निम्नलिखित संकुल स्पीशीज के चुंबकीय आघूर्णों के मान से आप क्या निष्कर्ष निकालेंगे?

उदाहरण	चुंबकीय आघूर्ण $(BM)$
$K _{4}\left[ Mn ( CN )_{6}\right.$	$2.2$
$\left[ Fe \left( H _{2} O \right)_{6}\right]^{2+}$	$5.3$
$K _{2}\left[ MnCl _{4}\right]$	$5.9$

Magnetic moment $\left( \mu  \right)$ is given as $\mu=\sqrt{n(n+2)}$

For value $n=1,$ $\mu=\sqrt{1(1+2)}=\sqrt{3}=1.732$

For value $n=2,$ $\mu=\sqrt{2(2+2)}=\sqrt{8}=2.83$

For value $n=3,$ $\mu=\sqrt{3(3+2)}=\sqrt{15}=3.87$

For value $n=4,$ $\mu=\sqrt{4(4+2)}=\sqrt{24}=4.899$

For value $n=5,$ $\mu=\sqrt{5(5+2)}=\sqrt{35}=5.92$

$(i)$ $K _{4}\left[ Mn ( CN )_{6}\right]$

For in transition metals, the magnetic moment is calculated from the spin-only formula. Therefore,

$\sqrt{n(n+2)}=2.2$

We can see from the above calculation that the given value is closest to $n=1$. Also, in this complex, Mn is in the $+2$ oxidation state. This means that $Mn$ has $5$ electrons in the $d$ orbital.

Hence, we can say that $CN ^{-}$ is a strong field ligand that causes the pairing of electrons.

$(ii)$ $\left[ Fe \left( H _{2} O \right)_{6}\right]^{2+}$

$\sqrt{n(n+2)}=5.3$

We can see from the above calculation that the given value is closest to $n=4$. Also, in this complex, $Fe$ is in the $+2$ oxidation state. This means that $Fe$ has $6$ electrons in the $d$ -orbital

Hence, we can say that $H _{2} O$ is a weak field ligand and does not cause the pairing of electrons.

$(iii)$ $K _{2}\left[ MnCl _{4}\right]$

$\sqrt{n(n+2)}=5.9$

We can see from the above calculation that the given value is closest to $n=5$. Also, in this complex, $M n$ is in the $+2$ oxidation state. This means that $M n$ has $5$ electrons in the $d$ -orbital.

Hence, we can say that $Cl ^{-}$ is a weak field ligand and does not cause the pairing of electrons.