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2 years ago

What is the hybridization and geometry of the carbonyl carbon in carboxylic acids and their derivatives?

Chemistry

1 answer:

OverLord2011 [107]2 years ago

5 0

Answer:

The carbonyl carbon of carboxylic acids and its derivatives has sp² hybridization and trigonal planar geometry.

Explanation:

Carboxylic acids and its derivatives such as acid chlorides, esters, amides etc, contain a carbonyl group (C=O), in which the carbon atom forms double bond with the oxygen atom. Also, the carbonyl carbon is bonded to two R-groups.

As the carbonyl carbon is bonded to three atoms, thus it has trigonal planar geometry.

In carboxylic acids and it derivatives, the carbonyl carbon is sp² hybridized, as it forms three σ-bonds and one π-bond.

The sp² hybridized carbon is formed by the mixing of the 2s orbital with two 2p orbitals. Whereas, one remaining 2p orbital remains unhybridised.

Therefore, the carbonyl carbon of carboxylic acids and its derivatives has sp² hybridization and trigonal planar geometry.

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riadik2000 [5.3K]

Answer : The energy removed must be, -67.7 kJ

Solution :

The process involved in this problem are :

$(1):C_6H_6(g)(425.0K)\rightarrow C_6H_6(g)(353.0K)\\\\(2):C_6H_6(g)(353.0K)\rightarrow C_6H_6(l)(353.0K)\\\\(3):C_6H_6(l)(353.0K)\rightarrow C_6H_6(l)(335.0K)$

The expression used will be:

$\Delta H=[m\times c_{p,g}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,l}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = heat released by the reaction = ?

m = mass of benzene = 125 g

$c_{p,g}$ = specific heat of gaseous benzene = $1.06J/g^oC$

$c_{p,l}$ = specific heat of liquid benzene = $1.73J/g^oC$

$\Delta H_{vap}$ = enthalpy change for vaporization = $33.9kJ/mole=33900J/mole=\frac{33900J/mole}{78.11g/mole}J/g=434.0J/g$

Molar mass of benzene = 78.11 g/mole

Now put all the given values in the above expression, we get:

$\Delta H=[125g\times 1.06J/g.K\times (353.0-(425.0))K]+125g\times -434.0J/g+[125g\times 1.73J/g.K\times (335.0-353.0)K]$

$\Delta H=-67682.5J=-67.7kJ$

Therefore, the energy removed must be, -67.7 kJ

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2 years ago

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Answer:

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Explanation:

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2. It will be melt when ice is warmed at a consistent pressure of 1 atm.

3. If water vapour pressure is continued to increase at a temperature of 100 C, it will be condense.

4. If water vapour pressure is continued to increase at a temperature of -50 C, it will be deposited.

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