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(B) To determine the highest paramagnetism,we calculate the number of unpaired electrons in each ion:$1$. $[Cr(H_2O)_6]^{3+}$: $Cr$ is in $+3$ oxidati

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$[Fe(H_2O)_6]^{2+}$

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(B) To determine the highest paramagnetism,we calculate the number of unpaired electrons in each ion:$1$. $[Cr(H_2O)_6]^{3+}$: $Cr$ is in $+3$ oxidation state. $Cr$ $(Z=24)$ is $[Ar] 3d^5 4s^1$. $Cr^{3+}$ is $[Ar] 3d^3$. Number of unpaired electrons $(n)$ = $3$.$2$. $[Fe(H_2O)_6]^{2+}$: $Fe$ is in $+2$ oxidation state. $Fe$ $(Z=26)$ is $[Ar] 3d^6 4s^2$. $Fe^{2+}$ is $[Ar] 3d^6$. Since $H_2O$ is a weak field ligand,electrons remain unpaired: $t_{2g}^4 e_g^2$. Number of unpaired electrons $(n)$ = $4$.$3$. $[Zn(H_2O)_6]^{2+}$: $Zn$ is in $+2$ oxidation state. $Zn$ $(Z=30)$ is $[Ar] 3d^{10} 4s^2$. $Zn^{2+}$ is $[Ar] 3d^{10}$. Number of unpaired electrons $(n)$ = $0$.$4$. $[Cu(H_2O)_6]^{2+}$: $Cu$ is in $+2$ oxidation state. $Cu$ $(Z=29)$ is $[Ar] 3d^{10} 4s^1$. $Cu^{2+}$ is $[Ar] 3d^9$. Number of unpaired electrons $(n)$ = $1$.Since $[Fe(H_2O)_6]^{2+}$ has the maximum number of unpaired electrons $(n=4)$,it exhibits the highest paramagnetism.

$A$. Tetrahedral $NiCl_2Br_2^{2-}$	$B$. Square planar $Pt(NH_3)_2Cl_2$
$C$. Octahedral $Co(NH_3)_3Cl_3$	$D$. Square planar $Pd(NH_3)_3Br^{+}$
$E$. Octahedral $Co(en)_3^{3+}$

Amongst the following ions,which one has the highest paramagnetism?

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Amongst $[Ni(H_{2}O)_{6}]^{2+}$, $[Ni(PPh_{3})_{2}Cl_{2}]$, $[Ni(CO)_{4}]$ and $[Ni(CN)_{4}]^{2-}$, the paramagnetic species are

Which of the following would exhibit ionisation isomerism?

Which of the following molecules/ions can exhibit isomerism:
$A$. Tetrahedral $NiCl_2Br_2^{2-}$ $B$. Square planar $Pt(NH_3)_2Cl_2$
$C$. Octahedral $Co(NH_3)_3Cl_3$ $D$. Square planar $Pd(NH_3)_3Br^{+}$
$E$. Octahedral $Co(en)_3^{3+}$

Here,$en = 1,2-\text{diaminoethane}$.

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Of the following,which is diamagnetic in nature?

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