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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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In which of the following cases $C — C$ bond length will be highest?

${CH}_{3} — {CF}_{3}$

${FCH}_{2} — {CH}_{2} F$

$F_{2} CH$

${CF}_{3} — {CF}_{3}$

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Explanation

(b) According to Bent's Rule more electronegative substituents attached to hybrid orbital that contains more p-character. Hence, more p-characters, longer will be bond length.

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Which of the following statement is correct in the context of the allene molecules, $C_{3} H_{4}$ ?

The central carbon is Sp hybridized

The terminal carbon atoms are $s p^{2}$ hybridized

The planes containing the $C H_{2}$ groups are mutually perpendicular to permit the formations two separate $π$ - bonds

All are correct

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Explanation

In allene (C3H4), the central carbon atom is sp hybridized with linear geometry. The terminal carbon atoms are sp2 hybridized, and the planes containing the CH2 groups are perpendicular, allowing the formation of two separate Ï€-bonds. All the given statements are correct for the allene molecule.

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Select the incorrect statement about $N_{2} F_{4} and N_{2} H_{4}$ :

(i) In $N_{2} F_{4}$ , d-orbitals are contracted by electronegative fluorine atoms, but d-orbitals contaraction is not possible by H-atom in $N_{2} H_{4}$

(ii) The N-N bond energy in $N_{2} F_{4}$ is lesser than N-N bond energy in $N_{2} H_{4}$

(iii) The N-N bond length in $N_{2} F_{4}$ is more than that in $N_{2} H_{4}$

(iv) The N-N bond length in $N_{2} F_{4}$ is less than that in $N_{2} H_{4}$

Choose the correct code:

(i),(ii) and (iii)

(i) and (iii)

(ii) and (iv)

(ii) and (iii)

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The shape of $X e {F_{3}}^{+}$ is :

Trigonal planar

Pyramidal

Bent T-shape

See- saw

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Which of the following shape are not possible for possible value of 'n' in $X e F n$ molecule?

Linear

Square planar

Trigonal planar

Capped octahedral

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Explanation

(c) In $X e F n,$ possible value of n is 2,4,6,8 , then compound should be Xe $F_{2}$ (linear), Xe $F_{4}$ (Square planar), Xe $F_{6}$ (Capped octahedral) . so in this case trigonal planar molecule does not possible .

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Which of the following compound has maximum "C $—$ C" single bond length?

$C H_{2} C H C C H$

$H C C C C H$

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

$C H_{2} C H C H C H_{2}$

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Explanation

(b) $H C \equiv \underset{s p}{C} — \underset{s p}{C} \equiv C H$

(d) $H_{2} C = \underset{s p^{2}}{C H} — \underset{s p^{2}}{C H} = C H_{2}$

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Which statement is true about the most stable Lewis structure for $C S_{2}$ ?

There are no lone pairs in molecule

All bonds are double bonds

The central atom does not have an octet of electrons

A sulfur atom must be the central atom for the stucture to be stable

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Minimum $F — S — F$ bond angle present in:

${SSF}_{2}$

${SF}_{6}$

${SF}_{2}$

$F_{3} SSF$

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$S b F_{5}$ reacts with $X e F_{4}$ and $X e F_{6}$ to form ionic compounds $[X e {F_{3}}^{+}]$ $[S b {F_{6}}^{-}]$ and $[X e {F_{5}}^{+}]$ $[S b {F_{6}}^{-}]$ then molecular shape of $[X e {F_{3}}^{+}]$ ion and $[X e {F_{5}}^{+}]$ ion respectively :

Square pyramidal, T-shaped

Bent T-shape, Square pyramidal

See- saw , square pyramidal

Square pyramidal , See-saw

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Explanation

In the reaction between SbF5 and XeF4/XeF6, the ionic compounds formed are [XeF3]+ [SbF6]- and [XeF5]+ [SbF6]-. Based on the VSEPR theory, the molecular shape of [XeF3]+ is bent T-shaped and that of [XeF5]+ is square pyramidal.

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In which of the following species maximum atom can lie in same plane?

$X e F_{2} o_{2}$

$P C l_{5}$

$A s {H_{4}}^{+}$

$X e F_{4}$

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