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Chemistry MCQs for NEET — Practice Questions with Answers

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Inorganic benzene reacts with HCl to form a compound $B_{3} N_{3} H_{9} {Cl}_{3}$ . The protonation occurs at:

B-atom

N-atom

First at B-atom then rearrangers into N-atom

First N-atom then rearrangers into B-atom

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Given the correct order of initials T or F for the following statements. Use T if staement is true and F if it is false:

(i) $({CH}_{3})_{2} P ({CF}_{3})_{3}$ is non-polar $({CH}_{3})_{3} P ({CF}_{3})_{2}$ is polar molecule

(ii) ${CH}_{3} \hat{P} {CH}_{3}$ bond angles are equal in $({CH}_{3})_{3} P ({CF}_{3})_{2}$ molecule

(iii) ${PF}_{3}$ will be more soluble in polar solvent than ${SiF}_{4}$

TTF

FFT

FFF

FTT

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Which substance has the strongest London dispersion forces?

$S i H_{4}$

$C H_{4}$

$S n H_{4}$

$G e H_{4}$

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Explanation

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Which of the following compounds has the lowest boiling point?

HCl

HBr

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Explanation

(b) HCl has least boiling point among all halogen acids because it contains V.W. force never contains H-bond HCl < HBr < HI < HF .

Only HF has hydrogen bonding rest all have V.W. force of attraction.

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The incorect order of boiling point is:

$H_{2} O > C H_{3} O H$

$N {(C H_{3})}_{3} > N H {(C H_{3})}_{2}$

$H_{3} P O_{4} > M E_{3} P O_{4}$

$C H_{3} N_{3} > H N_{3}$

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Explanation

(d) Correct order of B.P

$C H_{3} N_{3} < H N_{3}$

Among $C H_{3} N_{3}$ there are dipole-dipole interaction while among $H N_{3}$ intermolecular H- bonding occurs.

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When the substances Si, KCl, $C H_{3}$ OH and $C_{2} H_{6}$ are arranged in order of increasing melting point , what is the correct order?

Si, KCl , $C H_{3}$ OH , $C_{2} H_{6}$

$C H_{3} O H$ , $C_{2} H_{6}$ , Si ,KCl

KCl, Si, $C_{2} H_{6}$ , $C H_{3} O H$

$C_{2} H_{6}, C H_{3} O H, K C l, S i$

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Explanation

(d) $C_{2} H_{6} < C H_{3} - O H < K C L < S i$

( molecular lattice) ( molecular lattice ( Ionic lattice) (covalent lattice)

with H-bonding)

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Iodine molecules are held in the solid lattice by............

London forces

dipole-dipole interactions

Covalent bonds

coulombic force

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Explanation

Iodine molecules in the solid lattice are held together by London dispersion forces, which are weak intermolecular forces arising from the instantaneous dipole moments in non-polar molecules. These forces are responsible for the solid-state of iodine at room temperature.

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Nodal planes of $π - bond (s) in {CH}_{2} = C = C = {CH}_{2} are located in :$

all are in molecular plane

two in a molecular plane and in a plane peerpendicular to moleclar plane which contains $C - C σ - bond$

one in molecular plane and two in plane perpendicular to molecular plane which bisects $C - C σ - bonds$

two in a molecular plane and one in a plane perpendicular to molecular plane which bisects $C - C σ - bonds$ at right angle.

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Explanation

The nodal planes of Ï€-bonds in CH2=C=C=CH2 are located in two planes - one in the molecular plane containing the carbon atoms, and the other perpendicular to the molecular plane containing the C-C Ïƒ-bonds. This is due to the formation of Ï€-bonds by the sideways overlap of p-orbitals.

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The hybridisation of the central atom will change when:

$N H_{3}$ combines with $H^{+}$

$H_{3} B O_{3_{}}$ combines with $O H^{-}$

$N H_{3}$ forms $N {H_{2}}^{-}$

$H_{2} O$ combines with $H^{+}$

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Which bonds are formed by a carbon atom with $s p^{2}$ - hybridisation?

$4 π - b o n d s$

$2 π - b o n d s a n d 2 σ b o n d s$

$1 π - b o n d s a n d 3 σ - b o n d s$

$4 δ - b o n d s$

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Explanation

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Yes. You can attempt every Chemistry question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.