The increasing order of magnetic behaviour for following complex is

A. \([Ti(H_2O)_6]^{3+}\)

B. \([V(H_2O)_6]^{3+}\)

C. \([MnCl_4]^{2-}\)

D. \([MnF_6]^{3-}\)

E. \([Cr(H_2O)_6]^{3+}\)

Choose the correct answers from the options given below:

\(A < B < E < D < C\)

The correct answer is option 4. \(A < B < E < D < C\).

To arrange the given complexes in the increasing order of their magnetic behaviour we have find the values of magnetic moment of each of the given complexes.

We know the value of magnetic moment can be calculated using the formula:

\(\mu = \sqrt{n (n + 2)}\); where,  n = number of unpaired electrons.

A. \([Ti(H_2O)_6]^{3+}\)

Here, the central metal ion, \(Ti\) is in \(+3\) oxidation state. So, the configuration will be

\(Ti^{3+} = [_{18}Ar] 4s^03d^1\)

So, the number of unpaired electrons, \(n = 1\)

Thus the magnetic moment will be

\(\mu = \sqrt{1 (1 + 2)}\)

or, \(\mu = \sqrt{3}\)

or, \(\mu = 1.73 BM\)

B. \([V(H_2O)_6]^{3+}\)

Here, the central metal ion, \(V\) is in \(+3\) oxidation state. So, the configuration will be

\(V^{3+} = [_{18}Ar] 4s^03d^2\)

So, the number of unpaired electrons, \(n = 2\)

Thus the magnetic moment will be

\(\mu = \sqrt{2 (2 + 2)}\)

or, \(\mu = \sqrt{8}\)

or, \(\mu = 2.82 BM\)

C. \([MnCl_4]^{2-}\)

Here, the central metal ion, \(Mn\) is in \(+2\) oxidation state. So, the configuration will be

\(Mn^{2+} = [_{18}Ar] 4s^03d^5\)

So, the number of unpaired electrons, \(n = 5\)

Thus the magnetic moment will be

\(\mu = \sqrt{5 (5 + 2)}\)

or, \(\mu = \sqrt{35}\)

or, \(\mu = 5.92 BM\)

D. \([MnF_6]^{3-}\)

Here, the central metal ion, \(Mn\) is in \(+3\) oxidation state. So, the configuration will be

\(Mn^{3+} = [_{18}Ar] 4s^03d^4\)

So, the number of unpaired electrons, \(n = 4\)

Thus the magnetic moment will be

\(\mu = \sqrt{4 (4 + 2)}\)

or, \(\mu = \sqrt{24}\)

or, \(\mu = 4.90 BM\)

E. \([Cr(H_2O)_6]^{3+}\)

Here, the central metal ion, \(Cr\) is in \(+3\) oxidation state. So, the configuration will be

\(Cr^{3+} = [_{18}Ar] 4s^03d^3\)

So, the number of unpaired electrons, \(n = 3\)

Thus the magnetic moment will be

\(\mu = \sqrt{3 (3 + 2)}\)

or, \(\mu = \sqrt{15}\)

or, \(\mu = 3.87 BM\)

Comparing the magnetic moment values the increasing order will be as follow: \(A < B < E < D < C\) which corresponds to option 4.