Animated Solution for Chemistry - Periodicity in Properties: Outermost electronic configuration of a group-13 element $E$ is $4s^{2}4p^1$. The electronic configuration of an element of p-block period-five placed diagonally to element, $E$ is

The periodic table is not just a list of elements; it is a highly structured, geometric map of chemical behavior. When you understand how to navigate its rows and columns, deducing the properties and electronic configurations of elements becomes a logical game rather than an exercise in rote memorization. Let's embark on a journey to solve this classic JEE problem by decoding the map.

Element E

Our journey begins with a mysterious element, $E$. We are given two crucial pieces of intelligence: it belongs to Group 13, and its outermost electronic configuration is $4s^{2}4p^1$.

What does this tell us? The principal quantum number of the outermost shell is $n=4$. In the architecture of the periodic table, the principal quantum number of the valence shell directly corresponds to the period of the element. Therefore, element $E$ resides in the 4th period.

Combining this with the fact that it is in Group 13, we can pinpoint its exact location on our map. The element in Period 4, Group 13 is Gallium (Ga). We have successfully established our starting coordinate.

The Diagonal Shift

The problem now asks us to find the electronic configuration of an element placed diagonally to element $E$ in the p-block. But what exactly does a 'diagonal' move entail in this context?

Imagine standing on the square for Gallium. To move diagonally downwards and to the right, you must take one step down and one step to the right.

Taking one step down means moving to the next period. So, our period number increases by 1:

Taking one step to the right means moving to the next group. So, our group number increases by 1:

Our new coordinates are Period 5, Group 14. Looking at the periodic table, the element occupying this specific slot is Tin (Sn). We have found our target!

Now comes the final and most critical phase: writing the full electronic configuration for Tin (Sn). We must build this systematically to avoid common traps.

First, we identify the noble gas core. Since Tin is in the 5th period, its inner core of electrons will perfectly match the noble gas at the end of the previous period (Period 4). That noble gas is Krypton (Kr). So, our configuration begins with $[\text{Kr}]$.

Next, we determine the valence shell. Tin is in Group 14, which means it has 4 valence electrons. Because it is in the 5th period, these electrons will populate the $5s$ and $5p$ orbitals. Following the Aufbau principle, the $5s$ orbital fills first, followed by the $5p$ orbital, giving us a valence configuration of $5s^{2}5p^2$.

The Crucial Trap: It is incredibly easy to just write $[\text{Kr}]5s^{2}5p^2$ and move on. However, we must remember the d-block! Between the $5s$ and $5p$ orbitals, the periodic table houses the entire 4d transition series. Before an electron can ever enter the $5p$ subshell, the lower-energy $4d$ subshell must be completely filled with 10 electrons.

Therefore, we must insert the fully occupied $4d^{10}$ subshell into our configuration. Combining all these pieces, the complete and correct electronic configuration for Tin is:

This perfectly matches option (d). By trusting the geometric logic of the periodic table and carefully applying the Aufbau principle, we have flawlessly navigated from an abstract configuration to the exact identity and structure of a completely different element.