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:
a) and
:
There are no insoluble precipitate forms.
b) and
:
There are the insoluble precipitates of forms.
c) and
:
There are the insoluble precipitates of forms.
d) and
:
As is insoluble, therefore no precipitate forms.
e) and
:
There are no insoluble precipitates forms.
Explanation:
a)
Solubility rule suggests:- ⇒ soluble,
⇒ soluble.
KCl ⇒ soluble, ⇒ soluble.
There are no insoluble precipitate forms.
b)
Solubility rule suggests:- ⇒ soluble,
⇒ soluble.
⇒ insoluble,
⇒ soluble.
There are the insoluble precipitates of forms.
c)
Solubility rule suggests:- ⇒ soluble,
⇒ soluble.
⇒ soluble,
⇒ insoluble.
There are the insoluble precipitates of forms.
d)
Solubility rule suggests:- ⇒ soluble,
⇒insoluble.
As is insoluble, therefore no precipitate forms.
e)
Solubility rule suggests:- ⇒ soluble,
⇒ soluble.
⇒ soluble,
⇒ soluble.
There are no insoluble precipitates forms.