Explain briefly how the $+2$ oxidation state becomes more stable in the first half of the first-row transition elements as the atomic number increases.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The $+2$ oxidation state is achieved by the loss of two $4s$ electrons from the metal atoms. As we move from $Sc$ $(Z=21)$ to $Mn$ $(Z=25)$,the number of electrons in the $3d$ orbital increases from $1$ to $5$. The electronic configurations for the $+2$ ions are as follows:
$Sc^{2+}: 3d^1$
$Ti^{2+}: 3d^2$
$V^{2+}: 3d^3$
$Cr^{2+}: 3d^4$
$Mn^{2+}: 3d^5$
As the number of $d$ electrons increases,the $d$-orbital approaches a half-filled configuration. According to Hund's rule and the principle of exchange energy,a half-filled $d^5$ configuration is exceptionally stable. Therefore,the stability of the $+2$ oxidation state increases across the series,reaching a maximum stability at $Mn^{2+}$ due to its stable $d^5$ configuration.

Explore More

Similar Questions

Write the ground state electronic configurations of $Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu$.

Which one of the following sets correctly represents the increase in the paramagnetic property of the ions?

Transition metals are not characterized by

Which of the following pairs of ions are colored in their aqueous solution?

Which of the following is a colourless ion?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo