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(A) To determine the number of unpaired $d$ electrons,we analyze the electronic configuration of the central metal ions in the given octahedral comple

Vedclass · Accepted Answer

$2$

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(A) To determine the number of unpaired $d$ electrons,we analyze the electronic configuration of the central metal ions in the given octahedral complexes:$1$. $[V(H_2O)_6]^{3+}$: $V^{3+}$ is $3d^2$. Number of unpaired electrons $(n)$ = $2$ (Even).$2$. $[Cr(H_2O)_6]^{2+}$: $Cr^{2+}$ is $3d^4$. Since $H_2O$ is a weak field ligand,electrons fill according to Hund's rule: $t_{2g}^3 e_g^1$. Number of unpaired electrons $(n)$ = $4$ (Even).$3$. $[Fe(H_2O)_6]^{3+}$: $Fe^{3+}$ is $3d^5$. With a weak field ligand,$t_{2g}^3 e_g^2$. Number of unpaired electrons $(n)$ = $5$ (Odd).$4$. $[Ni(H_2O)_6]^{3+}$: $Ni^{3+}$ is $3d^7$. With a weak field ligand,$t_{2g}^5 e_g^2$. Number of unpaired electrons $(n)$ = $3$ (Odd).$5$. $[Cu(H_2O)_6]^{2+}$: $Cu^{2+}$ is $3d^9$. With a weak field ligand,$t_{2g}^6 e_g^3$. Number of unpaired electrons $(n)$ = $1$ (Odd).The complexes with an even number of unpaired electrons are $[V(H_2O)_6]^{3+}$ and $[Cr(H_2O)_6]^{2+}$.Therefore,the total number of such complexes is $2$.

Number of complexes from the following with an even number of unpaired $d$ electrons is . . . . . . . . . .
$[V(H_2O)_6]^{3+}, \quad [Cr(H_2O)_6]^{2+}, \quad [Fe(H_2O)_6]^{3+}, \quad [Ni(H_2O)_6]^{3+}, \quad [Cu(H_2O)_6]^{2+}$
[Given atomic numbers: $V=23, Cr=24, Fe=26, Ni=28, Cu=29$]

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Number of complexes from the following with an even number of unpaired $d$ electrons is . . . . . . . . . .$[V(H_2O)_6]^{3+}, \quad [Cr(H_2O)_6]^{2+}, \quad [Fe(H_2O)_6]^{3+}, \quad [Ni(H_2O)_6]^{3+}, \quad [Cu(H_2O)_6]^{2+}$[Given atomic numbers: $V=23, Cr=24, Fe=26, Ni=28, Cu=29$]