Chemical Bonding and Atomic Structure NEET Questions — Free MCQ Practice

Chemical Bonding and Atomic Structure MCQs for NEET — Chemistry Questions with Answers

Practice free Chemical Bonding and Atomic Structure (Chemistry) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

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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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Select the correct statement:

$H S O_{5}^{-}$ ion has one $S — O — H$ linkage

Number of $B — O — B$ linkages in Borax is equal to number of $P — O — P$ linkages in $P_{4} O_{10}$

Hybridization of both sulphur on $(H_{2} S_{2} O_{5})$ (pyrosulphurous acid) is name but oxidization state of both sulphur are different

Tetra -Polyphosphoric has four $P — O — P$ and no $P — P$ linkage

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The H-O-H bond angles in $H_{3} O^{+}$ are approximately $107^{°}$ . The orbitals used by oxygen in these bond are best describe as:

$p - o r b i t a l s$

sp-hybrid orbitals

$s p^{2} .$ hybrid orbital

$s p^{3}$ . hybrid orbital

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Which of the following molecular species is not linear?

$(CN)_{2}$

${OCN}^{-}$

${XeF}_{2}$

$S_{3}^{2 -}$

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Explanation

Among the given options, the molecular species S32- (trisulfur dianion) is not linear. It has a bent or V-shaped molecular geometry due to the presence of two terminal sulfur atoms bonded to the central sulfur atom with a bond angle less than 180Â°.

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The compound that has the highest ionic character assoicated with the X-Cl bond is:

$P C l_{5}$

$B C l_{3}$

$C C l_{4}$

$S i C l_{4}$

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Explanation

(d) In $S i C l_{4}$ difference between electronegativity of Si (1.8) and chlorine (3.0) is higher than in other given compounds

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In which of the folllowing transformations, the bond order has increased and the magnetic behaviour has changed?

$C_{2}^{+} \to C_{2}$

${NO}^{+} \to NO$

$O_{2} \to O_{2}^{+}$

$N_{2} \to N_{2}^{+}$

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Explanation

(a)

(a) $\underset{Paramagnetic}{\underset{BO = 1.5}{C_{2}^{+}}} \overset{}{\to} \underset{Diamagnetic}{\underset{BO = 2}{C_{2}}}$ (b) $\underset{Diamagnetic}{\underset{BO = 3}{{NO}^{+}}} \underset{}{\to} \underset{Paramagnetic}{\underset{BO = 2.5}{NO}}$

(c) $\underset{Paramagnetic}{\underset{BO = 2}{O_{2}}} \overset{}{\to} \underset{Paramagnetic}{\underset{BO = 2.5}{O^{+}}}$ (d) $\underset{Diamagnetic}{\underset{BO = 3}{N_{2}}} \underset{}{\to} \underset{Paramagnetic}{\underset{BO = 2.5}{N_{2}^{+}}}$

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The bond having the highest bond energy is :

C=C

C=S

C=O

P=N

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Explanation

(c) More effective axial and sideways overlapping between atomic orbitals of carbon and those of oxygen atom is higher due to smaller size of oxygen atom. oxygen is more electronegative than oxygen atom, due to these factor CO has highest bond energy .

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