B & Ga
B & Tl
Ti & B
B & In
To determine the elements of Group 13 with the highest and lowest first ionization enthalpies, we need to understand the concept of ionization enthalpy and the trends in the periodic table.
Ionization Enthalpy: Ionization enthalpy is the energy required to remove the outermost electron from an isolated gaseous atom to form a cation. In general, the ionization enthalpy depends on factors such as the atomic size, nuclear charge, and electron shielding effect.
Trends in Group 13: As we move down the group in the periodic table, the atomic size increases due to the addition of new electron shells, which leads to an increase in distance between the nucleus and the outermost electron. This generally results in a decrease in ionization enthalpy down the group. However, due to the poor shielding effect of d and f orbitals (present in heavier elements like Ga and Tl), there are some exceptions to this trend.
Let's analyze the elements of Group 13:
From this analysis, we can conclude that:
Therefore, the elements of Group 13 with the highest and lowest first ionization enthalpies are Boron (B) and Gallium (Ga), respectively.
The question asks about the elements of Group 13 with the highest and lowest first ionization enthalpies. Group 13 of the periodic table consists of the elements: Boron (B), Aluminium (Al), Gallium (Ga), Indium (In), and Thallium (Tl).
Ionization Enthalpy: It is the energy required to remove an electron from an isolated gaseous atom or ion.
The general trend for ionization enthalpy across a group is that it decreases as we move down the group because the electrons are further away from the nucleus, and the effect of shielding increases.
Therefore, B has the highest ionization enthalpy, and Ga has relatively higher ionization energy among the lower elements despite being lower in the group's sequence, consequently making option B & Ga correct.
Conclusion: The elements of Group 13 with the highest and lowest first ionization enthalpies are Boron (B) and Gallium (Ga), respectively. Thus, the correct answer is B & Ga.
The correct orders among the following are:
[A.] Atomic radius : \(B<Al<Ga<In<Tl\)
[B.] Electronegativity : \(Al<Ga<In<Tl<B\)
[C.] Density : \(Tl<In<Ga<Al<B\)
[D.] 1st Ionisation Energy :
In\(<Al<Ga<Tl<B\)
Choose the correct answer from the options given below :
Given below are two statements:
Statement I: $ H_2Se $ is more acidic than $ H_2Te $
Statement II: $ H_2Se $ has higher bond enthalpy for dissociation than $ H_2Te $
In the light of the above statements, choose the correct answer from the options given below.
If $ \theta \in [-2\pi,\ 2\pi] $, then the number of solutions of $$ 2\sqrt{2} \cos^2\theta + (2 - \sqrt{6}) \cos\theta - \sqrt{3} = 0 $$ is:
A thin transparent film with refractive index 1.4 is held on a circular ring of radius 1.8 cm. The fluid in the film evaporates such that transmission through the film at wavelength 560 nm goes to a minimum every 12 seconds. Assuming that the film is flat on its two sides, the rate of evaporation is:
The major product (A) formed in the following reaction sequence is
