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The highest oxidation state shown by element with atomic number 23 is:
A. +5
B. +4
C. +3
D. +6

Last updated date: 20th Jun 2024
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Hint: The oxidation state of an atom is the number of electrons lost or gained by the atom reacting with any species. Highest oxidation state corresponds to the maximum number of electrons an atom can gain or lose to form a stable compound with reacting species.

Complete step-by-step answer:
Electrons that are present in the outermost shell are generally known as valence electrons and the number of valence electrons determines the oxidation state of an atom. We can also say that it is the number of electrons gained or lost from an atom during a chemical reaction.
The oxidation state of the elements belonging to the s-block and the p-block of the periodic table are generally calculated as the number of outermost electrons or eight minus the number of outermost electrons. For the d-block and f-block elements, it is determined not only on the basis of outermost electrons but also on d and f orbital electrons.
Element having atomic number 23 has electronic configuration \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^3}\]. From this, we can see that its outermost electrons are in d-orbital so the element belongs to the 3d-block of periodic table. Five outermost electrons are there in the element of atomic number 23 and the element is named Vanadium.
It can lose maximum five electrons to attain stable electronic configuration and thus it has the highest oxidation number +5.

Hence, the correct option is (A).

Note: Each transition element can show a minimum oxidation state corresponding to the number of s-electrons and maximum oxidation state equal to the total number of electrons present in both s and d-orbitals. In between oxidation states also become possible such as foe Vanadium, +2,+3,+4 also possible other than +5.