What is the electronic arrangement of a metal | vedclass

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(C) In an octahedral complex,the $d$-orbitals split into $t_{2g}$ and $e_g$ sets.When $\Delta_{0} < P$ (where $\Delta_{0}$ is the crystal field splitt

Vedclass · Accepted Answer

$t_{2g}^3 e_g^1$

Vedclass · Answer

(C) In an octahedral complex,the $d$-orbitals split into $t_{2g}$ and $e_g$ sets.When $\Delta_{0} For a $d^4$ configuration,the first three electrons occupy the lower energy $t_{2g}$ orbitals singly according to Hund's rule.The fourth electron will occupy the higher energy $e_{g}$ orbital because the energy required to pair it in the $t_{2g}$ orbital is greater than the energy required to promote it to the $e_{g}$ orbital.Therefore,the electronic arrangement is $t_{2g}^3 e_g^1$.

List-$I$ (Complex)	List-$II$ ($CFSE$ in $\Delta_0$)
$A$. $[Ti(H_2O)_6]^{2+}$	$I$. $-1.2$
$B$. $[V(H_2O)_6]^{2+}$	$II$. $-0.8$
$C$. $[Mn(H_2O)_6]^{3+}$	$III$. $-0.6$
$D$. $[Fe(H_2O)_6]^{3+}$	$IV$. $0$

What is the electronic arrangement of a metal atom/ion in an octahedral complex with $d^4$ configuration,if $\Delta_{0} < P$ (pairing energy)?

Similar Questions

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

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

$A$. $[Ti(H_2O)_6]^{2+}$ $I$. $-1.2$

$B$. $[V(H_2O)_6]^{2+}$ $II$. $-0.8$

$C$. $[Mn(H_2O)_6]^{3+}$ $III$. $-0.6$

$D$. $[Fe(H_2O)_6]^{3+}$ $IV$. $0$

The number of unpaired electrons in the complex ion $[CoF_6]^{3-}$ is (Atomic number of $Co = 27$).

Crystal field splitting energy $(CFSE)$ for $[CoCl_6]^{4-}$ is $18000 \ cm^{-1}$. The crystal field splitting energy $(CFSE)$ for $[CoCl_4]^{2-}$ will be (in $cm^{-1}$)

On the basis of crystal field theory,explain why $Co(III)$ forms a paramagnetic octahedral complex with weak field ligands,whereas it forms a diamagnetic octahedral complex with strong field ligands.

The electronic spectrum of $[Ti(H_2O)_6]^{3+}$ gives a single broad peak with a maxima at $20300 \, cm^{-1}$. The crystal field stabilization energy is equal to .......... $kJ \, mol^{-1}$

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