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Test: Chemical Bonding & Molecular Structure - 1 - NEET MCQ


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25 Questions MCQ Test Chemistry Class 11 - Test: Chemical Bonding & Molecular Structure - 1

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Test: Chemical Bonding & Molecular Structure - 1 - Question 1

In which of the following species the underlined carbon is having sp− hybridization

Detailed Solution for Test: Chemical Bonding & Molecular Structure - 1 - Question 1

Only in CH3CH2OH, carbon has sp3 hybridisation. In other molecules, the carbon atom has multiple bonds.

Test: Chemical Bonding & Molecular Structure - 1 - Question 2

Among the following, the species having the smallest bond length is:

Detailed Solution for Test: Chemical Bonding & Molecular Structure - 1 - Question 2

The bond order of given molecules are:
NO = 2.5, NO+ = 3, O2 = 2, NO = 2
Larger the bond order, the smaller the bond length.
NO+ has the largest bond order 3.
Therefore, it will have the smallest bond length.
Hence option A is the answer.

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Test: Chemical Bonding & Molecular Structure - 1 - Question 3

Among the following the maximum covalent character is shown by which compound:

Detailed Solution for Test: Chemical Bonding & Molecular Structure - 1 - Question 3

The correct option is D.

  • The proportion of covalent character in ionic bond is decided by polarisability of the metal cation as well as the electronegativity of both the elements involved in bonding.
  • Polarisability is also decided by the density of +ve charge on the metal cation. 
  • AlCl3 shows maximum covalent character. This is because Al cation has 3 unit +ve charge and shows strong tendency to distort the electron cloud
Test: Chemical Bonding & Molecular Structure - 1 - Question 4

The hybridisation of orbitals of N atom in NO3, NO2+, NH4+ are respectively:

Detailed Solution for Test: Chemical Bonding & Molecular Structure - 1 - Question 4
  • In NO3, the central N atom has 3 bonding domains and zero lone pairs of electrons.
  • In NO2, the central N atom has 2 bonding domains and zero lone pairs of electrons.
  • In NH4, the central N atom has 4 bonding domains and zero lone pairs of electrons.
  • The Hybridization of N atom in NO3, NO2+, NH4+ are sp2, sp, sp3 respectively.

Hence option A is the answer.

*Multiple options can be correct
Test: Chemical Bonding & Molecular Structure - 1 - Question 5

Which of the following statements are not correct?

Detailed Solution for Test: Chemical Bonding & Molecular Structure - 1 - Question 5
  • Statement A: "In canonical structures there is a difference in the arrangement of atoms."
    Incorrect. In canonical (resonance) structures, only the arrangement of electrons changes, not the arrangement of atoms.

  • Statement B: "Hybrid orbitals form stronger bonds than pure orbitals."
    Correct. Hybrid orbitals, such as sp, sp², and sp³, tend to form stronger and more stable bonds compared to unhybridized (pure) orbitals because of better orbital overlap.

  • Statement C: "NaCl being an ionic compound is a good conductor of electricity in the solid state."
    Incorrect. In the solid state, NaCl does not conduct electricity because its ions are fixed in place within the crystal lattice. Only in the molten or dissolved state can NaCl conduct electricity, as ions are free to move.

  • Statement D: "VSEPR Theory can explain the square planar geometry of XeF₄."
    Correct. VSEPR (Valence Shell Electron Pair Repulsion) Theory can explain the square planar geometry of XeF₄, as it accounts for the lone pairs on the central atom, Xe, which arrange themselves to minimize repulsion and give XeF₄ a square planar shape.

Test: Chemical Bonding & Molecular Structure - 1 - Question 6

When a gas phase atom in its ground state gains an electron. This is called:

Detailed Solution for Test: Chemical Bonding & Molecular Structure - 1 - Question 6

 

  • Electron Affinity refers to the energy change that occurs when a neutral atom in the gas phase gains an electron to form a negative ion. It is typically an exothermic process for most elements, meaning energy is released.

  • Electron Gain Enthalpy (Option A) is closely related to electron affinity, but it specifically refers to the enthalpy change associated with gaining an electron. The terms are often used interchangeably, though electron affinity is more commonly used in this context.

  • Ionization Enthalpy (Option B) is the energy required to remove an electron from a neutral atom in the gas phase, not to gain one.

  • Lattice Enthalpy (Option D) is the energy required to separate one mole of an ionic solid into its gaseous ions, which is unrelated to an individual atom gaining an electron.

Test: Chemical Bonding & Molecular Structure - 1 - Question 7

Elements in which apart from 3s and 3p orbitals, 3d orbitals also available for bonding In a number of compounds of these elements there are more than eight valence electrons around the central atom. One such example is:

Detailed Solution for Test: Chemical Bonding & Molecular Structure - 1 - Question 7

In H2SO4, S has more than 8 electrons in the valence shell.

Test: Chemical Bonding & Molecular Structure - 1 - Question 8

The number of types of bonds between two carbon atoms in calcium carbide is:

Detailed Solution for Test: Chemical Bonding & Molecular Structure - 1 - Question 8

CaC2 can be divided into two ions: Ca2+, C22-

  • There is a triple bond, between two carbon.
  • Among three bonds one is sigma, bond and the remaining two are pi bond.
Test: Chemical Bonding & Molecular Structure - 1 - Question 9

The electronic configuration of the outer most shell of the most electronegative element is:

Detailed Solution for Test: Chemical Bonding & Molecular Structure - 1 - Question 9
  • The most electronegative element is F. Its atomic number is 9.
  • So, the electronic configuration is 2s22p5.
Test: Chemical Bonding & Molecular Structure - 1 - Question 10

Rank the bonds in the set C=O, C-O, C≡O, in the order of decreasing bond length:

Detailed Solution for Test: Chemical Bonding & Molecular Structure - 1 - Question 10
- The bond lengths decrease as the number of bonds between atoms increases.
- C-O is a single bond, which is the longest among the three.
- C=O is a double bond, shorter than a single bond but longer than a triple bond.
- C≡O is a triple bond, making it the shortest.
- Therefore, the correct order of decreasing bond length is C-O > C=O > C≡O.
- Hence, the correct option is 2.
Test: Chemical Bonding & Molecular Structure - 1 - Question 11

A qualitative measure of the stability of an ionic compound is provided:

Detailed Solution for Test: Chemical Bonding & Molecular Structure - 1 - Question 11
  • Lattice enthalpy is a measure of the energy released when oppositely charged ions in the gas phase come together to form an ionic solid.
  • It provides a qualitative measure of the stability of an ionic compound, as a higher lattice enthalpy indicates a more stable and strongly bonded ionic lattice.
  • The greater the lattice enthalpy, the more energy is required to break the ionic bonds, indicating a stable ionic compound.
Test: Chemical Bonding & Molecular Structure - 1 - Question 12

The shift in electron density is symbolized by the crossed arrow in the below diagram. It depicts:

Detailed Solution for Test: Chemical Bonding & Molecular Structure - 1 - Question 12

Arrows represent the direction of the dipole moment.

Test: Chemical Bonding & Molecular Structure - 1 - Question 13

Which one of the following is paramagnetic?

Detailed Solution for Test: Chemical Bonding & Molecular Structure - 1 - Question 13

The Lewis structure of NO molecule can be represented as:

  • It is observed that the total no. of unbonded electrons is odd.
  • Therefore, there must be an incompletely filled subshell.

Thus, NO is paramagnetic in nature.

Test: Chemical Bonding & Molecular Structure - 1 - Question 14

Using MO theory predicts which of the following species has the shortest bond length?

Detailed Solution for Test: Chemical Bonding & Molecular Structure - 1 - Question 14

Bond length is inversely proportional to bond order.

Test: Chemical Bonding & Molecular Structure - 1 - Question 15

Rank the bonds in the set C=O, C-O, C≡O in order of decreasing bond strength:

Detailed Solution for Test: Chemical Bonding & Molecular Structure - 1 - Question 15

Bond order is directly proportional to bond strength. The higher the bond order, the stronger the pull between the two atoms and the shorter the bond length.

Correct order is: C≡O > C=O > C-O

Test: Chemical Bonding & Molecular Structure - 1 - Question 16

Bond lengths are lower in elements having:

Detailed Solution for Test: Chemical Bonding & Molecular Structure - 1 - Question 16
  • Bond length is the distance between the nuclei of two bonded atoms. The more electrons that are shared between the atoms (as in multiple bonds), the closer the atoms are pulled together, resulting in a shorter bond length.
  • In general, triple bonds are shorter than double bonds, which are in turn shorter than single bonds.
  • This is because triple bonds involve the sharing of three pairs of electrons, creating a stronger attraction and pulling the bonded atoms closer together.

Thus, elements with triple bonds have the shortest bond lengths among the options given.

Test: Chemical Bonding & Molecular Structure - 1 - Question 17

In BF3, the molecule below the dipole moment is zero although the B-F bonds are oriented at an angle of 1200 to one another. Net dipole moment in BF3 molecule is:

Detailed Solution for Test: Chemical Bonding & Molecular Structure - 1 - Question 17
  • In BF₃ (boron trifluoride), the molecule has a trigonal planar structure, with the three B-F bonds oriented at 120° to each other.
  • Each B-F bond is polar due to the difference in electronegativity between boron and fluorine, which creates individual dipole moments.
  • However, because of the symmetrical trigonal planar shape, the dipole moments of the three bonds cancel each other out.
  • As a result, the net dipole moment of the BF₃ molecule is zero.

This means that although the individual B-F bonds are polar, the overall molecule is non-polar due to its symmetry.

Test: Chemical Bonding & Molecular Structure - 1 - Question 18

Hybridization of Cand Cof H3C − CH = C = CH − CHare:

Detailed Solution for Test: Chemical Bonding & Molecular Structure - 1 - Question 18


 

C2 has 1 double bond, so it is sp2 hybridized while C3 has 2 double bonds so it is sp hybridized.

Test: Chemical Bonding & Molecular Structure - 1 - Question 19

Which one of the following pairs of species has the same bond order?

Detailed Solution for Test: Chemical Bonding & Molecular Structure - 1 - Question 19
  • The species which have the same number of total electrons will have the same bond order.
  • CNand NO+ each have 14 electrons and they will have the same bond order.
Test: Chemical Bonding & Molecular Structure - 1 - Question 20

Quartz is very hard and melts at 1550C . Reason is:

Detailed Solution for Test: Chemical Bonding & Molecular Structure - 1 - Question 20

Quartz/silica have extended covalent bond network.

  • Quartz (SiO₂) is a network solid where each silicon atom is covalently bonded to four oxygen atoms in a tetrahedral structure.
  • These covalent bonds form a continuous three-dimensional network throughout the crystal, making it extremely hard and giving it a high melting point.
  • The strength of these covalent bonds across the entire structure requires a large amount of energy to break, which is why quartz has a high melting point of around 1550°C.

Options A, B, and C are incorrect because they misrepresent the bonding in quartz:

  • SiO₂ does not exist as discrete molecules in quartz but as a continuous covalent network.
  • The bonds are covalent rather than electrovalent (ionic) or coordinate bonds.
Test: Chemical Bonding & Molecular Structure - 1 - Question 21

H-O-H bond angle in water is:

Detailed Solution for Test: Chemical Bonding & Molecular Structure - 1 - Question 21
  • Water has a bent (or V-shaped) molecular geometry due to the sp³ hybridization of the oxygen atom.
  • Oxygen has two lone pairs and two bonded pairs of electrons.
  • The lone pairs exert greater repulsion than bonding pairs, which compresses the bond angle from the ideal tetrahedral angle of 109.5° to about 104.5°.
Test: Chemical Bonding & Molecular Structure - 1 - Question 22

The shape of the below molecule is:
 

Detailed Solution for Test: Chemical Bonding & Molecular Structure - 1 - Question 22

The order of Repulsion: lone pair-lone pair > lone pair-bond pair > bond pair-bond pair

Due to the extra lone pair electron, the shape becomes bent.

Test: Chemical Bonding & Molecular Structure - 1 - Question 23

The sp3d2 hybridization of central atom of a molecule would lead to

Detailed Solution for Test: Chemical Bonding & Molecular Structure - 1 - Question 23

It will lead to an octahedral structure assuming that there are no unpaired electrons in the central atom. In case there are any , you have to consider the shape that bonding pairs of electrons are in.

so sp3d2 hybridization will lead to octahedral geometry.

Test: Chemical Bonding & Molecular Structure - 1 - Question 24

CO is isoelectronic with

Detailed Solution for Test: Chemical Bonding & Molecular Structure - 1 - Question 24

CO , N2 and NO+ have same number of electrons = 14.

Test: Chemical Bonding & Molecular Structure - 1 - Question 25

Rank the following bonds in order of increasing polarity: H-N, H-O, H-C.

Detailed Solution for Test: Chemical Bonding & Molecular Structure - 1 - Question 25

Polarity in a bond depends on the difference in electronegativity between the two atoms involved. The greater the difference, the more polar the bond.

  1. H-O: Oxygen is highly electronegative, so the H-O bond is the most polar of the three.
  2. H-N: Nitrogen is less electronegative than oxygen but more electronegative than carbon, so the H-N bond is moderately polar.
  3. H-C: Carbon is relatively close to hydrogen in electronegativity, so the H-C bond is the least polar.

Therefore, the order of increasing polarity is: H-C < H-N < H-O, which matches option B.

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