Question

The correct order of acidic strength is:
A. $K_2O$ > $CaO $ > $MgO$
B. $CO_2$ > $N_2O_5$ > $SO_3$
C. $Na_2O$ > $MgO$ > $Al_2O_3$
D. $Cl_2O_7 $> $SO_3$ > $P_4O_10$

Answer 1

Hint: We should know that apart from oxidation state there are several other factors which determine the acidity of a compound. Some of the factors are electronegativity, anion size, resonance and hybridization.

Step by step answer:
OPTION A:
> \[{K_2}O\]: We know oxygen has got an oxidation number of -2. So the sum of the oxidation numbers of K and O in \[{K_2}O\] should sum up to zero. Therefore,
  \[2 \times \left( x \right){\text{ }} + {\text{ }}\left( { - 2} \right) = 0\] (where let x is the oxidation number of K in \[{K_2}O\])
\[\xrightarrow{{}}{\text{ }}x = {\text{ }} + 1\]
So, the oxidation state of K is =+1.
>CaO: We know oxygen has got an oxidation number of -2. So the sum of the oxidation numbers of Ca and O in CaO should sum up to zero.Therefore,
 \[1 \times \left( x \right){\text{ }} + {\text{ }}\left( { - 2} \right) = 0\] (where let x is the oxidation number of Ca in CaO)
\[\xrightarrow{{}}{\text{ }}x = + 2\]
So, the oxidation state of Ca is = +2
>MgO: We know oxygen has got an oxidation number of -2. So the sum of the oxidation numbers of Mg and O in MgO should sum up to zero.Therefore,
\[1 \times \left( x \right){\text{ }} + {\text{ }}\left( { - 2} \right) = 0\] (where let x is the oxidation number of Mg in MgO)
\[\xrightarrow{{}}{\text{ }}x = + 2\]
So, the oxidation state of Mg is = +2
Therefore, the acidic strength in option A is not in the correct order.

OPTION B
> \[C{O_2}\]: We know oxygen has got an oxidation number of -2. So the sum of the oxidation numbers of C and O in \[C{O_2}\] should sum up to zero.Therefore,
\[2 \times \left( x \right){\text{ }} + {\text{ }}4 \times 1{\text{ }} = {\text{ }}0\] (where let x is the oxidation number of C in \[C{O_2}\])
\[\xrightarrow{{}}x = {\text{ }} + 4\]
So the oxidation state of C is = +4
>\[{N_2}{O_5}\]: We know oxygen has got an oxidation number of -2. So the sum of the oxidation numbers of N and O in \[{N_2}{O_5}\] should sum up to zero.Therefore,
 \[2 \times \left( x \right){\text{ }} + {\text{ }}5 \times \left( { - 2} \right){\text{ }} = {\text{ }}0\] (where let x is the oxidation number of N in \[{N_2}{O_5}\])
\[\xrightarrow{{}}x = {\text{ }} + 5\] So the oxidation state of N is = +5

> \[S{O_3}\]: We know oxygen has got an oxidation number of -2. So the sum of the oxidation numbers of S and O in \[S{O_3}\]should sum up to zero.Therefore,
\[1 \times \left( x \right) + 3 \times \left( { - 2} \right) = 0\]
\[\xrightarrow{{}}x = + 6\]
So the oxidation state of S = +6
Therefore, the acidic strength in option B is not in the correct order.

OPTION C:
> \[N{a_2}O\] : We know oxygen has got an oxidation number of -2. So the sum of the oxidation numbers of Na and O should sum up to zero. Therefore,
 \[2 \times \left( x \right){\text{ }} + {\text{ }}\left( { - 2} \right){\text{ }} = {\text{ }}0\] (where let x is the oxidation number of Na in \[N{a_2}O\])
\[\xrightarrow{{}}x = + 1\]
So the oxidation state of Na =+1
>MgO: We know oxygen has got an oxidation number of -2. So the sum of the oxidation numbers of Mg and O in MgO should sum up to zero. Therefore,
 \[1 \times \left( x \right){\text{ }} + {\text{ }}\left( { - 2} \right) = 0\] (where let x is the oxidation number of Mg in MgO)
\[\xrightarrow{{}}x = + 2\]
So, the oxidation state of Mg is = +2
>\[A{l_2}{O_3}\]: We know oxygen has got an oxidation number of -2. So the sum of the oxidation numbers of Al and O in \[A{l_2}{O_3}\] should sum up to zero. Therefore,
\[2 \times \left( x \right){\text{ }} + {\text{ }}3 \times \left( { - 2} \right){\text{ }} = {\text{ }}0\] (where let x is the oxidation number of Al in \[A{l_2}{O_3}\])
\[\xrightarrow{{}}x = + 3\]
So, the oxidation state of Al is = +3.
Therefore, the acidic strength in option C is not in the correct order.

OPTION D:
> \[C{l_2}{O_7}\]: We know oxygen has got an oxidation number of -2. So the sum of the oxidation numbers of Cl and O in \[C{l_2}{O_7}\] should sum up to zero.Therefore,
\[2 \times \left( x \right){\text{ }} + {\text{ }}7 \times \left( { - 2} \right){\text{ }} = {\text{ }}0\] (where let x is the oxidation number of Cl in \[C{l_2}{O_7}\])
\[\xrightarrow{{}}x = + 7\]
So, the oxidation state of Cl is = +7
 \[S{O_3}\]: We know oxygen has got an oxidation number of -2. So the sum of the oxidation numbers of S and O in \[S{O_3}\] should sum up to zero.Therefore,
\[1 \times \left( x \right){\text{ }} + {\text{ }}3{\text{ }} \times \left( { - 2} \right){\text{ }} = {\text{ }}0\] (where let x is the oxidation number of S in \[S{O_3}\])
\[\xrightarrow{{}}x = + 6\]
So, the oxidation state of S is +6
 \[{P_4}{O_{10}}\]: We know oxygen has got an oxidation number of -2. So the sum of the oxidation numbers of P and O in \[{P_4}{O_{10}}\] should sum up to zero.Therefore,
 \[4 \times \left( x \right){\text{ }} + {\text{ }}10 \times \left( { - 2} \right){\text{ }} = {\text{ }}0\] (where let x is the oxidation number of P in \[{P_4}{O_{10}}\])
\[\xrightarrow{{}}x = + 5\]
So, the oxidation state of P is = +5
Therefore, the acidic strength in option D is in the correct order.
So we conclude that Option D is the correct answer.

Additional Information: While moving left to right across a period, the number of valence electrons of elements increases and varies between 1 to 8. But the valency of elements, when combined with H or O, first increases from 1 to 4 and then it reduces to zero.
As we move down in a group, the number of valence electrons does not change. Hence all the elements of one group have the same valency.

Note: Oxidation state determines the number of electrons that an atom can gain, lose or share while chemically bonding with another atom of another element.