Answer:
5 electron groups, see saw
Explanation:
During the formation of SF4, the sulfur atom usually bonds with each of four fluorine atoms where 8 of valence electrons are used. The four fluorine atoms have 3 lone pairs of electrons in its octet which will further utilize 24 valence electrons. In addition, two electrons are present as a lone pair on the sulfur atom. We can determine sulfur’s hybridization state by counting of the number of regions of electron density on sulphur (the central atom in the molecule). When bonding takes place there is a formation of 4 single bonds to sulfur and it has 1 lone pair. Looking at this, we can say that the number of regions of electron density is 5. The hybridization state is sp3d.
SF4 molecular geometry is seesaw with one pair of valence electrons. The molecule is polar. The equatorial fluorine atoms have 102° bond angles instead of the actual 120° angle. The axial fluorine atom angle is 173° instead of the actual 180° bond angle.
The _____melting point________ is the temperature at which a substance changes from solid to liquid; _______boiling point_________ is the temperature at which a substance changes from a liquid to as gas; _______vapourisation_________ is the process by which atoms of molecules leave a liquid and become a gas.
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Combustion Have a great day!
Answer:
HCO₂/H₂O is not the acid-base conjugate pair.
Explanation:
<em>Acid and conjugate base pairs differ by an H+ ion.</em>
Neither HCO₂ nor H₂O has lost or gained protons.
The conjugate acid of H₂O is H₃O⁺
The conjugate base of HCO₃⁻ is CO₃²⁻
[A conjugate acid has one more H⁺ than its base]