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butalik [34]
4 years ago
13

The diatomic molecule of chlorine, cl2, is held together by a(n)____ covalent bond.

Chemistry
1 answer:
kvasek [131]4 years ago
7 0

Answer : The diatomic molecule of chlorine, Cl₂, is held together by a SINGLE covalent bond.

Covalent Bond :

It is type of chemical bond, which is formed by sharing of electron between atoms . The covalent bond is formed between two non metals . The valance electron are shared to form the bond . The shared electrons are known as BONDING electron pair and the electron pair which do not take part in bonding are known as NON- BONDING electron pair. Example : O₂ , H₂ , H₂O, NH₃ etc .

Formation of covalent bond in Cl₂ :

Chlorine is present in group 17 in p block , It is a non metal .

The electronic configuration of Cl is : 1s² 2s² 2p⁶ 3s² 3p⁵ .

Since the outer shell is n= 3 , which has 7 electrons in it , hence Cl has 7 valence electrons in it .

From electronic configuration , it can be seen that Cl need 1 electron to complete its octet (3s² 3p⁶ ). Hence when two Cl atoms come close they share one-one electron with each other (as shown in image ) .

Now the octet of both the atom are complete and they are in stable state together .

When both Cl atom share one e⁻ , there is a bond formed between Cl atoms . One bond consists of 2⁻ . Hence in Cl₂ SINGLE covalent bond present .

Cl + Cl -> Cl₂

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12.371 g

Explanation:

Given :

m_{evaporating\ dish}=1.135\ g

m_{evaporating\ dish}+m_{Hydrate\ sample}=25.637\ g

m_{evaporating\ dish}+m_{First\ heated\ sample}=16.428\ g

m_{evaporating\ dish}+m_{Second\ heated\ sample}=13.266\ g

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m_{evaporating\ dish}=1.135\ g

m_{evaporating\ dish}+m_{Hydrate\ sample}=25.637\ g

m_{Hydrate\ sample}=25.637-m_{evaporating\ dish}\ g=25.637-1.135\ g=24.502\ g

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m_{Second\ heated\ sample}=m_{salt\ anhydrous}=13.266-m_{evaporating\ dish}\ g=13.266-1.135\ g=12.131\ g

Mass of water:

m_{water}=m_{Hydrate\ sample}-m_{salt\ anhydrous}=24.502-12.131\ g=12.371\ g

m_{water}=12.371\ g

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