Ionization Energy Formula: Valence Electrons, Types & Factors

Sample Questions

Ques. Among the second period elements, the actual ionization enthalpies are in the order Li<B

Explain why [5 marks]

1. Be has higher Ionization energy \(\triangle\)_iH than B?
2. O has lower \(\triangle\)_iH than N? 

Ans. As we all know, Ionization energy is increasing as we go from left to right within the period because, as the atomic number increases, the electrons are added in the shells.

1. First consider the electronic configuration of Be and B.

Boron(B) = 1s²,2s²,2p¹

Beryllium(Be) = 1s², 2s²

From the electronic configuration, it is clear that Be has completely filled orbitals and it is more stable because the electrons present in the outermost shell are tightly bound to the nucleus. On the other hand, B has one electron present in the p-subshell but it is half-filled which makes it less stable. Also, it is loosely bound to the nucleus.

So. less ionization energy is required to remove an electron from Boron. As a result, Beryllium has higher ionization energy than Boron.

2. Take the electronic configuration of the elements:

O=1s²,2s²,2p⁴

N=1s²,2s²,2p³ 

Oxygen has 4 electrons in its p-orbital and it is unstable while, Nitrogen has 3 electrons and it has stable half-filled p-orbital. The removal of electrons is more difficult in the case of Nitrogen rather than Oxygen because the force of attraction is more in Nitrogen. So, it requires more energy to escape the electron from the valence shell. That’s why Oxygen has lower Ionization Energy than N. 

Ques. The first ionization enthalpy values of the third-period elements, Na, Mg and Si are 496,737 and 786kJ/mol respectively. Predict whether the first \(\triangle\)_iH₁ value for Al will be more close to 575 or 760 kJ/mol. Justify your answers. [2 marks]

Ans. The electronic configuration of Al and Mg is below:

Al = [Ne] 3s²3p¹

Mg = [Ne] 3s² 

The first ionization energy is lower than that of Mg because in the outermost shell there's only one electron present in p=orbit. The force of attraction between the nucleus and p-orbital is less as compared to the s-orbital. So, it is easier to remove an electron from the p-orbital of Al than the s-orbital of Mg. Thus, the first ionization energy of Al is less as compared to Mg. The first ionization energy of Al will be close to 575 Kj/mol. 

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Ques.The electronic configuration for some neutral atoms are: 

• 1s², 2s²
• 1s², 2s² ,2p⁶,3s¹
• 1s², 2s² ,2p⁶
• 1s², 2s² ,2p³

Which of these electronic configurations would you expect to have 

1. the highest first ionization enthalpy?
2. the highest second ionization enthalpy? [3 marks]

Ans.

1. (c) will have the highest first ionization enthalpy because it has 8 electrons in its valence shell. Due to this, it has a stable noble gas electronic configuration. More ionization energy is required to remove the electron from the outermost shell.
2. (b) will have the second-highest ionization enthalpy because after releasing the electron, it forms a cation and the remaining electron present in the valence shell will be 8. After that, it acquires a noble gas configuration in which removal of an electron is difficult and also more ionization energy is required to remove the electron.

Ques. Why is the second ionization enthalpy of boron higher than the second ionization enthalpy of carbon in spite of the fact that the first ionization enthalpy of boron is less than carbon? [2 marks] 

Ans. Boron has lower first ionization enthalpy than Carbon because ionization energy increases as we go from left to right in a period. Also, the atomic size of carbon is more than boron. But, after removing one electron, boron acquires a completely filled s-orbital configuration. It is in a stable state and it is very difficult to remove the electron from the stable state. So, the second ionization enthalpy of boron is more than carbon because in the case of carbon after removing one electron, it still has one electron left in the p-orbital. 

Ques. Why do inert gases have higher ionization enthalpy than halogens? [2 marks]

Ans. Inert gases have higher ionization enthalpy than halogens because inert gases have the most stable electronic configuration. Also, the force of attraction between the outermost shell and nucleus is more as compared to halogens. So, it is very difficult to remove the electron from the outermost shell. As a result, inert gases have more ionization enthalpy than halogens. 

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Ques. The first ionization enthalpy(\(\triangle\)_iH₁) and the second ionization enthalpy(\(\triangle\)_iH₂) in kJ/mol of three elements I,II,III are given below: [3 marks]

Identify the elements which are likely to be

• a non-metal
• an alkali metal 
• an alkaline earth metal. 

Ans.

1. Non-metals have higher ionization enthalpies than metals. In the ‘III’ category both the enthalpies are high. So, these are non-metal. 
2. In Alkali metals, the first ionization enthalpy is lower while the second is the highest. In the ‘I’ category first ionization enthalpy is low as 403.0 while the second is highest as 2640. So, the ‘I’ category elements are Alkali metals.
3. In Alkaline metals, mostly two electrons are present in the valence shell. In the ‘II’ category the first and second ionization enthalpies are in the ratio of 1:2. So, these are alkaline metals. 

Ques. The first ionization enthalpy values (in kJ/mol) of group 13 elements are:

How would you explain this deviation from the general trend? [5 marks]

Ans. As we all know that first ionization enthalpy decreases as we move from top to bottom which is the general nature of the elements of the periodic table. This kind of similar behaviour is expected from the elements of group 13. It means the first ionization energy B >Al>Ga>In>Tl. This behaviour is clearly seen in the B and Al. But, the ionization energy of Ga is higher than Al which contradicts the statement.

Ga has electronic configuration is [Ar] 3d¹⁰, 4s², 4p¹ and Al has electronic configuration is [Ne] 3s²3p¹. The removal of electrons from the 4p-orbital is easy as compared to 3p-orbital. But, in the case of Ga electrons are also present in the d-orbital which have a poor screening effect. Due to the poor screening effect, the removal of electrons from the 4p-orbital requires a little bit more energy than Al. So, the first ionization energy of Ga is more than Al.

Moving towards Ga to In, the first Ionization energy is less than that of Ga. it is because of the inert gas configuration of In. It has configuration [Kr] 4d¹⁰, 5s², 5p¹ .Due to the presence of electrons in the 4d-orbital, the screening effect occurs. Due to the screening effect, Nuclear charge increases. As a result, Indium has less ionization enthalpy than Ga.

Moving Further from In to Tl, it is seen that again Tl has more first ionization enthalpy than In. It is because of the addition of electrons in the f-orbital. Tl has inert core configuration [Xe] 3f¹⁴, 5d¹⁰, 6s²,6p¹

Due to the addition of electrons in the f and d-subshell, the nuclear charge increased by 32 protons. This increase in nuclear charge overweighs the effect of screening by the 4f and 5d-subshell. Due to the poor screening effect, electrons are strongly bound to the nucleus and hence require more energy. That’s why Tl has higher first ionization energy than In.

Ques. Write two important limitations of the Rutherford nuclear model of the atom. (CBSE 2016) [2 marks]

Ans. Important limitations of Rutherford Model :

1. According to the Rutherford model, electrons orbiting around the nucleus continuously radiates energy due to the acceleration; hence the atom will not remain stable.
2. As the electron spirals inwards; its angular velocity and frequency change continuously; therefore it. will emit a continuous spectrum.

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Ques.

1. Using Bohr’s second postulate of quantization of orbital angular momentum show that the circumference of the electron in the n,h orbital state in a hydrogen atom is n times the de-Broglie wavelength associated with it.
2. The electron in a hydrogen atom is initially in the third excited state. What is the maximum number of spectral lines which can be emitted when it finally moves to the ground state? (CBSE 2012) [3 marks]

Ans.

1. According to the de-Broglie hypothesis, this electron is also associated with wave character.

Hence a circular orbit can be taken to be a stationary energy state only if it contains an integral number of de- Broglie wavelengths i.e. we must have

2πr = nλ

According to Bohr’s-second postulate

mvr = nh/ 2π

2πr = nh/ mv

2πr = nλ

Hence the circumference of the electron in the nth orbital state in a hydrogen atom is n times the de-Broglie wavelength associated with it.

2. For third excited state n = 4

For ground-state n = 1

Hence, possible transitions are :

ni = 4 to nf = 3,2,1

ni = 3 to nf = 2,1

ni = 2 to nf = 1

Total number of transitions = 6

Ques. Write two important limitations of the Rutherford model which could not explain the observed features of atomic spectra. How were these explained in Bohr’s model. of a hydrogen atom? [3 marks]

Use the Rydberg formula to calculate the wavelength of the Hα line.

(Take R = 1.1 × 107 m^-1).

Ans.

Limitations of Rutherford Model:

1. Electrons moving in a circular orbit around the nucleus would get, accelerated, therefore it would spiral into the nucleus, as it loses its energy.
2. It must emit a continuous spectrum.

The explanation according to Bohr’s model of hydrogen atom:

1. Electron in an atom can revolve in certain stable orbits without the emission of radiant energy.
2. Energy is released/absorbed only when an electron jumps from one stable orbit to another stable orbit. This results in a discrete spectrum.

The wavelength of Hα line:

Hα line is formed when an electron jumps from nf = 3 to ni = 2 orbits. It is the Balmer series.

\(\frac{1}{\lambda} = R \left(\frac{1}{2^2}-\frac{1}{3^2}\right)\)

\(\frac{1}{\lambda} = 1.1 \times 10^7 \left(\frac{1}{4}-\frac{1}{9}\right) or \lambda = 656.3 nm\)