The correct match for the complex with its ma | vedclass

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(B) $1$. For $[Zn(OH_2)_6]^{2+}$: $Zn^{2+}$ is $3d^{10}$,which has no unpaired electrons,so it is diamagnetic.$2$. For $[Co(NH_3)_6]^{3+}$: $Co^{3+}$

Vedclass · Accepted Answer

$[Co(NH_3)_6]^{3+}$; diamagnetic

Vedclass · Answer

(B) $1$. For $[Zn(OH_2)_6]^{2+}$: $Zn^{2+}$ is $3d^{10}$,which has no unpaired electrons,so it is diamagnetic.$2$. For $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ is $3d^6$. $NH_3$ is a strong field ligand,causing pairing of electrons. It forms $d^2sp^3$ hybridization with no unpaired electrons,making it diamagnetic.$3$. For $[CoF_6]^{3-}$: $Co^{3+}$ is $3d^6$. $F^-$ is a weak field ligand,so no pairing occurs. It has $4$ unpaired electrons,making it paramagnetic.$4$. For $[V(OH_2)_6]^{2+}$: $V^{2+}$ is $3d^3$,which has $3$ unpaired electrons,making it paramagnetic.Therefore,the correct match is $[Co(NH_3)_6]^{3+}$; diamagnetic.

The correct match for the complex with its magnetic behavior is:

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