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Explain briefly how $ + 2$ state becomes more and more stable in the first half of the first row transition elements with increasing atomic number ?

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Last updated date: 24th Jun 2024
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Answer
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Hint: The $ + 2$ oxidation state is more stable in the first half of the first row transition elements due to least repulsion and increase in nuclear charges as we move ahead in the series of transition elements.

Complete step by step answer: Transition elements are referred to those elements that have partially filled d-orbital . These are called transition elements because they are present between groups $2$ and $13$ in the main group elements .
The elements of the first transition series are those for which the $3d$ electron shell contains between one and nine electrons.
The first row of transition elements varies from scandium $\left( {Sc} \right)$ to manganese $\left( {Mn} \right)$.
The transition elements acquire greater horizontal similarities in the properties in comparison to the other main group elements. The reason behind this is the common $n{s^2}$ configuration of the outermost shell.
If we look up to the common oxidation states of the transition elements , it reveals that except scandium $\left( {Sc} \right)$ , the most common oxidation state of the first row transition elements is $ + 2$ that arises from the loss of two elements from $4s$ orbital. This infers that leaving scandium $\left( {Sc} \right)$, the d-orbitals become more stable than s-orbital. Moreover , the $ + 2$ state becomes more stable in the first half of first row transition elements with increasing atomic number because $3d$ orbitals have only one element in each of $5,3d$ orbitals. Also, the repulsions are least and nuclear charge increases.

Note: So, due to least repulsion as the nuclear change increases, the $ + 2$ state seems to be stable in the first half row of transition elements whereas in the next half the electrons start to pair up resulting in greater electron repulsions.