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.