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Chemistry MCQs for NEET — Practice Questions with Answers

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Which bond angle would result in the maximum dipole moment for the triatomic molecule $X Y_{2}$ shown below?

$90^{°}$

$120^{°}$

$150^{°}$

$180^{°}$

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Explanation

(a) $μ = \sqrt{{μ_{1}}^{2} + {μ_{2}}^{2} + μ_{1} μ_{2} \cos θ}$ , if $θ = 90^{°}$ $μ$ is maximum.

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The hydrogen bonding is strongest in:

$O — H ‐ ‐ ‐ S$

$S — H ‐ ‐ ‐ O$

$F — H ‐ ‐ ‐ F$

$F — H ‐ ‐ ‐ O$

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Explanation

$‐ ‐ ‐ H — F > ‐ ‐ ‐ H — O > ‐ ‐ ‐ H — N$

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The ionisation energy will be maximum for the process:

$B a \to B a^{2 +}$

$B e \to B e^{2 +}$

$C s \to C s^{+}$

$L i \to L i^{+}$

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Explanation

(b) smaller is atom, more is energy needed to remove electron , i.e. ionisation energy. Also removal of two electrons needs more energy.

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In which of the process, the bond order increases and magnetic behaviour changes?

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

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

$N O \to N O^{+}$

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

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Explanation

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Specify the coordination geometry around and hybridization of N and B atoms in a 1:1 complex of $B F_{3}$ and $N H_{3}$ :

N: tetrahedral, $s p^{3}$ ; B : tetrahedral, $s p^{3}$

N: pyramidal, $s p^{3}$ ; B : pyramidal, $s p^{3}$

N: pyramidal, $s p^{3}$ ; B: planar, $s p^{2}$

N: pyramidal, $s p^{3}$ ; B: tetrahedral, $s p^{3}$

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Explanation

(a) No doubt $N H_{3}$ and $B F_{3}$ have $s p^{3}$ (pyramidal) and $s p^{2}$ (coplanar) hybridization respectively having one lone pair of electron on N atom which is responsible for pyramidal shape of $N H_{3}$ inspite of $s p^{3}$ hybridization. However, as soon as it is coordinated to $B F_{3}$ , both attain tetrahedral geometry and acquire $s p^{3}$ - hybridization.

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When $O_{2}$ is adsorbed on a metallic surface, electron transfer occurs from the ,metal to $O_{2}$ . The wrong statements(s) regarding this adsorption is (are):

$O_{2}$ is physisorbed

Heat is released

Occupancy of $π * 2 p$ of $O_{2}$ is increased

Bond length of $O_{2}$ is increased

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Explanation

(a)

(a) :Incorrect : Because electronic transfer takes place and it is chemisorbtion.

(b) :correct: Adsorbtion is always exothermic.

(d) :correct: Since the electron enters into $π *$ orbital bond order decrease and bond length increases.

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The $B F_{3}$ is a planarn molecule where as $N F_{3}$ is pyramidal because:

$B - F$ bond is more polar than $N - F$ bond.

Boron atom is bigger than nitrogen atom.

Nitrogen is more electronegative than boron.

$B F_{3}$ has no lone pair but $N F_{3}$ has a lone pair of electrons.

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Explanation

The BF3 molecule has a planar structure due to the absence of lone pairs on the central boron atom. In contrast, the NF3 molecule adopts a pyramidal structure because the nitrogen atom has one lone pair of electrons, causing repulsion and distorting the molecular geometry.

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Which of the following statement is true for $I O_{2} {F_{2}}^{-}$ according to VSEPR theory?

The lone pair and two I-O double bonds occupy the equatorial positions of trigonal
bipyramid.

It has $s p^{3}$ hybridization and is T-shaped.

Its structure is analogous to $S F_{4}$ .

(A) and (C) both

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Explanation

According to VSEPR theory, the IO2F2- ion has a trigonal bipyramidal structure with the two lone pairs and two I-O double bonds occupying the equatorial positions, which is analogous to the structure of SF4. Therefore, both options (A) and (C) are correct.

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In the structure of $H_{2} C S F_{4}$ , to decide the plane in which C = S is present the following bond angle value are given

Axial FSF angle $(i d e a l i s e d = 180 °)$ $\Rightarrow 170 °$

Equatorial FSF angle $(i d e a l i s e d = 120 °)$ $\Rightarrow 97 °$

After deciding the plane of double bond, which of the following statements is correct?

Two C-H bonds are in the same plane of axial S-F bonds.

Two C-H bonds are in the same plane of equatorial S-F bonds.

Total five atoms are in the same plane.

Equatorial S-F bonds are perpendicular to plane of $π$ bond.

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According to Molecular orbital theory which of the following statement is incorrect?

LUMO level for $C_{2}$ molecule is $σ_{2 p x}$ orbital.

In $C_{2}$ molecules both the bonds $π$ are bonds.

$C_{2}$ is paramagnetic but ${C_{2}}^{2 -}$ is diamagnetic.

In ${C_{2}}^{2 -}$ ion there is one $σ$ and two $π$ bonds.

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Explanation

(A), (B) M.O. for C₂ = ${σ_{1}}^{s^{2}} < {σ_{1}}^{*_{s^{2}}} < {σ_{2}}^{s^{2}} < {σ_{2}}^{*_{s^{2}}} < \frac{π_{2} p^{2} y = π_{2} p^{2} z}{H O M O} < \frac{σ_{2} p_{x}}{L U M O},$ two $π$ molecular orbitals are involved in bonding .

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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.