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.
संकुल $\left[ M \left( H _{2} O \right)_{6}\right] Cl _{2}$ के लिए धातु आयनों का युग्म जो $3.9 \,BM$ का एक स्पिन मात्र चुम्बकीय आघूर्ण देता है, होगा
किस तत्व का क्लोराइड रंगीन होगा
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