The D-And F-Block Elements - Class 12 Medical Chemistry - Extra Questions

$$Hg^{2+}$$ salts are colourless. The electronic configuration of Hg is $$[Xe]4f^{14}5d^{10}6s^{2}$$. It loses 2 electrons to become $$Hg^{2+}$$ which has fully filled 5d-orbitals (i.e. there is no unpaired electron is 5d). Hence, no d−d transition can take place and so, it is colourless. Tranisition of electron from d orbital to s orbital will require high energy and it doesnot fall within the visible region.

The copper atom is in the +1 oxidation state in which it has a full (totally occupied) set of 'd' orbitals. There can be no transitions between the non-degenerate orbitals and so there can be no frequencies of light absorbed due to d-d transitions. That is true for CuCl. $$Cu : [Ar] 3d^{10} 4s^1$$$$Cu^+ : [Ar] 3d^{10}$$. Thus, copper is +1. 

The steady decrease in the size of lanthanide ions with the increase in atomic number is called lanthanide contraction.

Consequences:

• There is not regular trend in ionization potential in the group due to lanthanide contraction.
  
• Similarity of 2nd and 3rd transition group
  

$$Zr$$ and $$Hf$$ exhibit  similar properties due to lanthanide contraction. Electrons present in $$f$$-subshell dosen't have good shielding effect due to which with increase in atomic number effective nuclear charge, size gets constricted and thus size of $$Hf$$ and $$Zr$$ becomes almost equal and hence both have similar properties.

$$Ti^{3+}$$ is colored where as $$Sc^{3+}$$ is colorless because $$Ti^{3+}$$ has an unpaired electron in d-orbital whereas $$Sc^{3+}$$ has $$3d^0$$ i.e, no electron 

Silver bromide used in photography because silver bromide is photosensitive.It decomposes and is converted into metallic silver grains when light is incident on it.

Colour of compounds of transition elements depends upon the unpaired electrons present in $$d$$ orbitals of transition elements. If $$d$$ orbitals are completely vacant as in $${Sc}^{3+},{Ti}^{4+}$$ or completely filled as in $${Cu}^{+},{Zn}^{2+}$$, the compounds will be colourless. But if any unpaired electron is present in $$d$$-orbitals, the compound will be coloured due to $$d\rightarrow d$$ transition. 

Transition elements have vacant $$d-$$ orbitals which can accomodate the ligands. Hence they forms coordinate bonds. The tendency to form coordinate compounds increases with decrease in atomic size and increase in positive charge, in general.

Due to presence of unpaired electrons, transition elements generally form coloured compounds.

$$\displaystyle Sc^{3+}$$ has $$\displaystyle 3d^0$$ outer electronic configuration with zero unpaired electrons. Hence, $$\displaystyle Sc^{3+}$$ salts are colorless.

$$\displaystyle Cr^{3+}$$  has $$\displaystyle 3d^3$$ outer electronic configuration with 3 unpaired electrons. Hence, $$\displaystyle Cr^{3+}$$ salt are coloured.

Note: Colour in transition metal compounds is due to presence of unpaired electrons which cause d-d transitions.

The balanced equations in the manufacture of $$\displaystyle K_2Cr_2O_7$$ from chromite ore are as given below.

Step I : roasting- conversion of $$\displaystyle chromite$$ ore into $$\displaystyle sodium$$ $$\displaystyle chromate$$
$$\displaystyle 4FeO \cdot Cr_2O_3 + 8Na_2CO_3+7O_2 \rightarrow 2Fe_2O_3 + 8Na_2CrO_4+8CO_2 $$

Step II :  Conversion of $$\displaystyle sodium$$ $$\displaystyle chromate$$ into $$\displaystyle sodium$$ $$\displaystyle dichromate$$

$$\displaystyle 2Na_2CrO_4+H_2SO_4 \rightarrow Na_2Cr_2O_7+Na_2SO_4+H_2O$$

Step III :  Conversion of $$\displaystyle sodium$$ $$\displaystyle dichromate$$ into $$\displaystyle potassium$$ $$\displaystyle dichromate$$

$$\displaystyle Na_2Cr_2O_7+2KCl \rightarrow K_2Cr_2O_7+2NaCl$$

Potassium dichromate ($$K_2Cr_2O_7$$) is prepared from chromite ore $$FeCr_2O_4$$. The chromite ore is fused with sodium or potassium carbonate in free access of air.

$$4FeCr_2O_4+8Na_2CO_3+7O_2\rightarrow 8Na_2CrO_4+2FeO_3+8CO_2$$

The yellow solution of sodium chromate is filtered and acidified with sulfuric acid to give a solution from which orange sodium dichromate , $$Na_2Cr_2O_7. 2H_2O$$ can be crystallized.

$$2Na_2CrO_4+2H^{+}\rightarrow Na_2Cr_2O_7+2Na+H_2O$$

Sodium dichromate is more soluble than potassium dichromate. Hence, sodium dichromate when fused with KCl forms orange crystals of potassium dichromate.

$$Na_2Cr_2O_7+2KCl\rightarrow K_2Cr_2O_7+2NaCl$$

$$Sc$$ has generally $$+3$$ oxidation state so there is no unpaired electron, therefore, all the compounds of $$Sc$$ are colourless.

The regular decrease in the size of lanthanoid ions from $${La}^{3+}$$ to $${Lu}^{3+}$$ is known as lanthanoid contraction. This is regular decrease in size due to the greater effect of the increased nuclear charge as compared to screening effect.