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Question:
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Match the following
$List-I
(Complex)$
(A) $\left[\mathrm{CoF}_6\left[^{3-}\right.\right.$
(B) $\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$
(C) $\left[\mathrm{FF}_6\right]^{3-}$
(D) $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}$
$List-II
(Spin only Magnetic moment)$
(I) 0
(II) $\sqrt{24}$
(III) $\sqrt{8}$
(IV) $\sqrt{35}$
(V) $\sqrt{15}$
The correct answer is:
Options:
$List-I
(Complex)$
(A) $\left[\mathrm{CoF}_6\left[^{3-}\right.\right.$
(B) $\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$
(C) $\left[\mathrm{FF}_6\right]^{3-}$
(D) $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}$
$List-II
(Spin only Magnetic moment)$
(I) 0
(II) $\sqrt{24}$
(III) $\sqrt{8}$
(IV) $\sqrt{35}$
(V) $\sqrt{15}$
The correct answer is:
Solution:
2075 Upvotes
Verified Answer
The correct answer is:
A-II, B-I, C-IV, D-III
$\left[\mathrm{CoF}_6\right]^{3-}: \mathrm{Co}$ is present as $\mathrm{Co}^{3+}$ and $\mathrm{F}^{-}$is a weakfield ligand.
Thus, $\mathrm{Co}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^6$ with four unpaired electrons.
$$
\Rightarrow \mu=\sqrt{\mathrm{n}(\mathrm{n}+2)} \text { B.M. }=\sqrt{4(4+2)}=\sqrt{24}
$$
$\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}: \mathrm{Co}$ is present as $\mathrm{Co}^{3+}$ and $\mathrm{C}_2 \mathrm{O}_4{ }^{2-}$ is strong-field ligand.
Thus, $\mathrm{Co}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^6$ with all paired electrons.
$$
\mu=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{0(0+2)}=0
$$
$\left[\mathrm{FeF}_6\right]^{3-}: \mathrm{Fe}$ is present as $\mathrm{Fe}^{3+}$ and $\mathrm{F}^{-}$is a weak-field ligand.
Thus, $\mathrm{Fe}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^5$ with five unpaired electrons.
$$
\Rightarrow \mu=-\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{5(5+2)}=\sqrt{35}
$$
$\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}: \mathrm{Mn}$ is present as $\mathrm{Mn}^{3+}$ and $\mathrm{CN}^{-}$is a strongfield ligand with $\mathrm{CN}^{-}$pairing-up only two electrons. Thus, $\mathrm{Mn}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^4$ with two unpaired electrons.
$$
\mu=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{2(2+2)}=\sqrt{8}
$$
Thus, $\mathrm{Co}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^6$ with four unpaired electrons.
$$
\Rightarrow \mu=\sqrt{\mathrm{n}(\mathrm{n}+2)} \text { B.M. }=\sqrt{4(4+2)}=\sqrt{24}
$$
$\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}: \mathrm{Co}$ is present as $\mathrm{Co}^{3+}$ and $\mathrm{C}_2 \mathrm{O}_4{ }^{2-}$ is strong-field ligand.
Thus, $\mathrm{Co}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^6$ with all paired electrons.
$$
\mu=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{0(0+2)}=0
$$
$\left[\mathrm{FeF}_6\right]^{3-}: \mathrm{Fe}$ is present as $\mathrm{Fe}^{3+}$ and $\mathrm{F}^{-}$is a weak-field ligand.
Thus, $\mathrm{Fe}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^5$ with five unpaired electrons.
$$
\Rightarrow \mu=-\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{5(5+2)}=\sqrt{35}
$$
$\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}: \mathrm{Mn}$ is present as $\mathrm{Mn}^{3+}$ and $\mathrm{CN}^{-}$is a strongfield ligand with $\mathrm{CN}^{-}$pairing-up only two electrons. Thus, $\mathrm{Mn}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^4$ with two unpaired electrons.
$$
\mu=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{2(2+2)}=\sqrt{8}
$$
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