What is the electronic arrangement of a metal | vedclass

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(C) In an octahedral complex,the $d$-orbitals split into two sets: $t_{2g}$ (lower energy) and $e_g$ (higher energy).For a $d^4$ configuration,if $\De

Vedclass · Accepted Answer

$t^3_{2g} e^1_g$

Vedclass · Answer

(C) In an octahedral complex,the $d$-orbitals split into two sets: $t_{2g}$ (lower energy) and $e_g$ (higher energy).For a $d^4$ configuration,if $\Delta_o First,three electrons occupy the $t_{2g}$ orbitals singly.The fourth electron will occupy the $e_g$ orbital because the energy required to pair in $t_{2g}$ $(	ext{pairing energy})$ is greater than the energy required to promote the electron to the $e_g$ level $(\Delta_o)$.Thus,the configuration is $t^3_{2g} e^1_g$.

List-$I$ (Complex)	List-$II$ ($CFSE$ in $\Delta_0$)
$A$. $[Cu(NH_3)_6]^{2+}$	$I$. $-0.6$
$B$. $[Ti(H_2O)_6]^{3+}$	$II$. $-2.0$
$C$. $[Fe(CN)_6]^{3-}$	$III$. $-1.2$
$D$. $[NiF_6]^{4-}$	$IV$. $-0.4$

What is the electronic arrangement of a metal atom/ion in an octahedral complex with a $d^4$ configuration,if $\Delta_o < \text{pairing energy}$?

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Match List-$I$ with List-$II$:

List-$I$ (Complex) List-$II$ ($CFSE$ in $\Delta_0$)

$A$. $[Cu(NH_3)_6]^{2+}$ $I$. $-0.6$

$B$. $[Ti(H_2O)_6]^{3+}$ $II$. $-2.0$

$C$. $[Fe(CN)_6]^{3-}$ $III$. $-1.2$

$D$. $[NiF_6]^{4-}$ $IV$. $-0.4$

Choose the correct answer from the options given below:

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