JEE Main 2025 — D And F Block Elements Question with Solution

We need to determine the number of electrons each element has in its 4d orbitals and then add them together.

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1. Niobium $(\mathrm{Nb},\mathrm{Z}=41)$

The ground‐state electronic configuration of niobium is:

$[\mathrm{Kr}]4d^{4}5s^1.$

Hence, Niobium has 4 electrons in its 4d orbitals.

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2. Ruthenium $(\mathrm{Ru},\mathrm{Z}=44)$

The ground‐state electronic configuration of ruthenium is:

$[\mathrm{Kr}]4d^{7}5s^1.$

Hence, Ruthenium has 7 electrons in its 4d orbitals.

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3. Sum of 4d Electrons

$x=4(\text{for Nb}),y=7(\text{for Ru}).$

Therefore,

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

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Answer: 11