JEE Chemistry 2025 Question with Solution

Given: Niobium (Nb) and ruthenium (Ru) have $x$ and $y$ electrons in their respective $4d$ orbitals.

Find: The value of $x + y$.

Use the electronic configurations of the two elements.

For Nb with atomic number $41$:

$$[Kr] \, 4d^4 5s^1$$

So, $x = 4$.

For Ru with atomic number $44$:

$$[Kr] \, 4d^7 5s^1$$

So, $y = 7$.

Now add the number of $4d$ electrons:

$$x + y = 4 + 7 = 11$$

Therefore, the value of $x + y$ is $11$.

Given: $x$ is the number of electrons in the $4d$ orbitals of Nb, and $y$ is the number of electrons in the $4d$ orbitals of Ru.

Find: $x + y$.

Niobium (Nb) has configuration $[Kr] \, 4d^4 5s^1$, so the number of electrons in $4d$ is $4$.

Ruthenium (Ru) has configuration $[Kr] \, 4d^7 5s^1$, so the number of electrons in $4d$ is $7$.

Hence,

$$x + y = 4 + 7 = 11$$

Therefore, the required numerical value is $11$.