Chemistry

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (ii)

The electronic configurations and the bond orders of the given species are given below:

Bond order of O₂ ^- : σ1s² σ∗1s² σ2s² σ∗2s² σ2p²_z (π2p²_x = π2p²_y) (π∗2p¹_x= π∗2p¹_y)

Bond order = $\frac{1}{2}$ ( N_b- N_a)

BO= $\frac{1}{2}$ (10-7)= $\frac{3}{2}$ = 1.5

Bond order of O₂ ⁺ : σ1s² σ∗1s² σ2s² σ∗2s² σ2p²_z (π2p²_x = π2p²_y) (π∗2p¹_x= π∗2p⁰_y)

Bond order = $\frac{1}{2}$ (10-5)= $\frac{5}{2}$ =2.5

Bond order of O₂ : σ1s² σ∗1s² σ2s² σ∗2s² σ2p²_z (π2p²_x = π2p²_y) (π∗2p¹_x= π∗2p¹_y)

BO= $\frac{1}{2}$ (10-6) = $\frac{4}{2}$ =2

So, the correct order of bond order is (ii) O₂ ^– < O₂ < O₂ ⁺

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (iv)

In the molecules of boron ( B₂ ), carbon ( C₂ ) and nitrogen ( N₂ ) the energy of s2p_z  molecular orbital is greater than the energy of P2p_x and P2p_y molecular orbital.

So, the correct order of energies of molecular orbital of N2 is

  (π2p_y ) > (σ2p_z ) < (* 2p_x ) ≈ (π*2p_y )

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (ii)

The element B represents the electronic configuration of phosphorus atom and element C represents the electronic configuration of chlorine atom. Both phosphorus and chlorine are non-metals hence the bonding forming between them will be a covalent bond.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (iv)

The electronic configuration of B given as 1S² 2S²  2P⁶ 3S² 3P³ which represents phosphorous atom. The electronic configuration of C is given as 1S² 2S²  2P⁶ 3S² 3P⁵   which represents chlorine molecule.

So, the molecular formula representing B and C will be phosphorous trichloride ( PCl₃ ) which corresponds to BC₃ .

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (ii)

The electronic configuration of C is given 1S² 2S²  2P⁶ 3S² 3P⁵  which the electronic configuration of chlorine atom ( Cl ). It is unstable due to one unpaired electron in the p-subshell. Hence, the C₂ will be the stable form of C as it will form a dichloride molecule ( Cl₂ ).

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (i)

The electronic configuration of A 1S² 2S²  2P⁶  is having no unpaired electrons in the outer most shells. So, an atom having no unpaired electron in its valence shell is said to be stable.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (ii)

For a molecule having a hybridization of sp², its geometry in plane is trigonal planar. For a molecule having a trigonal planar geometry, its bond angle will be 120°

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (iv)

The electrons involved in the formation of any chemical bond are the valence shell electrons. In the given electronic configuration,1s² 2s² 2p⁶ 3s² 3d² 4s² the valence shell electrons are in d-orbital and s-orbital. So, the four electrons involved in the chemical bond formation will be 3d² 4s².

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (ii)

Hydrogen bond is the strongest in that molecules have a higher difference in electronegativity. Due to the small size of the oxygen atom, it has the highest electronegative character. So, H₂O molecules will have the strongest hydrogen bonding

HCl, HI and H₂Sdo not have hydrogen bonding between their atoms. So, H₂O will have the strongest hydrogen bonding.