Due to some backbonding by sidewise overlapping between $$d$$ - orbitals of metal and $$p$$ - orbital of carbon, the $$F–C$$   bond has $$\sigma $$  and $$\pi $$ character.

For $$\left[ {C{u^{II}}{{\left( {N{H_3}} \right)}_4}} \right]\left[ {P{t^{II}}C{l_4}} \right]$$     four isomers are possible which are $$\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]\left[ {PtC{l_4}} \right],$$     $$\left[ {CuC{l_4}} \right]\left[ {Pt{{\left( {N{H_3}} \right)}_4}} \right],\left[ {PtC{l_3}\left( {N{H_3}} \right)} \right]\left[ {Cu{{\left( {N{H_3}} \right)}_3}Cl} \right]$$          and $$\left[ {Pt{{\left( {N{H_3}} \right)}_3}Cl} \right]\left[ {Cu\left( {N{H_3}} \right)C{l_3}} \right]$$

The organometallic compounds having sigma bond between carbon and metal are sigma bonded organometallic. An example of a sigma bonded organometallic compound is Grignard's reagent.
$$\underbrace {R - Mg}_{\sigma - bond} - X$$
Whereas, ruthenocene, ferrocene and cobaltocene are not sigma bonded organometallic compound.

Key Idea In case of high spin complex, $${\Delta _o}$$  is small. than the pairing energy. That means, the energy required to pair up the fourth electron with the electrons of lower energy $$d$$ - orbitals would be higher than that required to place the electrons in the higher $$d$$ - orbital. Thus, pairing does not occur.
For high spin $${d^4}$$  octahedral complex,

∴ Crystal field stabilisation energy
$$\eqalign{ & = \left( { - 3 \times 0.4 + 1 \times 0.6} \right){\Delta _o} \cr & = \left( { - 1.2 + 0.6} \right){\Delta _o} \cr & = - 0.6\,{\Delta _o} \cr} $$

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

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

$$\mathop {\left[ {Co{{\left( {N{H_3}} \right)}_5}Br} \right]C{l_2}}\limits_{{\text{0}}{\text{.02 }}mole} + 2AgN{O_3}$$       $$ \to \left[ {Co{{\left( {N{H_3}} \right)}_5}Br} \right]{\left( {N{O_3}} \right)_2}$$      $$ + \mathop {2AgC{l_{\left( {ppt.} \right)}}\left( Y \right)}\limits_{0.02 \times 2\, = \,0.04{\text{ }}moles} $$
$$\mathop {\left[ {Co{{\left( {N{H_3}} \right)}_5}Cl} \right]S{O_4}}\limits_{{\text{0}}{\text{.02 }}moles} + BaC{l_2}$$       $$ \to \left[ {Co{{\left( {N{H_3}} \right)}_5}Cl} \right]C{l_2} + $$       $$\mathop {BaS{O_{4\left( {ppt.} \right)}}\left( Z \right)}\limits_{{\text{0}}{\text{.02 }}mole} $$

Ions I and IV are the same $$(trans),$$  with mirror plane through $$en$$  groups.

The electronic configuration of $$V\left( {23} \right) = \left[ {Ar} \right]4{s^2},3{d^3}$$
Let in $${\left[ {V{{\left( {gly} \right)}_2}{{\left( {OH} \right)}_2}{{\left( {N{H_3}} \right)}_2}} \right]^ + }$$     oxidation state of $$V$$  is $$x.$$
$$\eqalign{ & x + \left( { - 1} \right) \times 2 + \left( { - 1} \right)2 + \left( {0 \times 2} \right) = + 1 \cr & x = + 5 \cr} $$
$${V^{5 + }} = \left[ {Ar} \right]4{s^0},3{d^0}$$     ( no unpaired electrons )
The electronic configuration of $$Fe\left( {26} \right) = \left[ {Ar} \right]4{s^2},3{d^6}$$
Let the oxidation state of $$Fe$$  in $${\left[ {Fe\left( {en} \right)\left( {ppy} \right){{\left( {N{H_3}} \right)}_2}} \right]^{2 + }}$$     is $$x.$$
$$\eqalign{ & \left[ {x + \left( 0 \right) + \left( 0 \right) + \left( 0 \right) \times 2} \right] = + 2 \cr & x = + 2 \cr} $$
$$F{e^{2 + }} = \left[ {Ar} \right],3{d^6}$$     $$(\because $$  4 unpaired electron )
but, $$bpy, en$$   and $$N{H_3}$$  all are strong field ligands, so pairing occurs and thus, $$F{e^{2 + }}$$  contains no unpaired electron.
The electronic configuration of $$Co\left( {27} \right) = \left[ {Ar} \right]4{s^2},3{d^7}$$
Let the oxidation state of $$Co$$  in $${\left[ {Co{{\left( {ox} \right)}_2}{{\left( {OH} \right)}_2}} \right]^ - }$$    is $$x$$
$$\eqalign{ & x + \left( { - 2} \right) \times 2 + \left( { - 1} \right) \times 2 = - 1 \cr & x = + 5 \cr} $$
$$C{o^{5 + }} = \left[ {Ar} \right],3{d^4}$$    [ 4 unpaired electrons ]
$$ox$$  and $$OH$$  are weak field ligands, thus pairing of electron units does not occur.
The electronic configuration of $$Ti\left( {22} \right) = \left[ {Ar} \right]4{s^2},3{d^2}$$
Oxidation state of $$Ti$$  in $${\left[ {Ti{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$$    is 3.
$$T{i^{3 + }} = \left[ {Ar} \right]3{d^1}$$    ( one unpaired electron )
Hence, complex $${\left[ {Co{{\left( {ox} \right)}_2}{{\left( {OH} \right)}_2}} \right]^ - }$$    has maximum number of unpaired electrons, thus show maximum paramagnetism.

In $$CuS{O_4} \cdot 5{H_2}O,$$    water acts as ligand and as a result it causes crystal field splitting making $$d{\text{ - }}d$$  transitions possible in $$CuS{O_4} \cdot 5{H_2}O.$$    Hence, it is coloured. In anhydrous $$CuS{O_4},$$  due to absence of ligand crystal field splitting is not possible hence no colour is observed.