Chemistry - PERIODIC TABLE Question with Solution | TestHub

The incorrect order is $Y<La(I{E}_1)$. Due to lanthanide contraction, La has a smaller atomic radius and higher effective nuclear charge than expected, leading to a higher first ionization energy ($I{E}_1$) than Y. Thus, $Y(I{E}_1)>La(I{E}_1)$ is incorrect.

Explanation:

The first ionization energy ($I{E}_1$) is the energy required to remove the outermost electron from a gaseous atom. Generally, ionization energy decreases down a group and increases across a period. However, there are exceptions due to factors like electron configuration and shielding effects.

Let's analyze each option:

A. $Cd<Hg(I{E}_1)$

This order is correct. In Group 12, as we move from Cadmium (Cd) to Mercury (Hg), the atomic size increases, but the presence of filled 4f orbitals in Hg leads to poor shielding and a significantly higher effective nuclear charge. This results in a stronger attraction for the valence electrons, making it harder to remove them. Therefore, $I{E}_1$ of Hg is higher than Cd.

B. $Ag<Au(I{E}_1)$

This order is correct. Similar to the Cd-Hg trend, moving from Silver (Ag) to Gold (Au) in Group 11, the lanthanide contraction effect in Au causes its atomic radius to be unexpectedly small and its effective nuclear charge to be high. This leads to a higher $I{E}_1$ for Au compared to Ag.

C. $Y<La(I{E}_1)$

This order is incorrect. Yttrium (Y) is in Period 5, and Lanthanum (La) is in Period 6. Normally, $I{E}_1$ decreases down a group. However, due to the lanthanide contraction, the 4f orbitals are filled before La, leading to a significant increase in effective nuclear charge for La. This makes it harder to remove an electron from La than from Y. Therefore, $I{E}_1$ of La is higher than Y, meaning $La>Y(I{E}_1)$. The given order $Y<La(I{E}_1)$ is incorrect.

D. None of these

Final Answer is C.