Ncert Solutions Chemistry Class 12th

In octahedral complex the splitting of the d orbital will be such a way that the d_x²_-y² and d_z² orbitals which face towards the axes along the direction of the ligand will experience more repulsion and will be raised in the energy and the other three orbitals which are directed between the axes are lowered in energy.

4.12 2NH₃ (g)? N₂ (g) + 3H₂ (g)

Rate of zero order reaction is equal to rate constant. i.e. Rate = 2.5 * 10^-4mol L^-1sec^-1.

According to rate law,

-d [NH₃] / 2dt = d [N₂] / dt

2.5 * 10^-4mol L^-1sec^-1 = d [N₂] / dt

i.e. the rate of production of N₂ is 2.5 * 10^-4mol L^-1 sec^-1.

According to rate law,

-d [NH₃] / 2dt = d [H₂] / 3dt

d [H₂] / dt = -3 X d [NH₃] / 2dt

i.e. rate of formation of H₂ is 3 times rate of reaction = 3 * 2.5 * 10^-4mol L^-1sec^-1

= 7.5 * 10^-4mol L^-1sec^-1

Rate of formation of N₂ and H₂ is 2.5 * 10^-4 mol L^-1sec^-1 and 7.5 * 10^-4 mol L^-1sec^-1 respectively

(i) In the coordination entity iron exists in + 2 oxidation state. Overall charge balance:

X + 6 (-1) = -4 X = + 2.

Its electronic configuration is: 3d⁶

CN^- is strong field ligand so it causes pairing of the unpaired electron and undergoes hybridisation to form 6 d²sp³ hybrid orbitals to be filled by the six cyanide ions. It's geometry is octahedral with no unpaired electrons and hence is diamagnetic complex.

CuSO₄ + KCN? K₂ [Cu (CN)₄] + K₂SO₄

[Cu (H₂O)₄]²⁺ + 4CN^-? [Cu (CN)₄]^2- + 4H₂O

The coordination entity formed is K₂ [Cu (CN)₄] .

IUPAC name of the coordination entity is potassium tetracyanocuprate (II). It is a very stable complex. The copper atom present inside the coordination sphere does not separate out to form copper ions and cyanide ions due to strong bond between them.It does not ionize to give Cu²⁺ ions and hence on adding H₂S, since there are no copper ions present so no precipitate of copper sulfide is formed.

4.11 (a) Rate = k [A] [B]², Rate = 0 * 10^-6 [0.1] [0.2]²

Rate = 8 * 10^-9 mol L^-1sec^-1

(b) 2A +B = A₂B

=0.06+ 0.18 = 0.02

Situation when A remains 0.06 mol L^-1

Now, According to rate law, Rate = k [A] [B]²

Rate = 2 * 10^-6 [0.06] [0.18]²

i.e. Rate = 3.888 * 10^-9 mol L^-1sec^-1

Initial rate of reaction is 8 * 10^-9 mol L^-1sec^-1. and rate when concentration of A is 0.06 mol L^-1 is 3.888 * 10^-9 mol L^-1sec^-1.

Copper sulphate exists as [Cu (H₂O)₄]SO₄ . It is blue in colour due to presence of the [Cu (H₂O)₄]²⁺ ions.

When KF is added water is replaced by fluoride ion and green colour is due to [Cu (F)₄]^2- ions.

When KCl is added water is replaced by chloride ion and bright green colour is due to presence of [CuCl₄]^2- ions.

In both the cases water, weak field ligand is replaced by fluoride and chloride ions.

There are four different types of ligands present in the complex. So, by fixing the position of 2 ligands we get 2 geometrical isomers and by changing the position of fixed isomer we get one more geometrical isomer.

Since it is tetrahedral complex so it should be optically active . But however it has not been possible to resolve optically active d and l forms of such a complex due to its complicated nature.

Two types of ligand : chloride ion is unidentate ligand and -en (ethylenediamine) is bidentate ligand with two sites of attachment. The complexes in which two symmetrical bidentate chelating ligands AA and two monodentate ligands a, are coordinated to central metal atom M, exhibit the phenomenon of optical isomerism and can be resolved into their optical isomers.

An example of this type of complexes is given as shows both geometrical as well as optical isomerism. Its cis form is unsymmetrical, while the trans form is symmetrical because it contains a plane of symmetry. Hence, optical isomerism is shown by cis form.