Question

the complex with highest number of unpaired electrons is A) K4[Fe(CN)6] C) [Ti(H O) ] B) K4[FeF6] 3+ 2 D) [Cr(NH )

Answer 1

Answer:

Explanation:

Larger than the magnetic moment of the cyanide complex because there are more unpaired electrons in the fluoride complex

For [FeF

6

​

]

3−

and [Fe(CN)

6

​

]

3−

magnetic moment of the fluoride complex is expected to be larger than the magnetic moment of the cyanide complex because there are more unpaired electrons in the fluoride complex. Higher is the number of unpaired electrons, higher is the magnetic moment.

Answer 2

(A) The complex with the highest number of unpaired electrons is $\[{K_4}\left[ {Fe{{(CN)}_6}} \right]\]$.

Explanation:

In $\[{K_4}\left[ {Fe{{(CN)}_6}} \right]\]$, the oxidation state of the central metal is $\[ + 2\]$.

$\[{4\left( { + 1} \right) + X + 6\left( { - 1} \right) = 0}\]$

$\[{4 - 6 + x = 0}\]$

$\[{x = 6 - 4 = + 2}\]$

Here, iron contains 4 unpaired electrons.

IN $\[\left[ {Ti{{({H_2}O)}_6}} \right]\]$, the oxidation state of central metal is zero.

$\[{x + 6\left( 0 \right) = 0}\]$

$\[{x = 0}\]$

Here, titanium contains 2 unpaired electrons.

In $\[{K_4}{\left[ {Fe{F_6}} \right]^{3 + }}\]$, the oxidation state of central metal is $\[ + 5\]$.

$\[{4\left( { + 1} \right) + x + 6\left( { - 1} \right) = + 3}\]$

$\[{4 - 6 + x = + 3}\]$

$\[{x = 6 - 4 + 3 = + 5}\]$

Here, iron contains 3 unpaired electrons.

In $\[{\left[ {Cr{{(N{H_3})}_6}} \right]^{3 + }}\]$, the oxidation state of central atom is $\[ + 3\]$.

$\[x + 6(0) = + 3\]$

$\[x = + 3\]$

Here, chromium contains 3 unpaired electrons.