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# For $4{d^3}5{s^2}$ configuration belongs to which group?A. IIAB. IIBC. VBD. IIIB

Last updated date: 20th Jun 2024
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Answer
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Hint: The configuration $4{d^3}5{s^2}$is the outer electronic configuration of element Niobium which is represented by Nb. Its atomic number is 41. n is the number of electrons present in s orbital.

Complete answer:
The electronic configuration of the element niobium (Nb) is $[Kr]4{d^3}5{s^2}$.
The symbol [Kr] is the condensed electron configuration of the element krypton present in the ground state. It represents all the electrons present in this element. The ground state atom of this element will contain the electronic configuration $4{d^3}5{s^2}$. This element will be present just beneath the period in which krypton is present. Krypton is the last element of the fourth period. Therefore, the element is present in the fifth period of the periodic table. The value of n is 5.
The block of the element present in the periodic table is determined by the type of the occupied electron orbital with highest potential energy. The electron with the highest potential energy in the ground state atom of the element is present is 4d orbital. Therefore, the element is present in the d-block of the periodic table.
The group of the element is calculated by the formula as shown below.
$Group = ns + (n - 1)d{e^ - }$
$\Rightarrow Group = 2 + 3$
$\Rightarrow Group = 5$
Therefore, the group in which $4{d^3}5{s^2}$ configuration belongs is VB.
Therefore, the correct option is C.

Note:
The electron consisting of the highest potential energy in the ground state atom of the element is obtained using the Aufbau diagram. The actual electronic configuration of niobium is $[Kr]4{d^4}5{s^1}$ but the new electronic configuration is $[Kr]4{d^3}5{s^2}$which is stabilized with the Aufbau principle.