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3 years ago

What would be the bond order for He 2 2- molecule

Chemistry

1 answer:

LuckyWell [14K]3 years ago

8 0

Answer:

In He2 molecule,

Atomic orbitals available for making Molecular Orbitals are 1s from each Helium. And total number of electrons available are 4.

Molecular Orbitals thus formed are:€1s2€*1s2

It means 2 electrons are in bonding molecular orbitals and 2 are in antibonding molecular orbitals .

Bond Order =Electrons in bonding molecular orbitals - electrons in antibonding molecular orbitals /2

Bond Order =Nb-Na/2

Bond Order =2-2/2=0

Since the bond order is zero so that He2 molecule does not exist.

Explanation:

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Oxygen has a total of 8 electrons, does that satisfy the octet rule? Explain why or why not.

murzikaleks [220]

Answer:

Explanation:

The first 2 electrons are in the inner shell, and the other 6 are valence electrons. Oxygen needs 2 more electrons to fill its valence shell and satisfy the octet rule.

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3 years ago

Calculate the ph of the resulting solution if 29.0 ml of 0.290 m hcl(aq) is added to

Mama L [17]

We have the given reaction as;

$HCl_{(aq)} + NaOH_{(aq)}$ ----> $NaCl_{(aq)} + H_{2}O_{(l)}$ 

Answer A) The pH will be 12.36,

We have to convert the concentrations of HCl and NaOH into moles,

So we have, n(HCl) = (0.0290 L) X (0.290 mol/L) = 8.41X $10^{-3}$ moles 

and for NaOH we have, n(NaOH) = (0.0390 L)(0.290 mol/L) = 1.13X $10^{-2}$ moles

Now, it seems NaOH is in excess, so amount remaining will be;

1.13 X $10^{-2}$ - 8.41 X $10^{-3}$ = 2.89 X $10^{-3}$ moles

Now, the total volume will become as = 0.0390 + 0.0290 = 0.068 L 

So, the concentration of [ $OH^{-}$ ] = 2.89×10ˉ³ mol / 0.068 L = 4.25 X $10^{-2}$ M

pOH = - log [ $OH^{-}$ ] = -log (4.25× $10^{-2}$ ) = 1.37

Hence, pH = 14 - pOH = 14 - 1.37 = 12.6

So the pH of the solution will be 12.6 which is basic in nature. 

Answer B) The pH will be 1.68

Now, for the given concentration we need to find moles for HCl and NaOH also;

n(HCl) = (0.0290 L)(0.290 mol/L) = 8.41 X $10^{-3}$ mol 

n(NaOH) = (0.0190 L)(0.390 mol/L) = 7.41 X $10^{-3}$ mol 

here we can see, HCl is in excess amount so the remaining will be;

8.41X $10^{-3}$ - 7.41 X $10^{-3}$ = 1.0 X $10^{-3}$ mol

Here, the total volume will be = 0.0290 + 0.0190 = 0.0480 L

So the concentration of [HCl] = 1.0 X $10^{-3}$ mol / 0.0480 L = 2.08 X $10^{-2}$ M

Which is = [H⁺] 

So, the pH = - log [ $H^{+}$ ] = -log(2.08X $10^{-2}$ ) = 1.68

Hence, the pH will be 1.63 which is more acidic in nature.

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3 years ago

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silver (I) nitrate reacts with nickel (II) chloride to produce silver (I) chloride and nickel (II) nitrate wright the balanced c

Naya [18.7K]

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3 years ago