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algol [13]
3 years ago
15

Predict the best choice in each of the following. You may wish to review the chapter on electronic structure for relevant exampl

es.
(a) the most metallic of the elements Al, Be, and Ba
(b) the most covalent of the compounds NaCl, CaCl2, and BeCl2
(c) the lowest first ionization energy among the elements Rb, K, and Li
(d) the smallest among Al, Al+, and Al3+
(e) the largest among Cs+, Ba2+, and Xe
Chemistry
1 answer:
alisha [4.7K]3 years ago
6 0

Explanation:

(a)  It is known that both Be and Ba are group 2 elements. And, in metals more easily an element is able to lose valence electrons more will be its metallic character.

Elements of group 2 have more metallic character as compared to elements of group 3 (Al is a group 3 element).

Therefore, out of the given options Ba is larger in size so, it is most metallic in nature.

(b)   A covalent compound is defined as the compound in which combining atoms form a chemical bond due to sharing of electrons.

Whereas a compound in which combining atoms tend to form a chemical bond due to transfer of electrons is known as an ionic compound.

Both NaCl and CaCl_{2} are ionic in nature. On the other hand, in BeCl_{2} size of Be is smaller and it has high charge density due to which it will attract the valence electrons of chlorine atom.

Hence, sharing of atoms occur in BeCl_{2}. Hence,  BeCl_{2} is a covalent compound.

(c)  Ionization energy is defined as the heat or energy necessary to remove an electron from a gaseous atom or ion.

Larger is the size of an atom, less amount of energy will be needed by it to remove a valence electron. Hence, lower will be the value of ionization energy.

Therefore, out of the given options Rb has the lowest first ionization energy.

(d)    Higher is the positive charge on an atom smaller will be its size because a positive charge occurs due to loss of electrons.

As a result, size of an atom becomes small. Therefore, out of the given options Al^{3+} is the smallest.

(e)   When we move from left to right in a periodic table then there will be decrease in size of atom due to the addition of electrons into same shells.

Cs is placed at the bottom right corner and Xe is placed at the bottom left corner of a periodic table.

Hence, size of Cs^{+} is largest as compared to both Ba^{+} and Xe.

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In a 0.730 M solution, a weak acid is 12.5% dissociated. Calculate Ka of the acid.
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Approximately 1.30 \times 10^{-2}, assuming that this acid is monoprotic.

Explanation:

Assume that this acid is monoprotic. Let \rm HA denote this acid.

\rm HA \rightleftharpoons H^{+} + A^{-}.

Initial concentration of \rm HA without any dissociation:

[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}.

After 12.5\% of that was dissociated, the concentration of both \rm H^{+} and \rm A^{-} (conjugate base of this acid) would become:

12.5\% \times 0.730\; \rm mol \cdot L^{-1} = 0.09125\; \rm mol \cdot L^{-1}.

Concentration of \rm HA in the solution after dissociation:

(1 - 12.5\%) \times 0.730\; \rm mol \cdot L^{-1} = 0.63875\; \rm mol\cdot L^{-1}.

Let [{\rm HA}], [{\rm H}^{+}], and [{\rm A}^{-}] denote the concentration (in \rm mol \cdot L^{-1} or \rm M) of the corresponding species at equilibrium. Calculate the acid dissociation constant K_{\rm a} for \rm HA, under the assumption that this acid is monoprotic:

\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}.

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