Ionisation enthalpy $({\Delta _i}{{\rm H}_1}KJmo{l^{ - 1}})$ for the elements of group 13 follows the order:
A. B$\,>\,$Al$\,>\,$Ga$\,>\,$In$\,>\,$Tl
B. B$\,<\,$Al$\,>\,$Ga$\,<\,$In$\,>\,$Tl
C. B$\,<\,$Al$\,>\,$Ga$\,>\,$In$\,>\,$Tl
D. B$\,>\,$Al$\,<\,$Ga$\,>\,$In$\,<\,$Tl
Answer
629.1k+ views
Hint:Let us understand what ionisation enthalpy is, it is the energy required to remove the most loosely bound electron from the isolated gaseous atom. It depends on the size of the atom, charge on the nucleus, type of the electron evolved and the effective screening effect by the inner electron shells.
Complete answer:
Though when the other factors are equal the order of I.E in orbitals is s>p>d>f. But the increase in I.E is not quite smooth in the group-13 whose elements have been listed in the options. For ex. The first I.E for the group-13 where a p electron is removed is actually less than that for the adjacent group-2 where an s electron is being removed.
The values for B and Al are lower because removal of one electron leaves a stable filled s shell. In general the first I.E decreases in a regular way on descending the main groups. A departure from this trend occurs in group-13, where the expected decrease occurs between B and Al, but the values for the remaining elements do not continue the trend, and are irregular the reason for the change at Ga is that it is proceeded by ten elements of the first transition series where the 3d shell is being filled this makes Ga smaller than it would otherwise be. A similar effect is observed with the second and third transition series, and the presence of the three transition series not only has a marked effect on the values for Ga, In, Tl but the effect still shows in group 14,15. Tl has higher I.E than In due to poor shielding of 14f electrons in the inner shell.
Therefore from above we conclude the I.E trend for the transition series is B$\,>\,$Al$\,<\,$Ga$\,>\,$In$\,<\,$Tl and the correct option is D.
Note:
The factors listed above are interrelated. In a small atom the electrons are tightly held, whilst in a larger atom the electrons are less strongly held. Thus the ionization enthalpy decreases as the size of the atoms increases.
Complete answer:
Though when the other factors are equal the order of I.E in orbitals is s>p>d>f. But the increase in I.E is not quite smooth in the group-13 whose elements have been listed in the options. For ex. The first I.E for the group-13 where a p electron is removed is actually less than that for the adjacent group-2 where an s electron is being removed.
The values for B and Al are lower because removal of one electron leaves a stable filled s shell. In general the first I.E decreases in a regular way on descending the main groups. A departure from this trend occurs in group-13, where the expected decrease occurs between B and Al, but the values for the remaining elements do not continue the trend, and are irregular the reason for the change at Ga is that it is proceeded by ten elements of the first transition series where the 3d shell is being filled this makes Ga smaller than it would otherwise be. A similar effect is observed with the second and third transition series, and the presence of the three transition series not only has a marked effect on the values for Ga, In, Tl but the effect still shows in group 14,15. Tl has higher I.E than In due to poor shielding of 14f electrons in the inner shell.
Therefore from above we conclude the I.E trend for the transition series is B$\,>\,$Al$\,<\,$Ga$\,>\,$In$\,<\,$Tl and the correct option is D.
Note:
The factors listed above are interrelated. In a small atom the electrons are tightly held, whilst in a larger atom the electrons are less strongly held. Thus the ionization enthalpy decreases as the size of the atoms increases.
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