use simple structure and bonding models to account for each of the following:
A. the bond length between the two carbon atoms is shorter in C2h4 then in c2h6.
B.the bond lengths in SO3 are all identical and are shorter than a sufur oxygen single bond.
C. the N-O bonds in the NO2- ion are equal in length, whereas they are unequal in HNO2.
D. for sulfur, the fluorides SF2,SF4, and SF6, are known to exist, whereas for oxygen only OF3 is known to exist.
e. the I3- ion is linear.
PLEASE HELP!!
A. Ethane is H3C-CH3.
Ethene is H2C=CH2
The double bond is shorter than a single bond.
B. Draw the electron dot structure of SO3. You will find a S=O bond and two
S-O bonds. You can draw two more resonance structures with the double bond alternating (they don't actually move--these are just the structures we can draw). The actual structure, then, is a resonacne hybrid. The reason the bond lengths are the same is because of resonance (or you can say the hybrid structure is the true structure instead of any one of the resonance structures you drew above.)
Now, I suggest you reread your text and notes about resonance, hybrids, and electron dot strucutures, and try your hand at C, D, and E. We shall be happy to critique your thoughts.
use simple structure and bonding models to account for each of the following:
A. the bond length between the two carbon atoms is shorter in C2h4 then in c2h6.
B.the bond lengths in SO3 are all identical and are shorter than a sufur oxygen single bond.
C. the N-O bonds in the NO2- ion are equal in length, whereas they are unequal in HNO2.
D. for sulfur, the fluorides SF2,SF4, and SF6, are known to exist, whereas for oxygen only OF3 is known to exist.
e. the I3- ion is linear.
-o-n--c
c.) The NO2
-1 ion exhibits resonance, giving bonds of equal length, but in HNO2, the bonding of
the H atom fixes one oxygen atom’s bond to N as a single bond. The other oxygen atom is
bonded to the nitrogen with a double bond. There is no resonance in HNO2.
d) The central atom sulfur can form hybrid orbitals capable of forming equal bonds with fluorine
in SF4 and SF6, which oxygen cannot because oxygen electrons on n=2 have no d orbitals
available with which to hybridize. Recall that d orbitals start at n=3.
C. In NO2-, there are three resonance structures that can be drawn. By looking at the Lewis structures, we can see that there is a double bond between N and one of the O atoms, and a single bond with the other O atom. These bonds are identical, resulting in equal bond lengths. In HNO2, there is no resonance, so the N-O bonds are not equivalent and have different lengths.
D. For sulfur, the valence electron configuration is 3s2 3p4. Since sulfur can accommodate more than 8 valence electrons, it can form multiple bonds with fluorine atoms. In SF2, there is one lone pair on the sulfur atom and two bonding pairs, resulting in a bent molecular geometry. In SF4, there are two lone pairs and four bonding pairs, resulting in a see-saw molecular geometry. In SF6, there are six bonding pairs and no lone pairs, resulting in an octahedral molecular geometry.
On the other hand, oxygen has a valence electron configuration of 2s2 2p4 and can only accommodate 8 valence electrons. Therefore, it can only form a maximum of two bonds. OF3 is known to exist because oxygen can accommodate one lone pair and three bonding pairs, resulting in a trigonal pyramidal molecular geometry. There are no known oxygen fluorides with more than three fluorine atoms bonded to oxygen because it would exceed the limit of 8 valence electrons for oxygen.
E. The I3- ion is linear because it has a linear molecular geometry. This is because all three iodine atoms are bonded to the central iodine atom through single bonds, resulting in a linear arrangement.
C. In NO2- ion, the nitrogen atom is surrounded by three oxygen atoms. The nitrogen atom has a lone pair of electrons, which contributes to the overall electron distribution of the molecule. This leads to resonance, where the double bond character is distributed over both N-O bonds equally. As a result, the N-O bonds in NO2- are equal in length.
In HNO2, there is no resonance because there is no lone pair of electrons on the nitrogen atom. Therefore, the double bond character is localized between one of the N-O bonds. This unequal distribution of electron density results in different bond lengths for the N-O bonds in HNO2.
D. Sulfur (S) has a larger atomic size than oxygen (O). This means that sulfur can accommodate more fluorine (F) atoms around it due to a larger coordination number. Additionally, the presence of d-orbitals in the outermost energy level of sulfur allows it to form more covalent bonds with fluorine.
For oxygen, the maximum number of covalent bonds it can form is two. Oxygen can only form three covalent bonds (as in OF3) if it expands its octet and forms a molecular compound with a more electronegative element, such as fluorine.
E. The I3- ion consists of one central iodine (I) atom and two iodine ions (I-). The central iodine atom forms covalent bonds with the two I- ions, resulting in a linear geometry. This arrangement occurs because the lone pairs on the central iodine atom repel each other, causing the I-I-I bond angles to maximize at 180 degrees. Thus, the I3- ion adopts a linear shape.