15-18 When two atoms form a covalently-bonded diatomic molecule, the distance be

Question

15-18 When two atoms form a covalently-bonded diatomic molecule, the distance between nuclei at which the potential energy is at a minimum is called the bond energy. the bond length. the molecular diameter. the covalent radius the covalent diameter. Which reaction below represents the first ionization of O? O^+ (g) + e^- rightarrow O(g) O(g) + e^- rightarrow O^- (g) O^- (g) rightarrow O(g) + e^- O(g) rightarrow O^+ (g) + e^- O^- (g) + e^- rightarrow O^2- (g) Place the following in order of increasing IE. K Ca Rb Which ionization process requires the most energy? W(g) rightarrow w^+ (g) + e^- W^+ (g) rightarrow W^2 (g) + e^- W^2 + (g) rightarrow W^3+ (g) + e^- W^3 (g) rightarrow W^4 (g) + e^-

Explanation / Answer

(15) Bond energy is the energy required to break the bonds. So, it is not the correct choice since they are talking about the distance.

Bond length tells us about the strength of of a covalent bond between two atoms and hence it is the correct choice.

Molecular diameter is what we calculate by assume an atom a sphere. So, it is not what they are talking about in the question.

Covalent radius is like a measurement of the size of the atom that's going to take part in the formation of covalent bond. So, It's again not to be the right choice.

Covalent diameter is like double of covalent radius which is again not the right choice.

So, the only correct choice is the (B) bond length.

(16) Ionization energy is the energy required to remove electron from the valence shell of a gaseous atom. When an electron is removed then a cation is formed. the equation in general could be shown as...

X(g) -----------> X(g)^+ +e-

So, the correct choice is (D). O(g) ------------> O+(g) + e-

(17) It is hard to remove electron from a gaseous atom if size is small since it has more attraction towards its valence electrons. Since atomic size decreases in a period(left to right in periodic table) and increases in a group(top to bottom in periodic table). So, Ionization energy increases as we move left to right and decreases as we move top to bottom.

K and Rb are the elements of same group. Since Rb is below K in the first group so K is smaller in size as compared to Rb and so the ionization potential is greater for K than Rb. Now if we compare K and Ca then they are the elements of same period. Ca is right to K means size of Ca is smaller than K and so the ionization energy is greater for Ca than K.

So, the increasing oder is Rb < K < Ca. So, the correct choice is (D).

(18) W is Tungsten with atomic number 74. It's electronic configuration is....

[Xe] 4f14 5d4 6s2

6s is completely field where as the 5d is niether completely filled nor half field. We know that ionization energy is greater if the electron is removed frommore stable configuration.

When one electron is removed from 6s then we get W+(g) . Now 6s with one electron is again very stable since its half filled and also there is positive charge on W, so, it is much hard to remove the another electron from 6s, So, it takes up most energy.

Correct choice is (B) W+(g) --------> W2+(g) + e-

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