DIRECTIONS: These are Assertion-Reason type questions.Question contains two statements:Statement-1 (Assertion) and Statement-2 (Reason). Answer these questions from the following four options.
Statement-1 :Ice is less dense than liquid water.
Statement-2 :There are vacant spaces between hydrogen bonded water molecules in ice.

Statement- 1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement -1
Statement -1 is True, Statement -2 is True ; Statement-2 is NOT a correct explanation for Statement - 1
Statement - 1 is True,Statement- 2 is False
Statement -1 is False, Statement -2 is True

Solution

It is correct that during formation of ice from water there are vacant spaces between hydrogen bonded molecules of ice. Ice has a cage like structure. Due to this reason ice is less dense than liquid water. Hence both statement-1 and statement-2 are true and statement-2 is correct explanation of statement-1.

Solution

The bond angles HXH vary in the order : NH₃ > PH₃ AsH > SbH₃. In the formation of \(XH_{4}^{+}\), the 3bond angle HXH increases to 109º 28'. Hence, maximum increase in the bond angle will be for SbH₃.

The electronegativity difference between N and F is greater than that between N and H yet the dipole moment of NH₃(1.5 D) is larger than that of NF₃ (0.2D). This is because

in NH₃ the atomic dipole and bond dipole are in the same direction whereas in NF₃ these are in opposite directions
in NH₃ as well as NF₃ the atomic dipole and bond dipole are in opposite directions
in NH₃ the atomic dipole and bond dipole are in the opposite directions whereas in NF₃ these are in the same direction
in NH₃ as well as in NF₃ the atomic dipole and bond dipole are in the same direction

Solution

In X — H — Y, X and Y both are electronegative elements

thus electron density on X will increase and on H will decrease
in both electron density will decrease
in both electron density will increase and on H will decrease
thus electron density will decrease on X and will increase on H

Solution

In X — H - - - Y, X and Y both are electronegative elements(i.e attracts the electron pair) then electron density on X and Y will increase and on H will decrease.

The relationship between the dissociation energy of N₂ and \(N_{2}^{+}\)is :

Dissociation energy of \(N_{2}^{+}\) > dissociation energy of N₂
Dissociation energy of N2 = dissociation energy of \(N_{2}^{+}\)
Dissociation energy of N₂ > dissociation energy of \(N_{2}^{+}\)
Dissociation energy of N₂ can either be lower or higher than the dissociation energy of \(N_{2}^{+}\)

Solution

Dissociation energy of any molecule depends upon its bond order. Bond order in N₂ molecule is 3 while bond order in \(N_{2}^{+}\)is 2.5. Further we know that more the Bond order,more is the stability and more is the bond dissociation energy.

The hybridization of P is \(PO_{4}^{3-}\) is the same as of

S in SO₃
N in \(NO_{3}^{-}\)
I in \(ICl_{2}^{+}\)
I in ICl₄

Solution

Solution

Structure I has two resonance forms, which make the hydrogen bonding more stable. The II molecule has no such resonance stabilization.

The dipole moments of diatomic molecules AB and CD are 10.41D and 10.27 D, respectively while their bond distances are 2.82 and 2.67 Å, respectively. This indicates that