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Chemistry MCQs for NEET — Practice Questions with Answers

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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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The geometry of ammonia molecule can be best described as:

Nitrogen at one vertex of a regular tetrahedron, the other three vertices being occupied by three hydrogens

Nitrogen at the centre of the tetrahedron, three vertics being occupied by three hydrogens

Nitrogen at the centre of an equilateral triangle, three corners being occupied by three hydrogens

Nitrogen at the junction of a T, three open ends being occupied by three hydrogens

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Explanation

(b) In $\overset{. .}{N} H_{3}$ central atom is nitrogen which is $s p^{3}$ hybrdized hence, it will be at the centre of tetrahedron with H-atom at three vertices.

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Assuming pure 2s and 2p orbitals are carbon are used in forming $C H_{4}$ molecule, which of the following statements is false?

Three C $—$ H bonds will be at right angle

One C $—$ H bond will be weaker then other three C $—$ H bonds

The shape of molecule will be tetrahedral

The angle of C $—$ H bond formed by s-s overlapping will be uncertain with respect to other three bonds

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Which molecular geometry is least likely to result from a trigonal bipyramidal electron geometry?

Trigonal planar

See-saw

Linear

T-shaped

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Explanation

(a) In trigonal by pyramidal geometry lone pair electrons cannot occupy axial positions.

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Which of the following orbital can not form $π$ as well as $δ$ -bond?

$d_{x y}$

$d_{z^{2}}$

$d_{x^{2} - y^{2}}$

$d_{y z}$

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Explanation

(b) $d_{z^{2}}$ can not form $π$ as well as $δ$ bond but it can form $σ$ -bond.

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