If I understood correct, the high energy of atomisation for d-block element comes from the ability to make many metallic bonds due to half electron filled d-orbitals, but it seems here that copper has no free d-orbitals yet has high enthalpy of atomisation? Is there a mistake in my understanding or is there a reason for this anomaly?
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$\begingroup$ It's s orbital is half filled... $\endgroup$– MithoronNov 15, 2019 at 20:21
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$\begingroup$ yes but other elements in coppers row has half filled s orbital+many half filled subshells in d-orbitals yet it has lower enthalpy then coppers. I am of the understanding that more half filled orbitals=moer metallic bonding $\endgroup$– tryst with freedomNov 16, 2019 at 4:35
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1$\begingroup$ There is no filled $3d$ subshell in metallic copper. Metallic band structure makes combinations of initially filled and empty valence orbitals overlap each other in energy. $\endgroup$– Oscar LanziNov 16, 2019 at 11:35
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$\begingroup$ Copper, with a nearly ideal Fermi surface, contributes one electron per atom to the conduction band in the crystal state. I'm unclear what the question is about. $\endgroup$– Jon CusterNov 18, 2019 at 16:29
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$\begingroup$ Oh lol I asked this qsn after reading my high school text, in it says that transistion elements is able to create multiple bonds due to presence of half filled d -orbitals $\endgroup$– tryst with freedomNov 19, 2019 at 7:38
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1 Answer
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When we look up the Wikipedia articles for various elements, we find that the heat of fusion and heat of vaporization of copper are modestly below those quantities from iron, cobalt and nickel, but hugely exceeding zinc.
In the first-row transition elements when they're condensed as metals, we don't really have separate $3d$ and $4s$ subshells because the metallic bonding spreads these subshells into bands and the bands overlap in energy. This article in Chem Libretexts shows this overlap between $s$ and $p$ orbitals in sodium; a similar concept involving $s$, $d$ and even$p$ orbitals in transition metals. So we can't separately consider $3d$ and $4s$ subshells in a first-row transition element such as copper. With contributing orbitals from different subshells mixed together we have a significant number of filled and empty states mixed together in the valence band, leading to relatively strong metallic bonding as long as we keep that energy overlap. Among elements in this period, only when we get to zinc do we see the $3d-4s$ band overlap drop off and the nonbonding character of the filled $3d$ subshells emerge as a major factor. Therefore, metallic bonding and with it, enthalpy of atomization drop off sharply at zinc, not so much at copper.
answered Nov 17, 2019 at 2:05