NEET Chemistry questions with solutions

Q1. Among the following organic acids, the acid present in rancid butter is:

1. pyruvic acid
2. lactic acid
3. butyric acid
4. acetic acid

Answer: butyric acid

Butyric acid is a short-chain fatty acid responsible for the unpleasant smell of rancid butter due to its breakdown in fats.

Q2. A ketone has molar mass \( 86 . \) Which of the following cannot be IUPAC name of this ketone?

1. 2 -pentanone
2. 2-methyl butanone
3. 3 -methyl butanone
4. 3 -pentanone

Answer: 3 -methyl butanone

The molar mass of 3-methyl butanone exceeds 86, so it cannot be the correct IUPAC name for the ketone.

Q3. An ideal gas is compressed in a closed container its U?

1. Increases
2. Decreases
3. Remains same
4. Both (1) \& (2)

Answer: Increases

For an ideal gas, internal energy (U) depends only on temperature. Compression raises temperature, so U increases.

Q4. Which of the following will change the value of the equilibrium constant for the reaction? \( N_{2}(g)+O_{2}(g) \leftrightharpoons 2 N O(g) \)

1. Add more \( N_{2} \)
2. Increase of pressure
3. Use a smaller reaction vessel
4. Increase the temperature

Answer: Increase the temperature

The equilibrium constant is only affected by temperature changes because it is related to the reaction's enthalpy and Gibbs free energy.

Q5. Electronic configuration \( [\boldsymbol{K} \boldsymbol{r}] \boldsymbol{4} \boldsymbol{d}^{10} \) belongs to:

1. Period(4) Group(10) Block(d)
2. Period(5) Group(12) Block (d)
3. Period(5) Group(10) Block \( (d) \)
4. Period(4) Group(12) Block(d)

Answer: Period(5) Group(10) Block \( (d) \)

The configuration shows 4d^10, indicating the 5th period (n=5) and a full d-subshell (10 electrons), placing it in Group 10.

Q6. Internal resistance of a battery cell increases with:

1. increase in concentration of electrolyte
2. increase in distance between two electrodes
3. increase in area of the plates inside the electrolyte
4. increase in size of the electrodes

Answer: increase in distance between two electrodes

Internal resistance increases when the electrodes are farther apart because ions must travel a longer path through the electrolyte, causing more opposition to current. The other options generally reduce resistance by improving conductivity or increasing effective contact area.

Q7. Compare the properties of a typical metal and a non-metal on the basis of the following. Conductivity of heat and electricity:

1. Metals are good conductor of heat and electricity
2. Non metals are good conductor of heat and electricity
3. Metals are bad conductor of heat and electricity
4. Both B and C

Answer: Metals are good conductor of heat and electricity

Metals have delocalized electrons that move easily through the structure, so they conduct heat and electricity well. Non-metals generally lack these free-moving charge carriers, so they are usually poor conductors.

Q8. The plot shows the variation of \( -\ln \boldsymbol{K}_{\boldsymbol{p}} \) versus temperature for the two reactions. \( M(s)+\frac{1}{2} O_{2}(g) \rightarrow M O(s) \) and \( C(s)+\frac{1}{2} O_{2}(g) \rightarrow C O(s) \) Identify the correct statement?

1. \( A t T>1200 K, \) carbon will reduce \( M O(s) \) to \( M(s) \)
2. At \( T<1200 \mathrm{K} \), oxidation of carbon is unfavourable.
3. At \( T<1200 \) K, the reaction \( M O(s)+C(s) \rightarrow M(s)+ \) \( C O(g) \) is spontaneous
4. Oxidation of carbon is favourable at all temperatures

Answer: \( A t T>1200 K, \) carbon will reduce \( M O(s) \) to \( M(s) \)

At temperatures above 1200 K, the reaction becomes spontaneous, and carbon can reduce MO(s) to M(s). [AI-generated key — verify before high-stakes use]

Q9. Select the correct statement.

1. Carbon-carbon bond order in graphite is more than diamond
2. Carbon-carbon bond order in graphite is less than diamond
3. Hybridization of each carbon in diamonds is \( S p_{3} \)
4. Hybridization of each carbon in graphite is \( S p_{3} \)

Answer: Hybridization of each carbon in diamonds is \( S p_{3} \)

In diamond, each carbon forms four single bonds in a tetrahedral arrangement, which corresponds to sp3 hybridization. Graphite instead has sp2-hybridized carbons arranged in planar sheets with delocalized pi electrons.

Q10. \( \mathbf{2} \boldsymbol{C} \boldsymbol{a}_{3}\left(\boldsymbol{P} \boldsymbol{O}_{4}\right)_{2}+\boldsymbol{6} \boldsymbol{S} \boldsymbol{i} \boldsymbol{O}_{2}+\mathbf{1 0} \boldsymbol{C} \rightarrow \boldsymbol{P}_{4}+ \) \( \boldsymbol{C a} \boldsymbol{S i O}_{3}+\mathbf{1 0 C O} \) Which coefficient balances the reaction given above?

1. 2
2. 4
3. 5
4. 6

Answer: 2

The equation is balanced by matching each element’s atom count on both sides. With 2 Ca3(PO4)2, there are 4 P atoms and 6 Ca atoms, so the product must be 1 P4 and 6 CaSiO3, which also fits the oxygen and carbon counts already shown.

Q11. Diborane \( \left(B_{2} H_{6}\right) \) reacts independently with \( O_{2} \) and \( H_{2} O \) to produce :

1. \( H B O_{2} \) and \( H_{3} B O_{3} \)
2. \( H_{3} B O_{3} \) and \( B_{2} O_{3} \)
3. \( B_{2} O_{3} \) and \( H_{3} B O_{3} \)
4. \( B_{2} O_{3} \) and \( \left[B H_{4}\right]^{-} \)

Answer: \( H_{3} B O_{3} \) and \( B_{2} O_{3} \)

Diborane burns in oxygen to form boron(III) oxide, while hydrolysis with water produces boric acid. So the two independent reactions give B2O3 and H3BO3, respectively.

Q12. A beaker contains a saturated solution of copper(l) chloride. Copper(l) chloride is slightly soluble salt with a solubility product of \( 1.2 \times 10^{-6} \) Which of the following salts when added to the solution would precipitate copper(l) chloride?

1. Sodium chloride
2. Potassium bromide
3. Silver(I) nitrate
4. Lead(II) acetate E. Magnesium iodide

Answer: Silver(I) nitrate

Silver(I) nitrate provides Ag+ ions, which react with Cl− to form AgCl, an insoluble precipitate. Removing chloride from solution shifts the equilibrium for CuCl(s) formation, causing more copper(I) chloride to precipitate.

Q13. The state of equilibrium refers to:

1. State of rest
2. Dynamic state
3. Stationary state
4. State of inertness

Answer: Dynamic state

Equilibrium is called a dynamic state because microscopic processes still occur, but they balance each other so there is no net change. The system appears unchanged overall, even though activity continues internally.

Q14. Internal energy does not include

1. vibrational energy
2. rotational energy
3. energy arising by gravitational pull
4. nuclear energy

Answer: energy arising by gravitational pull

Internal energy is the sum of microscopic energies within a system, such as vibrational, rotational, and nuclear contributions. Energy arising from gravitational pull is a macroscopic potential energy due to position in an external field, so it is not part of internal energy.

Q15. The pH of \( D_{2} O \) and \( H_{2} O \) at \( 298 \mathrm{K} \) is:

1. 7.0, 7.0
2. 7.35, 7.0
3. 7.0,6.85
4. 6.85,7.35

Answer: 7.0, 7.0

The pH of D₂O and H₂O at 298 K is 7.0, as both are neutral solutions with equal concentrations of hydrogen and hydroxide ions. [AI-generated key — verify before high-stakes use]