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irina1246 [14]
3 years ago
11

Use the images to explain why carbon forms a bond with four chlorine atoms.

Chemistry
2 answers:
lina2011 [118]3 years ago
5 0

Explanation:

The Lewis dot diagram shows how electrons participate in a bond with Carbon and Chlorine. This is shown by the sticks and the 2 paired electrons near the carbon atom which represent the bonds. These electrons form these bonds because they form octets when they are bonded which most molecules and compounds follow

Hoped this helped, 2Trash4U

kati45 [8]3 years ago
5 0

Answer: Chlorine and Carbon exhibits covalent bonding which involves the sharing if electrons. Carbon is less electronegative than Chlorine so it is considered as the central atom on this compound. Carbon shares it 4 valence electrons on Cl to attain octet. Now Chlorine is more electronegative so it will receive the lone pairs based from the Lewis structure.

Since C os in group 4 and Cl in group 7 the number of electrons will be:

C = 4

Cl= 7 x 4

# of e- = 32 - 8 ( the number of bonds)

Total number of e-. = 26 e-

Distribute the electrons to the most electronegative atom which is Cl have 6 e- each to attain octet.

Explanation:

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Explanation:

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Example:-

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Calculate the standard reaction enthalpy for the reaction NO2(g) → NO(g) + O(g) given +142.7 kJ/mol for the standard enthalpy of
bulgar [2K]

Answer:

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

Explanation:

O_2(g) \rightarrow \frac{2}{3}O_3(g),\Delta H^o_{1}=142.7 kJ/mol..[1]

O_2(g) \rightarrow 2 O(g),\Delta H^o_{2}=498.4 kJ/mol..[2]

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NO_2(g)\rightarrow NO(g) + O(g),\Delta H^o_{4}=?..[4]

Using Hess's law:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

2 × [4] = [2]- (3 ) × [1] - (2) × [3]

2\times \Delta H^o_{4}=\Delta H^o_{2} -3\times \Delta H^o_{1}-2\times \Delta H^o_{3}

2\times \Delta H^o_{4}=498.4 kJ/mol-3\times 142.7 kJ/mol-2\times -200 kJ/mol

2\times \Delta H^o_{4}=470.3 kJ/mol

\Delta H^o_{4}=\frac{470.3 kJ/mol}{2}=235.15 kJ/mol

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

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