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3 years ago

Using molecular orbital theory, the bond order of the C-C bond in the C2 molecule is _____. g

Chemistry

1 answer:

In-s [12.5K]3 years ago

7 0

Answer:

The bond order for C2 molecule is 2.

Explanation:

Bond order can be defined as the half of the difference between the number of electrons in the bonding orbital and the number of electrons in the antibonding orbitals. It can be represented mathematically by; .

Bond order,n= [number of electrons in the bonding molecular orbitals(BMO) - the number or electrons in the anti-bonding molecular orbitals(AMO) ] / 2.

The electronic configuration of the C2 molecule is given below;

C2 = (1sσ)^2 (1s^*σ)^2 (2sσ)^2 (2s^*σ)^2 (2pπ)^4.

The ones with the (*) are known as the Anti-bonding molecular orbitals while the ones without (*) are known as the bonding molecular orbitals. Hence, we have 8 Electrons from the bonding molecular orbitals and 4 Electrons from the anti-bonding molecular orbitals.

So, from the formula given above, the bond order of C2 molecule is;

===> 8-4/2= 4/2.

===> 2.

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3 years ago

What is the molarity of a KF(aq) solution containing 3.0 mol of KF in 2.00L of solution?

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Answer : The molarity of $KF$ in the solution is 1.5 M.

Explanation : Given,

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Formula used :

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Now put all the given values in this formula, we get:

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3 years ago

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Salsk061 [2.6K]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

Calculate the ph of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pka ???? 4.76) of (a) 2:1

givi [52]

According to Hasselbach-Henderson equation:

$pH=pK_{a}+log\frac{[A^{-}]}{[HA]}$

Here, $[A^{-}]$ is concentration of conjugate base and [HA] is concentration of acid.

In the given problem, conjugate base is $CH_{3}COOK$ and acid is $CH_{3}COOH$ thus, Hasselbach-Henderson equation will be as follows:

$pH=pK_{a}+log\frac{[CH_{3}COOK]}{[CH_{3}COOH]}$ ...... (1)

(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{2}{1}=5.06$

Therefore, pH of solution is 5.06.

(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{3}=4.28$

Therefore, pH of solution is 4.28.

(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{5}{1}=5.45$

Therefore, pH of solution is 5.45.

(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{1}=4.76$

Therefore, pH of solution is 4.76.

(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{10}=3.76$

Therefore, pH of solution is 3.76.

4 0

3 years ago

The buffer solution is used to control the pH to insure that it does not become too high because excessively basic solutions cou

oksian1 [2.3K]

Answer:

pH = 12.7

Explanation:

First, we have to calculate the [Ca²⁺] in a solution of about 250 ppm CaCO₃.

$\frac{250mgCaCO_{3}}{L} .\frac{1gCaCO_{3}}{1000mgCaCO_{3}} .\frac{1molCa^{2+} }{100gCaCO_{3}} =2.5 \times 10^{-3} M$

Now, let's consider the dissolution of Ca(OH)₂ in water.

Ca(OH)₂(s) ⇄ Ca²⁺(aq) + 2 OH⁻(aq)

The solubility product Ksp is:

Ksp = [Ca²⁺] × [OH⁻]²

[OH⁻] = √(Ksp/[Ca²⁺]) = √(6.5 × 10⁻⁶/2.5 × 10⁻³) = 5.1 × 10⁻² M

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log (5.1 × 10⁻²) = 1.3

pH + pOH = 14 ⇒ pH = 14 - pOH = 14 - 1.3 = 12.7

5 0

3 years ago