Part a use valence bond theory to devise a hybridization and bonding scheme for co2. match the words in the left column to the a
ppropriate blanks in the sentences on the right. make certain each sentence is complete before submitting your answer.
1 answer:
Q1)
first lets draw the lewis structure for CO₂
C is the central atom with O atom on either side of C
O - C - O
the number of valence electrons around each atom
C- 4
O - 6
O - 6
total number of valence electrons - 16
number of electron pairs - 16/2 = 8
1 pair is shared between 1 O atom and C and another electron pair is shared between C and other O atom
that leaves us with 8 -2 = 6 pairs
add 3 pairs each for O atoms as lone pairs and octets for both O atoms are completed. But in C the octet is not complete yet. So we convert the 2 single bonds between C and O to 2 double bonds on either side. then octet of C is complete. number of lone pairs around O is reduced to 2 lone pairs.
the structure is as follows
.. . .
: O = C = O :
to determine the geometry lets use VSEPR
number of valence electrons around C - 4
number of bonds around C (2 O with 2 electrons contributed - 4
total number of electrons - 8
number of electron pairs - 4
VSEPR geometry for 4 pairs with no lone pairs is linear geometry
1. The lewis structure for CO₂ has a central carbon atom attached to oxygen atoms through two double bonds.
2. Carbon dioxide has a linear electron geometry
3. Carbon atom is sp hybridized.
4. Carbon dioxide has two <span>C(p)-O(p)</span> π bonds and two O(sp²)-C(sp) sigma bonds.
the geometry of CO₂ linear means its sp hybridised where each orbital has 50% s and 50% p character. One s orbital and one p orbital have mixed and form 2 sp orbitals. the remaining p orbitals take part in π bonds and sp orbitals take part in sigma bonding. So between C and O there's one sigma bond and one pi bond. Altogether 2 sigma bonds and 2 pi bonds.
O has 3 sp² orbitals with 2 of them being lone pair orbitals. One sp² orbital bonds with sp orbital of C forming a sigma bond. the other p orbital of O bonds with p orbital of C forming a pi bond.
first lets draw the lewis structure for CO₂
C is the central atom with O atom on either side of C
O - C - O
the number of valence electrons around each atom
C- 4
O - 6
O - 6
total number of valence electrons - 16
number of electron pairs - 16/2 = 8
1 pair is shared between 1 O atom and C and another electron pair is shared between C and other O atom
that leaves us with 8 -2 = 6 pairs
add 3 pairs each for O atoms as lone pairs and octets for both O atoms are completed. But in C the octet is not complete yet. So we convert the 2 single bonds between C and O to 2 double bonds on either side. then octet of C is complete. number of lone pairs around O is reduced to 2 lone pairs.
the structure is as follows
.. . .
: O = C = O :
to determine the geometry lets use VSEPR
number of valence electrons around C - 4
number of bonds around C (2 O with 2 electrons contributed - 4
total number of electrons - 8
number of electron pairs - 4
VSEPR geometry for 4 pairs with no lone pairs is linear geometry
1. The lewis structure for CO₂ has a central carbon atom attached to oxygen atoms through two double bonds.
2. Carbon dioxide has a linear electron geometry
3. Carbon atom is sp hybridized.
4. Carbon dioxide has two <span>C(p)-O(p)</span> π bonds and two O(sp²)-C(sp) sigma bonds.
the geometry of CO₂ linear means its sp hybridised where each orbital has 50% s and 50% p character. One s orbital and one p orbital have mixed and form 2 sp orbitals. the remaining p orbitals take part in π bonds and sp orbitals take part in sigma bonding. So between C and O there's one sigma bond and one pi bond. Altogether 2 sigma bonds and 2 pi bonds.
O has 3 sp² orbitals with 2 of them being lone pair orbitals. One sp² orbital bonds with sp orbital of C forming a sigma bond. the other p orbital of O bonds with p orbital of C forming a pi bond.
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Answer:
343.98 nm is the longest wavelength of radiation with enough energy to break carbon–carbon bonds.
Explanation:
A typical carbon–carbon bond requires 348 kJ/mol=348000 J/mol
Energy required to breakl sigle C-C bond:E
where,
E = energy of photon
h = Planck's constant =
c = speed of light =
= wavelength of the radiation
Now put all the given values in the above formula, we get the energy of the photons.
343.98 nm is the longest wavelength of radiation with enough energy to break carbon–carbon bonds.