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3 years ago

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What atomic or hybrid orbitals make up the sigma bond between c2 and c3 in propyne, chcch3 ? (c2 is the second carbon in the str

ucture as written.) orbital on c2 + orbital on c3 what is the approximate c1-c2-c3 bond angle ? ... °?

1 answer:

Anvisha [2.4K]3 years ago

4 0

Answer:

$\boxed{\text{$\sigma$(sp-sp$^{3}$); 180$^{\circ}$}}$

Explanation:

(a) Hybridization of orbitals

See the Lewis structure of propyne in the first diagram below.

C1 is directly bonded to two other atoms (H and C2) so it is sp hybridized.

C2 is directly bonded to two other atoms (C1 and C3) so it is sp hybridized.

C3 is directly bonded to four other atoms (C2 and 3 H) so it is sp³ hybridized.

(b) C2-C3 Sigma Bond

See the atomic and molecular orbitals in the second picture below.

C3 is using its hybrid atomic orbitals to form sigma molecular bonds. The C2-C3 sigma bond is formed by the overlap of the C2 sp atomic orbital with the C3 sp³ atomic orbital to make a σ(sp-sp³) molecular orbital.

(c) Bond angles

C1 and C2 are sp hybridized. Since the angle between sp orbitals is 180°, all atoms directly attached the C1 and C2 must be in a straight line. The C-C-C bond angle is 180°.

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Consider a hypothetical reaction in which a and b are reactants and c and d are products. if 25 grams of a completely reacts wit

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According to law of conservation of mass, the mass can neither be destroyed nor created in a chemical reaction. The mass of reactants and mass of products are equal in a chemical reaction.

The reaction between A and B (reactants) to form C and D (products) is given as:

$A+B\rightarrow C+D$

The mass of A = $25 g$ (given)

The mass of B = $26 g$ (given)

The mass of reactant = mass of A + mass of B

Substituting the values:

The mass of reactant = $25 g + 26 g = 51 g$

The mass of C = $14 g$

Let the mass of D = $X g$

So, the mass of product = mass of C + mass of D

Substituting the values:

the mass of product = $(14 +X) g$

According to law of conservation of mass:

the mass of reactant = the mass of product

Substituting the values,

$51 g = (14 +X) g$

$X = 51 -14 g$

$X = 37 g$

Hence, the amount of D produced in the reaction is $37 g$ .

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3 years ago

Which of the following statements are true? I. Models are always wrong unless they are proved by a theory. II. Elements, such as

bazaltina [42]

Answer:

The only option that has a true statement is hence only (ii) and (iv) are TRUE

Explanation:

The correct option is (ii) - Elements, such as lead, are made of tiny particles that mostly consist of open space.

for the first option (i) - Models are always wrong unless they are proved by a theory- This is not true because not all models are always wrong and they are not necessarily need to be proved by a theory to ascertain the veracity of the model, one such example is the Neil's bohr model of atom.

for the third option (iii) - The air you breathe is an example of a heterogeneous mixture. This is wrong because Air is a homogeneous mixture containing Nitrogen and Oxygen and other constituents as its composition.

for the last option (iv) - Because NH3 always contains the same relative numbers of atoms, it will always contain 4.6 g of nitrogen for every 1.0 g of hydrogen - this is also true as the amount contained is dependent on the number of atoms in the balanced equation for the production of ammonia.

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3 years ago

Write the electron configurations of these ions, potassium and magnesium

irinina [24]

Answer:

a

Explanation:

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3 years ago

4. If 175 undecayed nuclei remain from a sample of 2800 nuclei, how many half-

Irina18 [472]

Answer:

X = 4

Explanation:

Start 2800

End 175

175 = 2800x(0.5)^X

175/2800 = 0.5^X

0.0625 = 0.5^X

log(0.5 x) = log(0.0625)

x · log(0.5) = log(0.0625)

-0.301x = -1.204

x = -1.204/-0.301

x = 4

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3 years ago

The following pairs of soluble solutions can be mixed. In some cases, this leads to the formation of an insoluble precipitate. D

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Answer:

a) $K_{2} S$ and $NH_{4} Cl$ :

There are no insoluble precipitate forms.

b) $Ca Cl_{2}$ and $(NH_{4} )_{2} Co_{3}$ :

There are the insoluble precipitates of $CaCo_{3}$ forms.

c) $Li_{2}S$ and $MnBr_{2}$ :

There are the insoluble precipitates of $MnS$ forms.

d) $Ba(No_{3} )_{2}$ and $Ag_{2} So_{4}$ :

As $Ag_{2} So_{4}$ is insoluble, therefore no precipitate forms.

e) $Rb_{2}Co_{3}$ and $NaCl$ :

There are no insoluble precipitates forms.

Explanation:

a)

Solubility rule suggests:- $K_{2} S$ ⇒ soluble, $NH_{4} Cl$ ⇒ soluble.

KCl ⇒ soluble, $(NH_{4})_{2} S$ ⇒ soluble.

There are no insoluble precipitate forms.

b)

Solubility rule suggests:- $Ca Cl_{2}$ ⇒ soluble, $(NH_{4} )_{2} Co_{3}$ ⇒ soluble.

$CaCo_{3}$ ⇒ insoluble, $NH_{4} Cl$ ⇒ soluble.

There are the insoluble precipitates of $CaCo_{3}$ forms.

c)

Solubility rule suggests:- $Li_{2}S$ ⇒ soluble, $MnBr_{2}$ ⇒ soluble.

$LiBr$ ⇒ soluble, $MnS$ ⇒ insoluble.

There are the insoluble precipitates of $MnS$ forms.

d)

Solubility rule suggests:- $Ba(No_{3} )_{2}$ ⇒ soluble, $Ag_{2} So_{4}$ ⇒insoluble.

As $Ag_{2} So_{4}$ is insoluble, therefore no precipitate forms.

e)

Solubility rule suggests:- $Rb_{2}Co_{3}$ ⇒ soluble, $NaCl$ ⇒ soluble.

$RbCl$ ⇒ soluble, $Na_{2} Co_{3}$ ⇒ soluble.

There are no insoluble precipitates forms.

6 0

3 years ago