Sodium reacts with chlorine to form the compound sodium $2Na + Cl_2 -> 2NaCI$. How would you describe, in terms of electron arrangement, the type of bonding in a molecule of chlorine?

Md Ariful Islam

0 like 0 dislike
1 views
Share with your friends

1 Answers

Md Ariful Islam

Call

The chlorine molecule is represented as $Cl-Cl$, i.e. $Cl_2$. Between the chlorine atoms, 2 electrons overlap to form a region of high electron to which the positively charged chlorine nuclei are attracted, such that internuclear repulsion is negated and a net attractive force results. Because the electrons are shared between the nuclei, we conceive that each atom has 8 valence electrons.

Of course, on reaction with sodium, the sodium reduces the chlorine molecule to give $2xxCl^-$. The resultant bond between $Na^+$ and $Cl^-$ is ionic and a non-molecular substance results.

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
The Cl-Cl bond energy is 243 kJ/mol. Therefore the formation of a single bond between chlorine atoms ? A) should require the absorption of 243 kJ per mole of $Cl_2$ formed. B) should result in the release of 243 kJ per mole of $Cl_2$ formed.

The bond energy, or bond , expresses the energy needed to break a chemical bond, that's why it will always be listed as a positive value. In your case, 243...

1 Answers 1 views
Sodium chloride reacts with copper sulfate to produce sodium sulfate and copper chloride. $2NaCI (aq) + CuSO_4 (aq) -> Na_2SO_4 (aq) + CuCl_2 (s)$. What type of reaction is this?

Aqueous sodium chloride, $"NaCl"$, will not react with aqueous copper(II) sulfate, $"CuSO"_4$, because the two potential products are soluble in aqueous solution. The chemical equation given to you is...

1 Answers 1 views
What are the grams of sodium chloride produced when 10.0 g of sodium react with 10.0 g of chlorine gas in the equation $2Na+Cl_2 -> 2NaCl$?

1) We need to know which of reagents is excess, so we calculate it by reagents' mass. $M (Na)=23$ $frac\{g}{mol}$ $m (Na)=n×M=2×23=46$ g $M (Cl_2)=35.5×2=71$ $frac\{g}{mol}$ $m (Cl_2)=n×M=1×71=71$ g $\frac{10}{46}:\frac{x}{71}$...

1 Answers 1 views
What are the spectator ions in the equation $2Na^+(aq) + SO_4 ^(2-) (aq) + Ba^(2+)(aq) + 2Cl^(-)(aq) → BaSO_4(s) + 2Na^+(aq) + 2Cl^(-) (aq)$?

$Ba^(2+) + cancel(2Na^(+)) + SO_4^(2-) + cancel(2Cl^(-)) rarr BaSO_4(s) + cancel(2Cl^(-)) +cancel(2Na^(+))$ TO GIVE, $Ba^(2+) + SO_4^(2-) rarr BaSO_4(s)darr$ Which is the $"net ionic equation."$ The spectator ions are simply...

1 Answers 1 views
For the reaction $2Na(s) + Cl_2(g) -> 2NaCl(s)$, how many grams of $NaCl$ could be produced from 103.0 g of Na and 13.0 L of $Cl_2$ (at STP)?

You have to search the between Na and $Cl_2$ You have $ n= "mass"/"molecular mass" = (103,0 g)/(23,0 (g/(mol))) = 4,47 mol $ of Na. As 1 mole occupies 22,4...

1 Answers 1 views
Chlorine reacts with hot sodium hydroxide solution. A sample of a compound formed in the reaction was found to contain 10.8 g of sodium atoms, 16.7 g of chlorine atoms and 22.5 g of oxygen atoms. What is the empirical formula of the compound formed?

$"Moles of chlorine"=(16.7*g)/(35.45*g*mol^-1)=0.47*mol$ $"Moles of sodium"=(10.8*g)/(22.99*g*mol^-1)=0.47*mol$ $"Moles of oxygen"=(22.5*g)/(16.00*g*mol^-1)=1.41*mol$ We divide thru by the smallest quantity, that of sodium, to give an empirical formula.....of $NaClO_3$, i.e. $"sodium chlorate"$.

1 Answers 1 views
In the reaction $2Na+Cl_2 -> 2NaCl$, how many moles of $Cl_2$ react with 100 moles of sodium to form sodium chloride?

Remember that is a NUMERICAL (or extensive) quantity. I could say just as correctly that 2 dozen sodium atoms react with 1 dozen chlorine molecules to form 2 dozen sodium...

1 Answers 1 views
How many moles of chlorine gas are needed to make 0.6 moles of sodium chloride given the reaction: $2Na + Cl_2 -> 2NaCl$?

as, 1 mole of Cl$rarr$2 moles of NaCl so, 1 mole of NaCl$rarr 1/2$moles of Cl now, 0.6 mole NaCl$rarr1/2*0.6$ moles of Cl i.e. =$1/2*0.6=0.5*0.6$ =$0.3$ moles of Cl

1 Answers 1 views
Sodium and chlorine react to form sodium chloride: 2Na (s) + Cl2 (g) ----> 2NaCl (s) For the reaction of 44.0g Na with 69.9 g Cl2, what is the limiting reactant and what is the theoretical yield of sodium chloride in grams?

This is a limiting reactant problem. We know that we will need a balanced equation with masses, molar masses, and moles of the involved. 1. Gather all the information...

1 Answers 1 views
What mass of sodium chloride, $NaCl$, is made when $200$ $L$ of chlorine gas, $Cl_2$, reacts with excess sodium metal?

One mole of an ideal gas has a volume of $22.4$ $L$ at STP. Chlorine is not a perfectly ideal gas but can be treated as one at this temperature...

1 Answers 1 views
X