Chemistry MCQs for NEET — Practice Questions with Answers

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Which of the following does not possesses bond order as CO ?

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Explanation

The bond order of CO is 3. The bond order of NO+ is also 3, NO has a bond order of 2.5, N2 has a bond order of 3, and CN- has a bond order of 3. Therefore, NO (neutral) does not have the same bond order as CO.

Which of the following is the correct order for lone pair and bonding pair electrons ? Lp = Lone pair and Bp = Bonding pair

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Explanation

In VSEPR (Valence Shell Electron Pair Repulsion) theory, the repulsion between lone pairs (Lp) of electrons is greater than the repulsion between lone pair and bonding pair (Bp) of electrons, which in turn is greater than the repulsion between bonding pairs of electrons. Hence, the correct order is Lp-Lp > Lp-Bp > Bp-Bp.

Which theory is useful to determine geometrical structure of molecules ?

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Explanation

The VSEPR (Valence Shell Electron Pair Repulsion) theory is primarily used to determine the geometrical structure of molecules. It predicts the shape of individual molecules based on the extent of electron-pair electrostatic repulsion.

Which of the following has maximum bond angle ?

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Which of the following have equal bond order ?

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Explanation

The bond order of $O_2$ is 2, CN is 3, and $NO^+$ is also 3. Therefore, CN and $NO^+$ have equal bond orders.

The type of bond present in $CuSO_4 .5H_2O$

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$ CO_2$ is isostructural with

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$ NH_3$ has a higher boiling point than expected because

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Explanation

$NH_3$ has a higher boiling point than expected due to the presence of intermolecular hydrogen bonds. In ammonia, the nitrogen atom is highly electronegative and forms hydrogen bonds with hydrogen atoms of neighboring ammonia molecules. These hydrogen bonds require more energy to break, resulting in a higher boiling point.

The molecule with zero dipole moment is

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Explanation

Carbon tetrachloride (CClâ‚„) has a tetrahedral geometry with four chlorine atoms symmetrically arranged around the central carbon atom. This symmetry causes the dipole moments of the C-Cl bonds to cancel each other out, resulting in a net dipole moment of zero.

Molecular shaper of $SF_4 , CF_4, XeF_4$ are

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Explanation

The molecular shapes and lone pairs of electrons for the given molecules are as follows:

  • SFâ‚„ has a 'see-saw' shape with 1 lone pair of electrons.
  • CFâ‚„ has a tetrahedral shape with 0 lone pairs of electrons.
  • XeFâ‚„ has a square planar shape with 2 lone pairs of electrons. Thus, they have different shapes with 1, 0, and 2 lone pairs of electrons respectively.

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