The melting points of $Cu$, $Ag$ and $Au$ follow the order
Updated On: Jul 7, 2022
$Cu > Ag > Au$
$Cu > Au > Ag$
$Au > Ag > Cu$
$Ag > Au > Cu$
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The Correct Option isB
Solution and Explanation
All have $d _{10}$ configuration. There is only $1 e ^{-}$for metallic bonding. Now size of atoms or kernels increases down the group. Hence, metallic bonding weakness due to less attraction of electrons involved in metallic bonding. $Ag < Au < Cu$
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