What group elements of 4 electrons to form covalent bond?

Answer 1

* CH4, for the central C : neutral counting: C contributes 4 electrons, each H radical contributes one each: 4+4(1) = 8 valence electrons : ionic counting: C4- contributes 8 electrons, each proton contributes 0 each: 8 + 4(0) = 8 electrons. : Similar for H: : neutral counting: H contributes 1 electron, the C contributes 1 electron (the other 3 electrons of C are for the other 3 hydrogens in the molecule): 1 + 1(1) = 2 valence electrons. : ionic counting: H contributes 0 electrons (H+), C4- contributes 2 electrons (per H), 0 + 1(2) = 2 valence electrons : conclusion: Methane follows the octet-rule for carbon, and the duet rule for hydrogen, and hence is expected to be a stable molecule (as we see from daily life) * H2S, for the central S : neutral counting: S contributes 6 electrons, each hydrogen radical contributes one each: 6+2(1) = 8 valence electrons : ionic counting: S2- contributes 8 electrons, each proton contributes 0: 8+2(0) = 8 valence electrons : conclusion: with an octet electron count (on sulfur), we can anticipate that H2S would be pseudotetrahedral if one considers the two lone pairs. * SCl2, for the central S : neutral counting: S contributes 6 electrons, each chlorine radical contributes one each: 6+2(1) = 8 valence electrons : ionic counting: S2+ contributes 4 electrons, each chloride anion contributes 2: 4+2(2) = 8 valence electrons : conclusion: see discussion for H2S above. Notice that both SCl2 and H2S follow the octet rule - the behavior of these molecules is however quite different. * SF6, for the central S : neutral counting: S contributes 6 electrons, each fluorine radical contributes one each: 6+6(1) = 12 valence electrons : ionic counting: S6+ contributes 0 electrons, each fluoride anion contributes 2: 0+6(2) = 12 valence electrons : conclusion: ionic counting indicates a molecule lacking lone pairs of electrons, therefore its structure will be octahedral, as predicted by VSEPR. One might conclude that this molecule would be highly reactive - but the opposite is true: SF6 is inert, and it is widely used in industry because of this property. * TiCl4, for the central Ti : neutral counting: Ti contributes 4 electrons, each chlorine radical contributes one each: 4+4(1) = 8 valence electrons : ionic counting: Ti4+ contributes 0 electrons, each chloride anion contributes two each: 0+4(2) = 8 valence electrons : conclusion: Having only 8e (vs. 18 possible), we can anticipate that TiCl4 will be a good Lewis acid. Indeed, it reacts (in some cases violently) with water, alcohols, ethers, amines. * Fe(CO)5 : neutral counting: Fe contributes 8 electrons, each CO contributes 2 each: 8 + 2(5) = 18 valence electrons : ionic counting: Fe(0) contributes 8 electrons, each CO contributes 2 each: 8 + 2(5) = 18 valence electrons : conclusions: this is a special case, where ionic counting is the same as neutral counting, all fragments being neutral. Since this is an 18-electron complex, it is expected to be isolable compound. * Ferrocene, (C5H5)2Fe, for the central Fe: : neutral counting: Fe contributes 8 electrons, the 2 cyclopentadienyl-rings contribute 5 each: 8 + 2(5) = 18 electrons : ionic counting: Fe2+ contributes 6 electrons, the two aromatic cyclopentadienyl rings contribute 6 each: 6 + 2(6) = 18 valence electrons on iron. : conclusion: Ferrocene is expected to be an isolable compound.