The correct order of $E_{M^{2 +} / M}$ values with negative sign for the four successive elements $C r$ , $M n$ , $F e$ and $C o$ is?

M n > C r > F e > C o

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C r > F e > M n > C o

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F e > M n > C r > C o

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C r > M n > F e > C o

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Solution

The correct option is C $M n > C r > F e > C o$
${E_{M}}^{2 +} / M$
$_{25} M n ⟶ 4 s^{2} 3 d^{5}$ ( $E_{M} n^{2 +} / M n$ will be more negative than expected because of its' extra stability due to half filled $d -$ orbital)
$M n_{+} ⟶ 4 s^{1} 3 d^{5}$
$M n^{2 +} ⟶ 4 s^{0} 3 d^{5}$

$_{24} C r ⟶ 4 s^{1} 3 d^{5}$
$C r^{+} ⟶ 4 s^{0} 3 d^{5}$
$C r^{2 +} ⟶ 4 s^{0} 3 d^{4}$

$_{26} F e ⟶ 4 s^{2} 3 d^{6}$
$F e^{+} ⟶ 4 s^{1} 3 d^{6}$
$F e^{2 +} ⟶ 4 s^{0} 3 d^{6}$

$_{27} C o ⟶ 4 s^{2} 3 d^{7}$
$C o^{+} ⟶ 4 s^{1} 3 d^{7}$
$C o^{2 +} ⟶ 4 s^{0} 3 d^{7}$

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Similar questions

E^{0} M^{2 +} / M

For the four successive transition elements (Cr, Mn, Fe and Co), what is the order of stability of their +2 states?

(At. Nos. Cr = 24, Mn = 25, Fe = 26, Co = 27)

E_{M^{2 +} / M}^{o}

C r, M n, F e

C o

E^{0} M^{2 +} / M

E_{M^{2 +} / M}^{0}

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