An element has (Ar) 3d1 configuration in its +2 oxidation state. Its position in the periodic table is

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Solution

Dear Student,

Electronic configuration of the element in +2 oxidation state is [Ar] 3d¹.
Thus the atomic number of the element should be = atomic number of Ar+1+2= 18+1+2=21
Thus the element is Scandium. It is a d block element
The electronic configuration of Sc= [Ar] 3d¹ 4s². Here, n=4 therefore it belongs to the 4th period.
Group=total number of valence shell electrons=3
So, the element belongs to group 3, period 4.

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An element has $[A r] 3 d^{4}$ configuration in its +3 oxidation state. The atomic number of element is:

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[A r] 3 d^{4}

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1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}

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