Class 11th

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (ii)

In a molecular structure if a central atom has the same hybridization as the central atom of another molecular species, then the two structures are known as isostructural species.

(i) In NF₃ the central nitrogen atom is sp³ hybridized and in BF₃ the central boron atom is sp² . So, these two molecules are not isostructural pairs.

(ii) In BF₄ the central boron atom is sp³ hybridized and in NH₄ ⁺ the central nitrogen atom is also sp³ So, these two molecules are identified as isostructural pairs.

(iii) In BCl₃ the central boron atom is sp² hybridized and in BrCl₃ the central bromine atom is sp³d hybridized. So, these two molecules are not isostructural pairs.

(iv) In NH₃ the central nitrogen atom is sp³ and in NO₃ the central nitrogen atom is  sp² So, these two molecules are not isostructural pairs.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Ans: The average bond enthalpy is defined as the ratio of total bond dissociation enthalpy to the number of bonds broken in the structure.

The identical O- H bonds in water molecule does not have the same bond enthalpies. According to the structure of a water molecule, there are two O- H bonds, but there is a change in the breaking of the first O- H bond than the second because of the different charge.

Hence in water molecule the average bond enthalpy will be:

The average O-H bond enthalpy = $\frac{502+427}{2}$ = 464 mol^-1

In ethanol C₂H₅OH  the bond enthalpy of O-H is different because the electronic environment around oxygen atom is different. In ethanol the O-H is attached to the carbon atom and in water molecule the O-H is attached to the hydrogen atom.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Ans: (i) According to the electronic configuration, a total of 8 electrons must be present in the valence shell of an element. Element X has 4 valence electrons, so it will share the remaining 4 electrons for the formation of the bond, the molecular formula will be XH₄ . The element Y has 5 valence shell electrons, so it will form 3 bonds and the formula will be YH₃ . The element Z has 7 valence shell electrons, so it will form one bond with hydrogen and has the molecular formula H-Z .

(ii) Elements X, Y and Z having, 5 and 4 7 valence electrons respectively belongs to the second period and group 14^th ,15^th and 17^th . The electronegativity of the elements increases on moving across the period and the dipole moment is calculated by the difference in the electronegativity of the atoms in a molecular structure. Higher the difference in the electronegativity will be the dipole moment. So, the dipole moment of H - Z will be the highest.