The paramagnetic complex ion which has no unp | vedclass

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(D) The correct answer is $[Ni(NH_3)_6]^{2+}$.$1$. $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $3d^6$. $CN^-$ is a strong field ligand,so all $6$ electrons are pa

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$[Ni(NH_3)_6]^{2+}$

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(D) The correct answer is $[Ni(NH_3)_6]^{2+}$.$1$. $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $3d^6$. $CN^-$ is a strong field ligand,so all $6$ electrons are paired in $t_{2g}$ orbitals $(t_{2g}^6 e_g^0)$. It is diamagnetic.$2$. $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ is $3d^5$. $CN^-$ is a strong field ligand,so $t_{2g}^5 e_g^0$. It has $1$ unpaired electron in $t_{2g}$ orbitals.$3$. $[Zn(NH_3)_6]^{2+}$: $Zn^{2+}$ is $3d^{10}$. All orbitals are fully filled. It is diamagnetic.$4$. $[Ni(NH_3)_6]^{2+}$: $Ni^{2+}$ is $3d^8$. $NH_3$ is a weak field ligand for $Ni^{2+}$. The crystal field splitting results in $t_{2g}^6 e_g^2$ configuration. Here,all $6$ electrons in $t_{2g}$ orbitals are paired,and there are $2$ unpaired electrons in $e_g$ orbitals,making it paramagnetic.

List-$I$ (Molecules/ion)	List-$II$ (Hybridisation of central atom)
$A. PF_5$	$I. dsp^2$
$B. SF_6$	$II. sp^3d$
$C. Ni(CO)_4$	$III. sp^3d^2$
$D. [PtCl_4]^{2-}$	$IV. sp^3$

The paramagnetic complex ion which has no unpaired electrons in $t_{2g}$ orbitals is

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Match the List-$I$ with List-$II$
List-$I$ (Molecules/ion) List-$II$ (Hybridisation of central atom)
$A. PF_5$ $I. dsp^2$
$B. SF_6$ $II. sp^3d$
$C. Ni(CO)_4$ $III. sp^3d^2$
$D. [PtCl_4]^{2-}$ $IV. sp^3$
Choose the correct answer from the options given below $:$

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