Chemistry MCQs for NEET — Practice Questions with Answers

The NCERT text under 9.3 Alkenes states: 'If there is one double bond between two carbon atoms in alkenes, they must possess two hydrogen atoms less than alkanes. Hence, general formula for alkenes is $C_nH_{2n}$'.

The NCERT text states: 'The double bond is shorter in bond length (134 pm) than the C–C single bond (154 pm).' Therefore, a C-C single bond is longer than a C=C double bond.

The NCERT problem 9.9 provides examples of isomers for $C_5H_{10}$ including Pent-1-ene ($CH_2=CH-CH_2-CH_2-CH_3$) and Pent-2-ene ($CH_3-CH=CH-CH_2-CH_3$). These compounds have the same carbon chain but differ in the position of the double bond, which is a characteristic of position isomerism. The section on structural isomerism also defines: 'Position isomerism: When two or more compounds differ in the position of substituent atom or functional group on the carbon skeleton, they are called position isomers'.

The NCERT text defines position isomerism as compounds differing 'in the position of substituent atom or functional group on the carbon skeleton'. Propan-1-ol ($CH_3CH_2CH_2OH$) and Propan-2-ol ($CH_3CH(OH)CH_3$) have the same carbon skeleton and functional group (-OH), but the position of the -OH group is different. But-1-ene and 2-Methylprop-1-ene are chain isomers (or skeletal isomers). Diethyl ether and Butan-1-ol are functional group isomers. n-Butane and Isobutane are also chain isomers.

Corrosion involves the surface of metals being coated with oxides or other salts due to oxidation. Rusting of iron, tarnishing of silver, and the green coating on copper (patina) are all examples of corrosion. Electrorefining of copper is an industrial process used to purify copper, where impure copper is oxidized at the anode and pure copper is deposited at the cathode; it is not a degradation process like corrosion. (Refer to the section 'Corrosion' and 'Electrorefining' in the context).

The context states, 'Electrons released at anodic spot move through the metal and go to another spot on the metal and reduce oxygen in the presence of H+...'. Therefore, H+ ions are essential for the cathodic reaction where oxygen is reduced. Oxidation of iron occurs at the anodic spot, and oxygen is reduced at the cathodic spot. Rust ($\text{Fe}_2\text{O}_3\cdot x\text{H}_2\text{O}$) is formed by the further oxidation of $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$ by atmospheric oxygen. (Refer to the detailed explanation of iron corrosion chemistry).

The text provides the standard electrode potential for the anode reaction ($\text{Fe}^{2+}/\text{Fe}$) as $E^o = -0.44 V$ and for the cathode reaction ($ \text{O}_2|\text{H}_2\text{O} \text{, } \text{H}^+$) as $E^o = 1.23 V$. The overall reaction is $\text{2Fe(s)} + \text{O}_2\text{(g)} + \text{4H}^+\text{(aq)} \longrightarrow \text{2Fe}^{2+}\text{(aq)} + \text{2H}_2\text{O(l)}$. The standard cell potential is $E^o_{cell} = E^o_{cathode} - E^o_{anode} = 1.23 V - (-0.44 V) = 1.67 V$. (Refer to the 'Corrosion' section and the equations for $E^o_{cell}$).

The text states: 'Electrons released at anodic spot move through the metal and go to another spot on the metal and reduce oxygen in the presence of H+... Cathode: O2(g) + 4 H+(aq) + 4 e– ¾® 2 H2O (l)'. This clearly indicates that $\text{H}^+$ ions are consumed to facilitate the reduction of oxygen at the cathode. The context also mentions 'further production of hydrogen ions' when ferrous ions are oxidized to ferric ions, which means they are not simply 'produced' but rather involved in a cycle. (Refer to the detailed explanation of iron corrosion chemistry).

The text explicitly states: 'The ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust in the form of hydrated ferric oxide ($\text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O}$)'.

For a metal to act as a sacrificial electrode, it must be more easily oxidized than iron. This means it should have a more negative (or less positive) standard reduction potential than iron. From the appendix and context, iron has a standard reduction potential for $\text{Fe}^{2+}/\text{Fe}$ = -0.44 V. Comparing the options: Cu (+0.34 V), Ni (-0.23 V), Zn (-0.76 V), Ag (+0.80 V). Zinc (-0.76 V) has the most negative reduction potential among the choices, indicating it is most easily oxidized, and thus would be the most effective sacrificial electrode. (Refer to the 'Prevention of corrosion' section and Appendix III for standard potentials).