Question

The inner octahedral complex [CO(NH3)6]3+ has magnetic moment equal to
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Answer 1

$\large\underline{\bigstar \: \: {\sf Given-}}$

Inner octahedral complex ${\sf [Co(NH_3)_6]^{+3}}$

$\large\underline{\bigstar \: \: {\sf To \: Find -}}$

• Magnetic Moment (μ)

$\large\underline{\bigstar \: \: {\sf Solution-}}$

${\sf [Co(NH_3)_6]^{+3}}$

Electronic Configuration of ${\sf _{27}Co=[Ar]3d^7\:4s^2}$

So ,

${\sf Co^{+3}= [Ar] \: 3d^6}$

Due to presence of strong lig-and electrons get paired up so metal ion provides vacant '3d' orbital for hybridization.

• This is diamagnetic complex

______________________________

★ Magnetic Moment (μ)

$\large\implies\underline{\boxed{\sf \mu=\sqrt{n(n+2)}}}$

Here , n = number of unpaired electron.

As the complex is diamagnetic their is no unpaired electron.

Therefore ,

$\implies{\sf \mu=\sqrt{0(0+2)} }$

$\implies{\bf \mu = 0 }$

$\large\underline{\bigstar \: \: {\sf Answer-}}$

Magnetic Moment is zero.