Question:

Which of the following compounds is coloured?

Updated On: Aug 1, 2022
  • $Ti Cl_3$
  • $FeCl_3$
  • $CoCl_2$
  • All of these
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The Correct Option is D

Solution and Explanation

The compounds having metal ions with unpaired electrons, are coloured. Thus, in $TiCl _{3}$ $Ti ^{3+}=[ Ar ] 3 d^{1}$ (One unpaired electron) in $FeCl _{3}$ $Fe ^{3+}=[ Ar ] 3 d^{5}$ (Five unpaired electrons) in $CoCl _{2}$ $Co ^{2+}=[ Ar ] 3 d^{7}$ (Three unpaired electrons) Hence, all compounds are coloured.
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Concepts Used:

Properties of D Block Elements

  • Multiple oxidation states- The oxidation states of d block elements show very few energy gaps; therefore, they exhibit many oxidation states. Also, the energy difference between s and d orbital is very less. Therefore both the electrons are involved in ionic and covalent bond formation, which ultimately leads to multiple oxidation states.
  • Formation of complex compounds- Ligands show a binding behavior and can form so many stable complexes with the help of transition metals. This property is mainly due to:
    • Availability of vacant d orbitals.
    • Comparatively small sizes of metals.
  • Hardness- Transition elements are tough and have high densities because of the presence of unpaired electrons.
  • Melting and boiling points- Melting and boiling points of transition are very high because of the presence of unpaired electrons and partially filled d orbitals. They form strong bonds and have high melting and boiling points.
  • Atomic radii- The atomic and ionic radius of the transition elements decreases as we move from Group 3 to group 6. However, it remains the same between group 7 and group 10, and from group 11 to group 12 increases.
  • Ionization enthalpy- The ionization enthalpies of the transition elements are generally on the greater side as compared to the S block elements