	-	-	-	-	-	-	84078	92038	23070	25496	25661	27465
	9th	-	-	-	-	-	-	-	-	106434	115380	28932	31653	31853
	10th	-	-	-	-	-	-	-	-	-	131432	141362	35458	38473
	11th	-	-	-	-	-	-	-	-	-	-	159075	169988	42647
	12th	-	-	-	-	-	-	-	-	-	-	-	189368	201266

2.    Explain how the successive (first, second, third, etc) ionization potentials change in each atom.

Answer :- Let an atom is denoted by " M " .

Now the successive ionization potentials of M is :-

M(g) ------------> M⁺(aq) + 1e^- , I.E₁

M⁺ (aq) ------------> M²⁺(aq) + 1e^- , I.E₂

M²⁺(aq) ------------> M³⁺(aq) + 1e^- , I.E₃

here I.E₁ = First Ionisation potential

I.E₂ = Second Ionisation potential and

I.E₃ = Third Ionisation potential .

The decreasing order of Successive ionization potential for " M " is :

I.E₃ > I.E₂ > I.E₁  

I.E₃ is greater than I.E₂ and I.E₂ is greater than I.E₁ because effective nuclear charge per electron increases from M ( i.e I.E₁)  to M²⁺ ( i.e I.E₃) .

NOTE :-

IONISATION POTENTIAL :- The amount of energy required to remove an electron from outer most shell of atom in its isolated gaseous states.

FACTOR FAVORING IONISATION POTENTIAL :-

(1). Nuclear cahrge :- as the nuclear charge increases ionisation potential also increases .

(2) .Atomic or ionic radii :- Ionisation potential is inversely proportional to the atomic or ionic radii  .

(3) . Noble gas electronic configuration :- Ionisation enthalpy is more when an atom or ion have noble gas or stable electronic configuration .

3.     For most atoms, there are one or two huge jumps in the ionization potential.   Explain why these occur. In aluminum, explain these huge increases using the valence electrons.

Answer :- The huge jumps in the ionisation potential are due to the involement of Noble gas electronic configuration .

For e.g:-

Al(g) ----------> Al⁺(aq) + 1e^- , I.E₁ = 578

1s² 2s² 2p⁶ 3s² 3p¹   1s² 2s² 2p⁶ 3s²

Al⁺(g) ----------> Al²⁺(aq) + 1e^- , I.E₂ = 1817

1s² 2s² 2p⁶ 3s² 3p¹ 1s² 2s² 2p⁶ 3s¹

Al²⁺(g) ----------> Al³⁺(aq) + 1e^- , I.E₃ = 2745

1s² 2s² 2p⁶ 3s² 3p¹   1s² 2s² 2p⁶ ( Nobe gas or stable electronic cong. of Ne )

Al³⁺(g) ----------> Al⁴⁺(aq) + 1e^- , I.E₄ = 11577   

1s² 2s² 2p⁶   1s² 2s² 2p⁵   

here huge jumps in ionisation potential from I.E₃ (i.e 2745) to I.E₄ (i.e 11577) because of the involement of Noble gas electronic configuration of Ne i.e 1s² 2s² 2p⁶ . Due to this stable electronic configuration it becomes more difficult to remove the electron from outer most shell of Al³⁺ , therefore value of 4^th ionisation potential increases sharply .

4.      Explain any trend you observe when looking at only the first ionization energies of each of the 13 elements.

Answer :-

IONIC RADII TREND:- Ionisation energy is inversely proportional to the size or radii of the atom or ion i.e as the size increases then ionisation enthalpy decreases and vice-versa . we know for 13 elements in the given table ionisation energy increases down the group therefore size of the ions decreases .  

Hence Ionic radii or size decreases down the group for 13 elements in the given table .