Question

The melting points of $Cu$ , $Ag$ , and $Au$ follow the order
A.$Cu\rangle Ag\rangle Au$
B.$Cu\rangle Au\rangle Ag$
C.$Au\rangle Ag\rangle Cu$
D.$Ag\rangle Au\rangle Cu$

Answer 1

Hint: We have to know that, more prominent is the size of component, lesser will be the power of fascination and consequently simpler will be to liquefy that component thus dissolving point will be low. Because of lanthanide withdrawal the size of $4d$ and $5d$ arrangements are around something similar.

Complete answer:
We have to know that the given components are copper, silver, and gold individually. They all has a place with the very gathering that is bunch number $11$ of the cutting edge intermittent table and are a $d$ square or progress metal components with valence shell electronic setup as$n{s^1}\left( {n - 1} \right){d^{10}}$ .
Copper has a place with $3d$ arrangement, silver has a place with $4d$ arrangement and gold has a place with $5d$ arrangement. Metals have metallic holding, so strength of metallic bonds will decide the softening mark of the component. As we probably are aware down, the gathering size of components increases and more prominent is the size lesser will be the strength of metallic connections between them. Assuming the strength of metallic bonds diminishes, it will be simpler to soften that component.
The size diminishes from copper to silver, however as we move from silver to gold, the size remains practically steady because of the lanthanide constriction that happens because of the presentation of $f$ orbital. This $f$ orbital has helpless safeguarding and consequently power of fascination among core and valence electron increments and nuclear size of the component diminishes. Thus the liquefying direct abatements from copper toward silver, however the softening mark of gold turns out to be considerably higher than silver because of its higher thickness that occurred because of constriction.

The correct order is option (b) $Cu\rangle Au\rangle Ag$ .

Note:
We have to know that the electron that is available in the internal shell makes an obstacle between the core and external electron. Because of this less electron thickness arriving at the valence shell that we anticipate that it should be because of quality or electrons in inward orbital, this impact is known as protecting impact.