$$\left[ {Ti{{\left( {{H_2}O} \right)}_6}} \right]C{l_4}$$
Coordination number 6 $$ \Rightarrow $$  octahedral complex
$$Ti$$  is in $$+4$$  oxidations state $$ \Rightarrow $$  no unpaired electrons
$$ \Rightarrow $$  magnetic moment $$= 0 B.M.$$

The complex gives three ions in aqueous solution.
∴ The complex should be $$\left[ {CO{{\left( {N{H_3}} \right)}_5}N{O_2}} \right]C{l_2}.$$
It will give three ions on dissociation as follows :
$$\left[ {Co{{\left( {N{H_3}} \right)}_5}N{O_2}} \right]C{l_2} \to $$     $$\mathop {{{\left[ {Co{{\left( {N{H_3}} \right)}_5}N{O_2}} \right]}^{2 + }}}\limits_{1\,mol} + $$    $$\mathop {2C{l^ - }}\limits_{2\,mol} {\text{(Counter ion)}}$$
$${\text{2}}C{l^ - } + 2AgN{O_3} \to \mathop {2AgCl}\limits_{{\text{Silver ppt}}} + 2NO_3^ - $$

$$\eqalign{ & 2Au + 4C{N^ - } + {H_2}O + \frac{1}{2}{O_2} \to 2\mathop {{{\left[ {Au{{\left( {CN} \right)}_2}} \right]}^ - }}\limits_{'X'} + 2O{H^ - } \cr & 2{\left[ {Au{{\left( {CN} \right)}_2}} \right]^ - } + Zn \to \mathop {{{\left[ {Zn{{\left( {CN} \right)}_4}} \right]}^{2 - }}}\limits_{'Y'} + 2Au \cr} $$

$${\left[ {Cu{{\left( {N{H_3}} \right)}_5}} \right]^{2 + }} \to $$    geometry with $$C.N. = 5$$   is $$sq.$$  pyramidal, $$ds{p^3}$$  hybridisation and $${d_{{x^2} - {y^2}}}$$  orbital is used in hybridisation.
$$\eqalign{ & \left( {\text{A}} \right)Fe{\left( {CO} \right)_5} \to {d_{{z^2}}}s{p^3} \cr & \left( {\text{C}} \right){\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{2 + }} \to {d^2}s{p^3} \cr & \left( {\text{D}} \right){\left[ {Ir{F_6}} \right]^{3 - }} \to {d^2}s{p^3} \cr} $$

In $${\left[ {Mn{{\left( {CN} \right)}_6}} \right]^{3 - }},$$   oxidation state of $$Mn = + 3,M{n^{3 + }} = 3{d^4}$$

It has two unpaired electrons.
$$\eqalign{ & \mu = \sqrt {n\left( {n + 2} \right)} \cr & \,\,\,\,\, = \sqrt {2\left( {2 + 2} \right)} \cr & \,\,\,\,\, = \sqrt 8 \cr & \,\,\,\,\, = 2.82\,B.M. \cr} $$

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{No}}{\text{. of unpaired electron }} \cr & \left( {\text{A}} \right)C{o^{3 + }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4 \cr & \left( {\text{B}} \right)F{e^{3 + }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \cr & \left( {\text{C}} \right)M{n^{3 + }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4 \cr & \left( {\text{D}} \right)C{r^{3 + }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3 \cr} $$
The effective magnetic moment is given by the number of unpaired electrons in a substance, the lesser the number of unpaired electrons lower is its magnetic moment in Bohr — Magneton and lower shall be its paramagnetism

No explanation is given for this question. Let's discuss the answer together.

The complex $$\left[ {Cr{{\left( {N{H_3}} \right)}_6}} \right]C{l_3}$$    involves $${d^2}s{p^3}$$  hybridisation as it involves $$(n -1)d$$   orbitals for hybridisation. It is an inner orbital complex.

Tetrahedral complexes generally involve $$s{p^3}$$  hybridisation.

The chlorophyll molecule plays an important role in photosynthesis, contain porphyrin ring and the metal $$Mg\,{\text{not}}\,Ca.$$