Of the following complex ions,which is diamag | vedclass

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Question

(B) The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$. In $[Ni(CN)_4]^{2-}$,$Ni$ is in the $+2$ oxidation state,so its configuration is $Ni^{2+

Vedclass · Accepted Answer

$[Ni(CN)_4]^{2-}$

Vedclass · Answer

(B) The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$. In $[Ni(CN)_4]^{2-}$,$Ni$ is in the $+2$ oxidation state,so its configuration is $Ni^{2+} = 3d^8$.Since $CN^-$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbitals.This results in $d^8$ configuration with all electrons paired,making the complex diamagnetic.In contrast,$[NiCl_4]^{2-}$,$[CuCl_4]^{2-}$,and $[CoF_6]^{3-}$ contain unpaired electrons and are paramagnetic.

List-$I$	List-$II$
$a. [Fe(CN)_{6}]^{3-}$	$i. 5.92 \ BM$
$b. [Fe(H_{2}O)_{6}]^{3+}$	$ii. 0 \ BM$
$c. [Fe(CN)_{6}]^{4-}$	$iii. 4.90 \ BM$
$d. [Fe(H_{2}O)_{6}]^{2+}$	$iv. 1.73 \ BM$

Compound	Total stereoisomers
$A$. $[Co(en)_2Cl_2]^+$	$P$. $2$
$B$. $[Co(en)_3]Cl_3$	$Q$. $3$
$C$. $[Pt(NH_3)_2Cl_2]$	$R$. $4$
$D$. $[Pt(NH_3)_3Cl_3]^+$	$S$. $5$

Of the following complex ions,which is diamagnetic in nature?

Similar Questions

Which one of the following complexes exhibits chirality?

The number of geometric isomers that can exist for the square planar complex $[Pt(Cl)(py)(NH_3)(NH_2OH)]^+$ is ($py=$ pyridine):

Match List-$I$ with List-$II$.

List-$I$ List-$II$

$a. [Fe(CN)_{6}]^{3-}$ $i. 5.92 \ BM$

$b. [Fe(H_{2}O)_{6}]^{3+}$ $ii. 0 \ BM$

$c. [Fe(CN)_{6}]^{4-}$ $iii. 4.90 \ BM$

$d. [Fe(H_{2}O)_{6}]^{2+}$ $iv. 1.73 \ BM$

Choose the correct answer from the options given below.

Compound Total stereoisomers
$A$. $[Co(en)_2Cl_2]^+$ $P$. $2$
$B$. $[Co(en)_3]Cl_3$ $Q$. $3$
$C$. $[Pt(NH_3)_2Cl_2]$ $R$. $4$
$D$. $[Pt(NH_3)_3Cl_3]^+$ $S$. $5$

Match the compounds with their total number of stereoisomers and select the correct code:

$[Co(NH_3)_4Cl_2]NO_2$ and $[Co(NH_3)_4Cl.NO_2]Cl$ are $........$ isomers.

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