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3 years ago

What kind of bond will form between the atoms?

Chemistry

1 answer:

just olya [345]3 years ago

6 0

Answer:- (i) phosphorous and chlorine is polar covalent.

(ii) potassium and oxygen is ionic.

(iii) fluorine and fluorine is nonpolar covalent.

(iv) copper and aluminum is metallic.

(v) carbon and fluorine is polar covalent.

(vi) carbon and hydrogen is nonpolar covalent.

(vii) Aluminum and oxygen is ionic.

(viii) silver and copper is metallic.

Explanations:- An ionic bond forms between a metal and a non metal by the transfer of metal atom valence electrons to the non metal. In (ii) and (vii) the bond is between a metal and a non metal. Potassium is a metal and oxygen is non metal, so the bond between them is ionic. Similarly, Aluminum is a metal and oxygen is a non metal, so the bond between them is also ionic.

Metallic bond is formed between metals only. In (iv) we have metals copper and aluminum and in (viii) we have metals silver and copper. So, there would be metallic bond only.

Covalent bond is formed by the sharing of electrons between non metals. If the bond is between two similar non metals for example H and H, Cl and Cl, N and N etc then it is known as nonpolar covalent as the shared electrons are between both the non metals.

If the bond is formed between two different non metals like C and F, H and Cl etc then it is known as polar covalent as the shared electrons are more towards the more electron negative atom and due to which the highly electron negative atom has partial negative and the less electron negative atom has partial positive charge.

Note:- Even if the bond is between two different non metals but the electron negativity difference is too low then it is considered as nonpolar covalent.

Electron negativity difference between C and H is very low and so the C-H bond is considered as nonpolar covalent.

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Please explain to me how to answer no.2 questions

Marina CMI [18]

Answer: -

First Ionization energy IE 1 for element X = 801

Here X is told to be in the third period.

So principal quantum number n = 3 for X.

For 1st ionization energy the expression is

IE1 = 13.6 x Z ^2 / n^2

Where Z =atomic number.

Thus Z =( n^2 x IE 1 / 13.6)^(1/2)

Z = ( 3^2 x 801 / 13.6 )^ (1/2)

= 23

Number of electrons = Z = 23

Nearest noble gas = Argon

Argon atomic number = 18

Number of extra electrons = 23 – 18 = 5

a) Electronic Configuration= [Ar] 3d34s2

We know that more the value of atomic radii, lower the force of attraction on the electrons by the nucleus and thus lower the first ionization energy.

So more the first ionization energy, less is the atomic radius.

X has more IE1 than Y.

b) So the atomic radius of X is lesser than that of Y.

c) After the first ionization, the atom is no longer electrically neutral. There is an extra proton in the atom. Due to this the remaining electrons are more strongly pulled inside than before ionization. Hence after ionization, the radii of Y decreases.

3 0

3 years ago

Calculate the pH and fraction of dissociation ( α ) for each of the acetic acid ( CH 3 COOH , p K a = 4.756 ) solutions. A 0.002

marysya [2.9K]

Answer:

The degree of dissociation of acetic acid is 0.08448.

The pH of the solution is 3.72.

Explanation:

The $pK_a=4.756$

The value of the dissociation constant = $K_a$

$pK_a=-\log[K_a]$

$K_a=10^{-4.756}=1.754\times 10^{-5}$

Initial concentration of the acetic acid = [HAc] =c = 0.00225

Degree of dissociation = α

$HAc\rightleftharpoons H^++Ac^-$

Initially

At equilibrium ;

(c-cα) cα cα

The expression of dissociation constant is given as:

$K_a=\frac{[H^+][Ac^-]}{[HAc]}$

$1.754\times 10^{-5}=\frac{c\times \alpha \times c\times \alpha}{(c-c\alpha)}$

$1.754\times 10^{-5}=\frac{c\alpha ^2}{(1-\alpha)}$

$1.754\times 10^{-5}=\frac{0.00225 \alpha ^2}{(1-\alpha)}$

Solving for α:

α = 0.08448

The degree of dissociation of acetic acid is 0.08448.

$[H^+]=c\alpha = 0.00225M\times 0.08448=0.0001901 M$

The pH of the solution ;

$pH=-\log[H^+]$

$=-\log[0.0001901 M]=3.72$

3 0

4 years ago

Problem page gaseous ethane ch3ch3 will react with gaseous oxygen o2 to produce gaseous carbon dioxide co2 and gaseous water h2o

iVinArrow [24]

$m(\text{CO}_2) = 2.24 \; \text{g}$

Ethene react with oxygen at a $2 : 7$ molar ratio:

$2\; \text{C}_2 \text{H}_6 (g) + 7\; \text{O}_2 (g) \to 6\; \text{H}_2{O} (g) + 4\; \text{CO}_2 (g)$

Convert the quantity of each reactant supplied to number of moles of particles:

$n(\text{C}_2\text{H}_6) = 0.60 \; \text{g} / 28.05 \; \text{g} \cdot \text{mol}^{-1} = 0.0214 \; \text{mol}$
$n(\text{O}_2) = 3.27 \; \text{g} / 32.00 \; \text{g} \cdot \text{mol}^{-1} = 0.102 \; \text{mol}$

The question stated not whether both reactants were used up in this process. Thus start by testing the assumption that e.g., ethene was used up while some oxygen gas were left unreacted (ethene as the <em>limiting </em>reagent.) Under this assumption, the relative availability of the two species, $n(\text{C}_2 \text{H}_6) /2$ and $n(\text{O}_2) /7$ (as seen in the balanced chemical equation) shall satisfy the relationship

$n(\text{O}_2) / 7 - n(\text{C}_2 \text{H}_6) / 2 > 0$

In other words,

$n(\text{O}_2)/7 > n(\text{C}_2 \text{H}_6)/2$

$n(\text{C}_2 \text{H}_6) / n(\text{O}_2) < 2/7 \approx 0.286$

Evaluating the expression $n(\text{C}_2 \text{H}_6) / n(\text{O}_2)$ with data given in the question yields approximately $0.210 < 0.286$ , which does satisfy the relationship. Hence the assumption holds and ethene is the limiting reactant.

The quantity of a reactant produced in a chemical reaction is related to its stoichiometric (of relating to proportions) relationship with the limiting reactant (or any of the reactants in case of more than one limiting reactant.) For this scenario, given the molar ratio $n(\text{C}_2\text{H}_6) : n( \text{CO}_2) = 2:4$ ,