Redox Reactions MCQs for NEET — Chemistry Questions with Answers

To determine the oxidation number of phosphorus in each compound, we follow these steps:

For $H_4P_2O_7$ (pyrophosphoric acid):

• Hydrogen has an oxidation number of +1.
• Oxygen has an oxidation number of -2.
• The overall charge of the compound is 0.
• Let the oxidation number of phosphorus be x. $4(+1) + 2(x) + 7(-2) = 0$ $4 + 2x - 14 = 0$ $2x - 10 = 0$ $2x = 10$ $x = 5$

For $H_5P_3O_{10}$ (penta phosphoric acid):

• Hydrogen has an oxidation number of +1.
• Oxygen has an oxidation number of -2.
• The overall charge of the compound is 0.
• Let the oxidation number of phosphorus be x. $5(+1) + 3(x) + 10(-2) = 0$ $5 + 3x - 20 = 0$ $3x - 15 = 0$ $3x = 15$ $x = 5$

For $(HPO_3)_3$ (triphosphoric acid):

• Hydrogen has an oxidation number of +1.
• Oxygen has an oxidation number of -2.
• The overall charge of the compound is 0.
• Let the oxidation number of phosphorus be x. $3(+1) + 3(x) + 9(-2) = 0$ $3 + 3x - 18 = 0$ $3x - 15 = 0$ $3x = 15$ $x = 5$

Thus, the oxidation numbers are +5 for $H_4P_2O_7$, +5 for $H_5P_3O_{10}$, and +5 for $(HPO_3)_3$.

To determine the oxidation number of chlorine in each compound, we follow these steps:

For $HClO$ (hypochlorous acid):

• Hydrogen has an oxidation number of +1.
• Oxygen has an oxidation number of -2.
• Let the oxidation number of chlorine be x. $1(+1) + x + 1(-2) = 0$ $1 + x - 2 = 0$ $x - 1 = 0$ $x = +1$

For $HClO_2$ (chlorous acid):

• Hydrogen has an oxidation number of +1.
• Oxygen has an oxidation number of -2.
• Let the oxidation number of chlorine be x. $1(+1) + x + 2(-2) = 0$ $1 + x - 4 = 0$ $x - 3 = 0$ $x = +3$

For $HClO_3$ (chloric acid):

• Hydrogen has an oxidation number of +1.
• Oxygen has an oxidation number of -2.
• Let the oxidation number of chlorine be x. $1(+1) + x + 3(-2) = 0$ $1 + x - 6 = 0$ $x - 5 = 0$ $x = +5$

For $HClO_4$ (perchloric acid):

• Hydrogen has an oxidation number of +1.
• Oxygen has an oxidation number of -2.
• Let the oxidation number of chlorine be x. $1(+1) + x + 4(-2) = 0$ $1 + x - 8 = 0$ $x - 7 = 0$ $x = +7$

Thus, the oxidation numbers are +1 for $HClO$, +3 for $HClO_2$, +5 for $HClO_3$, and +7 for $HClO_4$.

Halogens are elements in Group 17 of the periodic table, and they can exhibit various oxidation states. The maximum positive oxidation state of a halogen is +7, which is seen in compounds like potassium permanganate (KMnO_4) where chlorine can be +7 in perchlorates (ClO_4^-). This is because halogens can lose up to seven electrons from their valence shell.

Chalcogens are elements in Group 16 of the periodic table, which includes oxygen, sulfur, selenium, tellurium, and polonium. The maximum positive oxidation state of chalcogens is +6. This is seen in compounds like sulfur hexafluoride (SF_6) and sulfuric acid (H_2SO_4), where sulfur exhibits an oxidation state of +6.

To find the oxidation number of nitrogen in each compound, we can use the rule that the sum of oxidation numbers in a neutral compound is zero and in a polyatomic ion is equal to the charge of the ion. For $N_3H$ (hydrazoic acid), let the oxidation number of nitrogen be x. The oxidation number of hydrogen (H) is +1. The equation is: $3x + 1 = 0 \Rightarrow x = -1/3$. For $H_2N_2O_2$ (hyponitrous acid), let the oxidation number of nitrogen be y. The oxidation numbers of hydrogen and oxygen are +1 and -2 respectively. The equation is: $2(+1) + 2y + 2(-2) = 0 \Rightarrow 2y - 2 = 0 \Rightarrow y = +1$. For $HNO_3$ (nitric acid), let the oxidation number of nitrogen be z. The oxidation numbers of hydrogen and oxygen are +1 and -2 respectively. The equation is: $+1 + z + 3(-2) = 0 \Rightarrow z - 5 = 0 \Rightarrow z = +5$. Therefore, the correct oxidation numbers are: –1/3, +1, +3, +5.

Zeolites are aluminosilicate minerals and tremolite is a type of inosilicate. In zeolite $Na_2Al_2Si_4O_{12}$, each silicon (Si) atom typically has an oxidation state of +4. In tremolite $Ca_2Mg_5(OH)_2(Si_4O_{11})_2$, each silicon (Si) atom also has an oxidation state of +4. Therefore, the correct answer is +4, +4.

In the given compound $AlF_xOy^n$, aluminum (Al) has a fixed oxidation number of +3. The oxidation number of fluorine (F) is -1, and oxygen (O) is -2. According to the rules for oxidation states, the sum of the oxidation numbers must equal the charge of the ion or molecule. The equation can be set up as: $+3 + x(-1) + y(-2) = n$. Simplifying, we get $3 - x - 2y = n$. Therefore, the value of n is $+3 - x - 2y$.