Question 14

The pair of ions having same electronic configuration is.............

(a) $C r^{3 +}, F e^{3 +}$ (b) $F e^{3 +}, M n^{2 +} +$ (c) $F e^{3 +}, C o^{3 +}$ (d) $S C^{3 +}, C r^{3 +}$

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Solution

$_{24} C r = [A r] 3 d^{5}, 4 s^{1}$ $_{24} C r^{3 +} = [A r] 3 d^{3}$
$_{26} F e = [A r] 3 d^{6}, 4 s^{2}$ $_{26} F e^{3 +} = [A r] 3 d^{5}$
$_{25} M n = [A r] 3 d^{5}, 4 s^{2}$ $_{26} M n^{2 +} = [A r] 3 d^{5}$
$_{27} C o = [A r] 3 d^{7}, 4 s^{2}$ $_{27} C O^{3 +} = [A r] 3 d^{6}$
$_{21} S o = [A r] 3 d^{1}, 4 s^{2}$ $_{21} S c^{3 +} = [A r]$

Thus, $F e^{3 +}$ and $M n^{2 +}$ have the same electronic confifuratoin.

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Similar questions

F e^{2 +}, F e^{3 +}, C r^{2 +}, C r^{3 +}, M n^{2 +}, M n^{3 +}

F e^{2 +}, F e^{3 +}, C r^{2 +}, C r^{3 +}, M n^{2 +}, M n^{3 +}

C r^{3 +}

F e^{3 +}

The oxidation half reaction for following reaction is

Fe²⁺(aq) + Cr₂O₇^2-(aq) → Fe³⁺ (aq) + Cr³⁺(aq)

F e^{3 +}

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