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401.

Bonding in A: CCl4 B: H2 ; C:HCl; D:NH,CI – involves coordinate bonding. 

Answer»

D: NH4Cl (Ammonium chloride). The bond formed between the nitrogen atom in ammonia and the chloride ion is a coordi nate bond.

402.

The molecule which contains a triple covalent bond is:(A) ammonia(B) methane(C) water(D) nitrogen

Answer»

(D) nitrogen

403.

At what temperature water has maximum density?

Answer»

4°C is maximum density.

404.

Arrange the following in order of increasing strengths of hydrogen bonding O, F, S, CI, N.

Answer»

CI < S < N < O < F.

405.

Name two types of hydrogen bonding.

Answer»

Inter molecular and Intra molecular hydrogen bond.

406.

Explain why a σ-bond is stronger than a π-bond?

Answer»

Orbitals can overlap to a greater extent in a σ-bond due to their axial orientation, so σ-bond is quite strong. On the other hand in a π-bond sideways overlapping is not to an appreciable extent due to the already present σ-bond which restricts the distance between the involved atoms. Hence σ-bonds is much stronger than a π-bond.

407.

Why is σ-bond stronger than π-bond?

Answer»

It is because extent of overlapping is more in σ-bond than π-bond.

408.

Why does He2 not exist?

Answer»

He2 does not exist because repulsive forces dominate attractive forces and bond order is 0 for He2.

409.

The bond present in HCl is ……………. A) Ionic B) Covalent C) Polar covalent D) Dative

Answer»

Correct option is  A) Ionic

Polar Covalent Bond
As there is not complete sharing of electron,there is slight +ve and -ve charge on both the atom.
So,option (C) is correct.
410.

Write the increasing order of energies of molecular orbitals in various diatomic molecules of second row elements.

Answer»

The increasing order of energies of molecular orbitals for molecules (except O2 and F2) is :

σ1s < σ*1s < σ2s < σ*2s < (π2px = π2py) < σ2pz < (π*2px = π*2py) < σ*2pz

The increasing order of energy of molecular orbitals for diatomic molecules like O2 and F2 is :

σ1s < σ*1s < σ2s < σ*2s < σ2pz < (π2px = π2py) < (π*2p = π*2py) < σ*2pz

411.

Define the bond length.

Answer»

Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule.

412.

How do you express the bond strength in terms of bond order?

Answer»

Bond strength represents the extent of bonding between two atoms forming a molecule. The larger the bond energy, the stronger is the bond and the greater is the bond order.

413.

How do you express the bond strength in terms of bond order?

Answer»

The amount of energy required to break one mole of bonds of a particular type so as to separate them into gaseous atoms is called bond energy.

Bond energy is directly proportional to the bond order. Thus, more the value of bond order, greater will be the bond energy and vice-versa.

414.

Between a neutral atom and its cations which has bigger size? Why?

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1. A neutral atom has bigger in size than its cation.

2. A cation has more protons and nucleus of the cation attracts the electrons in the outermost orbital more. 

3. Hence, the radius of the anion decreases. It means size of the anion decreases. 

4. So, a neutral atom is bigger than its cation.

415.

Write the favourable factors for the formation of ionic bonds.

Answer»

Ionic bond is a type of chemical bond which involves the transfer of one or more electrons from one ion to another ions of opposite charges.The ion which gains an electron is called as anions, whereas the ion which has loose an electron is called as cation.Mostly the cations are metals & anions are non metal. For eg

Na + Cl → Na+ + Cl → NaCl

(i) Low ionization enthalpy of element forming cation.

(ii) More negative electron gain enthalpy of element forming anion.

(iii) High lattice enthalpy of the ionic compound formed.

416.

Perform the experiments arranging the apparatus as shown in figure.Record your observations and identify what type of compounds sodium chloride and sugar are

Answer»

When electricity is passed through sodium chloride solution in which carbon rod is immersed hydrogen and chlorine gas are produced. It is an ionic compound. In the second when electricity is passed through sugar solution there is no charge. 

Hence it belongs to covalent compound.

417.

Give one property of hydrogen chloride which agrees with it being a covalent compound.

Answer»

Hydrogen chloride is a gas at room temperature and in dry state it is a bad conductor of electricity.

418.

Write a short note on bond length.

Answer»

Bond length :

  • Bond length is defined as the equilibrium distance between the nuclei of two covalently bonded atoms in a molecule.
  • Each atom of the bonded pair contributes to the bond length.
  • Bond length depends upon the size of atoms and multiplicity of bonds. It increases with increase in size of atom and decreases with increase in multiplicity of bond. 
  • e.g. C – C single bond is longer than C ≡ C triple bond.
  • Bond lengths are measured by X-ray and electron diffraction techniques.
419.

In case of ionic substances, a more appropriate term is formula weight, rather than ‘molecular weight’. Why?

Answer»

Molecules are not present in ionic substances. Only ions are present. The formula of an ionic substance represents the simpler ratio of ions in one mole of crystal. Hence a more appropriate term is ‘formula weight’.

420.

Which property of the above compound agrees with the being of a covalent compound ?

Answer»

Hydrogen chloride is a gas. In dry state, it is bad conductor of electricity. Hence, it is a covalent compound.

421.

Write a short note on bond angles.

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It is the average angle between two adjacent atoms bonded to the central atom in a molecule. Molecules with larger bond angles are more stable than those with smaller angles.

422.

Which one of the following is diamagnetic? (a) O2(b) O22-(c) O22+ (d) None of these

Answer»

(b) O22-

O22- is diamagnetic. Additional two electrons are paired in anti-bonding molecular orbits π*2py and π*2pz

423.

Name two compounds that are covalent when taken pure but produce ions when dissolved in water.

Answer»

Hydrogen chloride and hydrogen bromide.

424.

Sulphur dioxide is a diamagnetic molecule. Explain.

Answer»

Sulphur dioxide (SO2) is a diamagnetic molecule because it has all electrons paired (no free electrons).

425.

What is an ion?

Answer»

An ion is an electrically charged atom (or group of atoms).

426.

Elements X, Y and Z have atomic number 6, 9 and 12 respectively. Which are :(i) forms an anion(ii) forms a cation ?(iii) has four electrons in its valence shell ?ElementAtomic No.Electronic configurationX62,4Y92,7Z122,8,2

Answer»

(i) Y will form an anion

(ii) Z forms a cation

(iii) X has four electrons in its valence shell.

427.

What is a Coordination number?

Answer»

The number of ions of opposite charge that surround a given ion of given charge is known as coordination number of that given ion.

428.

What factors affect or which factors influence the formation cation or anion?

Answer»

Factors affecting formation of anion or cation are : 

1. Atomic size, 

2. Ionization potential, 

3. Electron affinity, 

4. Electronegativity.

429.

Two elements X and Y have the following configurations. X = 1s2 2s2 2p6 3s23p64s2 Y = 1s2 2s2 2p6 3s2 3p5What is the formula of the compound?

Answer»

The electronic configuration of X is 1s2 2s2p6 3s23p64s2  

So its valency is 2.

The electronic configuration of Y is  1s2 2s2 2p6 3s2 3p5

So its valency is 1. 

∴ The formula of the compound is XY2

430.

The electronegativities of two elements are 1 and 3. What type of bond is formed between the two elements? Why?

Answer»

The electronegativity difference of elements = 3 – 1 = 2. So the electronegativity difference is more than 1.9. Therefore the bond formed between the elements is ionic in nature.

431.

Enlist the characteristic features of hybrid orbitals.

Answer»

Characteristic features of hybrid orbitals :

  • Number of hybrid orbitals formed is exactly the same as the participating atomic orbitals.
  • They have same energy and shape.
  • Hybrid orbitals are oriented in space in such a way that there is minimum repulsion and thus are directional in nature.
  • The hybrid orbitals are different in shape from the participating atomic orbitals, but they bear the characteristics of the atomic orbitals from which they are derived.
  • Each hybrid orbitals can hold two electrons with opposite spins.
  • A hybrid orbital has two lobes on the two sides of the nucleus. One lobe is large and the other small.
  • Covalent bonds formed by hybrid orbitals are stronger than those formed by pure orbitals, because the hybrid orbital has electron density concentrated on the side with a larger lobe and the other is small allowing greater overlap of the orbitals.
432.

List the important conditions required for hybridization.

Answer»

Conditions for hybridization :

  • Orbitals belonging to the same atom can participate in hybridization.
  • Orbitals having nearly same energy can undergo hybridization.

[Note : 2s and 2p orbitals of the same atom undergo hybridization but 3s and 2p orbitals of the same atom do not.] 

433.

Write the formula to calculate bond order of molecule.

Answer»

Bond order of a molecule = \(\frac{N_b-N_a}{2}\) 

Where,

Nis the number of electrons present in bonding MOs and Na is the number of electrons present in antibonding MOs.

434.

Mention the steps involved in Hybridization.

Answer»

The steps involved in hybridization are : 

  • formation of the excited state and 
  • mixing and recasting of orbitals.
435.

Define the term hybridization.

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Hybridization is defined as the process of mixing of valence orbitals of same atom and recasting them into equal number of new equivalent orbitals (hybrid orbitals).

436.

Match the following:List-IList-IIA.C-C1.1.20ÅB.C=C2.1.43ÅC.C\(\equiv\)C3.1.54ÅD.C-N4.1.33ÅCode:ABCD(a)3412(b)1324(c)2143(d)4231

Answer»

The correct option is-(a) 3 4 1 2

437.

Match the following:List-I (Molecule)List-II (Bond order)ANitric oxide13BCarbon monoxide22COxygen31DHydrogen42.5Code:ABCD(a)4123(b)3214(c)1342(d)2431

Answer»

The correct option is- (a) 4 1 2 3

438.

Match the following: List-IList-IIAHgCl21BentBNO3-2LinearCSO23TetrahedralDSO42-4Trigonal planarCode:ABCD(a)1324(b)2413(c)3241(d)4132

Answer»

The correct option is- (b) 2 4 1 3

439.

Which one of the following has linear shape?(a) I3-(b) ICl4-(c) BrF5(d) IOF5

Answer»

The correct answer is- (a) I3-

440.

Which of the following has see saw shape?(a) PCl5(b) IO2F2-(c) SOF4(d) ClO33

Answer»

The correct option is- (b) IO2F2-

441.

Define Sigma bond. Give an example.

Answer»

A covalent bond formed due to the overlapping of orbitals of the two atoms along the line joining the two nuclei (orbital axis) is called sigma (σ) bond. 

For example, the bond formed due to s-s and s-p, p-p overlapping along the orbital axis are sigma bonds.

442.

What is Pi- bond? 

Answer»

A covalent bond formed by the side wise overlapping of p- or d orbitals of two atoms is called as pi (π) bond. 

For example, the bond formed due to the sideways overlapping of the two p- orbitals is a pi- bond.

443.

H3PO3 can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing H3PO3? If not, give reasons for the same.

Answer»

The given structures cannot be taken as the canonical forms of the resonance hybrid of H3PO3 because the positions of the atoms have changed.

444.

Which does not show resonance CO32- ,BO33- , SO42- ?

Answer»

BO33− does not show resonance.

445.

H3PO3 can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical form of the resonance hybrid representing H3PO3? If not, give reasons for the same.

Answer»

The formula 1 and 2 do not describe the canonical forms of the resonance hybrid representing H3PO3 because

(i) Position of one of the H nuclei is different.

(ii) The number of bonding and non-bonding pairs is different.

446.

Give two difference between Ϭ &amp; π bonds.

Answer»
δ- Bondπ- Bond
Ϭ- Bonds are formed by the overlapping of orbitals at the internuclear axis.π- Bonds are formed by the overlapping of orbitals perpendicular to the internuclear axis.
Ϭ- Bonds are stronger than π-Bonds.π-Bonds are weaker than Ϭ-Bonds.

447.

S – orbital does not show any preference for direction. Why?

Answer»

S – Orbital does not show any preference for direction because it is spherically symmetrical.

448.

Elements X, Y, and Z have atomic numbers 6, 9 and 12 respectively. Which one:(a) forms an anion,(b) forms a cation,(c) State type of bond between Y and Z and give its molecular formula.

Answer»

(a) Y

(b) Z

(c) X

449.

What are the conditions necessary for the formation of covalent molecules?

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(i) Both atoms should have four or more electrons in their outermost shells, i.e., non-metals.

(ii) Both the atoms should have high electronegativity.

(iii) Both the atoms should have high electron affinity and high ionisation potential.

(iv) Electronegativity difference between the two atoms should be zero or negligible.

(v) The approach of the atoms towards one another should be accompanied by decrease of energy.

450.

What is the difference between:(a) ionic compounds and polar covalent compounds,(b) ionic compounds and covalent compounds,(c) a polar covalent compound and a non-polar covalent compound?

Answer»

(a) Ionic compounds are formed as a result of transfer of one or more electrons from the atom of a metallic electropositive element to an atom of a non-metallic electronegative element. A polar covalent compound is the one in which there is an unequal distribution of electrons between the two atoms.

(b) Ionic compounds, made up of ions, are generally crystalline solids with high melting and boiling points. They are soluble in water and good conductors of electricity in aqueous solution and molten state. Covalent compounds, made up of molecules, can exist as soft solids or liquids or gases with low melting and boiling points. They are generally insoluble in water and poor conductors of electricity.

(c) Polar covalent compounds are formed between 2 non-metal atoms that have different electro negativities and therefore have unequal sharing of the bonded electron pair. Non-polar compounds are formed when two identical non-metals equally share electrons between them.