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choli [55]
3 years ago
7

What atomic or hybrid orbitals make up the bond between c2 and o in acetaldehyde, ch3cho?

Chemistry
2 answers:
SSSSS [86.1K]3 years ago
8 0

Acetaldehyde is an organic compound (a compound containing C atoms) composed of a carbonyl group. On the other hand, a carbonyl group is a functional group containing C = O. The hybrid orbitals of a compound determines the number pi and s orbitals in the electronic configuration. For a single bond, there are two s orbitals. For double bonds, on the other hand, the number of s orbital bond is 1 while the number of pi bonds is 2. For triple bonds, there are three pi bonds present in the cloud.

Thus for a c = O bond, the atomic orbital configuration is sp3 containing 1 s orbital and 2 pi bonds. 

Margaret [11]3 years ago
7 0

Answer:

In terms of central atom carbon:

three sp2 hybridized orbitals

one "2p" orbital.

one sp2 hybridized orbital and one "2p" orbital bonds with oxygen atom.

Explanation:

In acetaldehyde we have two carbons.

One carbon is simple alkyl carbon

The other carbon is carbonyl carbon

The carbonyl carbon is bonded to the alkyl carbon and hydrogen by sigma bonds (one with each) and to the oxygen atom by one sigma and one pi bond.Which means that there are three sigma bonds and one pi bond.

The hybridisation of carbonyl carbon is sp2.

The atomic orbitals involved are:

three "2p" orbitals

one "2s" orbital

Out of the three "2p" orbitals : two are involved in hybridization.

The third one is showing side ways overlapping with the "2p" orbital of oxygen atom.

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Answer : The correct option is, (A) AlCl_3

Explanation :

Van't Hoff factor : It is defined as the ratio of the experimental value of the colligative property to the calculated value of the colligative property.

We can determine the van't Hoff factor by the association and dissociation of the compound.

(A) AlCl_3

It is an electrolyte that dissociates to give aluminum ion and chloride ion.

The dissociation of AlCl_3 will be,

AlCl_3\rightarrow Al^{3+}+3Cl^{-}

So, Van't Hoff factor = Number of solute particles = Al^{3+}+3Cl^{-} = 1 + 3 = 4

(B) KI

It is an electrolyte that dissociates to give potassium ion and iodide ion.

The dissociation of KI will be,

KI\rightarrow K^{+}+I^{-}

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(C) CaCl_2

It is an electrolyte that dissociates to give calcium ion and chloride ion.

The dissociation of CaCl_2 will be,

CaCl_2\rightarrow Ca^{2+}+2Cl^{-}

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(D) MgSO_4

It is an electrolyte that dissociates to give magnesium ion and sulfate ion.

The dissociation of MgSO_4 will be,

MgSO_4\rightarrow Mg^{2+}+SO_4^{2-}

So, Van't Hoff factor = Number of solute particles = Mg^{2+}+SO_4^{2-} = 1 + 1 = 2

(E) Non-electrolyte

The dissociation non-electrolyte is not possible. So, the Van't Hoff factor will always be, 1.

Hence, the highest van't Hoff factor of solute is, AlCl_3

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