Question

The melting point of \[{\text{Cu}},{\text{Ag, and Au}}\] follow the order:
A.\[{\text{Cu > Au > Ag}}\]
B.\[{\text{Cu > Ag > Au}}\]
C.\[{\text{Au > Ag > Cu}}\]
D.\[{\text{Ag > Au > Cu}}\]

Answer 1

Hint: Greater is the size of element, lesser will be the force of attraction and hence easier will be to melt that element and so melting point will be low. Due to lanthanide contraction the size of 4 d and 5 d series are approximately the same.

Complete step by step solution:
The element with symbol \[{\text{Cu}},{\text{Ag, and Au}}\] is copper, silver and gold respectively. They all belongs to the same group that is group number 11 of the modern periodic table and are a d block or transition metal elements with valence shell electronic configuration as \[{\text{n}}{{\text{s}}^1}\left( {{\text{n}} - 1} \right){{\text{d}}^{10}}\]
Copper belongs to 3 d series, silver belongs to 4 d series and gold belongs to 5 d series. Metals have metallic bonding, so the strength of metallic bonds will determine the melting point of the element.
As we know down the group the size of elements increases and greater is the size lesser will be the strength of metallic bonds between them. If the strength of metallic bonds decreases then it will be easier to melt that element. The size decreases from copper to silver, but as we move from silver to gold, the size remains almost constant due to the lanthanide contraction that occurs due to introduction of f orbital. F orbital have poor shielding and hence force of attraction between nucleus and valence electron increases and atomic size of the element decreases.
Hence the melting point decreases from copper to silver, but the melting point of gold becomes even higher than silver due to its higher density that happened due to contraction. So the correct order is \[{\text{Cu > Au > Ag}}\].

The correct option is A.

Note: The electron that is present in the inner shell creates hindrance between the nucleus and outer electron. Due to this less electron density reaching the valence shell that we expect it to be due to presence or electrons in inner orbital, this effect is known as shielding effect.